Contents
1 Information and Disclaimers
4
2 Exam 1 Material
2.1 Monday 9 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Examples of Non-Commutative Rings . . . . . . . . . . . . . . . . .
2.1.2 Noetherian & Artinian Modules . . . . . . . . . . . . . . . . . . . . .
2.2 Wednesday 11 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Noetherian & Artinian Rings . . . . . . . . . . . . . . . . . . . . . .
2.3 Friday 13 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Noetherian & Artinian Rings . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Simple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Wednesday 18 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Example of a ring which is left Noetherian but not right Noetherian
2.4.2 Simple & Semisimple rings & modules . . . . . . . . . . . . . . . . .
2.5 Friday 20 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Semisimple Modules . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Monday 23 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 Composition Series & Length . . . . . . . . . . . . . . . . . . . . . .
2.7 Wednesday 25 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.1 More on Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Friday 27 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.1 Semisimple Rings - Headed towards proving Artin-Wedderburn . . .
2.9 Monday 30 January 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9.1 Continuing to work towards Artin-Wedderburn . . . . . . . . . . . .
2.10 Wednesday 1 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10.1 Rieffel’s Theorem & Most of proof of Artin-Wedderburn . . . . . . .
2.11 Friday 3 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.11.1 Artin-Wedderburn Modulo Uniqueness . . . . . . . . . . . . . . . . .
2.11.2 The map λ : R → R00 (M ) . . . . . . . . . . . . . . . . . . . . . . . .
2.12 Monday 6 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.12.1 End of proof of Artin-Wedderburn . . . . . . . . . . . . . . . . . . .
2.12.2 More about λ : R → R00 . . . . . . . . . . . . . . . . . . . . . . . . .
2.13 Wednesday 8 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.13.1 Finishing off information about λ : R → R00 . . . . . . . . . . . . . .
2.13.2 The Jacobsen Radical . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14 Friday 10 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.1 Semiprimitive Rings . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.2 Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.3 Wedderburn Radical . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15 Monday 13 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.1 Artinian rings are Noetherain . . . . . . . . . . . . . . . . . . . . . .
2.15.2 Commutative Algebra . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16 Wednesday 15 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16.1 History and Applications of Artin-Wedderburn . . . . . . . . . . . .
2.16.2 Commutative Algebra . . . . . . . . . . . . . . . . . . . . . . . . . .
2.17 Friday 17 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.17.1 More Commutative Algebra . . . . . . . . . . . . . . . . . . . . . . .
2.18 Wednesday 22 February 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
2.18.1 Modules over Artinian ring . . . . . . . . . . . . . . . . . . . . . . .
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3 Exam 2 Material
3.1 Wednesday 22 February 2012 . . . . . . . . . . . . . . . . . . . . .
3.1.1 Split Exact Sequences . . . . . . . . . . . . . . . . . . . . .
3.2 Friday 24 February 2012 . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Projective Modules . . . . . . . . . . . . . . . . . . . . . . .
3.3 Monday 27 February 2012 & Wednesday 29 February 2012 . . . . .
3.3.1 Exam Review . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Wednesday 29 February 2012 . . . . . . . . . . . . . . . . . . . . .
3.4.1 Projective Modules . . . . . . . . . . . . . . . . . . . . . . .
3.5 Friday 2 March 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 von Neumann Regular Rings . . . . . . . . . . . . . . . . .
3.5.2 Maschke’s Theorem . . . . . . . . . . . . . . . . . . . . . .
3.6 Monday 5 March 2012 . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.1 Semisimple Algebras over a Field . . . . . . . . . . . . . . .
3.7 Wednesday 7 March 2012 . . . . . . . . . . . . . . . . . . . . . . .
3.7.1 Semisimple Group Rings . . . . . . . . . . . . . . . . . . . .
3.7.2 Starting Representation Theory . . . . . . . . . . . . . . . .
3.8 Friday 9 March 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8.1 Representation Theory Vocab . . . . . . . . . . . . . . . . .
3.8.2 Starting Examples of Representations . . . . . . . . . . . .
3.9 Monday 12 March 2012 . . . . . . . . . . . . . . . . . . . . . . . .
3.9.1 Representations of Cyclic Groups and Permutation Groups
3.10 Wednesday 14 March 2012 . . . . . . . . . . . . . . . . . . . . . . .
3.10.1 Trace & Characters . . . . . . . . . . . . . . . . . . . . . .
3.11 Friday 16 March 2012 . . . . . . . . . . . . . . . . . . . . . . . . .
3.11.1 Review/Clarification . . . . . . . . . . . . . . . . . . . . . .
3.11.2 Equality of Characters . . . . . . . . . . . . . . . . . . . . .
3.11.3 Class Functions . . . . . . . . . . . . . . . . . . . . . . . . .
3.12 Monday 26 March 2012 . . . . . . . . . . . . . . . . . . . . . . . .
3.12.1 Useful Character Theory Formulas . . . . . . . . . . . . . .
3.13 Wednesday 28 March 2012 . . . . . . . . . . . . . . . . . . . . . . .
3.13.1 Character Tables . . . . . . . . . . . . . . . . . . . . . . . .
3.14 Friday 30 March 2012 . . . . . . . . . . . . . . . . . . . . . . . . .
3.14.1 Characters and Inner Products . . . . . . . . . . . . . . . .
3.14.2 Localization . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.15 Monday 2 April 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.15.1 Facts about Localization . . . . . . . . . . . . . . . . . . . .
3.16 Wednesday 4 April 2012 . . . . . . . . . . . . . . . . . . . . . . . .
3.16.1 Localization of Modules . . . . . . . . . . . . . . . . . . . .
3.17 Friday 6 April 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.17.1 More on Localization . . . . . . . . . . . . . . . . . . . . . .
3.18 Monday 9 April 2012 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.18.1 Useful facts about localization . . . . . . . . . . . . . . . .
4 Post-Exam 2 Material
4.1 Monday 9 April 2012 . . . . . . . . . . . . . . . .
4.1.1 Tensor Products . . . . . . . . . . . . . .
4.2 Wednesday 11 April 2012 . . . . . . . . . . . . .
4.2.1 Tensor Product Properties . . . . . . . . .
4.3 Friday 13 April 2012 . . . . . . . . . . . . . . . .
4.3.1 More on Tensor Products . . . . . . . . .
4.3.2 Functors on Module Categories . . . . . .
4.4 Monday 16 April 2012 . . . . . . . . . . . . . . .
4.4.1 Tensor Product is a Right Exact Functor
4.4.2 Flat Modules . . . . . . . . . . . . . . . .
4.5 Wednesday 18 April 2012 . . . . . . . . . . . . .
4.5.1 More Tensor Products . . . . . . . . . . .
4.6 Friday 20 April 2012 . . . . . . . . . . . . . . . .
4.6.1 Tangent in Commutative Algebra . . . . .
4.7 Monday 23 April 2012 . . . . . . . . . . . . . . .
4.7.1 Covariant Hom Functor . . . . . . . . . .
4.7.2 5 Lemma . . . . . . . . . . . . . . . . . .
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4.8
4.9
Wednesday 25 April 2012 . . . . . . . . . .
4.8.1 5 Lemma Consequences . . . . . . .
4.8.2 Contravariant Hom Functor . . . . .
Friday 27 April 2012 . . . . . . . . . . . . .
4.9.1 Adjointness of Hom and Tensor . . .
4.9.2 Projective Dimension, Regular Local
. . . .
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. . . .
Rings
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67
Chapter 1
Information and Disclaimers
Information. These are class notes for the second year graduate level algebra course at UNL (Math 902) as taken
in class and later typed by Kat Shultis. The notes are from the Spring of 2012, and that semester the course was taught
by Tom Marley. During the fall semester we covered the following topics:
B Ring Theory: Noetherian/Artinian, Semisimple, Simple, Artin-Wedderburn Theory
B Representation Theory: Representations of Group Rings, Characters
B Commutative Algebra: Localization, Tensor Products, Projective and Injective Modules
For each class day, I’ve indicated the topic at the top of that day’s notes.
Disclaimer. I created these notes in order to help me study, and so I’ve expanded proofs where I find it useful,
shortened things I was comfortable with before this course, and changed at least one proof to one that I like better
than the one presented in class. These notes are not meant to be a substitute for your own notes. They have been
proof-read, but are not guaranteed to be without errors. If you find errors, please email Kat at the following address:
s-kshulti1 “at” math.unl.edu
Notation. I’m tired of deciding whether or not 0 ∈ N and so I’ve adopted the following for notational ease. For any
n ∈ Z, set Nn := {a ∈ Z | a ≥ n}. So, 0 ∈ N0 , but 0 ∈
/ N1 .
4
Chapter 2
Exam 1 Material
2.1
Monday 9 January 2012
2.1.1
Examples of Non-Commutative Rings
Comment. In this class, all rings will have a multiplicative identity.
Example 1. Here we’ll give a few examples of how to construct non-commutative rings.
1. Matrix Rings: Let R be any ring. Let Mn (R) = {n × n matrices with entries in R}. Then Mn (R) is a noncommutative ring for n ≥ 2.
2. Endomorphism Rings: Let R be a commutative ring, and M an R−module. Then
EndR (M ) = {f : M → M |f is R − linear}
is a ring under addition and function composition. It is typically non-commutative.
3. Group Rings: Let G be a group. Let R be a commutative ring, and let R[G] be the free R−module with basis {g|g ∈
G}. That is, every element
as r1 g1 + . . . + rn gn with ri ∈ R and gi ∈ G. The multiplication
 in R[G]
 is written uniquely
!
X
X
X
to make this a ring is 
rg g 
sh h =
rg sh gh. In particular, if G =< a | a2 = 1 >, and R = k is
g∈G
h∈G
g,h∈G
a field, then k[G] = {r0 + r1 a | r0 , r1 ∈ k}. Note here that R[G] is non-commutative if and only if G is non-abelian.
4. Skew-Polynomial Rings: Let R be a commutative ring. Let σ : R → R be a ring homomorphism, and let
R[x; σ] be the R−module R[x]. That is, every element has the form an xn + . . . + a1 x + a0 , with ai ∈ R. Here, the
multiplication is given as follows for monomials, and extended linearly: an xn · bm xm = an σ n (bm )xn+m . Note here that
if σ is the identity homomorphism, then this is just R[x]. Also, R[x; σ] is non-commutative when σ is not the identity.
2.1.2
Noetherian & Artinian Modules
Definition 2. Let R be a ring, and M a left R−module. Then M is called Noetherian (resp. Artinian) if M satisfies
ACC (resp. DCC) on left submodules. Recall here what ACC and DCC say: ACC (resp. DCC) is if given any ascending
(resp. descending) chain N0 ⊆ N1 ⊆ . . . (resp. N0 ⊇ N1 ⊇ . . .) of submodules of M , there exists a k ∈ N such that
Nk = Nk+i for all i ≥ 0.
Proposition 3. Let R be a ring, and M a left R−module. TFAE:
a. M is Noetherian (resp. Artinian),
b. every non-empty set of submodules of M has a maximal (resp. minimal) element,
In the Noetherian case, we also have:
c. every submodule of M is finitely generated.
Proof. We’ll only do the proof in the Noetherian case, but the Artinian case proof follows mutatis mutandis1 .
(a. ⇒ b.) Let Λ be a nonempty set of submodules of M . Let M1 ∈ Λ. By way of contradiction, assume Λ does not
have a maximal element. Then there exists some M2 ∈ Λ so that M1 ( M2 . This process can be continued, that is,
we can find some M3 ∈ Λ so that M2 ( M3 , etc. This gives a violation of ACC, so we have a contradiction, and that
completes this part of the proof.
(b. ⇒ c.) Let N be a submodule of M and let Λ = {finitely generated submodules of N }. Note that 0 ∈ Λ so that
Λ is non-empty. By assumption, then Λ has a maximal element, say N 0 . If N 0 ( N , then there is some x ∈ N \ N 0 ,
and we can set N 00 = N 0 + Rx, so that N 00 is finitely generated, and N 00 ) N 0 . This is a contradiction as N 00 ∈ Λ but
N 0 was maximal. Hence N 0 = N , so that N ∈ Λ, and N is finitely generated.
∞
[
(c. ⇒ a.) Let N0 ⊆ N1 ⊆ . . . be an ascending chain of submodules of M , and set N =
Ni . Note that N is a
i=0
1 See
Hungerford page xv. This is latin for (roughly) “by changing the things which (obviously) must be changed (in order that the
argument will carry over and make sense in the present situation).”
5
submodule of M since the submodules are nested, and so N is finitely generated. Say N = Rx1 + . . . + Rx` and choose
t so that xi ∈ Nt for all i. Then Nt ⊆ N ⊆ Nt so that N = Nt = Nt+j for all j ≥ 0. Hence, the chain terminates and
this completes the proof.
Comment. Let R = k be a field. Then every R−module is a k−vector space, so that any k−vector space V is
Noetherian if and only if it is Artinian which in turn is true if and only if dimk V < ∞. Similarly, if R is a division
ring, then every left R−module has a basis, and a well defined notion of dimension and we again get that Noetherian,
Artinian and finite dimensional are all equivalent for vector spaces over division rings.
Example 4. Let R = k[x] where k is a field. Then R is Noetherian but not Artinian. The non Artinian part can
be seen with the following descending chain (x) ) (x2 ) ) (x3 ) ) . . ..
na
o
Example 5. Let M =
|a ∈ Z, n ≥ 0 , and consider M as a Z−module. Note here that Z ⊆ M ⊆ Q. Also, let
n
2
a
a
T = M/Z. Then T is Artinian but not Noetherian. We’ll show this, and for notational purposes, let n denote n + Z.
2
2
1
For n ≥ 0 let Nn = Z · n and notice that Nn ⊆ T .
2
First, we show that Nn ( Nn+1 which shows that T is not Noetherian as it provides an ascending chain that doesn’t
1
1
1
1
stabilize. Since n = 2 · n+1 , then for any a ∈ Z, we have a · n = 2a · n+1 which gives the containment. To
2
2
2
2
1
1
1
a
show the containment is strict, suppose that n+1 = a · n for some a ∈ Z. Then we get n+1 − n ∈ Z but that
2
2
2
2
1
a
1 − 2a
− n = n+1 ∈
/ Z so we have a contradiction, and hence the containment is strict.
2n+1
2
2
Now, we claim that every proper Z−submodule of T is Nn for some n. If we can show this, then any descending
chain of submodules of T must start with Nn for some n, and the only submodules of Nn are Ni where 0 ≤ i < n and
so the chain will stabilize, and we’ll get that T is Artinian. So let A ( T be a proper submodule. Choose n to be the
1
1
largest integer so that n ∈ A. Such an element exists as A is a proper submodule, and 0 ∈ A. So we claim that
2
2
a
a
A = Nn . We clearly have that Nn ⊆ A, so let m ∈ A and suppose that m ∈
/ Nn . Without loss of generality, we
2
2m
may assume that a is odd, and we also have that m ≥ n + 1. Since gcd(a, 2 ) = 1, then there exist integers x, y so
a
1
a
a
1
that ax + 2m y = 1. Hence, x m + y = m . Modding out by Z gives x · m = m and since m ∈ A this gives that
2
2
2
2
2
1
∈ A. This is a contradiction as m ≥ n + 1 and n was chosen to be maximal. Hence, A = Nn .2
2m
2.2
Wednesday 11 January 2012
2.2.1
Exact Sequences
Definition 6. A sequence of left R−modules and R−homomorphisms
...
Ni−1
fi−1
Ni
fi
Ni+1
...
is exact if image(fi−1 ) = ker(fi ) for all i.
Definition 7. An exact sequence is a Short Exact Sequence (SES) if it is of the form
0
f
A
g
B
C
0.
(2.1)
Equivalently, this is if f is injective, g is surjective, and image(f ) = ker(g).
Comment. Let N be a submodule of M . Then the following sequence is always exact where i and π are the usual
inclusion and projection maps:
0
2 We
N
i
M
π
M/N
0.
did most of this example on Wednesday, but we at least tried to start it today, and it fits better here.
6
In fact, if one considers any SES, it has this form (up to isomorphism). This is because (using the generic notation
from the SES in 2.1) we have that f is an injection so that A ∼
= f (A) ⊆ B. Similarly, as g is a surjection, then by
the first isomorphism theorem we have C = image(g) ∼
= B/ ker(g) = B/f (A).
Example 8. The sequence below is a short exact sequence. Note that the map from Z to Z is being given by
multiplication by 2.
0
2
Z
Z/2Z
Z
0.
Definition 9. Let M1 , M2 be R−modules and note that the following is a SES:
0
M1
f
M1 ⊕ M2
g
M2
0.
where f (m1 ) = (m1 , 0) and g(m1 , m2 ) = m2 . A short exact sequence of this form is called split exact.
Proposition 10. Let R be a ring and
0
A
B
C
0.
a SES of left R−modules. Then B is Noetherian (resp. Artinian) if and only if both A and C are Noetherian (resp.
Artinian).
Proof. We’ll only show this in the Artinian case, but the Noetherian case follows mutatis mutandis. Also, without loss
of generality, we’ll assume that A is a submodule of B and that C = B/A.
(⇒) Let B be Artinian, and let N0 ⊇ N1 ⊇ . . . be a descending chain of submodules of A. Then this is also a descending
chain of submodules of B since A ⊆ B and all submodules of A are also submodules of B. Hence the chain terminantes
and A is Artinian.
Next, let B be Artinian, and let N0 ⊇ N1 ⊇ . . . be a descending chain of submodules of C. Then for all i we have
Ni = Bi /A where Bi is a submodule of B. We then have that B0 ⊇ B1 ⊇ . . . is a descending chain of submodules
of B and so there is some n ∈ N such that Bn = Bn+i for all i ≥ 0. Thus, Nn = Bn /A = Bn+i /A = Nn+i for all i
and so C is Artinian as well.3
(⇐) Suppose that A and B/A are Artinian. Let N1 ⊇ N2 ⊇ . . . be a descending chain in B. Then N1 ∩ A ⊇ N2 ∩ A ⊇ . . .
is a descending chain in A and A is Artinian, and so there exists a k ∈ N so that Nk ∩ A = Nk+i ∩ A for i ≥ 0. Also,
N1 + A
N2 + A
⊇
⊇ . . . is a descending chain in B/A and B/A is Artinian, and so there exists an ` ∈ N so that
A
A
N` + A
N`+i + A
=
for i ≥ 0. Let t = max(k, `). Then we claim that Nt = Nt+i for i ≥ 0. To show this it is sufficient
A
A
(by induction) to show that Nt ⊆ Nt+1 . So let n ∈ Nt ⊆ Nt + A = Nt+1 + A and note that this last equality is due
to the correspondence theorem, and the fact that t ≥ `. Then n = n0 + a where n0 ∈ Nt+1 and a ∈ A. Then, we get
n − n0 ∈ A ∩ Nt = A ∩ Nt+1 ⊆ Nt+1 , and since n0 ∈ Nt+1 this gives that n ∈ Nt+1 as well.
Corollary 11. Let M1 , . . . , Mn be R−modules. Then ⊕ni=1 Mi is Noetherian (resp. Artinian) if and only if every Mi
is Noetherian (resp. Artinian).
Proof. The proof of this follows from the proposition applied to a split exact sequence and induction.
2.2.2
Noetherian & Artinian Rings
Definition 12. Let R be a ring. Then R is both a left and a right R−module. Note that the left (resp. right)
submodules of R are left (resp. right) ideals. We say R is left (resp. right) Noetherian (resp. Artinian) if it is Noetherian
(resp. Artinian) as a left (resp. right) R−module.
Proposition 13. Let R be a left Noetherian (resp. Artinian) ring. Then every finitely generated left R−module is
Noetherian (resp. Artinian).
Proof. We’ll only show this in the Artinian case, but the proof for Noetherian follows mutatis mutandis. Let R be
a left Artinian ring. Then Rn is a left Artinian R−module for all n. If M is a finitely generated left R−module, then
there is some n ∈ N and some module homomorphism ϕ : Rn → M so that ϕ is surjective. Thus, M ∼
= Rn / ker ϕ. Since
quotients of left Artinian modules are also Artinian, then M is Artinian.
3 In
class we said this direction is trivial, but I feel it is worth stating the actual argument.
7
2.3
2.3.1
Friday 13 January 2012
Noetherian & Artinian Rings
Definition 14. Let R be a ring. R is Noetherian (resp. Artinian) if it is both left and right Noetherian (resp. Artinian).
Remark 15. Let φ : R → S be a ring homomorphism. Then S is a left R−module via r · s = φ(r)s. So every left
S−module is also a left R−module. Suppose S is finitely generated as a left R−module. That is, S = Ru1 + . . . + Rut
for some u1 , . . . , ut ∈ S. If R is left Noetherian (resp. Artinian) as a ring, then S is left Noetherian (resp. Artinian)
as a ring. This is because as S is finitely generated as a left R−module, then S is Noetherian as a left R−module.
Example 16. 1. Consider the ring homomorphism φ : R → Mn (R) given by r 7−→ rIn , and let Eij be the matrix
with 1 in the i,
0 elsewhere. Then Mn (R) is generated as an R−module (left or right) by the Eij ’s. That
Pj entry and P
is, Mn (R) = i,j REij = i,j Eij R. Thus, if R is left Noetherian (resp. Artinian), so is Mn (R). Here, a special
case is that if D is a division ring, then Mn (D) is both Noetherian and Artinian.
2. Let G be a group, and R a ring. Consider φ : R → R[G] where R[G] = ⊕g∈G Rg = ⊕g∈G gR, and the map is given by
r 7−→ reG . If |G| < ∞, then R[G] is a finitely generated R−module (on either side), so if R is left Noetherian (resp.
Artinian), so is R[G]. Here, a special case is that if F is a field, and |G| < ∞, then F [G] is both Noetherian and Artinian.
Comment. Many times we’ll only say something on the left, but it is also true on the right.
Theorem 17 (Hilbert Basis Theorem). Let R be a left Noetherian ring, and x and indeterminate. Then R[x] is
also left Noetherian.
Remark 18. Subrings of Artinian (or Noetherian) rings are not necessarily Artinian (or Noetherian).
Example 19. 1. We have Z ⊆ Q and Q is both Noetherian and Artinian, but Z is not Artinian.
2. We can adjoin infinitely many indeterminants to get k[x1 , . . .] ⊆ k(x1 , . . .) where k is a field and k(x1 , . . .) is
Noetherian but k[x1 , . . .] is not Noetherian.
3. If k is a field then R = k[x, y] is Noetherian and contains S = k[x, xy, xy 2 , xy 3 , . . .] but S is not Noetherian.
Example 20. Let R be a ring, and S = Mn (R). For k ∈ {1, . . . , n} let Ik = {[aij ] ∈ S | aij = 0 whenever j 6= k}.
These are columns in the matrices. Then Ik is a left ideal, but not a right ideal for n ≥ 1.
Comment. If I is a left ideal of R, then R/I is a left R−module, but it is not necessarily a ring (unless I is a 2−sided
ideal). An ideal of R is a left ideal which is also a right ideal. If I is an ideal, then R/I is a ring. Also, (0) and R
are the trivial ideals.
2.3.2
Simple Rings
Definition 21. A ring R is called simple if it has no nontrivial ideals.
Comment. 1. If R is a commutative ring, then R is simple if and only if it is a field.
2. Division rings are simple.
3. There exist simple rings which aren’t division rings (and even ones that aren’t left or right Artinian)!
Example 22. Let F be a field, and σ : F → F a non-zero ring homomorphism which is not surjective, and consider
R = F [x; σ]. Recall that this means we set x · a = σ(a)x for a ∈ F. Then R is left Noetherian but not right Noetherian.
Proof. We can use the Euclidean algorithm on R = F [x; σ]. That is, for f, g ∈ R with g 6= 0, there exist q, r ∈ R such
that f = qg + r and deg(r) < deg(g). If I =
6 (0) is a left ideal of R, then choose g ∈ I \ {0} of smallest degree. Then
I = Rg, which means that every left ideal of R is finitely generated, and hence R is left Noetherian.
It remains to show that R is not right Noetherian. So choose b ∈ F \ σ(F ). Then for all n ≥ 1 we claim that
xn bx ∈
/ xn−1 bxR + . . . + x0 bxR. This will give an infinite properly ascending chain of right ideals. We show this by
contradiction. So suppose that xn bx = xn−1 bxfn−1 (x)+. . .+bxf0 (x) with f0 (x) ∈ R. Then we get by rearranging things
that bxf0 (x) = xg(x) for some g(x) ∈ R as all the other terms start on the left with an x. So we have f0 (x) = cr xr +. . .+c0
and g(x) = ar xr + . . . + a0 since both f0 (x) and g(x) must have the same degree. The leading term of bxf0 (x) is thus
bxcr xr = bσ(cr )xr+1 and the leading term of xg(x) is xar xr = σ(ar )xr+1 . The coefficients of these terms must be equal
so we get bσ(cr ) = σ(ar ) which implies that b = σ( acrr ) ∈ σ(F ). This is a contradiction and so completes the proof.
2.4
Wednesday 18 January 2012
2.4.1
Example of a ring which is left Noetherian but not right Noetherian
Q Q
q1 q2
Example 23. The ring R =
=
: q1 , q2 ∈ Q, a ∈ Z is left Noetherian but not right Noethe0 Z
0 a
rian.
8
Proof. We’ll
The
here is if thereis an
first case show R
is left Noetherian by showing any left ideal I is finitely generated.
0 aq
q1 q2
q1 q2
0 q
element
=
∈ R such that a 6= 0. Then for any q ∈ Q we have that
∈ I.
0 a
0 q
0 0
0 0
0 1
Let J = R
. Note that J is a finitely generated left ideal which lives in I and consists of all elements of R
0 0
whose
element is in the upper right hand corner of the matrix. Define a ring homomorphism φ : R → Q × Z
only nonzero
q1 q2
7−→ (q1 , a). By noticing that φ is surjective and that ker(φ) = J, we see that R/J is a ring, and that
by
0 a
R/J ∼
= Q × Z as rings. Thus, R/J is left Noetherian, so the left ideal I/J is finitely generated and hence I is finitely
generated because J is finitely
generated.
Q Q
q1 q2
The second case is if I ⊆
= W. Note that W is a left ideal of R and so let r =
∈ R and
0 0
0 a
w1 w2
q1 w1 q1 w2
w=
∈ W. then rw =
∈ W. So we can consider W to be a 2−dimensional Q vector
0
0
0
0
qw1 qw2
w1 w2
. As I ⊆ W is a Q−subspace, then it
space where scalar multiplication is given by q
=
0
0
0
0
has a basis of at most 2 elements, and so is finitely generated as a Q−vector space. Also, a Q−basis for I will be a
finite generating set for I as a left ideal, so that I is finitely generated.
We’ll
show
R is not right Noetherianby producing
ascendingchain of right ideals. Let In =
an infinite
now
0 2an
0 2an
q1 q2
0 2abn
: a ∈ Z for any n ≥ 0. Then
=
so that since ab ∈ Z we get In is
0 a
0 0
0 0
0 0
1
1
0 2n
0 2n+1
actually a right ideal of R. In fact, In =
R. Since
∈ In+1 \ In then we get that this is an
0 0
0
0
infinite properly ascending chain I0 ( I1 ( . . . so that R is not right Noetherian.
2.4.2
Simple & Semisimple rings & modules
Definition 24. Let M be a nonzero left R−module. M is said to be simple if (0) and M are the only submodules of M .
Note. This means that (0) is not simple. Also, there are simple rings that aren’t simple modules, in fact M2 (Q) is
such an example.
Remark 25. A left R−module M is simple if and only if M ∼
= R/I where I is a maximal left ideal of R.
Proof. (⇒) First, let M be simple and choose x ∈ M \ {0}. Then Rx is a left submodule of M so M = Rx is cyclic. We
define an R−linear map ψ : R → M given by r 7−→ rx. Let I = ker ψ so that I is a left ideal of R. Note that ψ is surjective
and so R/I ∼
= M . By the correspondence theorem, and since M is simple, we get that I is a maximal left ideal of R.
(⇐) This direction is trivial.
Definition 26. Let M be an R−module. M is called semisimple if every submodule of M is a direct summand of
M . That is, if N ⊆ M is a submodule of M , then there exists a submodule N 0 ⊆ M such that N ⊕ N 0 = M. Here
⊕ is the internal direct sum, meaning that N + N 0 = M and N ∩ N 0 = (0).
Remark 27. The good news is that simple modules are semisimple. Unfortunately, there are simple rings which aren’t
semisimple.
Definition 28. A ring is called left (resp. right) semisimple if it is semisimple as a left (resp. right) module over itself.
Remark 29. If M is semisimple, then all submodules of M and all quotients of M are also semisimple.4
2.5
Friday 20 January 2012
2.5.1
Semisimple Modules
Example 30. Let D be a division ring. Then every D−module is semisimple.
Proof. Let V be a left D−module, and let W be a submodule of V . Let β be a basis for W . Then there exists a basis,
β 0 , of V which contains β. Let β1 = β 0 \ β, and let W1 be the span of β1 . Then W1 ⊕ W = V.
Lemma 31. Every nonzero semisimple module contains a simple module.
4 We
sort of proved this in class.
9
Proof. Let M 6= 0 be a semisimple module, and choose x ∈ M \ {0}. Then Rx 6= 0 is semisimple as it is a submodule
of M . Thus, we can assume M = Rx. Let Λ = {N : N is a submodule of M, x ∈
/ N }. By Zorn’s Lemma, Λ has a
maximal element, we’ll call it N . As M is semisimple, M = N ⊕ N 0 for some N 0 . Also, note that N 0 6= 0 as x ∈
/ N.
Then we claim that N 0 is simple. Let A 6= 0 be a submodule of N 0 . Since A * N, then N + A ) N. By maximality
of N , x ∈ N + A so that N + A = M = Rx. We wish to show that T
A = N 0 . Now, let n0 ∈ N 0 ⊂ M. Then n0 = a + n
0
0
for some a ∈ A and n ∈ N. Rearranging this gives n = n − a ∈ N
N = (0). Thus, n0 = a ∈ A so that A = N 0 .
Theorem 32. Let M be an R−module. TFAE:
1. M is semisimple,
P
2. M is a sum of simple submodules, that is, M = i∈I N
i where each Ni is simple, and
L
3. M is a direct sum of simple submodules, that is, M = i∈I Ni where each Ni is simple.
Proof. (1) ⇒ (3) If M = (0), then we’re done, so we may assume M 6= (0). Let T = {N | N is a simple submodule of M }.
By the lemma T 6= ∅, so let
(
)
X
M
Λ= J ⊆T |
E=
E .
E∈J
E∈J
Note that ∅ ∈ Λ so that λ 6= ∅. Let C be a totally ordered subset of Λ. We claim A =
[
J ∈ Λ. Suppose not. Then
J∈C
X
M
6=
E∈A
E so that there exist E1 , . . . , En ∈ A such that E1 + . . . + En 6= E1 ⊕ . . . ⊕ En . But there is some J ∈ C
E∈A
such that {E1 , . . . , En } ⊆ J. That forms a contradiction
X
M and so we have that A ∈ Λ. Thus, by Zorn’s Lemma, there
is some J ∈ Λ which is maximal. Let B =
E=
E. We claim that B = M. If not, then as M is semisimple,
E∈J
E∈J
there is some C 6= 0 submodule of M such that M = B ⊕ C. As C is semisimple, then by lemma 31 it contains some
E 0 which is simple. Let J 0 = J ∪ {E 0 }. Then J 0 ∈ Λ, which contradicts the maximality of J, and so B = M.
(3) ⇒ (2) Trivial.
(2) ⇒ (1) Let A be a submodule of M , let T = {N | N is a simple submodule of M }, and let
(
!
)
X
M
\ X
Λ= J ⊆T |
E=
E; A
E = (0) .
E∈J
E∈J
E∈J
Again,X
∅ ∈ Λ,M
so that Λ 6= ∅. Also, we have that Zorn’s Lemma applies, so let J ∈ Λ be maximal, and set
0
E=
E. Let B = A+A0 = A⊕A0 . Note that this is actually a direct sum as J ∈ Λ means that A∩A0 = (0).
A =
E∈J
E∈J
So we now claim that B = M. As M is a sum of simple submodules, it sufficies to show that every simple submodule of
0
M is contained in B. Suppose not, that is,!there is a simple !
submodule E 0 ⊆ M with E 0 * B. Then E 0 ∩!B = (0)
! as E is
X
X
M
\
L 0
simple. Let J 0 = J ∪ {E 0 }. Then
E + E0 =
E
E as E 0 ∩ A0 = (0) and A
E + E 0 = (0).5
E∈J
E∈J
E∈J
Hence, J 0 = J ∪ {E 0 } ∈ Λ which contradicts the maximality of J. Thus, B = A ⊕ A0 = M so that M is semisimple.
2.6
Monday 23 January 2012
2.6.1
Composition Series & Length
Definition 33. Let M be an R−module. A series (or filtration) for M is a finite sequence of submodules
(0) = Mn ⊆ Mn−1 ⊆ . . . ⊆ M0 = M.
The factors of the series is the set {Mi /Mi+1 }n−1
i=0 . The length of the series is the number of proper inclusions. A
refinement of a series is a series for M which contains the original series as a sub-series. A refinement is proper if the
length of the refinement is larger than the length of the original series. Two series are equivalent if they have the same
length and there is a bijection between the sets of factors such that corresponding factor modules are isomorphic.
Example 34. The sequence6 (0) ( (18) ( (3) ( Z is a series with factors {(18), Z/(6), Z/(3)}. This series has the
following series as a proper refinement: (0) ( (72) ( (18) ( (9) ( (3) ( Z with factors {(72), Z/(4), Z/(2), Z/(3), Z/(3)}.
An equivalent series is (0) ( (72) ( (24) ( (12) ( (4) ( Z since this series has factors {(72), Z/(3), Z/(2), Z/(3), Z/(4)}.
5 This
6 The
isn’t hard to show but isn’t what I would call obvious either. The details are in my written notes.
notation (n) is used for nZ = Zn.
10
Lemma 35. Let M be an R−module and A ⊆ A0 , B ⊆ B 0 submodules of M. Then
A + (A0 ∩ B 0 ) ∼ B + (A0 ∩ B 0 )
.
=
A + (A0 ∩ B)
B + (A ∩ B 0 )
Proof. The trick here7 is to how that both modules are isomorphic to
A0 ∩ B 0
.
(A0 ∩ B) + (A ∩ B 0 )
Because of symmetry, it is sufficient to do this for only one of the modules in the statement. By defining
ψ : A + (A0 ∩ B 0 ) →
(A0
A0 ∩ B 0
∩ B) + (A ∩ B 0 )
via a + x 7−→ x we can show that this is a surjective homomorphism with kernel equal to A + (A0 ∩ B) so that by the
first isomorphism theorem we’re done.
Theorem 36 (Shreier Refinement Theorem). Any two series for a module M have equivalent refinements.
Proof. Let
(0) = Mn ⊆ Mn−1 ⊆ . . . ⊆ M0 = M
(2.2)
(0) = Np ⊆ Np−1 ⊆ . . . ⊆ N0 = M
(2.3)
and
be two series for M . For i ∈ {0, . . . , n − 1} and j ∈ {0, . . . , p} let Mij = Mi+1 + Mi ∩ Nj . Then for each i, j, we have
Mij ⊆ Mij−1 . Also, Mip = Mi+1 and Mi0 = Mi . This gives a refinement of (2.2) as
(0) ⊆ Mn−1p ⊆ Mn−1p−1 ⊆ . . . ⊆ Mn−10 = Mn−2p ⊆ . . . ⊆ Mn−20 ⊆ . . . ⊆ M00 = M.
(2.4)
Similarly, for i ∈ {0, . . . , n}, and j ∈ {0, . . . , p − 1} let Nji = Nj+1 + Mi ∩ Nj . Again, we have for each i, j that
Nji ⊆ Nji−1 , that Njn = Nj+1 , and Nj0 = Nj . So we again get a refinement, but this time of (2.3):
(0) = Np−1n ⊆ Np−1n−1 ⊆ . . . ⊆ Np−10 ⊆ Np−2n ⊆ . . . ⊆ Np−20 ⊆ . . . ⊆ N00 = N.
(2.5)
Now, the nonzero factors of (2.4) are {Mij /Mij+1 } and the nonzero factors of (2.5) are {Nji /Nji+1 }. Then by the lemma,
we get that
M ij
Mi+1 + Mi ∩ Nj ∼ Nj+1 + Mi ∩ Nj
Nj i
=
=
=
Mij+1
Mi+1 + Mi ∩ Nj+1
Nj+1 + Mi+1 ∩ Nj
Nji+1
showing that the two refinements are equivalent.
Definition 37. A composition series is a series with no proper refinements. In particular, in a composition series, all
of the nonzero factor modules are simple.
Corollary 38 (Jordan-Holder Theorem). If M has a composition series, then any two composition series for M
are equivalent. In particular, they have the same length.
Definition 39. If an R−module M has a composition series, then the length of M , which is denoted λR (M ) is the
length of any composition series for M . If M has no composition series, then we define λR (M ) = ∞.
Remark 40. 1. If M is a module of finite length, then any series can be refined to a composition series.
2. λR (M ) = 0 implies that M = 0.
3. λR (M ) = 1 implies that M is a simple module.
Proposition 41. If λR (M ) < ∞ and N is a submodule of M , then λR (N ) ≤ λR (M ) with equality holding if and only
if N = M .
Proof. Consider the series 0 ⊆ N ⊆ M . As M has finite length, we can refine this to a composition series
0 = M0 ⊆ M1 ⊆ . . . ⊆ Mn ⊆ M . Since this is a refinement, then for some t ∈ {0, . . . , n} we have Mt = N .
Thus, λR (N ) = t ≤ n = λR (M ) and we clearly get that n = t if and only if N = M .
7 I’m
not going to go through all the details of this proof here.
11
2.7
2.7.1
Wednesday 25 January 2012
More on Length
Proposition 42. Let M be a left R−module. Then M has finite length if and only if M is both Noetherian and Artinian.
Proof. (⇐) Any ascending or descending chain of submodules of M can be refined to a composition series. Thus, any
such chain has at most λR (M ) strict containments, so that any ascending or descending chain stabilizes.
(⇒) If M = 0, then M clearly has length 0 so we may assume M 6= 0. Let Λ = {N | N ⊆ M ; N 6= 0}. As M is
Artinian, and M ∈ Λ, then there exists some M1 ∈ Λ which is minimal, and hence simple. If M = M1 , then λR (M ) = 1.
If not, then choose M2 which is minimal among all submodules of M which properly contain M1 . Again, M2 /M1 is
clearly simple. Proceeding in this way we get 0 = M0 ( M1 ( M2 ( . . . where Mi /Mi−1 is simple for all i. As M is
Noetherian, we eventually have M = Mn for some n so that we have a composition series, and hence finite length.
Notation. If we are thinking of R as a left R−module we write R R and if we are thinking of R as a right R−module
then we write RR .
Corollary 43. Suppose λR (R R) < ∞. Then any finitely generated left R−module, M , has finite length.
Proof. Since λR (R R) < ∞ then R is left Noetherian and left Artinian. Thus, any finitely generated left R−module,
M is Artinian and Noetherian, so that λR (M ) < ∞.
Proposition 44. Let 0 → L → M → N → 0 be a short exact sequence of R−modules. Then λR (M ) = λR (L) + λR (N ).
Proof. Since M is Noetherian (resp. Artinian) if and only if L and N are, then λR (M ) < ∞ if and only if both λR (L)
and λR (N ) are finite. So we may assume λR (M ) < ∞, so that L and N also both have finite length, and set ` = λR (L),
and n = λR (N ). Without loss of generality, we may assume L ⊆ M and N = M/L. We have composition series
0 = L` ( L`−1 ( . . . ( L0 = L
(2.6)
0 = Mn /L ( Mn−1 /L ( . . . ( M0 /L = N.
(2.7)
L = Mn ( Mn−1 ( . . . ( M0 = M
(2.8)
and
This gives that
with Mi /Mi+1 ∼
which is simple because (2.7) is a composition series. Combining (2.6) with (2.8) gives a
=
composition series for M of length ` + n:
Mi /L
Mi+1 /L
0 = L` ( L`−1 ( . . . ( L0 = L = Mn ( Mn−1 ( . . . ( M0 = M.
Corollary 45. By applying the proposition to the short exact sequence 0 → A1 → A1 ⊕ A2 → A2 → 0 we get
λR (A1 ⊕ A2 ) = λR (A1 ) + λR (A2 ).
Ln
Pn
Corollary 46. By induction we then have λR ( i=1 Ai ) = i=1 λR (Ai ).
∼ Ln D
generated D−module. Since M has a finite basis, M =
Example 47. Let D be a division ring, and M
i=1
Lna finitely P
n
where n = dimD (M ). Thus, λD (M ) = λD ( i=1 D) = i=1 λD (D) = n = dimD (M ).
Example 48. Let R = Z and M = Z/(12). Then the following series: 0 ⊂ (6)/(12) ⊂ (3)/(12) ⊂ Z/(12) = M is a
composition series for M so that λR (M ) = 3.
Note. Since Z is Noetherian but not Artinian, we have that λZ (Z) = ∞.
Example 49. Let k be a field, R = k[x], and M = k[x]/(x2 ) = k · 1 ⊕ k · x. Consider the map ϕ : R → (x)/(x2 )
given by 1 7−→ x + (x2 ). This map is a surjection and ker(ϕ) = (x) so that (x)/(x2 ) ∼
= R/(x) and R/(x) ∼
= k is simple.
R/(x) ∼
2
2
Also, (x)/(x2 ) = R/(x) by the isomorphism theorems. Thus, 0 ( (x)/(x ) ( R/(x ) = M is a composition series so
that λR (M ) = 2 = dimk (M ).
Example 50. Let R = R[x] and M = R[x]/(x2 + 1) = R · 1 ⊕ R · x. As (x2 + 1) is a maximal ideal, then M is simple
and λR (M ) = 1. So here we have λR (M ) = 1 2 = dimR (M ).
Example 51. Let R = C[x] and M = C[x]/(x2 + 1) = C · 1 ⊕ C · x. Then 0 ( (x − i)/(x2 + 1) ( C[x]/(x2 + 1) = M
is a composition series, so λR (M ) = 2 = dimC (M ).
12


a1

..  | a ∈ D . Then V is a left R−module. In fact,
i
.

an
V is a simple R−module. It is enough here to show that if v ∈ V \ {0} then Rv = V , and this isn’t hard to show as D is a
division ring. So now, fix i ∈ {1, . . . , n} and let Ii = {[ak` ]n×n | aki = 0, ak` ∈ R}. This is the set of matrices where the ith
column is nonzero, but
other columns are zero. Thus, for all i we have Ii is a left ideal of R and Ii ∼
= V as left modules.
Lall
n
It is clear that R = i=1 Ii ∼
= V n , so that λR (R) = λR (V n ) = nλR (V ) = n. However, note that dimD (R) = n2 .


Example 52. Let D be a division ring, R = Mn (D) and V = 

2.8
2.8.1
Friday 27 January 2012
Semisimple Rings - Headed towards proving Artin-Wedderburn
Recall. A ring R is called left semisimple if R R is semisimple.
Example 53. 1. Any division ring D is left semisimple as D D is simple.
2. If D is a division ring then R = Mn (D) is left semisimple. This can be seen by letting V be the set of n×1 (i.e. column)
matrices over D. Then V is a simple left R−module and R R ∼
= V n which proves that R is left semisimple. Symmetrically, RR is isomorphic to W n where W is the set of 1 × n (i.e. row) matrices over D so that RR is also semisimple.
Corollary 54. Let D1 , . . . , Dk be division rings and n1 , . . . , nk be positive integers. Then Mn1 (D1 ) × . . . × Mnk (Dk )
is left (and right) semisimple.
Comment. The following theorem is going to be our goal to prove for the next several class periods.
Theorem 55 (Artin-Wedderburn Theorem). Let R be a left semisimple ring. Then there exist unique division
rings D1 , . . . , Dk and integers n1 , . . . nk such that R ∼
= Mn1 (D1 ) × . . . × Mnk (Dk ).
Note. Note that by the previous Corollary, the Artin-Wedderburn Theorem implies that if R is left semisimple then
R is right semisimple.
Proposition 56. Let R be left semisimple. Then,
1. R R = I1 ⊕ . . . ⊕ In where each Ij is a simple left ideal of R.
2. λR (R R) < ∞ (which then implies that R is left Noetherian and left Artinian)
3. Every simple left R−module is isomorphic to Ij for some j, where the Ij ’s are as in the first statement of the theorem.
L
Proof. 1. Since we know that R R is semisimple, we have that R R = j∈Λ Ij where each Ij is a simple left ideal of
R. As 1 ∈ R we can write 1 = xj1 + xj2 + . . . + xj` where each xji ∈ Iji . Now, given any r ∈ R we have that
L`
r = r · 1 = rxj1 + . . . + rxj` and each rxji ∈ Iji as each Iji is a left ideal of R. Thus, R R ⊆ i=1 Iji so that
R R = Ij1 ⊕ . . . ⊕ Ij` and we can reindex these so we have R R = I1 ⊕ . . . ⊕ In .
2. For each i we have that λR (IP
i ) = 1 since each Ii is simple. Thus by the additivity of length we get that
n
λR (R R) = λR (I1 ⊕ . . . ⊕ In ) = j=1 λR (Ij ) = n.
3. First, let J be a simple left ideal. Consider 0 ⊆ J ⊆ R as a series for R R. This can be refined to a composition
series since the length of R as a left module over itself is finite. So we get:
0 = Jn ⊆ Jn−1 ⊆ . . . ⊆ J1 ⊆ J0 = R
where Jn−1 = J since J is simple. We also have the following as a composition series for R:
0 ⊆ I1 ⊆ I1 ⊕ I2 ⊆ . . . ⊆ I1 ⊕ . . . ⊕ In = R.
∼ Ii for some i since the factors of the second series here
By the Jordan Holder Theorem (Corollary 38) we get that J =
are the Ii ’s. Now, let M be a simple left R−module, and let x ∈ M \ {0}. As M is simple, we have that M = Rx.
Now, define φ : R → M via r 7−→ rx. This is a surjective homomoprhism of left R−modules. Let K = ker φ. Thus,
R/K ∼
= M as left R−modules by the first isomorphism theorem. As R R is semisimple, we also have that R = K ⊕ I
∼
∼
for some left ideal I. Thus, M ∼
= R/K ∼
= K⊕I
K = I and I = Ii for some i by the previous argument. Thus, every
simple left R−module is isomorphic to one of the Ii ’s.
Lemma 57. Let R be a ring, I a simple left ideal, and M a simple left module. Then either I ∼
= M or IM = 0.
Proof. Suppose IM 6= 0. Then let x ∈ M so that Ix 6= 0. Note that Ix is a left submodule of M , so as M is simple, we
must have that Ix = M. Define, φ : I → M by a 7−→ ax. Note that φ is a surjective homomorphism of left R−modules
since M = Ix. Now, if ker φ = I, then Ix = 0, which is a contradiction, and since I is simple, we then must have that
ker φ = 0 so that φ is an isomorphism.
13
Lemma 58 (Schur’s Lemma). Let R be a ring and M a simple left R−module. Then
EndR (M ) = {f : M → M | f is a left R − linear map}
is a division ring.
Proof. Let f ∈ EndR (M ) be chosen so that f 6= 0. Then image(f ) 6= 0 so that we must have image(f ) = M since M
is a simple module. Similarly, we cannot have ker(f ) = M since f 6= 0 so we must have that ker(f ) = 0. Thus, f is
an isomorphism, and so has an inverse in EndR (M ).
2.9
Monday 30 January 2012
2.9.1
Continuing to work towards Artin-Wedderburn
Proposition 59. Let R be a left semisimple ring.P
Let {I1 , . . . , Ik } be the set of distinct (up to isomorphism) simple
left ideals of R.8 For each i ∈ {1, . . . , k}, let Ri =
L where the sum is taken over all left ideals, L, of R, such that
L∼
= Ii . Then
1. Ri is a ring with identity for all i.
2. Ri is left semisimple for all i.
3. Ri has one distinct (up to isomorphism) simple left ideal for each i.
4. Ri is simple for all i.
5. R ∼
= R1 × . . . × Rk as rings.
Proof. Recall from Lemma 57 that if I, J are simple left ideals, then I J implies that IJ = 0. Hence, Ri Rj = 0 for
i 6= j and each Ri is a left ideal of R. Since R is semisimple, then R = I1 + . . . + Ik ⊆ R1 + . . . + Rk ⊆ R by proposition
56. Fix some x ∈ R and write x = x1 + . . . + xk with xi ∈ Ri for each i. We also have 1 = e1 + . . . + ek where ei ∈ Ri .
Thus, xi = xi · 1 = xi e1 + . . . + xi ek = xi ei = (x1 + . . . + xk )ei = xei since xj e` = 0 whenever j 6= `. This means
that each xi is uniquely determined so that R = R1 ⊕ . . . ⊕ Rk . Since each Ri is a left ideal, they’re closed under
addition and multiplication, but we also want to show that they have identities. In particular, for any r ∈ Ri , then
r = r · 1 = r(e1 + . . . + ek ) = rei and similarly ei r = (e1 + . . . + ek )r = 1 · r = r so that ei is the identity element of Ri so
that Ri is a ring with identity. We can easily9 check that the map φ : R → R1 × . . . × Rk defined by x 7−→ x(e1 , . . . , ek )
is an isomorphism, which completes the proof of parts 1 and 5.
Clearly, if J ⊆ Ri is a left ideal of R, then J is a left ideal of Ri . Conversely, if J ⊆ Ri is a left ideal of Ri , then
J is also a left ideal of R since RJ ⊆ RRi = (R1 + . . . + Rk )Ri = Ri Ri ⊆ Ri . Thus, the left ideals of Ri are precisely
the left ideals of R which are contained in Ri . Now, since Ri is a left ideal of R, then by definition of Ri we get that
Ri is left semisimple as an R−module, so it is a direct sum of simple left R−submodules. Thus, Ri is left semisimple
as a ring, which completes the proof of 2.
Next, let J ⊆ Ri be a simple left ideal. Then, using the same fact about the left ideals of Ri , we get that J ∼
= Ij
for some j. Hence, J ∼
= Ii , which gives 3.
Finally, let J be a nonzero (2-sided) ideal of Ri . In order to show Ri is simple, it is enough to show that J ⊇ K for every
left ideal K of R such that K ∼
= Ii . This is because Ri is defined to be the sum of all such ideals, so if J contains all of
them, then J = Ri . So let K be any left ideal of R such that K ∼
= Ii . Since J 6= 0, and J is left semisimple, then it must
contain a simple left ideal, say L. Clearly, we must have that L ∼
= Ii and since K ∼
= Ii , then we know that L ∼
= K. Since
0
0
Ri is semisimple, Ri = L⊕L where L is a left ideal of Ri . This gives that ei = `1 +`2 where `1 ∈ L and `2 ∈ L0 . For any
` ∈ L we have ` = `ei = ``1 + ``2 so that ``2 = ` − ``1 ∈ L ∩ L0 so that ``2 = 0 and hence ` = ``1 and L = L`1 . Now, let
ψ : L → K be a left R−module isomorphism. Then K = ψ(L) = ψ(L`1 ) = Lψ(`1 ) ⊆ Jψ(`1 ) ⊆ J since J is an ideal.
Corollary 60. Let R be a left semisimple ring. The following are equivalent:
1. R is simple.
2. R has a unique (up to isomorphism) simple left module.
Corollary 61. If R is a simple ring which is also left semisimple, then R is left Artinian.
2.10
Wednesday 1 February 2012
2.10.1
Rieffel’s Theorem & Most of proof of Artin-Wedderburn
Definition 62. Let R be a ring. We define here the opposite ring of R, which is denoted Rop . As a set Rop is in
bijective correspondence with R and we write ao as the element in Rop which corresponds to a ∈ R. The operations
on Rop are as follows:
ao + bo = (a + b)o and ao · bo = (ba)o .
Remark 63. For any ring R, we get EndR (R) ∼
= Rop . This isomorphism can be shown using the map f : EndR (R) →
op
o
R given by φ 7−→ (φ(1)) .
8 We
9I
know this set is finite by Proposition 56.
didn’t carefully write the details, but I thought through them, and they’re not hard.
14
Example 64. Let M be a left R−module. Then M n = M ⊕ . . . ⊕ M where there are n copies of M in the direct
sum. Also, if we define Mi to be Mi = {(0, . . . , m, . . . , 0) | m ∈ M } ⊆ M n where the m is in the ith spot for a given
i, then we have the usual inclusion and projection maps: ρi : Mi ,→ M n and πi : M n Mi . For ψ ∈ EndR (M n ) let
ψij : M → M be the composition of the following maps:
M∼
= Mi
ρi
Mn
ψ
Mn
πj
Mj ∼
=M
ψij
so that ψij ∈ EndR (M ) for all i, j. It is a fact that the map EndR (M n ) → Mn (EndR (M )) given by ψ 7−→ [ψij ]n×n = [ψ]
is a ring homomorphism.10 A special case is when M =R R; then EndR (Rn ) ∼
= Mn (EndR (R)) ∼
= Mn (Rop ).
Notation. Let M be a left R−module. We define
R0 = R0 (M ) := EndR (M )
and note here that M is a left R0 −module where we set φ · x = φ(x) for any φ ∈ R0 and x ∈ M . We also define
R00 = R00 (M ) := EndR0 (M ).
For a given a ∈ R, we have the map `a : M → M by x 7−→ ax and we note that for any a ∈ R, φ ∈ R0 and x ∈ M then
`a (φ · x) = `a (φ(x)) = aφ(x) = φ(ax) = φ · ax = φ · `a (x) since φ is an R−linear map. Hence, `a ∈ R00 (M ). Similarly
if I is a left ideal of R, and a ∈ I then ra : I → I given by x 7−→ xa is an element of R0 (I) since ra (bx) = bxa = bra (x)
for any b ∈ R and x ∈ I.
Remark 65. The map λ : R → R00 (M ) given by a 7−→ `a is a ring homomorphism.
Proof. Given any r, s ∈ R we have λ(r + s) = `r+s which is the map from M to M given by left multiplication by r + s
and this is equal to `r + `s = λ(r) + λ(s) since M is a module. Also, λ(rs) = `rs is left multiplication by rs. We have
for any x ∈ M that `rs (x) = rsx = r(sx) = `r (`s (x)) = `r `s (x) so that λ(rs) = λ(r)λ(s) and we do indeed have a ring
homomorphism.
Theorem 66 (Rieffel). Let R be a simple ring, and I a nonzero left ideal of R. Then λ : R → R00 (I) given by
a 7−→ `a is an isomorphism.
Proof. Since R is a simple ring, we know that ker(λ) = 0 or R. However, λ(1) = `1 6= 0 so we must have that ker(λ) = 0
and λ is injective. Now, we claim that λ(I) is a left ideal of R00 = R00 (I). To show this, let f ∈ R00 = EndR0 (I) and
let `a ∈ λ(I). We have f `a (x) = f (ax) = f (rx (a)) = f (rx · a) = rx f (a) = f (a)x = `f (a) (x) since rx ∈ R0 (I) and so
we get that f `a = `f (a) ∈ λ(I) and since it is clear that λ(I) is an additive subgroup of R00 (I), we get that λ(I) is a left
ideal of R00 (I). Consider the (2-sided) ideal IR of R.11 Since R is simple and IR =
6 0 we must have that IR = R. Thus,
λ(R) = λ(IR) = λ(I)λ(R). Finally, using the fact that 1R00 = `1 ∈ λ(R) and that λ(I) is a left ideal of R00 we get that
R00
=
=
⊆
=
R00 λ(R)
R00 λ(I)λ(R)
λ(I)λ(R)
λ(R)
so that λ is surjective.
Theorem 67. Let R be a simple ring. The following are equivalent:
1. R is left semisimple,
2. R is left Artinian, and
3. R ∼
= Mn (D) for some division ring D and n ∈ N.
Proof. (3 ⇒ 1) We’ve already shown this. See Example 52.
(1 ⇒ 2) We’ve already shown this too. See Corollary 61.
(2 ⇒ 3) Since R is left Artinian, it contains a simple left ideal I. (We can do this by choosing a minimal element of the
set of nonzero left ideals). By Rieffel’s Theorem (Theorem 66), we know that R ∼
= R00 (I) = EndR0 (I). By Schur’s Lemma
10 The additive part is easy enough, and I think the multiplicative part just comes from being careful about indices and matrix multiplication,
but I haven’t gotten it P
to work quite right.
11 Recall that IR = { n
k=1 ik rk | ik ∈ I, rk ∈ R, n ∈ N}.
15
(Lemma 58) we know that R0 = EndR (I) is a division ring since I is simple, so we’ll set D = R0 . We now claim that I is
finitely generated as an R0 module and prove it by contradiction. If I is not finitely generated, then there is an infinite set
S = {e1 , e2 , e3 , . . .} ⊆ I such that S is linearly independent over R0 . For each n, let Jn = {f ∈ R00 | f (e1 ) = . . . = f (en ) =
0}. Note that Jn is itself a left ideal of R00 for each n. Also, Jn ⊇ Jn+1 for all n. In fact, we have strict containment since
there is an element g ∈ R00 such that g(ei ) = 0 for all i =
6 n + 1 and g(en+1 ) = 1 which is also an element of Jn \Jn+1 . Since
R00 ∼
= R this is a contradiction since we know R is left Artinian. Hence, for some n ∈ N we have I ∼
= Dn as left D−modules.
This means that R ∼
= EndD (Dn ) = Mn (Dop ). Note that Dop is also a division ring to complete the proof.
2.11
2.11.1
Friday 3 February 2012
Artin-Wedderburn Modulo Uniqueness
Corollary 68. Let R be a ring. The following are equivalent:
1. R is left semisimple
2. R is right semisimple
3. R ∼
= Mn1 (D1 ) × . . . × Mnk (Dk ) where each Di is a division ring and ni ∈ N for each i.
Proof. (1 ⇒ 3) We showed in Proposition 59 that R ∼
= R1 × . . . × Rk where each Ri is simple and left semisimple. By
Rieffel’s Theorem (Theorem 66), we have that each Ri ∼
= Mni (Di ) for some ni ∈ N and division ring Di .
(3 ⇒ 2) Each Mni (Di ) is right semisimple and a finite product of right semisimple rings is right semisimple. We proved
this on the left on homework #2 but the proof on the right follows by mutatis mutandis.12
(2 ⇒ 1) Since R is right semisimple, then Rop is left semisimple. Using the implication (1 ⇒ 3) we have that
Rop ∼
= Mn1 (D1 ) × . . . × Mnk (Dk ) for ni ∈ N and Di a division ring for each i. Now,
op
op
R = (Rop )op ∼
= (Mn1 (D1 ) × . . . × Mnk (Dk ))op ∼
= Mn1 (D1 ) × . . . × Mnk (Dk )
and since Diop is also a division ring for each i. Then by the same problem in homework #2 and since Mn (D) is left
semisimple for any n ∈ N and any division ring D, we get that R is left semisimple .
Remark 69. Suppose that R is semisimple. Then λR (R R) < ∞ and λR (RR ) < ∞.
Exercise. If R ∼
= Mn1 (D1 ) × . . . × Mnk (Dk ) where each ni ∈ N and each Di is a division ring, then λR (R R) =
λR (RR ) = n1 + . . . + nk .
Question. If λR (R R) and λR (RR ) are both finite, are they necessarily equal?13
Remark 70. We will prove the uniqueness part of Artin-Wedderburn, but we need some more machinery first.
2.11.2
The map λ : R → R00 (M )
Remark 71. Given a map f : E → E, then f ∈ R00 (E) if and only if f is both additive and f ◦ φ = φ ◦ f for all
φ ∈ R0 (E). This is since f (φ(x)) = f (φ · x) = φ · f (x) = φ(f (x)) for all f ∈ R00 (E) and φ ∈ R0 (E).
Remark 72. Let f ∈ EndR0 (E) = R00 (E), and let n ∈ N.
(f (x1 ), . . . , f (xn )). Then f (n) ∈ R00 (E n ) = EndR0 (E n ) (E n ).
Define f (n) : E n → E n by (x1 , . . . , xn ) 7−→
Proof. It is clear that f (n) is additive, so we merely need to show that f (n) ψ = ψf (n) for any ψ ∈ R0 (E n ). We can write
this using matrix notation as ψ = [ψij ]n×n where ψij ∈ EndR (E) = R0 (E) and f (n) = f In . Thus, since f ∈ R00 (E)
we have f ψij = ψij f for all i, j and hence f (n) ψ = f In [ψij ] = [f ψij ] = [ψij f ] = [ψij ]f In = [ψij ]f (n) .
Lemma 73. Let E be a semisimple left R−module, f ∈ R00 (E) and x ∈ E. Then there exists some a ∈ R such that
f (x) = `a (x) = ax.14
Proof. As E is semisimple, then we have E = Rx ⊗ N for some submodule N . Let π : E → E be defined by
π(rx + n) = rx. Then π is R−linear, meaning that π ∈ R0 (E) so that f (x) = f (π(x)) = π(f (x)) ∈ Rx.
Theorem 74 (Jacobsen Density Theorem). Let R be a ring, and E a semisimple R−module. Let f ∈ R00 (E) and
x1 , . . . , xn ∈ E. Then there exists some a ∈ R such that f (xi ) = `a (xi ) for all i.
Proof. As we noted before, f (n) ∈ R00 (E) and E n is semisimple. So let x = (x1 , . . . , xn ) ∈ E n . By Lemma 73, there is
some a ∈ R such that f (n) (x) = `a (x). Rewriting this gives (f (x1 ), . . . , f (xn )) = f (n) (x) = `a (x) = ax = (ax1 , . . . , axn ).
Thus, f (xi ) = `a (xi ) for each i = 1, . . . , n.
12 The statement in the homework was: Let R , . . . , R be rings and R = R × · · · × R . Prove that R is left semisimple if and only
n
n
1
1
if Ri is left semisimple for i = 1, . . . , n.
13 Tom Marley seemed unsure if this was true or not, but didn’t claim it was an open question.
14 Note here that a will depend on x.
16
Corollary 75. If R is a ring, E is a semisimple R−module, and E is finitely generated over R0 (E), then λ : R → R00 (E)
given by a 7−→ `a is surjective.
Proof. Let E = R0 x1 + . . . + R0 xn and let f ∈ R00 (E). Then the Jacobsen Density Theorem (Theorem 74) says that
there is some a ∈ R such that f (xi ) = `a (xi ) for each i. Then since f and `a are R0 −linear, we get that f = `a = λ(a)
so that λ is surjective.
Corollary 76. If R is a simisimple ring, and E is a free left R−module, then λ : R → R00 (E) is an isomorphism.
Proof. As R is semisimple, so is E. Let e ∈ E be part of an R−basis for E. Then E = R0 e since we can send a basis element
anywhere. Hence E is finitely generated over R0 and so by Corollary 75, we have that λ is surjective. If a ∈ ker λ then
`a (E) = 0 so that ae = 0. This means that a = 0 as e is in a basis for E. Hence, λ is injective and thus an isomorphism.
Example 77. Let A be the Weyl algebra (the long example from homework #1). Recall that A is simple but not
left or right Artinian. Let E be any simple left A−module and consider the map λ : A → A00 (E). As A is simple,
and ker λ is an ideal, then λ is injective. Since E is simple, we get that E = Ax for any x ∈ E \ {0} so that E is a
finitely generated A−module. If we were to have that E was finitely generated as an A0 −module, then the first corollary
implies that λ is surjective. This then gives that A ∼
= A00 (E) = EndA0 (E). Note that by Schur’s Lemma (Lemma 58),
0
A is a division ring since E is simple. Then E = (A0 )n so that A ∼
= EndA0 ((A0 )n ) ∼
= Mn ((A0 )op ) which means that
A is semisimple. This is a contradiction, and so we conclude that E is not finitely generated as an A0 −module.
Remark 78. The arguments above give a different way to prove all but the uniqueness of the Artin-Wedderburn Theorem.
2.12
Monday 6 February 2012
2.12.1
End of proof of Artin-Wedderburn
Corollary 79. Let D be a division ring, and V be a D−module. Then λ : D → EndD0 (V ) is an isomorphism, where
D0 = EndD (V ).15
Proof. V is free and semisimple because D is a division ring, and so the hypotheses of Corollary 76 are satisfied and
hence λ is an isomorphism.
Corollary 80. Let D1 and D2 be division rings. Suppose that Mn1 (D1 ) ∼
= Mn2 (D2 ). Then n1 = n1 and D1 ∼
= D2 .
op
Proof. We’ll first show D1 ∼
= D2 . Recall from Example 64 that Mni (Di ) ∼
= EndDiop ((Di )n ) for i = 1, 2. So we have that
op n ∼
op n
EndD1op ((D1 ) ) = EndD2op ((D2 ) ). If we can show that for any division rings D1 and D2 and natural numbers n1 and
op
op
n2 we have that EndD1 (D1n1 ) ∼
= EndD2 (D2n2 ) implies that D1 ∼
= D2 then we’ll know that D1 ∼
= D2 and hence D1 ∼
= D2 .
n1
So let R = EndD1 (D1 ). Note that R is simple by Schur’s Lemma (Lemma 58), and semisimple by part 2 of Example
53. Also, D1n1 and D2n2 are simple left R−modules since any single element of either can be extended to give a basis, and
basis elements can be sent anywhere by endomorphisms. Now, since R is Artinian and simple, then it has precisely one
simple module, up to isomorphism, and so D1n1 ∼
= EndR (D2n2 ) ∼
= D2 .
= D2n2 as R−modules. Hence, D1 ∼
= EndR (D1n1 ) ∼
∼
Now, we have that Mn1 (D1 ) = Mn2 (D1 ) but that holds only if n1 = n2 , which completes the proof.
Lemma 81. Suppose φ : A1 × . . . × Ak → B1 × . . . × B` is an isomorphism where Ai and Bj are simple rings for
each i and j. Then k = ` and after reordering Ai ∼
= Bi for each i.
Proof. Let I1 = A1 × (0) × . . . × (0). Then φ(I1 ) is an ideal of B1 × . . . × B` . So after rearranging, we get
φ(I1 ) ∼
= B1 × . . . × Bt × (0) × . . . × (0). As rings, we have A1 ∼
= φ(I1 ), so φ(I1 ) has exactly two ideals, and hence t = 1.
Now, by modding out by I1 and φ(I1 ) respectively, we have that A2 × . . . × Ak ∼
= B2 × . . . × B` and using induction
on this argument completes the proof.
Corollary 82 (Uniqueness of Artin Wedderburn Theorem). If Mn1 (D1 ) × . . . × Mnk (Dk ) ∼
= Mm1 (D10 ) × . . . ×
Mm` (D`0 ) where Di and Di0 are division rings, and ni and mi are positive integers. Then k = `, and (after rearranging)
we have Di ∼
= Di0 , and ni = mi for each i.
2.12.2
More about λ : R → R00
Definition 83. Let φ : R → S be a ring homomorphism such that φ(R) ⊆ Z(S) = {s ∈ S | sx = xs for all x ∈ S}.
then S is an R-algebra. We say S is a finitely generated R−algebra if S is finitely generated as a ring over φ(R).
Remark 84. We frequently assume (by modding out by ker φ), that R ⊆ S and S = R[u1 , . . . , un ] for some ui ∈ S,
but that the ui ’s do not commute.
Remark 85. Let R be a ring, and E an R−module.
15 Recall
that this map λ is given by λ(a) = `a and `a is left multiplication by a.
17
1. If r ∈ Z(R), then `r ∈ R0 (E) since `r (ax) = r(ax) = a(rx) = a`r (x). Thus, Z(R) · `1 ⊆ R0 (E).
2. If E is finitely generated as a Z(R)−module, then E is finitely generated as an R0 (E)−module.
Proof. Let E = Z(R)u1 + . . . + Z(R)un for some u1 , . . . , un ∈ E. Then, we have
E
=
=
⊆
⊆
Z(R)u1 + . . . + Z(R)un
Z(R)`1 u1 + . . . + Z(R)`1 un
R0 u1 + . . . + R0 un
E
since E is also an R0 = R0 (E) module.
Proposition 86. Suppose R is finitely generated as a Z(R)−module, and that E is a finitely generated, semisimple
R−module. Then λ : R → R00 (E) is surjective.
P
Proof. We have R = Z(R)u1 + . . . + Z(R)un and E = Rw1 + . . . + Rwm . Then E =
Z(R)ui wj , so that E is
finitely generated as a Z(R)−module. By the second remark in Remark 85, we have that E is finitely generated as
an R0 (E)−module. Corollary 75 then tells us that λ is surjective.
Corollary 87. Let R be a finite dimensional k−algebra, where k is a field. That is, k ⊆ Z(R) and dimk (R) < ∞.
Also, let E be a finitely generated, semisimple R−module. Then λ : R → R00 (E) is surjective.
Theorem 88 (Burnside). Let k be an algebraically closed field, and V a finite dimensional k−vector space. Let R
be a subring of Endk (V ) with k ⊆ R.16 If V is a simple R−module, then k = EndR (V ) and R = Endk (V ).
Proof. As k ⊆ Z(R), then k ⊆ EndR (V ) ⊆ Endk (V ). Since dimk Endk (V ) < ∞, then dimk EndR (V ) < ∞. Since V
is a simple R−module, then EndR (V ) is a division ring. Let α ∈ EndR (V ), and consider the (commutative) ring k[α].
As k[α] ⊆ EndR (V ), then k[α] is a domain because by Schur’s Lemma (Lemma 58), we have that EndR (V ) is a division
ring since V is simple. Also, k[α] is finite dimensional over k since EndR (V ) is a finite dimensional k−vector space.
Hence, k[α] is a field, and algebraic over k, so we get α ∈ k and hence k = EndR (V ). Next, since V is simple, then
V is also semisimple. Because, V is a finitely generated R0 −vector space (since R0 = k), we have by Corollary 75, that
λ : R → R00 (V ) = EndR0 (V ) = Endk (V ) is surjective. Also, λ is the inclusion map, so we must have R = Endk (V ).17
2.13
Wednesday 8 February 2012
2.13.1
Finishing off information about λ : R → R00
Proposition 89. Suppose that E is an R−module and λ : R → R00 (E) is an isomorphism. Then Z(R0 ) = Z(R)`1 .
Proof. We’ve already seen that Z(R)`1 = {`r | r ∈ Z(R)} ⊆ R0 (E) in the first part of Remark 85. Let f ∈ R0 and
r ∈ Z(R). Then for any e ∈ E, we have (f ◦ `r )(e) = f (re) = rf (e) = (`r ◦ f )(e), so that f `r = `r f and `r ∈ Z(R0 ) and
hence Z(R)`1 ⊆ Z(R0 ). Now, if g ∈ Z(R0 ), then for all f ∈ R0 , we have f (g(e)) = g(f (e)). Equivalently, g(f · e) = f · g(e)
so that g is R0 −linear and g ∈ R00 = λ(R) = {`r | r ∈ R}. Hence, g = `r for some r ∈ R. We must now show that
r ∈ Z(R). We know that for all s ∈ R, that rs · e = g(s · e) = s · g(e) = sr · e for all e ∈ E. Thus, (rs − sr)E = 0 so
that `rs−sr = 0. As λ is an isomorphism, then rs − sr = 0 so that r ∈ Z(R).
Corollary 90. If D is a division ring, and V is a D−vector space, then by setting R = D and V = E, we then have
Z(Mn (D)) = Z(D)In = {rIn | r ∈ Z(D)}. Also, Mn (D) ∼
= EndDop (V ) where V = (Dop )n .
2.13.2
The Jacobsen Radical
Remark 91. Let R be a ring. Every proper left ideal of R is contained in a maximal left ideal of R. (The proof of
this is by Zorn’s Lemma).
T
Definition 92. Let R be a ring. The Jacobsen radical of R, denoted J(R), is J(R) = m where the intersection is
taken over all maximal left ideals of R.
Lemma 93. Let R be a ring, and y ∈ R. The following are equivalent:
1. y ∈ J(R),
2. 1 − xy is left invertible for every x ∈ R, and
3. yM = 0 for every simple left R−module, M .
16 Here,
17 Some
we are identifying an element, a, of k with the endomorphism given by multiplication by a.
of this is from Wednesday’s class, but it is nicer not to break up the proof in the notes.
18
Proof. (1 ⇒ 2) Let x ∈ R. Suppose that 1 − xy is not left invertible. Then R(1 − xy) is a proper left ideal. Let m
be a maximal left ideal containing R(1 − xy). Then 1 − xy ∈ m. But y ∈ J(R) implies that y ∈ m so that xy ∈ m
and hence 1 ∈ m which is a contradiction and so we must have that 1 − xy is left inevertible.
(2 ⇒ 3) Let M be a simple left R−module, and suppose yM 6= 0. Then there is some u ∈ M such that yu 6= 0. Since M
is simple, then Ryu = M , and there is some r ∈ R such that ryu = u. Thus, (1 − ry)u = 0 but as 1 − ry is left invertible
by hypothesis, then we must have that u = 0. This is a contradiction as we couldn’t have had u = 0 and so yM = 0.
(3 ⇒ 1) Let m be a maximal left ideal. Then R/m is simple and by assumption we have yR/m = 0 so we must have
y ∈ m and so y ∈ J(R) as m was an arbitrary choice of maximal left ideal of R.
Definition 94. Let M be a left R−module. The annihilator of M is AnnR (M ) = {r ∈ R | rM = 0}.
Remark 95. 1. AnnR (M ) is always a (2-sided) ideal. This is because if r ∈ AnnR (M ) and s ∈ R, then (rs)M =
r(sM ) ⊆ rM = 0, and (sr)M = s(rM ) = s · 0 = 0.
2. Warning!!! If x ∈ M , then AnnR (x) = {r ∈ R | rx = 0} is a left ideal, but it is not necessarily a 2-sided ideal.
T
Corollary 96. Let R be a ring. Then J(R) = AnnR (M ) where the the intersection is taken over all simple left
R−modules, M . Hence, J(R) is a 2−sided ideal.
Proof. This is from 1 ⇔ 3 in the lemma.
Proposition 97. Let R be a ring, and y ∈ R. The following are equivalent:
1. y ∈ J(R), and
2. 1 − xyz is a unit for all x, z ∈ R.
Proof. (1 ⇒ 2) Let y ∈ J(R), and x, z ∈ R. Then yz ∈ J(R) so 1 − xyz is left invertible by Lemma 93. Let u be a
left inverse of 1 − xyz, so that u is right invertible. We have 1 = u(1 − xyz) = u − uxyz so that u = 1 − (−u)(xyz).
Since xyz ∈ J(R), then u is left invertible. Thus, u is a unit and u−1 = 1 − xyz is also a unit.
(2 ⇒ 1) Let z = 1. Then 1 − xy is a unit and hence left invertible for all x ∈ R, so that y ∈ J(R) by Lemma 93
T
Corollary 98. J(R) = m where this time the intersection is over all maximal right ideals, m, of R.
2.14
Friday 10 February 2012
2.14.1
Semiprimitive Rings
Definition 99. A ring R is called semiprimitive if J(R) = 0. Apparently some people call this Jacobsen-semisimple,
or simply J-semisimple.18
Example 100. 1. Any semisimple ring is semiprimitive.
Proof. Let R be a semisimple ring. Then we know R = I1 ⊕ . . . ⊕ In where each Ii is a simple left ideal of R. This
means we can write 1 = u1 + . . . + un where ui ∈ Ii for each i. For any y ∈ J(R) we then have yui = 0 for all i
since Ii is a simple left R−module. Thus, y · 1 = yu1 + . . . + yun = 0 so that y = 0, and hence J(R) = 0.
2. Any simple ring is semiprimitive because J(R) cannot equal R.
3. Z is semiprimitive since the maximal ideals are {pZ | p prime}.
4. If F is a field, then F [x] is semiprimitive because the maximal ideals are generated by prime elements, and there
are infinitely many of them so that only the 0 polynomial can divide all of them.
Proposition 101. Let R be a ring. The following are equivalent:
1. R is semisimple, and
2. R is left Artinian and J(R) = 0.
Proof. (1 ⇒ 2) We’ve already seen that semisimple rings are left Artinian from Theorem 67, and semiprimitive from
the first part of Example 100.
(2 ⇒ 1) To show this, we start by claiming that any simple left ideal, I, of R is a direct summand of R. Since I 6= 0
we know that I 6⊂ J(R) = 0. Thus, there is a maximal left ideal m such that I 6⊂ m. By the maximality of m, we get
I + m = R, and I ∩ m = (0) since I is simple. Thus, R = I ⊕ m as desired. Now, since R is left Artinian, then any
nonzero left ideal, L, contains a simple left ideal. This is easily seen by choosing a minimal element of the nonempty set
Λ = {nonzero left ideals contained in L}. Let R1 be a simple left ideal of R. By what we’ve already shown, we have
that R = I1 ⊕ A1 for some left ideal A1 . If A1 = (0), then we’re done. So we assume A1 =
6 (0). By the same reasoning, A1
contains a simple left ideal, I2 . We have R = I2 ⊕A2 and now, A1 = (A1 ∩R) = A1 ∩(I2 ⊕A2 ) = I2 ⊕(A1 ∩A2 ) so that R =
I1 ⊕I2 ⊕(A1 ∩A2 ). This gives a descending sequence of left ideals: A1 ⊇ A1 ∩A2 ⊇ . . . and since R is Artinian, this sequence
must stabilize, and so we must eventually have A1 ∩ . . . ∩ An = 0. Hence, R = I1 ⊕ . . . ⊕ In and R is hence semisimple.
18 No
pun intended.
19
2.14.2
Nilpotence
Definition 102. Let R be a ring, and I a left ideal. We say I is nil if every element of I is nilpotent. We say I is
nilpotent if I n = 0 for some n ∈ N.19
Remark 103. Nilpotent left ideals are nil, but nil left ideals are not necessarily nilpotent.
Example 104. Let k be a field, and set R = k[x1 , x2 , x3 , . . .]/(x21 , x32 , x43 , . . .). Then the ideal (x1 , x2 , . . .) is nil, but
not nilpotent.
Remark 105. In the commutative case, the sum of two nilpotent elements is nilpotent. Indeed, if xn = y m = 0,
then (x + y)m+n−1 = 0.20 This tells us that the set of nilpotent
elements form
is
an ideal.
However, this fails
if R 0 0
0 1
0
0
non-commutative. Indeed in M2 (Z) we have that if A =
and B =
, then A2 = B 2 =
.
1 0
0 0
0 0
0 1
1 0
However, A + B is a unit. Indeed, A + B =
and (A + B)2 =
. This is because the left ideal
1 0
0 1
generated by A contains non-nilpotent elements.
Exercise. A finite sum of nilpotent left ideals is nilpotent.
Conjecture. The sum of two nil left ideals is nil.
2.14.3
Wedderburn Radical
Definition 106. Let R be a ring. The Wedderburn radical of R (if it exists), denoted W (R), is defined to be the
largest nilpotent left ideal of R.21
Proposition 107. Every nil left ideal is contained in J(R).
Proof. Let I be a nil left ideal, y ∈ I, and x ∈ R. Then xy ∈ I and hence is nilpotent. Note that if u is nilpotent with
un = 0, then (1−u)−1 = 1+u+u2 +. . .+un−1 . Thus, since xy is nilpotent, then 1−xy is a unit and hence y ∈ J(R).
Corollary 108. It is immediate that when W (R) exists, it is contained in J(R).
Theorem 109. If R is a left (or right) Artinain ring, then J(R) is nilpotent.
Proof. Let J = J(R) and consider the descending chain J ⊇ J 2 ⊇ . . .. Since R is left Artinian, there exists some k
so that J k = J k+1 = I. If we can show that I = 0, then we’ll have that J(R) is nilpotent. Suppose not. Then consider
the set Λ = {L | L is a left ideal and IL 6= 0}. As we’re assuming I 6= 0, then R ∈ Λ so that Λ 6= ∅. Since R is Artinian,
there is some L ∈ Λ which is minimal. Since we have IL 6= 0 then there exists some y ∈ L such that Iy 6= 0. As Iy is a
left ideal, we must have Iy ⊆ L, since I(Iy) = I 2 y = Iy 6= 0. Hence, Iy ∈ Λ and Iy ⊆ L which means by the minimality
of L that Iy = L. Now, we get that y = iy for some i ∈ I. Thus, (1 − i)y = 0 but i ∈ I ⊆ J so that 1 − i is a unit,
and hence y = 0. This is a contradiction as we couldn’t have had that y = 0 and so we must have that I = 0.
Corollary 110. It is immediate that for left (or right) Artinian rings, R, then W (R) exists.
2.15
Monday 13 February 2012
2.15.1
Artinian rings are Noetherain
Lemma 111. Let R be a semisimple ring, and M an R−module. The following are equivalent:
1. M is finitely generated,
2. M is Artinian,
3. M is Noetherian, and
4. λR (M ) < ∞.
Proof. Since free modules over a semisimple ring are semisimple and quotients of semisimple
modules are semisimple,
L
then any module over a semisimple ring is semisimple. Hence, we have that M = j∈Λ Ij where each Ij is simple.
Note that |Λ| < ∞ if and only if M has a composition series, which is true if and only if λR (M ) < ∞ which is true
if and only if M is both Noetherian and Artinian. Also, as R is both Noetherian and Artinian, then if M is finitely
generated, we have that M is both Noetherian and Artinian.
19 Recall
P
here that I n is the left ideal generated by product of n elements from I, i.e. element of the form m
j=1 aj1 . . . ajn where aj i ∈ I.
easily seen by writing out the product using the binomial coefficients since each term of the sum will have either xn or
20 This can be
y m as a factor.
21 That
is, it contains all nilpotent left ideals, and W (R) is nilpotent.
20
Remark 112. If M is an R−module, and I is an ideal of R such that IM = 0, then M is naturally an R/I−module
by setting rm = rm for any r ∈ R/I.
Note. If I is an ideal, and I ⊆ J(R), then J(R/I) = J(R)/I. In particular, J(R/J(R)) = 0.
Theorem 113. If R is a left Artinian ring, then R is left Noetherian.
Proof. Let J = J(R). Since R is left Artinian, then J n = 0 for some n. Also, J(R/J) = 0, and R/J is left Artinian.
Thus, R/J is semisimple by Proposition 101. If λR (R/J i ) < ∞ for all i, then λR (R R) = λR (R/0) = λR (R/J n ) < ∞
so that R is left Noetherian as well. We’ll prove now that λR (R/J i ) < ∞ for all i by induction. Note that since
R/J is Artinian, then by Lemma 111, we have λR (R/J) < ∞ which takes care of the base case. Suppose now that
λR (R/J i−1 ) < ∞ for some i > 1, and consider the module J i−1 /J i . Since J i−1 is an ideal of R, then it is left Artinian as
an R−module and hence J i−1 /J i is left Artinian. Also, J i−1 /J i is an R/J−module and is Artinian as an R/J−module.
By Lemma 111, we then have that λR (J i−1 /J i ) = λR/J (J i−1 /J i ) < ∞. Now, consider the short exact sequence
0 → J i−1 /J i → R/J i → R/J i−1 → 0
where the maps are the obvious choices. Then λR (R/J i ) = λR (J i−1 /J i ) + λR (R/J i−1 ) < ∞ and hence we get the
desired result.
2.15.2
Commutative Algebra
Definition 114. Let R be a commutative ring. An ideal P 6= R is prime if whenever ab ∈ P then a ∈ P or b ∈ P .
Equivalently, P is prime if and only if R/P is a domain.
/ S ⊂ R. We say S is multiplicatively closed if s1 s2 ∈ S whenever
Definition 115. Let R be a commutative ring, and 0 ∈
s1 , s2 ∈ S. For convenience, we usually assume 1 ∈ S.
Example 116. There are two main examples of multiplicatively closed sets:
1. If P is a prime ideal, then R \ P is multiplicatively closed.
2. If x ∈ R \ {0}, the S = {xn | n ≥ 0} is a multiplicatively closed set.
Proposition 117. Let R be a commutative ring, S 6= ∅ a multiplicatively closed set of R with 0 ∈
/ S. Then there is
a prime ideal P of R such that P ∩ S = ∅.
Proof. Let Λ = {I | I is an ideal of R, and I ∩ S = ∅}. Note that (0) ∈ Λ so that Λ 6= ∅. By Zorn’s lemma, there
exists a maximal element of Λ, say P . We need to show that P is prime. Note that P 6= R as S 6= ∅, and suppose
that P is not prime. Then there exist elements a, b ∈ R \ P such that ab ∈ P . Then we have (P, a) = P + Ra ) P
and (P, b) ) P . Hence, (P, a) ∩ S 6= ∅ and (P, b) ∩ S 6= ∅. Let s1 = x1 + r1 a and s2 = x2 + r2 b where xi ∈ P and
ri ∈ R for i = 1, 2. Then s1 s2 = (x1 + r1 a)(x2 + r2 b) = x1 x2 + x1 r2 b + x2 r1 a + r1 r2 ab ∈ P so that P ∩ S 6= ∅. This
is a contradiction and so we must have that P is prime.
p
n
Theorem 118
T (Krull). Let R be a commutative ring. Let Nilrad(R) = {r ∈ R | r = 0 for some n} = (0). Then,
Nilrad(R) = P where the intersection is taken over all prime ideals of R.
p
Proof. (⊆) Let r ∈ p
(0) and P be a prime ideal. Then rn = 0 ∈ P for some n and hence r ∈ P as P is prime.
(⊇) Suppose that r ∈
/ (0), and let S = {rn | n ≥ 0} so that 0 ∈
/ S and S is multiplicatively closed. Then by Proposition
117, there exists some prime ideal P such that P ∩ S = ∅ and hence r ∈
/ P.
2.16
2.16.1
Wednesday 15 February 2012
History and Applications of Artin-Wedderburn
Note. We discussed some history of algebra in class today, but it seems not worth typing up. The dates in question
were 1843-1965, but as was said in class...“History goes on.” –T.M.
Theorem 119. Let R be a simple, left Artinian ring. Then R ∼
= Mn (D) for some division ring D.
Proof. Let M be a simple left R−module, and let D = R0 (M ) = EndR (M ) which is a division ring. Now, consider the map
λ : R → R00 (E) = EndD (M ). It is clear that ker(λ) = AnnR (M ) which is a 2-sided ideal and so ker(λ) = (0). If we can
show that M is finitely generated over D, then we’ll have that λ is surjective by Corollary 75. Suppose that dimD (M ) = ∞.
Then choose {e1 , e2 , . . .} to be a countable D−linearly independent set in M . Let In = {r ∈ R | rei = 0 for i = 1, . . . , n}
and note that In is a left ideal of R with In ⊇ In+1 for each n. Also, for each n, there exists an f ∈ EndD (M ) such that
f (ei ) = 0 for i = 1, . . . , n but f (en+1 ) 6= 0. By the Jacobsen Density Theorem (Theorem 74), there is some r ∈ R such that
f (ei ) = rei for i = 1, . . . , n+1. This gives that r ∈ In \In+1 so that the containment is always strict. This violates the fact
that R is left Artinian, and so we must have that dimD (M ) < ∞. This means that M ∼
= Dn for some n and λ is an isomorn ∼
op
op
∼
phism. Hence, R = EndD (D ) = Mn (D ) and since D is a division ring whenever D is, that completes the proof.
21
Theorem 120 (Special Case of Wedderburn’s Theorem). Let R be a finite dimensional k−algebra, where k is
an algebraically closed field. Then R is semisimple if and only if R ∼
= Mn1 (k) × . . . × Mn` (k).
Proof. (⇐) This direction is clear by Example 53.
(⇒) Since R is semisimple, then by Artin-Wedderburn (Theorem 55) we have that R ∼
= Mn1 (D1 ) × . . . × Mn` (D` ), where
each Di = EndR (Ei ) where E1 , . . . , E` are the distinct up to isomorphism left simple modules of R. Since dimk (R) < ∞,
then Ei = Rxi for any xi ∈ Ei \ {0} so that dimk (Ei ) < ∞. Also, it is clear that Di = EndR (Ei ) ⊆ Endk (Ei ) ∼
= Mn (k)
where n = dimk (Ei ). Thus, dimk (Di ) < ∞ for all i. Let αi ∈ Di . Then k[αi ] is a finite dimensional k−vector space,
and also a commutative domain. Hence, k[αi ] = k since k is algebraically closed, and so Di = k for all i.
2.16.2
Commutative Algebra
Remark 121. For the time being, R will be a commutative ring.
Definition 122. Let I be an ideal of R. A prime ideal P of R is said to be minimal over I if I ⊆ P and whenever
Q is also prime with I ⊆ Q ⊆ P then P = Q.
Remark 123. By Zorn’s lemma, every ideal I 6= R has a prime minimal over it.
Notation. We set Spec(R) = {P | P is a prime ideal of R} and minR (R/I) = {P ∈ Spec(R) | P is minimal over I}.
Definition 124. If P ∈ Spec(R), then the height of P is
ht(P ) = sup{n | there exists a chain of primes P = Pn ) Pn−1 ) . . . ) P0 }.
Also, the dimension of a ring R is dim R = sup{ht(P ) | P ∈ Spec(R)}.
Example 125. Let F be a field.
1. dim(F ) = 0 since Spec(F ) = {(0)}.
2. dim(F [x]) = 1 since Spec(F [x]) = {(0), (f ) | f is irreducible }.
3. dim(Z) = 1 since Spec(Z) = {(0), (p) | p is prime }.
4. dim(Z/(6)) = 0 since Spec(Z/(6)) = {(2), (3)}.
Remark 126. We have dim(R) = 0 if and only if every prime of R is minimal over (0) which is true if and only if
every prime of R is maximal over (0).
2.17
2.17.1
Friday 17 February 2012
More Commutative Algebra
√
Definition 127. If I is an ideal, then the radical of I is I = {r ∈ R | rn ∈ I for some n}.
√
T
Exercise. I = P where the intersection is taken over all primes P ∈ minR (R/I).
Fact. If R is Noetherian, then ht(P ) < ∞ for all P ∈ Spec(R), however, there exist Noetherian rings with dim(R) = ∞.
Exercise. If P ∈ Spec(R), and P ⊇ I1 I2 . . . In , then P ⊇ Ij for some j.
Proposition 128. Let R be Noetherian. Then minR (R/I) is a finite set for every ideal I of R.
Proof. Suppose not, and consider the set Λ = {I | minR (R/I) is an infinite set}. Since R is Noetherian, and
λ 6= ∅, then there exists a maximal element of Λ, say I. Since minR (R/I) is an infinite set, then P ∈
/ Spec(R).
Thus, there exist some a, b ∈ R \ I such that ab ∈ I. Let J1 = (I, a) = I + Ra and J2 = (I, b). Note that
J1 , J2 ) I. Thus, by the maximality of I, we have that minR (R/Ji ) is finite for i = 1, 2. But J1 J2 ⊆ I since
(i1 + r1 a)(i2 + r2 b) = i1 i2 + r1 ai2 + r2 bi1 + r1 r1 ab ∈ I for all i1 , i2 ∈ I and r1 , r2 ∈ R. Let P ∈ Spec(R). Then P ⊇ I
if and only if P ⊇ J1 or P ⊇ J2 so minR (R/I) ⊆ minR (R/J1 ) ∪ minR (R/J2 ) which is a contradiction, and so we must
have that minR (R/I) is a finite set for every ideal I of R.
Theorem 129. Let R be a commutative ring. The following are equivalent:
1. R is Artinian, and
2. R is Noetherian with dim(R) = 0.
Proof. (1 ⇒ 2) We’ve already shown that since R is Artinian, then R is Noetherian in Theorem 113. So we claim
first that R has only finitely many maximal ideals. If not, then we can let {m1 , m2 , . . .} be an infinite set of distinct
maximal ideals of R. By considering the descending chain m1 ⊇ (m1 ∩ m2 ) ⊇ (m1 ∩ m2 ∩ m3 ) ⊇ . . . we see that
m1 ∩ . . . ∩ mi = m1 ∩ . . . ∩ mi+1 for some i. Thus, mi+1 ⊇ m1 ∩ . . . ∩ mi ⊇ m1 m2 . . . mi . Since mi+1 is maximal, then it
is also prime, and so for some j ∈ {1, . . . , i} we have mj ⊆ mi+1 by maximality of mj this means that mj = mi+1 which
22
contradicts the choice of the set of maximal ideals. Thus, R has only finitely many maximal ideals. Now, T
we’ve shown in
t
Theorem 109 that since R is Artinian, then J(R) is nilpotent, say J(R)n = 0. Also, we have that J(R) = i=1 mi where
t
n n
n
t
n
n
{mi }i=1 is all of the maximal ideals of R. Then, m1 m2 . . . mt ⊆ ∩i=1 mi = J(R) = 0, and so for any P ∈ Spec(R),
we have P ⊇ mn1 . . . mnt and hence P ⊇ mr for some r. Thus, every prime is maximal and so dim(R) = 0.
(2 ⇒ 1) As R is Noetherian,
p then by Theorem 128, we have Spec(R) = minR (R) = minR (R/(0)) = {P1 , . . . , Pt }.
p Thus,
J = J(R) = ∩ti=1 Pi = (0).
Since
R
is
Noetherian,
then
J
is
finitely
generated,
and
each
element
of
J(R)
=
(0) is
p
nilpotent, by definition of (0). Then because J is finitely generated, we can let n be the maximal power needed to get
each generator of J to be zero, and then J n = 0 follows so that J is nilpotent. We now claim that λR (R/J i ) < ∞ for all i,
which would imply that R is Artinian. So, when i = 1, we have that R/J = R/(P1 ∩ . . . ∩ Pt ) = R/P1 × . . . × R/Pt by the
Chinese Remainder Theorem since mi + mj = R for all i 6= j. Thus, R/J is semisimple by Artin-Wedderburn, and hence
λR/J (R/J) = λR (R/J) < ∞ by Proposition 56. For the inductive step, suppose that λR (R/J i−1 ) < ∞. Then J i−1 /J i is
an R/J−module, and as R is Noetherian, then J i−1 is Noetherian, and hence J i−1 /J i is Noetherian as an R/J−module.
Since R/J is semisimple, then λR (J i−1 /J i ) = λR/J (J i−1 /J i ) < ∞ by Lemma 111. Then, the short exact sequence:
0 → J i−1 /J i → R/J i → R/J i−1 → 0
gives us that λR (R/J i ) = λR (J i−1 /J i ) + λR (R/J i−1 ) < ∞. Thus, λR (R) = λR (R/(0)) = λR (R/J n ) < ∞ by
Proposition 42.
Note. If R is any commutative ring, then Spec(R/I) = {P/I | P ∈ Spec(R), P ⊇ I}. Also, P/I = Q/I if and only
if P = Q.
Example 130. Let k be a field.
1. If R = k[x, y, z]/(x3 , y 3 , z 3 ) then R is Notherian by the Hilbert Basis Theorem. Also, Spec(R) = {(x, y, z)} so
dim(R) = 0, and R is Artinian.
2. If R = Z[x]/(12, x3 − x), then R is Noetherian by the Hilbert Basis Theorem. Also,
(2, x)
(3, x)
(2, x − 1)
(3, x − 1)
(2, x + 1)
(3, x + 1)
Spec(R) =
,
,
,
,
,
(12, x3 − x) (12, x3 − x) (12, x3 − x) (12, x3 − x) (12, x3 − x) (12, x3 − x)
and so dim(R) = 0 and R is Artinian.
2.18
Wednesday 22 February 2012
2.18.1
Modules over Artinian ring
Proposition 131. 22 Let R be a left Artinian ring, and M a left R−module. The following are equivalent:
1. M is finitely generated,
2. M is Noetherian,
3. M is Artinian,
4. λR (M ) < ∞.
Proof. (4 ⇒ 2) This is true by Proposition 42.
(2 ⇒ 1) This is clear by definition of Noetherian modules.23
(1 ⇒ 3) Finitely generated modules over an Artinian ring are all Artinian, since they’re isomorphic to a quotient of
a free module.
(3 ⇒ 4) Let J = J(R). Then J n = 0 for some n by Theorem 109 since R is Artinian. Also, R/J is semisimple by
the same argument as in part two of the proof of Theorem 129. We then claim that λR (M/J i M ) < ∞ for all i ≥ 1.
For the base case, we know that M/JM is an R/J−module and since M is Artinian, then M/JM is Artinian and
so by we have that λR (M/JM ) = λR/J (M/JM ) < ∞ by Lemma 111. So consider the R/J−module J i−1 M/J i M .
This module is also Artinian, and so λR/J (J i−1 M/J i M ) < ∞ by Lemma 111. Then the additivity of length on exact
sequences gives that λR (M/J i M ) < ∞ for all i since 0 → J i−1 M/J i M → M/J i M → M/J i−1 M → 0 is an exact
sequence. Thus, λR (M ) = λR (M/(0)) = λR (M/J n M ) < ∞.
22 Tom
23 Or
was sick on Monday 20 February 2012.
by Proposition 3 if you’re so inclined.
23
Chapter 3
Exam 2 Material
3.1
3.1.1
Wednesday 22 February 2012
Split Exact Sequences
Proposition 132. Let R be a ring and
0
f
g
A
B
C
0
a short exact sequence of left R−modules. The following are equivalent:
1. There is an R−module homomorphism j : C → B such that gj = 1C .
2. There is an R−module homomorphism i : B → A such that if = 1A .
3. There are R−module homomorphisms i : B → A and j : C → B such that 1B = f i + jg.
Proof. First note that gf = 0 since the sequence is exact. Also, recall that when g is surjective, then αg = βg implies
that α = β. This is a consequence of surjective maps being epimorphisms in the category of sets. Also, when f is
injective then f α = f β implies that α = β because injective maps are monomorphisms in the category of sets. For
a reminder of what epimorphisms and monomorphisms are, see 901 notes page 2.
(3 ⇒ 1) By the hypothesis here, we have that 1B = f i + jg. Composing with g on the left (i.e. as the last map applied)
gives 1C g = g = gf i + gjg = gjg since gf = 0. Hence we get that 1C = gj because g is surjective.
(1 ⇒ 2) Let K = ker(g) = image(f ) ∼
= A. Also, let ρ : K → A be an isomorphism given by ρ(k) = f −1 (k). Also, let b ∈ B
and note that g(b − jg(b)) = g(b) − gjg(b) = g(b) − g(b) = 0 so b − jg(b) ∈ K. Now, 1B − jg : B → K is an R−module
homomorphism. Let i : B → A be defined by ρ ◦ (1B − jg). Then, if = ρ(1 − jg)f = ρf − ρjgf = ρf = 1A since gf = 0.
(2 ⇒ 3) Define j : C → B by j(g(b)) = b − f i(b). Since g is surjective, then every element c ∈ C is of the form
g(b) for some b ∈ B, so this definition actually makes sense as long as we can show that it is well defined. So
suppose that g(b) = g(b0 ). Then, b − b0 ∈ ker(g) = image(f ), and we can choose α ∈ A so that b − b0 = f (α). Then
j(g(b)) − j(g(b0 )) = g(b − b0 ) = b − b0 − f i(b − b0 ) = b − b0 − f i(f (α) = b − b0 f (α) = 0 since if = 1A and hence j is
well defined. Now, (f i + jg)(b) = f i(b) + jg(b) = f i(b) + b − f i(b) = b so that we have f i + jg = 1B as desired.
Definition 133. A short exact sequence is split if it satisfies the equivalent conditions in the previous proposition.
Moreover, when this happens, B ∼
= A ⊕ C.
3.2
3.2.1
Friday 24 February 2012
Projective Modules
Definition 134. An R−module, P , is called projective if given any diagram of the form below (with exact row),
P
h
g
f
M
0
N
then there exists a map h : P → M such that the diagram commutes.
Definition 135. An R−module, E, is called injective if given any diagram of the form below (with exact row),
E
h
g
0
f
M
24
N
then there exists a map h : N → E such that the diagram commutes.
Proposition 136. Any free module is projective.
Proof. Let F be a free R−module, and consider any diagram of the form below with exact row where M and N are
R−modules.
F
g
f
M
0
N
Let S be a basis for F . For each x ∈ S, choose mx ∈ M so that f (mx ) = g(x). Note that this is possible since f
is surjective. Then, define a map h : F → M by setting h(x) = mx for all x ∈ S. Then we have that f h = g by
construction, and so F is projective.
Proposition 137. Let P be an R−module. The following are equivalent:
1. P is projective, and
2. there exists an R−module, Q, such that P ⊕ Q is a free R−module.
Proof. (1 ⇒ 2) Note that any R−module is the homomorphic image of a free module. So let F be a free R−module
such that there exists a surjective map f : F → P . Let Q = ker(f ). Then the sequence 0 → Q → F → P → 0 is exact.
Since P is projective, there exists a map h : P → F such that the diagram below commutes.
0
f
Q
F
0
P
idP
h
P
∼ P ⊕ Q.
Thus, f h = idP , meaning that the sequence splits, and hence F =
(2 ⇒ 1) Now, suppose that P ⊕ Q ∼
= F where F is a free R−module. Let π : F → P be the usual projection map,
and i : P → F the usual injection so that πi = idP . Consider the diagram below where M and N are any R−modules,
and the row is exact.
F
π
i
P
g
M
f
N
0
Since F is projective, there is a map e
h : F → M such that f e
h = gπ. Thus, f e
hi = gπi = g so that the map h = e
hi
will give f h = g.
Example 138. Let R = Z/(6) ∼
= (2) ⊕ (3) so that (2) and (3) are projective R−modules, but are not free R−modules.
Exercise. Let R be a commutative ring, and I an ideal of R. Then I is free if and only if I = 0 or I = (a) for some
non-zerodivisor a ∈ R.
Example 139. Let R = Mn (D) for some division ring D, and let I be the matrices whose only nonzero entries are
in the first row. Note that I is then an ideal of R and that R = I ⊕ J where J is the matrices whose first row is
all zero. Thus, both I and J are projective, but neither is free. Note here that n = dimD (I) < dimD (R) = n2 and
n(n − 1) = dimD (J) < dimD (R) = n2 .
25
√
√
Remark 140. 1. Let R = Z[ −5] and I = (2, 1 + −5). Then I is projective, but not free.
2. A ring R is a PID if and only if every ideal is free.
3. Let R be a commutative ring, and e a nontrivial idempotent. Then R = Re ⊕ R(1 − e) so that Re and R(1 − e)
are projective but not free.
4. Let S = R[x, y, z]/(x2 + y 2 + z 2 − 1) and define φ : S 3 → S via (a, b, c) 7−→ xa + yb + zc. Let K = ker(φ). Note
that φ is surjective since φ(x, y, z) = x2 + y 2 + z 2 = 1. Then the diagram below is commutative with exact row
so that the sequence splits, and K ⊕ S ∼
= S3.
0
f
S3
K
0
S
idS
S
1
Thus K is a projective S−module, but it is not free.
5. In 1955, Serre asked if there exist non-free projective modules over k[x1 , . . . , xn ] where k is a field. The answer is
no, and was proved by Quillen-Suslin in the mid 1970’s.
3.3
Monday 27 February 2012 & Wednesday 29 February 2012
3.3.1
Exam Review
Note. For most of these two days we reviewed for the exam. I’m not typing up all that was said, but I’m not skipping
any new material.
Theorem 141. Let R be a ring, and M a semisimple module. The following are equivalent:
1. M is finitely generated,
2. λR (M ) < ∞,
3. M is Artinian,
4. M is Noetherian, and
5. M = E1 ⊕ . . . ⊕ Es where each Ei is simple.
3.4
Wednesday 29 February 2012
3.4.1
Projective Modules
Proposition 142. Let R be a ring. The following are equivalent:
1. R is semisimple, and
2. every R−module is projective.
Proof. (1 ⇒ 2) Let M be an R−module and φ : F → M a surjective module homomorphism where F is a free module.
Then the sequence below is exact where K = ker(φ).
0
K
i
φ
F
M
0
Since R is semisimple, F is also semisimple, and so F = K ⊕ N for some R−module, N . Thus, there exists a map
π : F → K such that π(x) = x for all x ∈ K. Hence, the sequence splits, F ∼
= K ⊕ M and hence M is projective.
(2 ⇒ 1) Now, let I be an ideal of R and consider the short exact sequence 0 → I → R → R/I → 0. This sequence
splits since R/I is projective, and so the map i : I → R splits, meaning that I is a direct summand of R and hence
R is semisimple.
Fact. If k is a field, then every projective module of k[x1 , . . . , xn ] is free.
Definition 143. A projective module, P , is called stably free if P ⊕ Rn ∼
= Rm for some n, m ∈ N. Here, the rank of
P is m − n.
Fact. In the early 1970s, Suslin proved the following statement: Let R be a smooth finitely generated k−algebra, where k is
an algebraically closed field of dimension d and characteristic 0. If P is a stably free projective of rank ≥ d, then P is free.2
Fact. There exist stably free projective modules of rank d − 2 which are not free. This was proved by Murthy.
Fact. In 2011, Jean Fasel3 proved something about the rank d − 1 case.4
1 This
is HARD to show!
must be that the dimension of R is d, since there are no nonzero prime ideals in a field, and we aren’t talking about vector space
dimension either. My notes are unclear on this.
3 Who interviewed recently for the tenure-track job.
4 Tom was unspecific about exactly what he proved, and I believe we were running out of time, so I just scribbled something down and left.
2 This
26
3.5
Friday 2 March 2012
3.5.1
von Neumann Regular Rings
Definition 144. A ring, R, is called von Neumann regular if for every a ∈ R there is some x ∈ R such that axa = a.
Example 145. 1. Any division ring is von Neumann regular.
2. Products of von Neumann regular rings are von Neumann regular.5
3. Arbitrary products of division rings are von Neumann regular.
6
4. If R is a von Neumann regular ring, and I is an ideal of R, then
Q R/I is von Neumann regular as well.
5. Let F be a field, and take Λ to be an infinite set. Let R = i∈Λ F , and let I = ⊕i∈Λ F . Note that I is an ideal
of R, and R is von Neumann regular, so R/I is von Neumann regular, but we showed in the homework that R/I
is not a product of fields.
Proposition 146. Let R be a ring. The following are equivalent:
1. R is von Neumann regular,
2. every finitely generated left ideal of R is generated by an idempotent, and
3. every finitely generated left ideal of R is a direct summand of R.
Proof. (2 ⇒ 3) Let I be a finitely generated left ideal of R. Then I = Re by assumption, where e2 = e. Then
R = Re ⊕ R(1 − e) = I ⊕ R(1 − e) so that I is a direct summand of R as desired.
(3 ⇒ 1) Let a ∈ R. Then by assumption, R = Ra ⊕ J for some left ideal J of R. Let 1 = ra + j for some r ∈ R and j ∈ J.
Then a = ara + aj, so that aj = a − ara ∈ Ra ∩ J and hence aj = 0 and a = ara so that R is von Neumann regular.
(1 ⇒ 2) For this direction we induct on the number of generators of I, which we’ll denote by n. For n = 1, we have that
I = Ra. Then there exists some x ∈ R such that axa = a since R is von Neumann regular. Set e = xa. We have that
e2 = xaxa = xa = e so that e is an idempotent. We clearly have Re ⊆ Ra. Also, a = axa = ae ∈ Re so Re = Ra = I
and I is generated by an idempotent. Now suppose that n > 1 and we have I = Ra1 + . . . + Ran . By induction, there
exist idempotents e1 , and e2 such that Ra1 + . . . + Ran−1 = Re1 and Ran = Re2 . So we have that I = Re1 + Re2 .
Note that I = Re1 + Re2 = Re1 + Re2 (1 − e1 ) since e2 = e2 e1 + 1e2 (1 − e1 ) and e2 (1 − e1 ) = −e2 e1 + 1e2 . Now, by
the n = 1 case, there is an idempotent f ∈ R such that Rf = Re2 (1 − e1 ). This gives that I = Re1 + Rf . Note that
f e1 = 0 since f = re2 (1 − e1 ) for some r ∈ R so that f e1 = re2 (1 − e1 )e1 = 0. Thus, f (f + e1 ) = f 2 + f e1 = f 2 = f .
Now, since f, e1 ∈ I, then R(f + e1 ) ⊆ I. Also, f = f (f + e1 ) ∈ R(f + e1 ) so that e1 = f + e1 − f ∈ R(f + e1 ) and
hence I = Re1 + Rf ⊆ R(f + e1 ) and hence we’re done by the n = 1 case.
Corollary 147. Let R be a ring. The following are equivalent:
1. R is semisimple, and
2. R is von Neumann regular and left Noetherian.
Proof. (1 ⇒ 2) We know that semisimple rings are left Noetherian, so let I be a (finitely generated)7 left ideal of R.
Then I is a direct summand of R since R is semisimple. By proposition 146, R is von Neumann regular.
(2 ⇒ 1) Let I be a left ideal of R. Then I is finitely generated since R is left Noetherian. By proposition 146, I is
a direct summand of R, so R is semisimple.
Note. Since semisimple was equivalent to semiprimitive and left Artinian (see proposition 101), then a ring is von
Neumann regular and left Noetherian if and only if it is semiprimitive and left Artinian.
3.5.2
Maschke’s Theorem
Theorem 148 (Maschke’s Theorem, 1899). Let G be a finite group, and k a field. Then R = k[G] is semisimple
if and only if the characteristic of the field does not divide the order of the group (char(k) - |G|).
Proof. (⇒) We proceed by contrapositive,
so suppose that char(k) | |G|. Letting
the identity of the field,
P
P1k denoteP
k, we have that |G| · 1k = 0. Let x = g∈G g ∈ k[G], and let h ∈ G. Then hx = g∈G hg = g∈G g = x since as sets
P
P
P
hG = G. Similarly, xh = x. Thus, x ∈ Z(k[G]). Note that x2 = x( g∈G g) = g∈G xg = g∈G x = |G| · x = 0. Thus,
Rx is a nilpotent ideal and Rx 6= 0. Hence Rx ⊆ J(R) 6= 0 which by proposition 101 means that R is not semisimple.
(⇐) Now, let I be a left ideal of R = k[G]. We need to show that I is a direct summand of R, and note that it is enough
to show that the injective map i : I → R splits. So we want to find an R−module homomorphism ρ : R → I such that
ρi = idI , that is, we need ρ(a) = a for all a ∈ I. However, since I is a k−subspace of R, then there exists a k−linear
map π : R → I which fixes I. As I is a left ideal of R, then for all g ∈ G, gπg −1 : R → I given by r 7−→ g · π(g −1 r)
is also k−linear. So define ρ : R → I by
1 X
ρ=
gπg −1 .
|G|
g∈G
5 This
is shown by doing the obvious thing to try. Details are in my handwritten notes.
is shown by doing the obvious thing to try. Details are in my handwritten notes.
7 All left ideals are finitely generated since R is left Noetherian.
6 This
27
We need to show ρ is R−linear, and that ρi = idI . We already have that ρ is k−linear, so it is enough to show that
for all h ∈ G, and a ∈ R, that ρ(ha) = hρ(a). Indeed, we have
1 X
ρ(ha) =
gπg −1 (ha)
|G|
g∈G
=
1 X
gπ((h−1 g)−1 a)
|G|
=
1 X
h(h−1 g)π((h−1 g)−1 a)
|G|
=
1 X
h
gπg −1 (a)
|G|
g∈G
g∈G
g∈G
= hρ(a)
since h−1 G = G as sets. It remains then to show that
ρ(a) = a. Recall that π(a) = a. then
P for all−1a ∈ I that
1
1
gπg
(a)
=
|G|a
= a. Hence, ρ is a splitting map for
gπg −1 (a) = g · π(g −1 a) = gg −1 a = a, so ρ(a) = |G|
g∈G
|G|
8
i, and hence I is a direct summand of R.
3.6
3.6.1
Monday 5 March 2012
Semisimple Algebras over a Field
Recall. Let R be a semisimple ring.
. . ⊕ I where I is a simple left ideal. Also, if we set D = EndR (I), then
B If R is simple, then R ∼
= nI = |I ⊕ .{z
}
n copies
R∼
= EndD (I) = R00 (I) = Mr (Dop ) where r = dimD (I).
B In general, R ∼
= n1 I1 ⊕ . . . ⊕ nt It where each Ij is a simple left ideal, and Ii Ij whenever i 6= j. On
Pthe homework,
we introduced the notation that B(Ij ) is the sum off all left ideals isomorphic to Ij , that is, B(Ij ) = J ∼
J∼
= nj Ij .
=Ij
Thus, we have that R ∼
= B(I1 ) × . . . × B(It ). Also, if we set Dj = EndR (Ij ) = EndB(Ij ) (Ij ), then B(Ij ) ∼
= EndDj (Ij )
by the simple case. Note also that these expressions for R are unique up to isomorphism.
Theorem 149. Let R be a semisimple ring which is also a finite dimensional k−algebra where k is an algebraically
closed field. We write R = n1 I1 ⊕ . . . ⊕ nt It where Ii Ij whenever i 6= j. Then,
1. EndR (Ij ) = k,
2. nj = dimk (Ij ), and
Pt
3. dimk (R) = j=1 n2j .
Proof. 1. Let Dj = EndR (Ij ), and note that this is a division ring. Also note that the map k → Dj given by a 7−→ `a is
an embedding since k ⊆ Z(R), so that `a ∈ EndR (Ij ). So consider k ⊆ EndR (Ij ) = Dj . In fact, we have k ⊆ Z(Dj )
since for all f ∈ EndR (Ij ), then f ◦ `a = `a ◦ f . We also have that mj = dimk (Ij ) ≤ dimk (R) < ∞ and that
Dj = EndR (Ij ) ⊆ Endk (Ij ) ∼
= Mmj (k). Thus, dimk (Dj ) ≤ m2j < ∞. Let α ∈ Dj . Then k[α] is a commutative
domain which is finite dimensional over k. Thus, α is algebraic over k and so α ∈ k and Dj = k.
2. Now, nj Ij ∼
= B(Ij ) = EndDj (Ij ) = Endk (Ij ) ∼
= Mmj (k). Taking dimensions of the two ends over k gives nj mj = m2j
and so we have mj = nj as desired.
Pt
Pt
3. Now it is easy to see that dimk (R) = j=1 nj dimk (Ij ) = j=1 n2j .
Proposition 150.
P Let G be a finite group, and k a field. Let C1 , . . . , Cr be the distinct conjugacy classes of G. For
each i, let zi = g∈Ci g ∈ k[G]. Then {zi }ri=1 forms a k−basis for Z(k[G]).
P
Proof. For any g ∈ G, and for any i ∈ {1, . . . , r}, we clearly have gCi g −1 = Ci . Thus, gzi g −1 = g( h∈Ci h)g −1 =
P
−1
= zi . Hence, gzi = zi g for all g ∈ G, meaning that zi ∈ Z(k[G]). Since Ci ∩ Cj = ∅ whenever
h∈Ci ghg
i 6= j and the elements of G are linearly independent over k, we seeP
that {z1 , . . . , zr } is a linearly independent
set over k. It remains so show that they span Z(k[G]). Let x =
g∈G ag g ∈ Z(R) and let h ∈ G. Then
P
P
−1
−1
x = hxh−1 =
a
hgh
=
a
g.
By
the
linear
independence
of the elements of G, we get that
g∈G g
g∈G h gh
ag = ah−1 ghPfor all g ∈ G and h ∈ H. thus, if g1 , g2 ∈ Ci for some i, then ag1 = ag2 . So for each i, choose gi ∈ Ci .
r
Then, x = i=1 ag1 zi ∈ span{z1 , . . . , zr } and so the set {z1 , . . . , zr } forms a basis for Z(k[G]).9
8 Part of this proof was actually done in class on March 5th, but it is easiest to read when it is put into one day rather than split amongst
more than one.
9 Part of the proof was done on Wednesday, but as always, the proof reads easier if it is not broken up.
28
3.7
3.7.1
Wednesday 7 March 2012
Semisimple Group Rings
Recall. A field k is algebraically closed if and only if whenever F is a field, with k ⊆ F , and F is algebraic over k,
then F = k.
Exercise. If k ⊆ R where k is a field, R is a commutative domain, and dimk (R) < ∞, then R is a field, and R is
algebraic over k.
Proposition 151. Let R = k[G], where G is a finite group, and k an algebraically closed field such that char(k) - |G|.
Then the number of conjugacy classes of G is the number of distinct simple left ideals of R.
Proof. Let t be the number of distinct simple left ideals of R. Then R = n1 I1 ⊕ . . . ⊕ nt It where each Ij is
simple and Ii Ij whenever i 6= j. We also have R ∼
= B(I1 ) × . . . × B(It ) ∼
= Mn1 (k) × . . . × Mnt (k) where
B(Ij ) = Endk (Ij ) = Mnj (k) is simple and semisimple. Thus, Z(R) = Z(Mn1 (k))×. . .×Z(Mnt (k)) ∼
= k·idn1 ×. . .×k·idnt .
Thus, dimk (Z(R)) = dimk (k · idn1 × . . . × k · idnt ) = t is the number of conjugacy classes by Proposition 150.
Corollary 152 (Summary). Let R = k[G], where G is a finite group, and k an algebraically closed field such that
char(k) - |G|. Write R = n1 I1 ⊕ . . . ⊕ nt It where each Ij is a simple left ideal and Ii Ij whenever i 6= j. Then,
1. nj = dimk (Ij ),
2. k = EndR (Ij ),
Pt
2
3.
j=1 nj = |G|, and
4. t is the number of conjugacy classes of G.
Corollary 153. Let R = k[G], where G is a finite group, and k an algebraically closed field such that char(k) - |G|.
Then G is abelian if and only if dimk (M ) = 1 for all simple left R−modules, M .
Proof. G is abelian if and only if the number of conjugacy classes of G is |G| so that t = |G|. This is true if and only
if nj = 1 for all j by corollary 152 which is true if and only if dimk (Ij ) = 1 for all j. In turn, this is true if and only
if dimk (M ) = 1 for all simple left R−modules M since such an M will be congruent to Ij for some j.
Example 154. Let G = S3 . Then, we know that t = 3. The conjugacy classes are: {(1)}, {(12), (13), (23)}, and
{(123), (132)}. Also, we must have that n21 + n + 22 + n23 = 6 so we have n1 = n2 = 1 and n3 = 2. This gives that
I1 ∼
= k as k−vector spaces, but notice that I1 I2 as R−modules. Also, I3 ∼
= k 2 . Our goal is to describe these
= k, I2 ∼
better! And so we’ll start studying representation theory to do this.
3.7.2
Starting Representation Theory
Idea. Whenever G acts on something, we get a “representation” of G.
Definition 155. Let G act on the set X. For g ∈ G, define σg : X → X by a 7−→ g · a. Then σg is a bijection, so
σg ∈ S(X) = {bijective maps X → X}. Then φ : G → S(X) given by g 7−→ σg is a group homomorphism. When
|X| = n, then S(X) ∼
= Sn . We say that φ is a permutation representation of G.
Example 156. 1. If X = G, we can consider the action of left multiplication or the action of conjugation.
2. If X = G/H, we can act by translation, i.e. g · aH = (ga)H
3. If X is the set of Sylow p−subgroups of G, then we can act on X by conjugation.
Definition 157. Let k be a field, and V a k−vector space. We say G acts k−linearly if
1. G acts on V ,
2. g · (αv) = α(g · v) for all v ∈ V , α ∈ k, g ∈ G, and
3. g · (v1 + v2 ) = g · v1 + g · v2 for all g ∈ G, v1 , v2 ∈ V .
Remark 158. Given such a linear action, for each g ∈ G, we get a map φg in EndR (V )∗ , the group of units of
EndR (V ), defined by φg (v) = g · v for all v ∈ V . This is in the group of units since (φg )−1 = φg−1 . Also, note that
GLk (V ) = EndR (V ). So we get a group homomorphism ρ : G → GLk (V ) given by g 7−→ φg .
Definition 159. Let G be a group, and k a field. Then a k−linear representation of G is a homomorphism
ρ : G → GLk (V ) for some k−vector space V .
Remark 160. Let ρ : G → GLk (V ) be a k−linear representation of G. This gives rise to a k[G]−module
structure
P
on V as follows. We already have that V is an abelian group under addition, so, given v ∈ V and g∈G ag g ∈ k[G],
P
P
we set ( g∈G ag g)v = g∈G ag ρ(g)(v) ∈ V . Alternately, one could define gv = ρ(g)(v) and extend the G−action by
linearity. We’ll denote V by Vρ when we think of it as a k[G]−module.
29
3.8
3.8.1
Friday 9 March 2012
Representation Theory Vocab
Remark 161. If M is a k[G]−module, let k M be the underlying k−vector space. Then, given g ∈ G, the map
φg :k M →k M given by m 7−→ gm is an invertible k−endomorphism of k M and hence ρM : G → GLk (k M ) given
by g 7−→ φg is a group homomorphism.
Remark 162. We have constructed a bijection between k−linear representations of V and k[G]−modules.
Definition 163. Let G be a group, k a field, and ρ : G → GLk (V ) a k−linear representation of G.
1. A subrepresentation of ρ is a representation ψ : G → GLk (W ) where W is a subspace of V and for all g ∈ G, then
ρ(g)|W = ψ(g).
2. The zero representation is the map ρ : G → GLk (0). That is, as a k[G]−module, 0ρ = 0.
3. A representation, ρ, is called irreducible if it is not the zero representation and every subrepresentation is the zero
representation or itself.
4. We define to representations ρ1 : G → GLk (V1 ) and ρ2 : G → GLk (V2 ) to be isomorphic if (V1 )ρ1 ∼
= (V2 )ρ2 as
k[G]−modules.
5. The degree of ρ is dimk (V ).
ρ1 (g)
0
6. Given ρi : G → GLk (Vi ) for i = 1, 2, then we define ρ1 ⊕ ρ2 : G → GLk (V1 ⊕ V2 ) by g 7−→
.
0
ρ2 (g)
7. The representation ρ : G → k ∗ = GLk (k) ∼
= M1 (k) given by g 7−→ 1k is called the trivial representation.
8. Let R = k[G]. Then R is a left k[G]−module. Recall that ρG : G → GLk (k R) is given by g 7−→ φg . We call this
the regular representation.
Remark 164. Let G be a group, k a field, and ρ : G → GLk (V ) a k−linear representation of G.
1. If ψ : G → GLk (W ) is a subrepresentation of ρ, then we get that Wψ is a k[G]−submodule of Vρ .
2. The zero representation is a subrepresentation of every representation.
3. If ρ is an irreducible representation, then Vρ is a simple k[G]−module.
4. If ρ1 and ρ2 are isomorphic, with φ : (V1 )ρ1 → (V2 )ρ2 a k[G]−module isomorphism. Then φ is a k−linear isomorphism
and φ(g · v) = g · φ(v) for all g ∈ G and v ∈ V1 . So, (φ ◦ ρ1 (g))(v) = (ρ2 (g) ◦ φ)(v) for all g ∈ G and v ∈ V1 . That
is, ρ1 (g) = φ−1 ρ2 (g)φ for all g ∈ G.
5. Degree 1 representations are irreducible.
6. The only degree 0 representation is the zero representation.
7. As k[G]−modules, (V1 ⊕ V2 )ρ1 ⊕ρ2 ∼
= (V1 )ρ1 ⊕ (V2 )ρ2 .
8. The trivial representation has degree 1 and is hence irreducible.
3.8.2
Starting Examples of Representations
Example 165. Let G = C3 =< a | a3 = 1 >, and R = k[G] = k · 1 ⊕ k · a ⊕ k · a2 . Then φa : R → R is given by
"
#
0 0 1
2
2
1 7−→ a, a 7−→ a , and a 7−→ 1. Thus, we set A = 1 0 0 ∈ GL3 (k). Then, we have that A3 = I3 , the 3-by-3
0 1 0
identity matrix, and ρR : C3 → GL3 (k) is given by 1 7−→ I3 , a 7−→ A and a2 7−→ A2 .10
Remark 166. Suppose that k[G] is semisimple, which recall means that G is a finite group and that char(k) - |G|.
Then every simple k[G]−module is isomorphic to a simple left ideal of k[G], which is a direct summand of k[G]. Also,
k[G] ∼
= n1 I1 ⊕ . . . ⊕ nt It where Ij is simple for each j and every simple k[G]−module is isomorphic to some Ij . Thus,
the regular representation ρk[G] = n1 ρI1 ⊕ . . . ⊕ nt ρIt where ρIj is the irreducible subrepresentation corresponding to
Ij for each j. Hence, every irreducible k−representation of G is a direct summand of the regular representation, and
is isomorphic to ρIj for some j.
Remark 167. The trivial
P representation is a subrepresentation of the regular representation for any group G. Indeed, set
R = k[G] and let u = g∈G g. Then gu = u for all g ∈ G. Let I = Ru = ku. Then ρI : G → GLk (k I) is given by g 7−→ 1
and ρI is the trivial representation and a subrepresentation of the regular representation since I is a simple k[G]−module.
3.9
3.9.1
Monday 12 March 2012
Representations of Cyclic Groups and Permutation Groups
Example 168. Let G = Cn =< a | an = 1 > and let k be an algebraically closed field of characteristic not
dividing n. Every irreducible k−linear representation of G has degree 1, so there must be n such representations. Let
ρ : G → GL1 (k) = k ∗ be given by a 7−→ λ ∈ k ∗ . Since an = 1, then ρ(a)n = λn = 1 so that λ must be an nth root of unity.
10 There
are more details for this example in the handwritten notes.
30
Since the characteristic of the field does not divide n, then xn − 1 has n distinct roots in k. In fact, the group of nth roots
of unity is cyclic, say generated by ω. then, ρi : G → GL1 (k) given by a 7−→ ω i gives all irreducible representations of G.
From a module point of view, we set R = k[Cn ] = k · 1 ⊕ k · a ⊕ . . . ⊕ k · an−1 since an = 1. We then have
k[Cn ] ∼
= k[x]/(xn − 1) ∼
= k[x]/(x − ω 0 ) × . . . × k[x]/(x − ω n−1 ) = Re0 ⊕ . . . ⊕ Ren−1 ,
where ei = 1 in k[x]/(x − ω i ). As k−vector spaces, Rei ∼
= k for each i so each Rei is a simple left ideal. The action
of G on Rei is given by a · ei = a · 1 = x = ω i · 1 = ω i ei .
Remark 169. Suppose H C G. Then any k−linear representation of G/H induces a k−linear representation for G.
Indeed, if ψ : G/H → GLk (V ) is a k−representation for G/H then composition with the quotient map will give a
k−representation for G:
ρ
GLk (V )
G
G/H
ψ
Here, if ψ is an irreducible k−representation for G/H, then ρ is also an irreducible representation (for G).
Example 170. Let G = Sn , and H = An . Then for any n, we have Sn /An ∼
= C2 =< σ > for any σ ∈ Sn \ An . If
k is a field of characteristic not equal to 2, then ψ : C2 → k ∗ given by σ 7−→ −1 is an irreducible, and nontrivial
representation. Composing with the quotient map, we see that ρ is a nontrivial irreducible representation of Sn where
ρ : Sn → k ∗ is given by σ 7−→ 1 if σ ∈ An , and σ 7−→ −1 if σ ∈
/ An . That is, σ 7−→ sgn(σ)
P · 1k . This representation
is called the sign representation, and corresponds to the simple left ideal kw where w = σ∈Sn sgn(σ)σ.
Remark
P 171. Let H ≤ G, and suppose [G : H] = n. Let C1 , . . . , Cn be the distinct left cosets of H in G. For each i, let
ui = g∈Ci g ∈ k[G]. Since for any g ∈ G, and any i, we have gCi = Cj for some j, then gui = uj for some j. Also, since
the cosets are disjoint, then the set {u1 , . . . , un } is linearly independent over k. Let I = ku1 + . . . + kun = ku1 ⊕ .P
. . ⊕ kun .
Since gI ⊆ I for each g ∈ G, then I is a left ideal of k[G]. Note that I contains the left ideal k(u1 +. . .+un ) = k( g∈G g)
so that I is not a simple left ideal unless H = G, that is, unless n = 1.
Example 172. Let G = S3 , and k be an algebraically closed field with char(k) - 6 (that is, char(k) 6= 2, 3). Our
goal is to find all irreducible k−representations of G. Since G has 3 conjugacy classes, there are 3 distinct irreducible
k−representations. Let’s call them ρ1 , ρ2 , and ρ3 and set ni = deg(ρi ) for i = 1, 2, 3. Recall that n21 + n22 + n23 = 6 so we
must have that n1 = n2 = 1 and n3 = 2. Let ρ1 be the trivial representation, and ρ2 be the sign representation. It remains
then to find ρ3 . Note that if Ii is the ideal of k[S3 ] corresponding to ρi , then k[S3 ] = I1 ⊕I2 ⊕2I3 . Let’s consider a left ideal
corresponding to a subgroup H of S3 . If |H| = 3, then the corresponding ideal will have dimension 2. However, this ideal
contains the trivial representation, and so decomposes into 2 simple ideals of dimension 1. So instead, we need to consider
H =< (12) >. Let C1 =PH, C2 = (23)H and C3 = (13)H, that is, as sets, C1 = {(1), (12)}, C2 = {(23), (132)}, and C3 =
{(13), (123)}. Let ui = g∈Ci g, and let I = ku1 +ku2 +ku3 , which is a left ideal of k[S3 ] of dimension 3. Let A = k(u1 +
u2 + u3 ) ⊆ I. so I = A ⊕ J, where dimk (J) = 2. If J is not simple, then J = J1 ⊕ J2 where dimk (Ji ) = 1 for i = 1, 2. But
k[S3 ] contains only 2 simple left ideals of dimension 1. Thus, J must be a 2 dimensional simple left ideal, and correspond
to the k−representation ρ3 . Consider the short exact sequence 0 → A → I → I/A → 0. Since I ∼
= A⊕
I/A, then
ku
+
ku
+
ku
1 −1
1
2
3
∼
J = I/A =
= ku1 ⊕ku2 and u3 = −u1 −u2 . Thus, ρ3 : S3 → GL2 (k) is given by (12) 7−→
0 −1
k(u
1 + u2 + u
3 )
−1 1
and (123) 7−→
. This is sufficient to define ρ3 since these two elements generate S3 . Note that (12)u1 = u1 ,
−1 0
(12)u2 = u3 = −u1 − u2 , (123)u1 = u3 = −u1 − u2 , and (123)u2 = u1 which is how the representation ρ3 was determined.
3.10
3.10.1
Wednesday 14 March 2012
Trace & Characters
Pn
Recall. If k is a field, and A = [aij ] ∈ Mn (k), then tr(A) = i=1 ai i ∈ k is called the trace of A. Note that the map
tr : Mn (k) → k is k−linear. Also, recall that tr(AB) = tr(BA) for all A, B ∈ Mn (k).
Now, if V is a finite dimensional k−vector space, and ρ ∈ Endk (V ), then tr(ρ) is defined as tr(ρ) = tr([ρ]β ) where [ρ]β
is the matrix representation of ρ with respect to the basis β. Since tr(P AP −1 ) = tr(A), then tr(ρ) is actually well
defined, that is, it does not depend on the basis chosen.
Definition 173. Let k be a field, R a finite dimensional k−algebra, and let M be a finitely generated left R−module.
For r ∈ R define rM : M → M by a 7−→ ra. Then, rM ∈ EndR (M ).
31
Note. If M is finitely generated over a finite dimensional k−algebra, R, then dimk (M ) < ∞.
Definition 174. Define the character χM of the R−module M to be the map χM : R → k given by χM (r) = tr(rM ).
Note. It is clear that χM is k−linear. Also, if {u1 , . . . , un } is a k−basis for R, then χM is uniquely determined by
χM (u1 ), . . . , χM (un ).
Definition 175. The degree of χM is dimk (M ).
Remark 176. χM (1) = tr(idM ) = dimk (M ) · 1k ∈ k.
Remark 177. Let ρ : G → GLk (V ) be a k−representation of G, where G is a finite group. Then ρ corresponds to the
k[G]−module Vρ . The character of ρ is χVρ : k[G] → k. In this case, we often write χρ to simplify notation, and consider
χρ : G → k since G is a basis for k[G], so χρ is uniquely determined by χρ (g) for g ∈ G. That is, χρ (g) = tr(ρ(g)).
Also, we say that χρ is a k−character of G.
Proposition 178. Let R be a finite dimensional k−algebra, and 0 → L → M → N → 0 a short exact sequence of
left R−modules. Then, χM = χL + χN .
Proof. Let r ∈ R. We have a commutative diagram as below since for any m ∈ M , then (rN ◦ g)(m) = rg(m) =
g(rm) = (g ◦ rM )(m), and for any ` ∈ L, then (rM ◦ f )(`) = rf (`) = f (r`) = (f ◦ rL )(`).
0
f
L
rL
0
L
g
M
rM
f
0
N
M
rN
g
N
0
As k−vector spaces, the rows are split exact. So we have that M ∼
= L ⊕ N as k−vector spaces. We then get
rL 0
χM (r) = tr(rM ) = tr(rL ⊕ rN ) = tr
= tr(rL ) + tr(rN ) = χL (r) + χN (r)
0 rN
since the following diagram commutes.11
M ∼
=L⊕N
rM
rL ⊕ rN
M ∼
=L⊕N
Corollary 179. Let R be a finite dimensional k−algebra.
1. If N ⊆ M be finitely generated R−modules, then χM = χN + χM/N .
2. If A and B are finitely generated R−modules, then χA⊕B = χA + χB .
3. If M ∼
= N , then χM = χN .
Note. The converse of the third part of the corollary is false in general. An example is below.
Example 180. Let k be a field, and R = k[x]/(x2 ) = k ⊕ kx. Also, let M = R/(x) ⊕ R/(x) ∼
= k ⊕ k. then, xM = 0,
but xR 6= 0. So M R even though dimk (M ) = dimk (R) = 2.
0 0
A k−basis for R is {1, x}. Also, χR (1) = dimk (R) · 1k = 2 · 1k , and χR (x) = tr
= 0. For M we have
1 0
0 0
= 0. Hence χR = χM .
χM (1) = 2 · 1k and χM (x) = tr
0 0
Example 181. Let R = k, where k is a field of characterisitc 2. Also, let M = R2 = k 2 . Then χM (1) = dimk (M ) =
2 · 1k = 0. Thus, χM = 0, but M 0.
11 This
was confusing this day, but is made clearer by an exercise and clarifiction near the beginning of Friday’s class.
32
3.11
3.11.1
Friday 16 March 2012
Review/Clarification
Exercise. Let
0
f
g
A
B
α
0
γ
β
A
B
f
0
C
g
C
0
be a commutative diagram of left R−modules such that the rows are split exact. then there exists an R−module
isomorphism φ : B → A ⊕ C such that the following diagram commutes.
φ
B
A⊕C
α⊕γ
β
φ
B
A⊕C
We’ve shown that there exist maps i : B → A and j : C → B such that 1B = f i + jg and if = 1A . This is shown
by defining φ(b) = (i(b), g(b)).
Note. In the proof of Proposition 178, the diagram below commutes due to the above exercise and φ is an isomorphism.
φ
M
rM
M
L⊕N
rL ⊕ rN
φ−1
L⊕N
Thus, rM = φ ◦ (rL ⊕ rN ) ◦ φ−1 and hence tr(rB ) = tr(φ ◦ (rL ⊕ rN ) ◦ φ−1 ) = tr(rL ⊕ rN ).
3.11.2
Equality of Characters
Proposition 182. Let R be a finite dimensional k−algebra, which is semisimple and char(k) = 0. Also, let M and
N be finitely generated R−modules. Then M ∼
= N if and only if χM = χN .
Proof. (⇒) This direction has been done, see Corollary 179.
(⇐) Since R is semisimple, we can write R ∼
=n
1 I1 ⊕ . . . ⊕ nt It where each Ij is simple and Ii Ij if i 6= j. We can also
P
write R ∼
J. We also have M ∼
= B(I1 ) ⊕ . . . ⊕ B(It ) where B(Ij ) = J ∼
= m1 I1 ⊕ . . . ⊕ mt It and N ∼
= r1 I1 ⊕ . . . ⊕ rt It .
=Ij
It is enough to show that mi = ri for all i. So we write, M = N1 ⊕ . . . ⊕ Nt where Ni ∼
= mi Ii for each i, and let
ei ∈ B(Ii ) ⊂ R be the identity element of B(Ii ) for each i. We certainly have ei Ij = 0 for all i 6= j so ei Nj = 0 for
i 6= j. However, ei n = n for all n ∈ Ni . Consider the map

 

(ei )N1
0
···
0
0 0
··· 0
0
(ei )N2 · · ·
0

  0 1Ni · · · 0 
= .
(ei )M = 
..
..
..
..
. .
..
..

  ..
.
. .. 
.
.
.
.
0
0
0 (ei )Nt
0 0
··· 0
Thus, χM (ei ) = tr(1Ni ) = dimk (Ni ) · 1k = mi dimk (Ii ) · 1k . Similarly, χN (ei ) = ri dimk (Ii ) · 1k . Since χM = χN , then
mi = ri because char(k) = 0. Thus, M ∼
= N.
Corollary 183. Let R be a semisimple, finite dimensional k−algebra, where k is a field of characteristic 0. If M is
a finitely generated R−module, then M = 0 if and only if χM = 0.
33
Definition 184. A character χM 6= 0 is said to be irreducible if whenever χM = χN + χN 0 for R−modules N , and
N 0 , then one of χN or χN 0 is 0.
Corollary 185. Let R be a semisimple, finite dimensional k−algebra, where k is a field of characteristic 0. An
R−module M is simple if and only if χM is irreducible.
Corollary 186 (Recap). Let R be a semisimple, finite dimensional k−algebra, where k is a field of characteristic
0. Then R ∼
= n1 I1 ⊕ . . . ⊕ nt It . Let M be a finitely generated R−module. Then M ∼
= m1 I1 ⊕ . . . ⊕ mt It . Denote χIi
by χi . Then,
1. χi is irreducible for all i,
2. χi 6= χj for i 6= j,
3. {χ1 , . . . , χt } is the complete set of irreducible characters for R, and
4. χM = m1 χ1 + . . . + mt χt .
5. If additionally, we have that k is algebraically closed, then,
(a) ni = χi (1),
(b) χR = χ1 (1)χ1 + . . . + χt (1)χt , and
Pt
(c) dimk (R) = χR (1) = i=1 χi (1)2 .
Recall. If ρ : G → GLk (V ) is a k−representation for G, then χρ = χVρ where Vρ is the k[G]−module V (associated to
ρ). If k has characterisitc 0 and G is a finite group, then the above apply to χρ (although 5 only applies if k is algebraically
closed). In this case, we usually consider χρ : G → k given by g 7−→ tr(ρ(g)), and call these (k−)characters of the group G.
Note. Let R be a semisimple, finite dimensional k−algebra, where k is a field of characteristic 0. Then χρ is an
irreducible character if and only if ρ is an irreducible representation.
3.11.3
Class Functions
Definition 187. A function f : G → A with G a group, and A a set, is called a class function if f (gxg −1 ) = f (x)
for all g, x ∈ G. Equivalently, f is constant on conjugacy classes.
Remark 188. Group characters are class functions since we have
χρ (gxg −1 ) = tr(ρ(gxg −1 )) = tr(ρ(g)ρ(x)ρ(g)−1 ) = tr(ρ(x)) = χρ (x).
3.12
3.12.1
Monday 26 March 2012
Useful Character Theory Formulas
Recall. Our standard notation is as follows:
B G is a finite group, k is an algebraically closed field with char(k) - |G|
B Then, k[G] is semisimple, and k[G] = B(I1 )×. . .×B(It ) ∼
6 j.
= n1 I1 ⊕. . .⊕nt It where each Ij is simple, and Ij Ii if i =
B We’ll set χi = χIi for i = 1, . . . , t, and φ = χk[G] , that is, φ is the character associated to the regular representation.
Also, recall that φ = n1 χ1 + . . . + nt χt .
B Let ei be the identity of B(Ii ).
B Let C1 , . .P
. , Ct be the distinct conjugacy classes of G and set mi = |Ci |.
B Let zi = g∈Ci g ∈ k[G].
B Recall that Z(k[G]) = ke1 × . . . × ket = kz1 ⊕ . . . ⊕ kzt .
|G|, if g = 1
Lemma 189. φ(g) =
0,
if g 6= 1.
Proof. Recall that gk[G] : k[G] → k[G] is given by left multiplication by g, that is, u 7−→ gu. Also, φ(g) = tr(gk[G] ).
Thus, φ(1) = tr(1k[G] ) = dimk (k[G]) = |G| since the matrix form of 1k[G] is an |G| by |G| identity matrix. If g 6= 1,
then gh 6= h and since the elements of G form a basis for k[G], then gk[G] permutes the basis elements and has no fixed
points, so the matrix form of gk[G] has zeros on the main diagonal, and hence φ(g) = tr(gk[G] ) = 0.
Theorem 190. 1. For i = 1, . . . , t,
ei =
ni X
χi (g −1 )g.
|G|
g∈G
2. For any g ∈ Ci ,
zi = mi
t
X
χj (g)
j=1
34
nj
ej .
P
Proof. 1. As ei ∈ k[G], we can write ei = g∈G ag g where ag ∈ k. Let h ∈ G, and consider φ(ei h−1 ). By Lemma 189,
and since characters are linear functions, we have


X
X
ag φ(gh−1 ) = ah |G|.
φ(ei h−1 ) = φ 
ag gh−1  =
g∈G
g∈G
On the other hand, we can compute the trace of (ei h−1 )k[G] to determine φ(ei h−1 ). Since ei B(Ij ) = 0 for i 6= j,
−1
then ei h−1 B(Ij ) = 0 for i 6= j. Also, ei |B(Ij ) is the identity map on B(Ij ). The matrix form of ei hk[G]
is therefore
block diagonal, where the only nonzero diagonal block is for B(Ii ). Thus,
−1
tr((ei h−1 )k[G] ) = tr((ei h−1 )B(Ii ) ) = tr(h−1
)
B(Ii ) ) = ni χi (h
−1
since B(Ij ) ∼
). Hence,
= ni Ii and χi (h−1 ) = tr(h−1
Ii ). Putting this together, we have that ah |G| = ni χi (h
ah =
ni
χi (h−1 )
|G|
for each h ∈ G, which
P gives the desired result.
2. By definition, zi = g∈Ci g. Thus, for any g ∈ Ci we have

χj (zi ) = χj 

X
g =
g∈Ci
X
χj (g) = |Ci |χj (g) = mi χj (g)
g∈Ci
Pt
since χj is a class function. We can also write zi = `=1 u` e` where u` ∈ k. Thus,
!
t
t
X
X
χj (zi ) = χj
u` e` =
u` χj (e` ) = uj χj (1) = uj nj
`=1
`=1
since dimk Ij = nj and due to the linearity of χj . Putting this together, we have that mj χj (g) = uj nj for any
g ∈ Ci . Hence,
χj (g)
uj = mi
nj
for any g ∈ Ci , which gives the desired result.
Note. Note that part 1 of Theorem 190 gives that char(k) - ni for all i.
Corollary 191. 1. For all g ∈ G and i, j ∈ {1, . . . , t}, then
X
χi (g)χj (g −1 ) = δij |G|
g∈G
1, if i = j
0, if i 6= j.
2. For all g, h ∈ G, then
where δij =
t
X
χi (g)χi (h−1 ) = ∆gh |CG (g)|
i=1
1, if g ∼ h
. We use the notation g ∼ h to mean that g and h are conjugate, that is, there exists
0, if g h
some x ∈ G such that xgx−1 = h. Lastly, as usual, CG (g) = {x ∈ G | gx = xg} is the centralizer of g in G.
3. If g 6= 1, then
t
X
χi (1)χi (g) = 0.
where ∆gh =
i=1
35
Proof. 1. By Theorem 190, we know
ni X
χi (g −1 )g.
|G|
ei =
g∈G
It is clear that χj (ei ) = 0 whenever i 6= j. Also, when i = j, we have that χi (ei ) = tr(1B(Ii ) ) = ni . Thus, applying
χj to the expression for ei gives us that
ni X
δij ni = χj (ei ) =
χi (g −1 )χj (g).
|G|
g∈G
Rearranging this gives the desired result.
2. By Theorem 190, we know that
zi = mi
t
X
χj (g)
nj
j=1
for any fixed choice of g ∈ Ci . By definition, zi =
190 into the first expression for zi we get
zi
= mi
x∈Ci
t
X
χj (g)
nj
j=1
x. By plugging in the expression for ei from Theorem
!
nj X
−1
χj (h )h
|G|
h∈G
t
X
X χj (g)nj
χj (h−1 )h
n
|G|
j
j=1 h∈G


t
mi X X
−1 
χj (g)χj (h ) h.
|G|
j=1
= mi
=
P
ej
h∈G
Recall that by definition
zi =
X
w.
(3.1)
w∈Ci
By comparing coefficientsP
of the two expressions for zi we obtain the following since the group elements are linearly indet
pendent. If h ∈
/ Ci , then j=1 χj (g)χj (h−1 ) = 0 since h does not appear in the sum in equation (3.1). However, if h ∈
m i Pt
−1
) = 1. Since |CG (g)| = |G|
Ci , then we must have that |G|
j=1 χj (g)χj (h
mi , then this establishes the desired formula.
3. If g 6= 1, then g −1 1, and so we have by the previous part that
t
X
χi (1)χi (g) = ∆1g−1 |CG (1)| = 0.
i=1
3.13
3.13.1
Wednesday 28 March 2012
Character Tables
Remark 192. Each row of a character table corresponds to an irreducible character and each column corresponds to a
conjugacy class in the group. The entry corresponding to the ith character and j th conjugacy class is χi (g) for some g ∈ Cj .
1. Since the number of irreducible characters equals the number of conjugacy classes, the tables are square.
2. Since characters are class functions, it does not matter which g ∈ Cj is chosen.
3. The identity element of the group is always in its own conjugacy class, and χi (1) is always the degree of the representation
(see Remark 176). This means that the first column of the table gets filled in with the degree of the given representation.
4. The character associated to the trivial representation is always at the top of the table, and the trivial representation
sends every group element to 1 in the field, and hence every entry in the top row is a 1.
5. From part 3 of Corollary 191, we know that the dot product of the first column with any other column must be 0.
Example 193. Today we did lots of examples of character tables (and all were over the field C.
1. Let G = Cn =< a | an = 1 >. Recall from Example 168 that the irreducible representations are {ρj | j = 0, . . . , n − 1}
where ρj (a) = ω j and ω = e2πi/n . Let χj be the character associated to the representation ρj . In particular, if n = 3,
then we have the following character table:
36
2.
3.
4.
5.
1 a
a2
χ0 1
1
1
χ1 1 ω ω 2
χ2 1 ω 2 ω
Let G = V4 = {1, a, b, c} ∼
= C2 × C2 . The group is abelian, and so there are 4 conjugacy classes, and hence 4
irreducible representations all of degree 1. Let’s call them ρ0 , ρ1 , ρ2 , and ρ3 . We know that ρj is determined by
ρj (a) and ρj (b) since ab = c. We have then ρj (a)2 = ρj (b)2 = ρj (1) = 1 and so ρj (a) = ±1 and ρj (b) = ±1 for
each j. There are 4 ways to choose between these, and so they must give our 4 irreducible representations. Let
ρ0 (a) = ρ0 (b) = 1, ρ1 (a) = −1 and ρ1 (b) = 1, ρ2 (a) = 1 and ρ2 (b) = −1, and finally ρ3 (a) = ρ3 (b) = −1. Note that
ρj (c) = ρj (ab) = ρj (a)ρj (b). Since these are all elements of the field, then they are also the traces of the images
of ρj (g) for g ∈ G. This gives the following character table:
1 a
b
c
χ0 1 1
1
1
χ1 1 -1 1 -1
χ2 1 1 -1 -1
χ3 1 -1 -1 1
Let G = S3 . From Example 172, we know that there are 3 irreducible representations, and also where they send
the group elements. Thus, we can quickly fill in the table, shown below. Alternately, using the fact that ρ1 is the
sign representation, we can get the first two rows, and fill in the third row using part 5 of Remark 192.
(1) (12) (123)
χ0
1
1
1
χ1
1
-1
1
χ2
2
0
-1
Let G = Q8 = {±1, ±i, ±j, ±k}. The conjugacy classes of this group are C1 = {1}, C2 = {−1}, C3 = {±i},
C4 = {±j}, and C5 = {±k}, so there must be 5 irreducible representations. We then have 8 = n20 + n21 + n22 + n23 + n24
and hence n0 = n1 = n2 = n3 = 1 and n4 = 2. Let H = {±1} and note that H C G. We also have that
2
2
2
G/H ∼
= C2 × C2 ∼
= V4 since the elements of G/H are 1, i, j, k and i = j = k = 1. Since any irreducible
representation of G/H gives an irreducible representation for G, by composing with the canonical map G/H this gives
us that the degree 1 representations of G match the degree 1 representations of G/H, taking into account the fact
that in G/H, C1 and C2 become the same conjugacy class. We can thus fill in all but the last row of the table using
what we already know, and then get the bottom column using part 5 of Remark 192. The complete table is below:
1 -1 i
j
k
χ0 1 1
1
1
1
χ1 1 1 -1 1 -1
1 -1 -1
χ2 1 1
χ3 1 1 -1 -1 1
χ4 2 -2 0
0
0
Let G = A4 . The 4 conjugacy classes are C1 = {(1)}, C2 = {(12)(34), (13)(24), (14)(23)}, C3 = (123)H and
C4 = (132)H where H = C1 ∪ C2 is a normal subgroup of G. Thus there are 4 irreducible representations of A4 .
Since G/H ∼
= C3 , there are at least 3 irreducible representations of degree 1. Since the sum of the squares of the
degrees must equal the order of the group, i.e. we must have n20 + n21 + n22 + n23 = 24 then n0 = n1 = n2 = 1 and
n3 = 3. In part 1 of this example, we found the character table for C3 , and we can fill in the remainder of the table
using part 5 or Remark 192. The complete character table is below where ω = e2πi/3 as before.
(1) (12)(34) (123) (132)
χ0
1
1
1
1
χ1
1
1
ω
ω2
χ2
1
1
ω2
ω
χ3
3
-1
0
0
Remark 194. On the homework we show that if χ is a complex character (meaning that the field in question is the
complex numbers) for a finite group G, then χ(g −1 ) = χ(g). The first part of Corollary 191 gives
X
χi (g)χj (g) = δij |G|.
g∈G
Now, let g1 , . . . , gt be representatives of the conjugacy classes of G, and mi be the number of elements in each conjugacy
class. Then we have
t
X
m` χi (g` )χj (g` ) = δij |G|.
`=1
37
3.14
3.14.1
Friday 30 March 2012
Characters and Inner Products
Definition 195. Let G be a finite group, and k a field. Let Fk (G) denote the set of class functions G → k.
Note. It is clear that Fk (G) is a k−vector space since (f + g)(x) = f (x) + g(x) and (cf )(x) = cf (x) for all c ∈ k,
x ∈ G and f, g ∈ Fk (G). Also, dimk Fk (G) is the number of conjugacy classes of G.
Definition 196. Let the inner product of FC (G) be given by:
hφ, ψi :=
1 X
φ(g)ψ(g)
|G|
g∈G
and note that hφ, ψi ∈ C for all φ, ψ ∈ FC (G).
Note. This is indeed an inner product. Here are the details:
B Additive in the first variable:
hφ1 + φ2 , ψi =
1 X
(φ1 + φ2 )(g)ψ(g)
|G|
g∈G
=
1 X
(φ1 (g) + φ2 (g))ψ(g)
|G|
=
1 X
1 X
φ1 (g)ψ(g) +
φ2 (g)ψ(g)
|G|
|G|
g∈G
g∈G
g∈G
= hφ1 , ψi + hφ2 , ψi
B Compatiblity with scalars:
hcφ, ψi =
1 X
(cφ)(g)ψ(g)
|G|
g∈G
1 X
= c
φ(g)ψ(g)
|G|
g∈G
= chφ, ψi
B Conjugate Symmetry:
hφ, ψi =
1 X
φ(g)ψ(g)
|G|
g∈G
=
1 X
φ(g)ψ(g)
|G|
g∈G
=
hψ, φi
B Positive Definiteness:
hφ, φi =
1 X
φ(g)φ(g)
|G|
g∈G
=
1 X
|φ(g)|2 ≥ 0
|G|
g∈G
P
If φ = 0, then it is clear that hφ, φ, i = 0. If hφ, φi = 0, then g∈G |φ(g)|2 = 0. Since each |φ(g)|2 is a non-negative
real number, then we must have that |φ(g)|2 = 0 for all g ∈ G and hence φ(g) = 0 for all g ∈ G.
Proposition 197. The irreducible characters form an orthonormal basis for FC (G). That is, if {χ1 , . . . , χt } are the
irreducible characters, then hχi , χj i = δij .
38
P
−1
Proof. Recall from Corollary 191 that
) = δij |G|. Recall also from the homework that when
g∈G χi (g)χj (g
−1
the field is C we have χj (g ) = χj (g) for all irreducible characters χj . Thus, we can rearrange to obtain
P
|G|δij = g∈G χi (g)χj (g) = |G|hχi , χj i which proves the proposition.
Remark 198. We thus have that for all φ ∈ FC (G), that φ = hφ, χ1 i + . . . + hφ, χt iχt .
Remark 199. Recall that φ is a character if and only if φ = m1 χ1 + . . . + mt χt where mi ∈ N0 . Thus, mi = hφ, χi i.
We then have that for any character φ, that hφ, φi = m21 + . . . + m2t .
Corollary 200. A character φ is irreducible if and only if hφ, φi = 1.
3.14.2
Localization
Remark 201. From now on, all rings are assumed to be commutative.
Definition 202. Let R be a commutative ring, and S ⊆ R a multiplicatively closed set. Define a relation on R × S
by (r1 , s1 ) ∼ (r2 , s2 ) when there exists some t ∈ S such that t(s2 r1 − s1 r2 ) = 0 or equivalently when ts2 r1 = ts1 r2 .
Proposition 203. The relation in the previous definition is an equivalence relation.
Proof. Let r ∈ R and s ∈ S. Then for any t ∈ S, we have tsr = tsr and so (r, s) ∼ (r, s). Also, if (r1 , s1 ) ∼ (r2 , s2 ),
then there is some t ∈ S such that t(s2 r1 − s1 r2 ) = 0. This also gives that t(s1 r2 − s2 r1 ) = 0 and so (r2 , s2 ) ∼ (r1 , s1 ).
Finally, suppose that (r1 , s1 ) ∼ (r2 , s2 ) and (r2 , s2 ) ∼ (r3 , s3 ). Then there exist t, t0 ∈ S such that ts1 r2 = ts2 r1
and t0 s2 r3 = t0 s3 r2 . Let t00 = tt0 s2 , and note that t00 ∈ S since t, t0 , s2 ∈ S and S is multiplicatively closed. Then
t00 s3 r1 = tt0 s2 s3 r1 = t0 ts2 r1 s3 = t0 ts1 r2 s3 = tt0 s3 r2 s1 = tt0 s2 r3 s1 = t00 s1 r3 and so (r1 , s1 ) ∼ (r3 , s3 ).
r
denote the equivalence class of (r, s). Then RS is used to denote the set of equivalence classes,
s
o
nr
| r ∈ R, s ∈ S .
and is called the localization of R at S. Note here that RS =
s
Definition 204. Let
Theorem 205. The localization of a commutative ring R at any multiplicatively closed set S is a commutative ring
with identity under the following operations:
r1 r2
r1 r2
·
=
s1 s2
s1 s2
r1
r2
s2 r1 + s1 r2
+
=
.
s1
s2
s1 s2
and
Proof. We need to show that these operations are well defined, and that they satisfy the ring axioms. First, assume
r1
r2
r3
r4
that
=
and
= . Then there exist t1 , t2 such that t1 s2 r1 = t1 s1 r2 and t2 s4 r3 = t2 s3 r4 . We need to show
s1
s2
s3
s4
r2 r4
s3 r1 + s1 r3
s4 r2 + s2 r4
r1 r3
=
and
=
. Let t = t1 t2 . Then we have
that
s1 s3
s2 s4
s1 s3
s2 s4
ts2 s4 r1 r3 = t1 s2 r1 t2 s4 r3 = t1 s1 r2 t2 s3 r4 = ts1 s3 r2 r4
so the first equality holds. We also have that
ts2 s4 (s3 r1 + s1 r3 )
= (t1 s2 r1 )t2 s4 s3 + (t2 s4 r3 )t1 s2 s1
= (t1 s1 r2 )t2 s4 s3 + (t2 s3 r4 )t1 s2 s1
= t1 t2 s1 s3 s4 r2 + t1 t2 s1 s3 s2 r4
= ts1 s3 (s4 r2 + s2 r4 )
r1 r2
s2 r1 + s1 r2
and
s1 s2
s1 s2
are elements of RS for any r1 , r2 ∈ R and any s1 , s2 ∈ S since S is multiplicatively closed and hence RS is closed
r
0
s0 r + s0
s0 r
r
under both addition and multiplication. Also, if r ∈ R and s, s0 ∈ S, then + 0 =
=
= since for any
0
0
s s
ss
ss
s
0
t ∈ S, we have ts(s0 r) = t(ss0 )r. Thus, 0 is the additive identity. Also, if r ∈ R and s ∈ S, then −r ∈ R and we have
s
r −r
sr − sr
0
−sr + sr
−r r
−r
r
+
=
=
=
=
+ and so
is the additive inverse of . Next, for any r, r0 ∈ R and
s
s
ss
ss
ss
s
s
s
s
so that the second equality holds and hence the operations are well defined. Note next that
39
r r0
s0 r + sr0
sr0 + s0 r
r0
+ 0 =
=
= 0+
0
0
s s
ss
ss
s
and s, s0 , s00 ∈ S. Then using that R is a ring, we have
r00
r r0
+ 0 + 00 =
s s
s
s, s0 ∈ S, we have
=
=
=
=
=
r
. In order to show addition is associative, let r, r0 , r00 ∈ R
s
s0 r + sr0
r00
+
ss0
s00
00 0
0
s (s r + sr ) + ss0 r00
ss0 s00
s0 s00 r + ss00 r0 + ss0 r00
ss0 s00
0 00
s s r + s(s00 r0 + s0 r00 )
ss0 s00
00 0
r s r + s0 r00
+
s s0 s00 r
r00
r0
+
+
s
s0
s00
and hence (RS , +) is
multiplication
is associative, let r, r0 , r00 ∈ R and
an abelian
00 group0 as 00desired.0 00To see that
0 00 0
0 00
r r
r
rr r
rr r
r rr
r
r r
s, s0 , s00 ∈ S. Then,
·
· 00 = 0 · 00 = 0 00 = · 0 00 = ·
·
using only the associativity of R.
s s0
s
ss s
ss s
s ss
s
s0 s00
r s0
rs0
r
s0 r
s0 r
Next, let s, s0 ∈ S and r ∈ R. Then for any t ∈ S, we have ts(rs0 ) = t(ss0 )r and hence · 0 = 0 = = 0 = 0 ·
s s
ss
s
ss
s s
r r0
rr0
r0 r
r0
t
0
0
so that t is the multiplicative identity of RS . For any r, r ∈ R and any s, s ∈ S, we have · 0 = 0 = 0 = 0 · rs
s s
ss
ss
s
so that multiplication is commutative. It only remains to show then that multiplication distributes over addition. So
let r, r0 , r00 ∈ R and s, s0 , s00 ∈ S. Then,
00
r r0
s0 r + sr0 r00
r
+ 0 · 00 =
· 00
s s
s
ss0
s
0
0 00
(s r + sr )r
=
ss0 s00
0 00
s rr + sr0 r00
=
ss0 s00
0 00 00
s s rr + ss00 r0 r00
=
ss00 s0 s00
r0 r00
rr00
+
=
ss00
s0 s00
00
r r
r0 r00
=
· 00 + 0 · 00 .
s s
s s
This completes the proof that RS is a commutative ring with identity under the given operations.
Remark 206. The following are easy, but it is worth going through the proofs once.12
B The ring RS = {0} if and only if 0 ∈ S.
r
0
1
0
Proof. If 0 ∈ S, then = for any s, t ∈ S and r ∈ R since 0tr = 0s0. Conversely, if RS = {0}, then = for
s
t
s
s
any s ∈ S and so there is some t ∈ S such that ts1 = ts0 so that ts = 0. Since S is multiplicatively closed, this
gives that 0 = ts ∈ S.
B Let S 0 = S ∪ {1}. Then RS ∼
= RS 0 and so without loss of generality, we usually assume 1 ∈ S.
r
r
r0
r
7−→ . This map is well defined since if = 0 in RS , then there is
s
s
s
s
r
r0
0
0
0
some t ∈ S such that tsr = ts r and since t ∈ S , then = 0 in RS 0 as well. Next, we’ll show that it is a ring
s
s
Proof. Let φ : RS → RS 0 be given by
12 I
hope! Definitely not worth going through them more than once!!!
40
homomorphism. Let r1 , r2 ∈ R and s1 , s2 ∈ S. Then,
r1 r2
r1 r2
r1 r2
r1 r2
r1
r2
φ
·
=φ
=
=
·
=φ
·φ
s1 s2
s1 s2
s1 s2
s1 s2
s1
s2
and similarly
φ
r1
r2
+
s1
s2
=φ
s2 r1 + s1 rs
s1 s2
s2 r1 + s1 rs
r1
r2
=
=
+
=φ
s1 s2
s1
s2
r1
s1
+φ
r2
s2
.
r
= 0, then there exists some t ∈ S 0 such that tr = 0. If t = 1, then r = 0 and = 0. If t 6= 1, then
s
s
r
r
r
r
r
0
= . If ∈ RS 0 and s = 1,
t ∈ S, and so = 0 in RS . Hence, φ is injective. If ∈ RS and s ∈ S, then φ
s
s
s
s
s
0
rs
rs0
r
r
0
then for any s ∈ S, we have 0 = and so φ
= so that φ is surjective and hence an isomorphism..
ss
s
ss0
s
Now, if φ
r
B There is a natural ring homomorphism φ : R → RS given by r 7−→
r
.
1
Proof. We only need to show that this is indeed a ring homomorphism. So let r, r0 ∈ R. Then φ(rr0 ) =
rr0
=
1
r r0
r + r0
r
r0
· = φ(r)φ(r0 ). Similarly, φ(r + r0 ) =
= + = φ(r) + φ(r0 ). Thus, φ is a ring homomorphism.
1 1
1
1
1
B The map φ in the previous remark is injective if and only if S does not contain any zero divisors.
t a
0
ta
= · =
Proof. (⇒) Suppose that t ∈ S such that ta = 0 for some a ∈ R. Then φ(ta) = 0, meaning that
1
1 1
1
1
0
1 t a
a
in RS . Since t ∈ S, then we multiply by to get = · · = = φ(a). Since we’re assuming that φ is injective,
t
1
t 1 1
1
this means that a = 0 so that t is not a zero divisor.
0
a
(⇐) Suppose that φ(a) = 0. Then = and so there is some t ∈ S such that t(a − 0) = 0 or equivalently that
1
1
ta = 0. Since S doesn’t include any zerodivisors, this then means that a = 0.
B φ(t) is a unit for all t ∈ S.
Proof. Since t ∈ S, then
1
1
1 t
t
∈ RS . Also, · φ(t) = · = which is the identity element of RS .
t
t
t 1
t
Note. Sometimes, and especially in older literature, the notation S −1 R is used instead of RS .
Example 207. These examples don’t require proofs.13
B If R is a domain, and S = R \ {0}, then RS = Q(R) is the field of fractions of R.
B More generally, if R is an arbitrary commutative ring, and S is the set of non-zerodivisors of R, then S is multiplicatively
closed and RS is called the total quotient ring of R.
B If S consists of only units, then RS ∼
= R.
n
B If x ∈ R, then
0 } is multiplicatively closed. We usually denote RS by Rx in this case. Note here
n r S = {x | n ∈ No
that Rx =
| r ∈ R, n ∈ N0 .
xn
na
o
1
B We thus have Z2 =
| a ∈ Z, n ∈ N0 = Z
.
2n
2
B Let R = k[x] where x is an indeterminate and k is a field. Then Rx = k[x, x−1 ].
Remark 208. If ab = 0 in R but a, b 6= 0, and we want to make a into a unit, i.e. so that φ−1 (a) exists, then we
must set φ(b) = 0 since 0 = φ(ab) = φ(a)φ(b).
13 At
least not at this point in my opinion.
41
3.15
3.15.1
Monday 2 April 2012
Facts about Localization
Remark 209. Let P ∈ Spec(R). Then S = R \ P is a multiplicatively closed set. Instead of writing RS , in this case
the notation RP is used.14
nr
o
Proposition 210. Let I be an ideal of R. Then, IS =
| r ∈ I, s ∈ S is an ideal of RS . Note however that we
s
r
r0
/ I, but r ∈ I.
can have = 0 with r0 ∈
s
s
a a0
a a0
s0 a − sa0
∈ IS since s0 a, −sa0 ∈ I, and hence s0 a − sa0 ∈ I due
, 0 ∈ IS , with a, a0 ∈ I. Then, − 0 =
s s
s
s
ss0
r
r a
ra
to I being an ideal. Also, if 00 ∈ RS , then 00 = 00 ∈ IS since ra ∈ I due to I being an ideal. Thus, IS is an ideal
s
s s
s s
of RS for every ideal I of R.
Proof. Let
Proposition 211. In fact, every ideal of RS is of the form IS where I is an ideal of R.
r
Proof. Let J be an ideal of RS and set I = φ−1 (J) where φ : R → RS is the ring homomorphism given by r 7−→ .
1
a
Then, I is an ideal since φ is a ring homomorphism. It remains to show that IS = J. First, let ∈ IS where a ∈ I
s
a
a
a 1
a
and s ∈ S. Then φ(a) = ∈ J by definition of I and so = · ∈ J since J is an ideal. Next, let ∈ J. Then
1
s
1 s
s
a
a s
since J is an ideal, we have = · ∈ J. Thus, a ∈ φ−1 (J) = I.
1
s 1
an
a1
,..., .
Proposition 212. If I is an ideal of R, generated by a1 , . . . , an , then IS is an ideal of RS generated by
1
1
In particular, if R is Noetherian, then RS is also Noetherian.
Proof. Since I is generated by a1 , . . . , an , then every element of I looks like r1 a1 + . . . + rn an where ri ∈ R for all
x
x
a
i. Thus, if ∈ IS , then = where a ∈ I. We then have that a = r1 a1 + . . . + rn an for some ri ∈ R and hence,
s
s
s
a
r1 a1
rn an
r1 a1
rn an
a1
an
x
= =
+ ... +
=
+ ... +
so that IS is now seen to be generated by
,..., .
s
s
s
s
s 1
s 1
1
1
Proposition 213. Let R be a commutative ring and S a multiplicatively closed subset of R. Then, IS = RS if and
only if I ∩ S 6= ∅.
1
a
1
Proof. (⇒) If ∈ IS , then = for some a ∈ I and s ∈ S. Thus, there exists a t ∈ S such that ts = ta ∈ S ∩ I
1
1
s
since ts ∈ S and ta ∈ I are clear since S is multiplicatively closed and a ∈ I with I an ideal.
1
a
(⇐) Suppose that I ∩ S 6= ∅. Then there is some a ∈ I ∩ S. Then, = ∈ IS since a ∈ I and a ∈ S and so
1
a
IS = RS .
Remark 214. If J ⊆ I are ideals of R and S is a multiplicatively closed subset of R, then JS ⊆ IS .
Example 215. Let R = Z, and S = {2n | n ∈ N0 }. Then note that (6) ( (3) as ideals in Z, but in RS , we have that
3
1 6
(3)S = (6)S since = · ∈ (6)S .
1
2 1
Proposition 216. Let R be a commutative ring, S a multiplicatively closed subset of R, and φ : R → RS the ring
r
homomorphism given by r 7−→ . Then there is a bijection
1
{P ∈ Spec(R) | P ∩ S = ∅} ←→ {P ∈ Spec(RS )}
given by P 7−→ PS and φ−1 (Q) ←−[ Q.
Proof. First, let P ∈ {P ∈ Spec(R) | P ∩ S = ∅}. We need to show that PS ∈ Spec(RS ). We already know that PS is
a b
a b
p
an ideal of RS and we know that PS ( RS by Proposition 213 since P ∩ S = ∅. Thus, let · 0 ∈ PS . Then · 0 = 00
s
s
s
s
s
where p ∈ P and s00 ∈ S. Then there exists some t ∈ S such that ts00 ab = tss0 p. Note then that ts00 ab ∈ P since P
14 Although
this is a special case of localization, it is the one used most frequently.
42
is an ideal. Since ts00 ∈ S, then ts00 ∈
/ P and so we must have ab ∈ P . Since P is prime, then a ∈ P or b ∈ P and hence
a
b
one of or 0 is in PS .
s
s
Next, it is clear that φ−1 (Q) ∈ Spec(R) for all Q ∈ Spec(RS ) since pre-images of prime ideals are prime. Also,
if Q ∈ Spec(RS ), then Q 6= RS . Consider φ−1 (Q) ∩ S. If a ∈ φ−1 (Q) ∩ S, then φ(a) = a1 ∈ Q and a1 ∈ RS and
so aa ∈ Q so that Q = RS . This would be a contradiction, and so we must have that φ−1 (Q) ∩ S = ∅ and hence
φ−1 (Q) ∈ {P ∈ Spec(R) | P ∩ S = ∅}.
a
a
Next, we need to show φ−1 (Q)S = Q. If ∈ Q, then a ∈ φ−1 (Q) and so ∈ φ−1 (Q)S and hence Q ⊆ φ−1 (Q)S .
s
s
b
a
a
b
−1
−1
On the other hand, if ∈ φ (Q)S , then we have = 0 where b ∈ φ (Q) and s0 ∈ S. Then, ∈ Q and since Q
s
s
s
1
a
b 1
1
b
is an ideal and 0 ∈ RS , we then have = 0 = · 0 ∈ Q and hence φ−1 (Q) = Q.
s
s
s
1 s
a
a
b
Finally, we need to show that φ−1 (PS ) = P . If a ∈ φ−1 (PS ), then ∈ PS so that = for some b ∈ P and s ∈ S.
1
1
s
Thus, there exists some t ∈ S such that tsa = tb. This element is in P since b ∈ P . Thus, since s, t ∈ S and hence
s, t ∈
/ P , we have a ∈ P . This shows that φ−1 (PS ) ⊆ P . For the reverse direction, if a ∈ P , then a1 ∈ PS and so
a ∈ φ−1 (PS ) is clear.
Example 217. Let R = Z and S = {2n | n ∈ N0 }. We have Spec(Z) = {(p) | p prime} ∪ {0} and Spec(RS ) =
Spec(Z2 ) = {(p)s | p an odd prime} ∪ {(0)2 }.
na
o
Example 218. If R = Z and S = R \ (2), then Z(2) =
| a, b ∈ Z, b ∈
/ 2Z . Also, Spec(RS ) = Spec(Z(2) =
b
{(0)(2) , (2)(2) }.
Definition 219. A commutative ring R is said to be quasi-local if R has a unique maximal ideal. A commutative
ring R is said to be local if it is quasi-local and Noetherian. If R is a quasi-local ring, with maximal ideal m, then the
residue field of R is R/m.
Remark 220. If R is a ring, and P ∈ Spec(R), then RP is a quasi-local ring with maximal ideal PP = P RP .
Proof. Indeed, if Q ∈ Spec(RP ), then Q = qP for some q ∈ Spec(R) and q ∩ (R \ P ) = ∅. Thus, q ⊆ P and so
Q = qP ⊆ PP .
3.16
3.16.1
Wednesday 4 April 2012
Localization of Modules
m
des
0 0
note the equivalence class of (m, s) ∈ M ×S where the equivalence
n m relation is given
o by (m, s) ∼ (m , s ) if and only if there is
some t ∈ S such that t(s2 m1 −s1 m2 ) = 0. Also, set MS =
| m ∈ M, s ∈ S . This is called the localization of M at S.
s
Theorem 222. The localization of M at S is an RS -module under the following operations:
Definition 221. Let M be an R−medule, and S a multiplicatively closed subset of R. For m ∈ M and s ∈ S, let
m m0
s0 m + sm0
+ 0 =
s
s
ss0
and
r m
rm
·
= 0
s s0
ss
Proof. The proof is very similar to the proof of Theorem 205.15
Remark 223. These remarks are easy, and the proofs are similar to the similar cases for rings, so the proofs will be
omitted.
m
B There is a natural R−module homomorphism α : M → MS given by m 7−→ .
1
B Given an R−submodule N of M , then NS is an RS −submodule of MS .
B If T is an RS −submodule of MS , then T = NS where T = α−1 (T ).
x1
xn
B If N = Rx1 + . . . + Rxn , then NS = RS
+ . . . + RS .
1
1
Note. The R−submodules of MS are potentially different than the RS −submodules of MS . For example, Z(0) = Q.
The Z−submodules of a module may be different than the Q−submodules of the same module.
Proposition 224. If M is a Noetherian (resp. Artinian) module over a ring R, and S is a multiplicatively closed
subset of R, then MS is a Noetherian (resp. Artinian) RS −module.
15 And
is no longer worth doing for me.
43
Proof. The Noetherian case is super similar to the Artinian case.16 Suppose T0 ⊇ T1 ⊇ . . . is a decreasing chain
of RS −submodules of MS . Then, α−1 (T0 ) ⊇ α−1 (T1 ) ⊇ . . . is a descending chain of R−submodules of M . Since
M is Artinian, then there exists some k ∈ N0 such that α−1 (Tk ) = α−1 (Tk+i ) for all i ∈ N1 . This means that
(α−1 (Tk ))S = (α−1 (Tk+i ))S for all i ∈ N1 . However taking the inverse image and then localizing gives the module you
started with, and so Tk = Tk+i for all i ∈ N1 .
Proposition 225. Let M be an R−module. The following are equivalent:
1. M = 0,
2. MP = 0 for all P ∈ Spec R, and
3. Mm = 0 for all maximimal ideals m.
Proof. (1 ⇒ 2) If M = 0, then it is clear that any localization of M is also zero, and in particular, MP = 0 for all
prime ideals P .
(2 ⇒ 3) If MP = 0 for every prime ideal P , then if m is a maximal ideal, it is also a prime ideal, and hence Mm = 0.
(3 ⇒ 1) Let x ∈ M , and let I = AnnR (x) = {r ∈ R | rx = 0}. Then I is an ideal of R since R is a commutative
ring. Note that x = 0 if and only if I = R, which is true if and only if I 6⊆ m for any maximal ideal m. Let m be a
x
0
maximal ideal. Since Mm = 0, then = in Mm . Thus, there is some t ∈ R \ m such that tx = t(1x − 10) = 0. Thus,
1
1
t ∈ AnnR (x) = I. Hence I 6⊂ m and so I = R and x = 0.
Example 226. Let F be a field, and R = F × F . Then Spec(R) = {F × (0), (0) × F } = {m1 , m2 }. Then Rmi ∼
=F
for i = 1, 2, which is a field, but R is not a domain.
Lemma 227. Let {Mi }i∈I be a collection of R−modules, and S a multiplicatively closed subset of R. Then
!
M
M
∼
Mi
(Mi )S
=
i∈I
where the map is given by
(mi )
7−→
s
m i
s
i∈I
S
.
L
(mi )
(m0i )
=
in
i∈I Mi S . Then
0
s 0
s
0
0
there exists some t ∈ S such that ts0 (mi )= ts(m
i ). Hence, for all i ∈ I, we have ts mi = tsmi . Thus, for all i ∈ I,
0
0
mi
mi
mi
m
= 0i in (Mi )S . Therefore,
=
. Next, we’ll show that this is a homomorphism of RS −modules. Let
s
s
s
s0
!
M
(mi ) (m0i )
, 0 ∈
Mi . Then,
s
s
Proof. We first need to show that this map is well defined. So suppose that
i∈I
φ
(mi )
+
s
and if
r
s0
(m0i )
s0
S
=φ
(s0 mi + sm0i )
ss0
s0 mi + sm0i
ss0
rm =
=
mi
m0
+ 0i
s
s
=
0 m m0 (mi )
(mi )
i
i
+
=
φ
+φ
s
s0
s
s0
∈ RS , then
φ
r (mi )
·
s0
s
=φ
(rmi )
s0 s
=
i
s0 s
=
r mi r
(mi )
=
φ
s0 s
s
s
sothat φis an RS −module
homomorphism. It only then remains to show that the map is bijective. So, suppose that
m0i (mi )
(m0i )
mi
φ
= s0 and hence there exists a finite set Γ ⊂ I such that mi = ni = 0 for all
=φ
.
Then
s
s
s0
Q
i ∈ I \ Γ. However, for each i ∈ Γ, we have that there exists some ti ∈ S such that ti s0 mi = ti sm0i . Let t = i∈Γ ti
and notice that t ∈ S since S is multiplicatively closed. Then, we have that ts0 (mi ) = ts(m0i )since ts0 mi = tsm0i for all
M
(mi )
(m0 )
mi
i ∈ Γ and ts0 mi = tsm0i = 0 for all i ∈
/ Γ. Thus,
= 0i and φ is injective. Finally, let
∈
(Mi )S . Then,
s
s
si
i∈I
Q
Q
mi
mi
=
6 0 for only finitely many i ∈ I. Let Γ = i ∈ I |
6= 0 , let s = i∈Γ si , and set ni =
j∈Λ\{i} sj mi for i ∈ Γ
si
si Q
(ni )
ni
mi
and set ni = 0 if i ∈
/ Γ. Then, φ
=
=
since for i ∈ Γ, then si ni = si
j∈Λ\{i} sj mi = smi and
s
s
si
ni
0
mi
for i ∈
/ Γ, then ni = 0 and there is some t ∈ S such that tmi = 0 or equivalently,
= =
and so φ is surjective.
s
s
si
16 I’m
only going to do the Artinian case.
44
Corollary 228. If F is a free R−module, then FS is a free RS −module.
Proof. Let F be a free R−module, so that F ∼
= ⊕i∈I R. Thus, FS = ⊕i∈I RS which is a free RS −module.
Corollary 229. If P is a projective R−module, then PS is a projective RS −module.
Proof. Since P is projective, then there is some R−module, Q, such that P ⊕ Q = F where F is a free R−module.
Then by Lemma 227, PS ⊕ QS ∼
= FS . By the previous corollary, FS is a free RS −module, and hence PS is a projective
RS −module.
Definition 230. Let µR (M ) := inf{n ∈ N0 | there exist x1 , . . . , xn ∈ M such that M = Rx1 + . . . + Rxn }. This is
the minimal number of generators of M .
Note. It is clear that µR (M ) = 0 if and only if M = 0.
Lemma 231 (Nakayama’s Lemma). Let M be a finitely generate R−module, and J the Jacobson radical of R. If
M = JM , then M = 0.
Proof. Suppose µR (M ) = n > 0. Say M = Rx1 + . . . + Rxn . Then, M = JM = J(Rx1 + . . . + Rxn ) = Jx1 + . . . + Jxn .
This gives that xn = j1 x1 + . . . + jn xn where jk ∈ J for j = 1, . . . , n. Hence, (1 − jn xn = j1 x1 + . . . + jn−1 xn−1 . Since
jn ∈ J, then 1 − jn is a unit. Hence, xn = (1 − jn )−1 j1 x1 + . . . + (1 − jn )−1 jn−1 xn−1 , so that Rxn ∈ Rx1 + . . . + Rxn−1 .
This contradicts the minimality of the number of generators, and hence we must have µR (M ) = 0 and thus M = 0
as previously noted.
Corollary 232. If (R, m) is quasi-local, then J(R) = m. So if M is a finitely generated R−module, and M = mM ,
then M = 0.
Corollary 233. Let M be a finitely generated R−module, N ⊂ M be a submodule, and J be the Jacobson radical of
R. If M = N + JM , then N = M .
M
N + JM
M
=
=J·
. Since M is finitely generated, then M/N is also finitely generated, and hence
N
N
N
M/N = 0 by Nakayama’s Lemma (Lemma 231). Thus, M = N as desired.
Proof. We have
Corollary 234. Suppose M is a finitely generated R−module and x1 , . . . , xn ∈ M . Then x1 , . . . , xn generate M if
and only if x1 , . . . , xn generate M/JM where xi = xi + JM . In particular, this means that µR (M ) = µR/J (M/JM ).
Proof. If x1 , . . . , xn generate M , then we always have x1 , . . . , xn generate M/JM . For the reverse direction, let
M
N + JM
N = Rx1 + . . . + Rxn . Then,
=
. This gives that M = N + JM and so by the previous corollary, we
JM
JM
have N = M and so M is generated by x1 , . . . , xn as desired.
Corollary 235. If (R, m) is quasi-local, and M is finitely generated, then
µR (M ) = µR/m (M/mM ) = dimR/m (M/mM ).
3.17
3.17.1
Friday 6 April 2012
More on Localization
Definition 236. Let M be a finitely generated R−module, and T ⊂ M a subset. We say T is a minimal generating
set for M if the elements of T generate M , but not proper subset of T generates M .
Remark 237. Let (R, m) be a quasi-local ring, M be an R−module, and T be a subset of M .
B
B
B
B
Then T is a minimal generating set for M if and only if T is a basis for M/mM .
Every minimal generating set for M has the same number of elements.
Every generating set contains a minimal generating set.
If x ∈ M \ mM , then x is contained in some minimal generating set.
Example 238. Let R = Z/(6)[x], and I = (2, 3x) = (2 − 3x). Note here that both {2, 3x} and {2 − 3x} are minimal
generating sets.
Example 239. Let R = Z(2) . Then R is a local ring with maximal ideal m = (2)(2) . Let M = Q. Then Q = mQ since
a
a
=2·
∈ mQ, but Q 6= 0. Thus, Q is not a finitely generated R−module.
b
2b
45
Definition 240. Let f : M → N be an R−module homomorphism and S a multiplicatively closed subset of R. Then
f
m
f (m) 17
define : MS → NS by
7−→
.
1
s
s
Remark 241. Localization is a functor from the usual category of R−modules to the usual category of RS modules.
f
The functor takes the object M to the object MS and the morphism f : M → N to the morphism : MS → NS .
1
Proposition 242. Localization preserves short exact sequences. That is, if 0 → A → B → C → 0 is a short exact
sequence of R−modules, and S ⊂ R is a multiplicatively closed set, then 0 → AS → BS → CS → 0 is a short exact
sequence of RS −modules.
Proof. Let
0
f
A
g
B
C
0
be a short exact sequence of R−modules and S ⊂ R a multiplicatively closed subset of R. Note that the maps in the
f
f
g
f
a
localization of the sequence are : AS → BS and : BS → CS . First, we’ll show that
is injective, so let ∈ ker .
1
1
1
s
1
f (a)
Then,
= 0 so that there exists some t ∈ S such that tf (a) = 0. Since f is an R−module homomorphism, then
s
a
f
we have f (ta) = 0 so that ta = 0 by f being injective. Hence = 0 in AS and
is injective. Next, we’ll show that
s
1
g
c
g b
g(b)
c
is surjective. Let ∈ CS . Since g is surjective, there exists some b ∈ B such that c = g(b). Then
=
=
1
s
1 s
s
s
f
g
g
g f
and is surjective. Next, we’ll show that image ⊆ ker . Note here that it is enough to show that ◦ = 0. Since
1
1
1
1 1
f
g
gf
g f
the original sequence is exact, then gf = 0, and so
= ◦ = 0 as well. Finally, we’ll show that ker ⊆ image
1
1 1
1
1
b
g
g(b)
so that the sequence will be exact at BS . Let ∈ ker . Then,
= 0 so that there exists some t ∈ S such that
s
1
s
tg(b) = 0. Since g is an R−module homomorphism, we then have that g(tb) = 0. Since ker(g) = image(f ), then there
f a f (a)
tb
b
f
exists some a ∈ A such that f (a) = tb. Thus,
=
= ∈ image which completes the proof.
=
1 st
st
st
s
1
Proposition 243. Let (R, m) be a quasi-local ring, and P a finitely generated projective R−module. Then P is a free
R−module.
Proof. Let n = µR (P ), and say P = Rx1 + . . . + Rxn for some x1 , . . . , xn ∈ P . Define a map φ : Rn → P by ei 7−→ xi .
Then φ is surjective, so we let K = ker φ which gives that the following is a short exact sequence of R−modules
0
K
Rn
φ
P
0.
Since P is projective, this sequence splits, and so Rn ∼
= P ⊕K. Then, modding out by m gives that (R/m)n ∼
= Rn /mRn ∼
=
n
(P ⊕ K)/m(P ⊕ K) ∼
P/mP
⊕
K/mK.
Looking
at
the
dimensions
of
these
over
R/m
gives
dim
(R/m)
= n and
=
R/m
dimR/m P/mP = n since µR (P ) = n (see Corollary 235). Thus, dimR/m K/mK = 0 and so K = mK. By Nakayama’s
Lemma (Lemma 231), we then have that K = 0 and hence P ∼
= Rn .18
Remark 244. Let P be a finitely generated projective R−module. We can define a map from Spec R to N0 by sending
the prime ideal q to rank Pq . This makes sense because by Corollary 229, Pq is projective, and by Proposition 243,
Pq is free. It can be shown that this map is continuous where the Zarisky topology is used on Spec R and the discrete
topology is used on N0 . Hence, if Spec R is connected, then the map is constant. Also, R has no nontrivial idempotents
if and only if Spec R is connected. Lastly, it is a fact that if R has no nontrivial idempotents, then the map is constant.
Theorem 245. Let R be a ring, S ⊂ R a multiplicatively closed set, and φ : R → RS the cannonical map. Then (RS , φ)
has the following universal property: Given any ring, B, and ring homomorphism f : R → B, such that f (s) is a unit in
B for every s ∈ S. Then there exists a unique ring homomorphism h : Rs → B such that the diagram below commutes:
17 One
18 It
could in theory show that this is a well defined RS −module homomorphism, but I believe that the details aren’t tricky.
is clear that K is finitely generated as Rn ∼
= P ⊕ K K so that K is a homomorphic image of Rn .
46
f
R
B
φ
h
RS
Furthermore, if (T, g) where g : R → T also has this property, then T ∼
= RS .
Proof. Given a ring homomorphism f : R → B, define h : RS → B by
r
7−→ f (r)f (s)−1 . We first check
s
r
r0
= 0 ∈ RS . Then there exists some t ∈ S such that ts0 r = tsr0 . We have
s
s
then that h(t) is a unit in B, and so f (t)f (s0 )f (r) = f (ts0 r) = f (tsr0 ) = f (t)f (s)f (r0 ) with f (t), f (s), and
f (s0 ) units. Thus, f (r) = f (s0 )−1 f (t)−1 f (t)f (s0 )f (r) = f (s0 )−1 f (t)−1 f (t)f (s)f (r0 ) = f (s0 )−1 f (s)f (r0 ) and so
f (r)f (s)−1 = f (s)−1 f (s0 )−1 f (s)f (r0 ) = f (r)f (s0 )−1 . This shows that h is well defined. We must next check that h
r r0
is a ring homomorphism, so let , 0 ∈ RS . Then,
s s
0
r r0
s r + sr0
h
+ 0
= h
s s
ss0
that h is well defined. So let
= f (s0 r + sr0 )f (ss0 )−1
=
[f (s0 )f (r) + f (s)f (r0 )] f (s)−1 f (s0 )−1
= f (r)f (s)−1 + f (r0 )f (s0 )−1
0
r
r
+h 0
= h
s
s
and h
r r0
·
s s
3.18
Monday 9 April 2012
r r0 = f (rr0 )f (ss0 )−1 = f (r)f (r0 )f (s)−1 f (s0 )−1 = h
· h 0 which shows that h is a ring
s
s
r
−1
homomorphism. Let r ∈ R. Then (h · φ)(r) = h(φ(r)) = h
= f (r)f (1) = f (r) which shows that the diagram
1
0
in question commutes. Now, if h : RS → B is also a well defined ring homomorphism such that h0 φ = f , then for
r
. Also, for any s ∈ S, then f (s) is a unit and φ(s) is a unit, and so
any r ∈ R, we have f (r) = h0 (φ(r)) = h0
1
s −1
s −1
1
s
−1
0
−1
0
0
0
f (s) = h (φ(s)) = h
=h
=h
since h0 is a ring homomorphism and is a unit in RS . Thus,
1
1
s 1
r
since h0 is a ring homomorphism, we must have that h0
= f (r)f (s)−1 and hence h is unique. The uniqueness of
s
RS is the same argument as it always is to see that objects in a category satisfying universal properties are unique.
For an explicit description, see the Aside near the top of page 2 of the 901 notes.
3.18.1
=h
rr0
ss0
Useful facts about localization
Proposition 246. If N ⊆ M are R−modules and S is a multiplicatively closed subset of R, then (M/N )S ∼
= MS /NS
as RS −modules.
Proof. We have an exact sequence 0 → N → M → M/N → 0 where the map from N to M is the inclusion map.
Localizing at S gives the exact sequence 0 → NS → MS → (M/N )S → 0 and hence MS /NS ∼
= (M/N )S .
Corollary 247. If I ⊆ R is an ideal and S is a multiplicatively closed subset of R, let S = {s = s + I | s ∈ S} and
∼ RS /IS as rings. The map here is given by r 7−→ r .
note that S ⊆ R/I is multiplicatively closed. Then (R/I)S =
s
s
∼
Remark 248. As a special case,let P ∈ Spec R. Then the residue field of RP is RP /PP = (R/P )(0) = Q(R/P ), which
is the field of fractions of R/P .
47
Chapter 4
Post-Exam 2 Material
4.1
Monday 9 April 2012
4.1.1
Tensor Products
Definition 249. Let M, N, A be R−modules. We say a map f : M × N → A is R−bilinear if for every m, m0 ∈ M ,
n, n0 ∈ N and r ∈ R, we have the following:
B f (m + m0 , n) = f (m, n) + f (m0 , n),
B f (m, n + n0 ) = f (m, n) + f (m, n0 ), and
B f (rm, n) = f (m, rn) = rf (m, n).
Example 250. These examples are easily seen to be bilinear.1
B Let S be an R−algebra. Then the multiplication
map · : S × S → S given by (a, b) 7−→ ab is R−bilinear.
r rm
B The map RS × M → MS given by
, m 7−→
is R−bilinear.
s
s
B The map R/I × M → M/IM given by (r, m) 7−→ rm is R−bilinear.
Theorem 251. Let M and N be R−modules. Then there exists an R−module T together with an R−bilinear map
f : M × N → T with the following universal property: Given any R−bilinear map g : M × N → A where A is also an
R−modules, then there exists a unique R−module homomorphism h : T → A such that the following diagram commutes:
g
M ×N
A
f
h
T
Furthermore, if f 0 : M × N → T 0 also has this property, then T ∼
= T 0.
Proof. Once we have the existence of such a module T , the uniqueness of T is the same argument as it always is to
see that objects in a category satisfying universal properties are unique. For an explicit description, see the Aside near
the top of page 2 of the 901 notes. For this proof, we’ll explicitly construct T . Let F be a free R−module with basis
S = {e(m,n) | (m, n) ∈ M × N }. Then S is in bijection with M × N and every element of F can be written uniquely
t
X
as
ri e(mi ,ni ) where ri ∈ R. Let U be the submodule of F which is generated by all elements of the following forms:
i=1
B
B
B
B
e(m+m0 ,n) − e(m,n) − e(m0 ,n) ,
e(m,n+n0 ) − e(m,n) − e(m,n0 ) ,
e(rm,n) − re(m,n) , and
e(m,rn) − re(m,n)
where m, m0 ∈ M , n, n0 ∈ N , and r ∈ R. Set T = F/U and let m⊗n denote e(m,n) +U ∈ T = F/U . Using the above definitions, we see that (m+m0 )⊗n = e(m+m0 ,n) +U = (e(m,n) +e(m0 ,n) )+U = (e(m,n) +U )+(e(m0 ,n) +U ) = m⊗n+m0 ⊗n.
Similarly, m ⊗ (n + n0 ) = m ⊗ n + m ⊗ n0 , and (rm) ⊗ n = r(m ⊗ n) = m ⊗ (rn). It is clear that T is an R−module since
it is a quotient of a free R−module. Define f : M × N → T by (m, n) 7−→ m ⊗ n. By construction of T it is clear that
f is R−bilinear. It remains then to check the universal property, so let g : M × N → A be an R−bilinear map. Define
an R−module homomorphism e
h : F → A by e(m,n) 7−→ g(m, n). This is well defined since F is a free R−module. Since
g is R−bilinear, then U ⊆ ker e
h. For example, e
h(e(m,n+n0 ) − e(m,n) − e(m,n0 ) ) = g(m, n + n0 ) − g(m, n) − g(m, n0 ) and
this is equal to 0 because g is R−bilinear. Therefore, e
h induces an R−module homomorphism h : T → A which is given
1 The
details aren’t worth doing.
48
by m ⊗ n 7−→ g(m, n) and this map is well defined. The diagram in question commutes by definition of h. So suppose
there is a second map h0 : T → A which also satisfies g = h0 f . Then h0 (m ⊗ n) = h0 f (m, n) = g(m, n) = h(m ⊗ n)
and so h is unique because the set {m ⊗ n | m ∈ M, n ∈ N } generates T .
Definition 252. The module T in the above theorem is called the tensor product of M and N over R and is denoted
M ⊗R N .
4.2
Wednesday 11 April 2012
4.2.1
Tensor Product Properties
Remark 253. If M = Ru1 + . . . + Rut and N = Rw1 + . . . + Rws , then M ⊗R N =
X
R(ui ⊗ wj ). More precisely,
i,j
elements of M are of the form m =
and n is m ⊗ n =
t
X
i=1
!
ri ui
t
X
ri ui and elements of N are of the form n =
s
X
rj0 wj so that the tensor of m
j=1
 i=1 
s
X
X
⊗
rj0 wj  =
ri rj0 (ui ⊗ wj ). Thus, µR (M ⊗R N ) ≤ µR (M )µR (N ).2 Thus, if M
j=1
i,j
and N are finitely generated, we have that M ⊗R N is also finitely generated.
Proposition 254. Let M, N, L, and Mi for i ∈ Λ be R−modules, S a multiplicatively closed subset of R, and I an
ideal of R. Then we have the following isomorphisms along with the maps giving the isomorphisms:
!
M
M
1. M ⊗R N ∼
3. RS ⊗R M ∼
= N ⊗R M
= MS
(Mi ⊗R N )
5.
M i ⊗R N ∼
=
rm
r
m ⊗ n 7−→ n ⊗ m
⊗ m 7−→
i∈Λ
i∈Λ
s
s
(mi ) ⊗ n 7−→ (mi ⊗ n)
2. R ⊗R M ∼
=M
4. R/I ⊗R M ∼
= M/IM
r ⊗ m 7−→ rm
6. (M ⊗R N ) ⊗R L ∼
= M ⊗R (N ⊗R L)
r ⊗ m 7−→ rm
(m ⊗ n) ⊗ ` 7−→ m ⊗ (n ⊗ `)
Proof. The proofs of these statements are all very similar in form. We’ll do all the details for 3 and the important
details for 4. Also, property 6 above will be proved in more generality in Proposition 280.
r rm
, m 7−→
. We first need to show that fe is well defined, so suppose that
3. Define a map fe : RS × M → MS by
s
s
r
r0
rm
r0 m
= 0 . Then there exists some t ∈ S such that ts0 r = tsr0 . Thus, ts0 rm = tsr0 m so that
= 0 which shows
s
s
s
s
r r0
0
e
e
that f is well defined. Next, we must show that f is bilinear, so let , 0 ∈ RS , m, m ∈ M , and a ∈ R. Then we
s s
have the following:
0
0 r r0
s r + sr0
(s0 r + sr0 )m
s0 rm + sr0 m
rm r0 m
e r , m + fe r , m ,
fe
+ 0 , m = fe
,
m
=
=
=
+
=
f
s s
ss0
ss0
ss0
s
s0
s
s0
r
r(m + m0 )
r r
rm rm0
fe , m + m0 =
=
+
= fe , m + fe , m0 , and
s
s
s
s
s
s
r ar arm
r
r ram
rm
fe a · , m = fe
,m =
=
= fe , am = a
= afe , m .
s
s
s
s
s
s
s
Thus, by definition of the tensor product, there is an R−module homomorphism f : RS ⊗R M → MS such that the
r
r
diagram below commutes where the map φ : RS × M → RS ⊗R M is the canonical map given by
, m 7−→ ⊗ m:
s
s
e
f
RS × M
MS
φ
f
RS ⊗R M
2I
believe that Tom said that we get equality when R is local and M, N are finitely generated.
49
r
rm
⊗m =
by commutativity of the diagram. In order to show that f is an
s
s
m
1
isomorphism, we’ll show that it has an inverse. So, let g : MS → RS ⊗R M be given by
7−→ ⊗ m. First, we
s
s
m m0
have that for any , 0 ∈ MS and any r ∈ R, that
s s
0
m m0
s m + sm0
g
= g
+ 0
s
s
ss0
1
=
⊗ (s0 m + sm0 )
ss0
1
1
⊗ s0 m + 0 ⊗ sm0
=
0
ss
ss
s0
s
=
⊗ m + 0 ⊗ m0
ss0
ss
1
1
=
⊗ m + 0 ⊗ m0
s
s m
m0
+g
= g
s
s0
The map f then must act as f
and
m
rm 1
m
r
1
g r
=g
= ⊗ rm = ⊗ m = r
⊗ m = rg
,
s
s
s
s
s
s
so that g is an R−module homomorphism. Next,we need to
show that g is well defined. Since it is an R−module
m
m0
m m0
− 0 = 0 only if g is well defined, so it is sufficient to show
homomorphism, then if
=
, we’ll get that g
s s0
s
s
m
m
m
that if
= 0, then g
= 0. To that end, suppose that
= 0. Then by definition, there exists some t ∈ S
s
s
s
t
1
1
1
⊗m =
⊗ tm =
⊗ 0 = 0 so that g is indeed well defined. It then only
such that tm = 0. Now, ⊗ m =
s
st
st
st
remains to show that g and f are inverses. It is enough to show this on a generating set since they’re both R−module
r
m
homomorphisms. So, let ⊗ m ∈ RS ⊗R M and
∈ MS . Then, we have
s
s
r
r
rm 1
r
(g ◦ f )
⊗m =g f
⊗m =g
= ⊗ rm = ⊗ m
s
s
s
s
s
and
(f ◦ g)
m
s
m 1
m
=f g
=f
⊗m =
s
s
s
so that f ◦ g = idMS and g ◦ f = idRS ⊗S M as desired.
4. Define fe : R/I × M → M/IM by (r, m) 7−→ rm. It is easy to check that fe is well defined and R−bilinear. Thus,
it induces an R−module homomorphism f : R/I ⊗R M → M/IM where r ⊗ m 7−→ rm. Next, define a map
g : M → R/I⊗
homomorphism.
If x ∈ IM
Pn R M by m 7−→ 1⊗m. It is easy to check that g is an R−moduleP
P
P, then we can
write x = j=1 rj mj for some rj ∈ I and mj ∈ M . Then, g(x) = 1⊗x = 1⊗ rj mj = (rj ⊗mj ) = (0⊗mj ) = 0
so that IM ⊆ ker(g). Thus, we have an induced R−module homomorphism g : M/IM → R/I ⊗R M which is given
by m 7−→ 1 ⊗ m. It is easy to check that f and g are inverses.
Remark 255. If F is a free R−module of rank n and G is a free R−module of rank m, then
F ⊗R G ∼
= Rn ⊗R Rm ∼
=
n
M
(R ⊗R Rm ) ∼
=
i=1
n
M
Rm ∼
= Rmn
i=1
so that F ⊗R G is a free R−module of rank mn. More generally, if {ui }i∈I is a basis for F and {vj }j∈J is a basis for
G, then {ui ⊗ vj }(i,j)∈I×J is a basis for F ⊗R G.
50
4.3
4.3.1
Friday 13 April 2012
More on Tensor Products
Remark 256. If M and N are R−modules, and m ∈ M , n ∈ N , it is a hard question in general to determine when
m ⊗ n = 0 in M ⊗R N .
Lemma 257. Let M, N be R−modules and m ∈ M , n ∈ N . Then m ⊗ n = 0 if and only if f (m, n) = 0 for every
R−bilinear map f : M × N → A where A is an R−module.
Proof. (⇐) The canonical map φ : M × N → M ⊗R N given by (m, n) 7−→ m ⊗ n is R−bilinear, so if f (m, n) = 0
for every such map f , then m ⊗ n = φ(m, n) = 0.
(⇒) If f : M × N → A is R−bilinear, then there exists an R−module homomorphism h : M ⊗R N → A such that
the following diagram commutes.
f
M ×N
A
φ
h
M ⊗R N
Thus, f (m, n) = h(m ⊗ n) = h(0) = 0 as desired.
Example 258. Consider 2 ⊗ 1 in Z ⊗Z Z/2Z. We have
2 ⊗ 1 = 2 · 1 ⊗ 1 = 1 ⊗ 2 · 1 = 1 ⊗ 2 = 1 ⊗ 0 = 0.
Now, consider 2 ⊗ 1 in 2Z ⊗Z Z/2Z. Here, we have that 2 ⊗ 1 6= 0 and we can show this using the previous lemma.
Let f : 2Z × Z/2Z → Z/2Z be given by (2n, a) 7−→ na. It is easy to check that f is Z−bilinear. Also, f (2, 1) = 1 =
6 0
and so 2 ⊗ 1 6= 0 in 2Z ⊗Z Z/2Z.
Remark 259. The lesson to learn from the previous example is that if M 0 ⊆ M is a submodule, then M 0 ⊗R N cannot
generally be identified with a submodule of M ⊗R N since the ”inclusion” map may not be injective.
4.3.2
Functors on Module Categories
Notation. We will use the notation R − mod to denote the category of R−modules where the objects are R−modules
and the morphisms are R−module homomorphism.
Note. If M, N are R−modules, then HomR (M, N ) := {f : M → N | f is an R − module homomorphism} is an
R−module.
Recall. A covariant functor is a map F : R − mod → S − mod such that F (M ) is an S−module for every R−module
M , and given morphisms f : M → N and g : N → L in R − mod, then F (gf ) = F (g)F (f ). Also, we must have that
F (idM ) = idF (M ) .
Proposition 260. Let f : M → N and g : A → B be R−module homomorphisms. Then there is an R−module homomorphism f ⊗g : M ⊗R A → N ⊗R B such that m⊗a 7−→ f (m)⊗g(a). We say that f ⊗g is the tensor product of f and g.
Proof. Define f × g : M × A → N ⊗R B by (m, a) 7−→ f (m) ⊗ g(b). It is easy to show that f × g is R−bilinear. Thus,
there is an induced R−linear homomorphism f ⊗ g : M ⊗R A → N ⊗R B and it is given by m ⊗ a 7−→ f (m) ⊗ g(a).
Definition 261. Given an R−module, M , we can define a covariant functor FM : R − mod → R − mod by
FM (N ) = M ⊗R N and if g : N → L is a morphism in R − mod, then FM (g) = idM ⊗ g : M ⊗R N → M ⊗R L. The
typical notation for this functor is M ⊗R −.
Remark 262. The functor − ⊗R M can be similarly defined, and is also covariant.
Definition 263. A functor F : R − mod → S − mod is called additive if for R−modules M, N , and f, g ∈
HomR (M, N ), then F (f + g) = F (f ) + F (g).
Exercise. If F is an additive functor, then F (0) = 0 on both maps and objects.3
Note. The functors M ⊗R − and − ⊗R M are additive functors.
3 There
was some argument about whether or not this is actually true, and so I should do this exercise soon!
51
Notation. Let A be an R−module, and x ∈ R. Then µx,A is the map A → A given by a 7−→ xa.
Definition 264. A functor F : R − mod → S − mod is called multiplicative if given an R−module, M , and any
r ∈ R, then F (µr,M ) = µr,F (M ) .
Note. The functors M ⊗R − and − ⊗R M are multiplicative.
Proposition 265. Let R → S be a ring homomorphism, and let M be an S−module.4 For any R−module, N , then
M ⊗R N is an S−module via s · (m ⊗ n) := (sm) ⊗ n.
Proof. Let s ∈ S, and let fs : M × N → M ⊗R N be given by (m, n) 7−→ (sm) ⊗ n. This is R−bilinear and so it
induces an R−module homomorphism M ⊗R N → M ⊗R N .
Remark 266. If in the previous proposition, we choose M = S, then we see that S ⊗R − is an additive and multiplicative
functor. Moreover, S ⊗R M is an S−module.
Example 267. These are some of the most commonly used5 additive and multiplicative functors in commutative algebra.
B Modding out: Let I be an ideal of R. Then R/I ⊗R − : R − mod → R/I − mod is the functor which takes the
R−module, M to R/I ⊗R M ∼
= M/IM .
B Localization: Let S be a multiplicatively closed subset of R. Then RS ⊗R − : R − mod → RS − mod is the
functor which takes the R−module, M to RS ⊗R M ∼
= MS .
Definition 268. A covariant additive functor F on module categories is called right exact if whenever
f
L
g
M
0
N
is exact, then
F (L)
F (f )
F (M )
F (g)
F (N )
0
is also exact.
4.4
Monday 16 April 2012
4.4.1
Tensor Product is a Right Exact Functor
Theorem 269. The functor M ⊗R − is right exact.
Proof. Let
f
A
g
B
0
C
be an exact sequence of R−modules, and let M be an R−module. We need to show that
M ⊗R A
1⊗f
1⊗g
M ⊗R B
M ⊗R C
0
n
X
mi ⊗ci ∈ M ⊗R C. Since g is surjective, then for each i ∈
!
n
n
n
X
X
X
{1, . . . , n} there exists bi ∈ B such that g(bi ) = ci . Thus, (1 ⊗ g)
mi ⊗ bi =
mi ⊗ g(bi ) =
bi ⊗ ci as desired.
is exact. We’ll start by showing that 1⊗g is surjective. So let
i=1
i=1
i=1
i=1
Next, we have that gf = 0 since the original sequence was exact and so we have (1⊗g)(1⊗f ) = 1⊗(gf ) = 1⊗0 = 0 which
gives that image(1⊗f ) ⊆ ker(1⊗g). Finally, let E = image(1⊗f ) ⊆ M ⊗R B. As E ⊆ ker(1⊗g), we have an induced map
h := 1 ⊗ g : (M ⊗R B)/E → M ⊗R C which is given by m ⊗ b 7−→ m⊗g(b). Note that h is surjective by definition. Hence,
it suffices to show that h is an isomorphism, and so we’ll produce an inverse for h. To that end, define α : M ×C → (M ⊗R
B)/E by (m, c) 7−→ m ⊗ b where b ∈ B is chosen so that g(b) = c. This can be done since g is surjective, but we must show
that α is independent of the choice of b. If g(b) = g(b0 ) = c, then g(b−b0 ) = 0 which gives that b−b0 ∈ ker(g) = image(f ) so
there exists some a ∈ A such that f (a) = b−b0 . Thus, m⊗b−m⊗b0 = m⊗(b−b0 ) = m⊗f (a) = (1⊗f )(m⊗a) ∈ image(1⊗
f ) = E and hence m ⊗ b = m ⊗ b0 . It is easy to show that α is R−bilinear, and so there is an induced map α
e : M ⊗R C →
4 And
5I
so M will also be an S−module.
believe.
52
M ⊗R B/E such that α
e(m ⊗ c) = m ⊗ b for any b ∈ B with g(b) = c. If m ⊗ b ∈ M ⊗R B/E, then α
eh(m ⊗ b) = α
e(m ⊗
eh = idM ⊗R B/E on a generating set, and hence on all of M ⊗R B/E. Similarly, if m⊗c ∈ M ⊗R C,
g(b)) = m ⊗ b so that α
then he
α(m ⊗ c) = h(m ⊗ b) = m ⊗ g(b) = m ⊗ c for any b ∈ B such that g(b) = c. Thus, he
α = idM ⊗R C on a generating
set, and hence on all of the module. Thus, h is an isomorphism, and so image 1 ⊗ f = E = ker 1 ⊗ g as desired.
Remark 270. The functor M ⊗R − does not necessarily preserve injections!
Example 271. It is clear that the inclusion map i : 2Z → Z is injective. However, when we apply the functor
− ⊗Z Z/2Z we obtain the map 2Z ⊗Z Z/2Z → Z ⊗Z Z/2Z which takes 1 ⊗ 2 to 1 ⊗ 2. In Example 258 we showed that
1 ⊗ 2 is nonzero in 2Z ⊗Z Z/2Z but is equal to zero in Z ⊗Z Z/2Z and so this map is not injective.
4.4.2
Flat Modules
Definition 272. An R−module M is called flat (over R) if M ⊗R − preserves injections, or equivalently we can say
that it is an exact functor.
Lemma 273. Suppose the fullowing diagram commutes
A
α
β
B
g
f
A
0
C
α0
B
0
h
β0
C0
and f, g, h are isomorphism. Then the top row is exact if and only if the bottom row is exact.
Proof. Since f, g, h are isomorphisms, it is enough to prove that the bottom row is exact if the top row is exact. Let
b0 ∈ image α0 . Then there exists some a0 ∈ A0 such that α0 (a0 ) = b0 . Since f is an isomorphism, there is some a ∈ A
such that f (a) = a0 . Then as the top row is exact, we have that β(α(a)) = 0 ∈ C. As h is an isomorphism, then
h(β(α(a))) = 0. Now, by commutativity of the diagram, hβα = β 0 α0 f and so β 0 α0 f (a) = 0 as well. Note then that
β 0 (α0 (f (a))) = β 0 (α0 (a0 )) = β 0 (b0 ). Thus, b0 ∈ ker β 0 .
Next, let b0 ∈ ker(β 0 ). As g is an isomorphism, there is some b ∈ B such that g(b) = b0 . Also, as b0 ∈ ker β 0 ,
then β 0 (b0 ) = 0. As h is an isomorphism, then h−1 (β 0 (b0 )) = 0 as well. Note that h−1 β 0 g = β by the commutativity
of the right square, and so β(b) = h−1 (β 0 (g(b))) = h−1 (β 0 (b0 )) = h−1 (0) = 0. thus, b ∈ ker β = image α so there
exists some a ∈ A such that α(a) = b. Let a0 = f (a). Then, by the commutativity of the let square, we have that
α0 (a0 ) = α0 (f (a)) = g(α(a)) = g(b) = b0 so that b0 ∈ image α0 .
L
Lemma 274. Let {Ai }i∈Λ be a family of R−modules. Then Ai is flat for every i ∈ Λ if and only if i∈Λ Ai is flat.
Proof. Let
0
f
M
be exact, and consider the commutative diagram below:
!
M
1A ⊗ f
Ai ⊗ R M
0
i∈Λ
M
0
(Ai ⊗R M )
M
!
M
Ai
⊗R N
i∈Λ
1Ai ⊗ f 0
i∈Λ
where 1A is the identity map on
N
M
(Ai ⊗R N )
i∈Λ
Ai and 1Ai is the identity map on Ai . The vertical maps are the canonical
i∈Λ
isomorphisms as given in Proposition 254. Chasing the element (ai ) ⊗ m through the diagram gives:
53
(ai ⊗ m)
(ai ) ⊗ f (m)
(ai ⊗ m)
(ai ⊗ f (m))
M
so that the diagram commutes. Then we have that
Ai is flat if and only if the top row is exact, which by Lemma 273
i∈Λ
is if and only if the bottom row is exact. In turn, the bottom row is exact if and only if it is exact on each component,
and hence if and only if Ai is flat for all i ∈ Λ.
Proposition 275. Every projective R−module is flat.
Proof. Let P be a projective R−module. Then there exists some R−module, Q, and a free R−module, F , such that
F ∼
= P ⊕ Q. Thus, it is enough to show that every free module is flat. However, since every free R−module is just
a direct sum of some copies of R, then it is sufficient to show that R is flat. Let
f
0
M
N
be exact. Then consider the following diagram where the vertical maps are given by the canonical isomorphisms in
Proposition 254.
0
R ⊗R M
0
M
1⊗f
R ⊗R N
f
N
It is easy to show that the diagram commutes, and the bottom row is already exact. Thus, by Lemma 273, the top
row is exact and so R is flat, which completes the proof.
Proposition 276. Let S be a multiplicatively closed subset of R. Then RS is a flat R−module.
Proof. Let
f
0
M
N
be exact. Then consider the following diagram where the vertical maps are given by the canonical isomorphisms in
Proposition 254.
0
0
RS ⊗R M
1⊗f
f
1
MS
R S ⊗R M
NS
The diagram is easily shown to commute. Also, the bottom row is exact since localization is an exact functor. Thus,
the top row is exact, and so RS is a flat R−module.
Definition 277. A ring homomorphism φ : R → S is called a flat ring map if S is a flat R−module.
Example 278. The following ring homomorphisms are flat ring maps:
B The identity map. id : R → R.
B The canonical localization map. φ : R → RS .
B The inclusion map for adjoining an indeterminant. φ : R → R[x].
B The completion map. φ : R → R̂.
54
B In a local ring (R, m) of characteristic p > 0, the Frobenius map f : R → R given by r 7−→ rp is flat if and only
if R is a regular local ring.6 Which happens if and only if µR (m) = dim R.
4.5
Wednesday 18 April 2012
4.5.1
More Tensor Products
Example 279. How many distinct elements are in the Z−module Z45 ⊗Z Z3 ⊗Z Z/100Z? This question is easily
answered by using the isomorphisms in Proposition 254. Note that in general (R/I)S ∼
= R/I ⊗R RS ∼
= RS /IS . We have
Z45 ⊗Z Z3 ⊗Z Z/100Z ∼
= Z3 ⊗Z (Z45 ⊗Z Z/100Z)
∼
= Z3 ⊗Z (Z45 /100Z45 )
∼
= Z3 ⊗Z (Z45 /4Z45 )
∼
= Z3 ⊗Z (Z/4Z)45
∼
= Z3 ⊗Z (Z/4Z)
∼
= (Z ⊗Z Z/4Z)3
∼
= (Z/4Z)3
92
1
= 2 so 52 is a unit in Z45 , and also that 45 ≡ 1 in Z/4Z. Hence,
452
5
Z45 ⊗Z Z3 ⊗Z Z/100Z has 43 = 64 distinct elements.
using the fact that 100 = 52 · 4, that
Note. If R → S is a ring homomorphism, A is an R−module, and B is an S−module, then A ⊗R B is an S−module
via s · (a ⊗ b) = a ⊗ (sb).
Proposition 280. Let R → S be a ring homomorphism, A be an R−module, and B, C be S−modules. Note that B, C
are also R−modules via the homomorphism. Then
A ⊗R (B ⊗S C) ∼
= (A ⊗R B) ⊗S C
as S−modules and the isomorphism is given by a ⊗ (b ⊗ c) 7−→ (a ⊗ b) ⊗ c.
Proof. Fix some a ∈ A and define a map fa : B × C → (A ⊗R B) ⊗S C which is given by (b, c) 7−→ (a ⊗ b) ⊗ c. It is
easy to show that fa is S−bilinear and so there is an induced S−module homomorphism fea : B ⊗S C → (A ⊗R B) ⊗S C
under which b ⊗ c 7−→ (a ⊗ b) ⊗ c. Now, define a map g : A × (B ⊗S C) → (A ⊗R B) ⊗S C by (a, x) 7−→ fea (x). In
particular g(a, b ⊗ c) = (a ⊗ b) ⊗ c. It is easy to see than g is R−bilinear, and so there is an induced R−module
homomorphism ge : A ⊗R (B ⊗S C) → (A ⊗R B) ⊗S C under which a ⊗ (b ⊗ c) 7−→ (a ⊗ b) ⊗ c. Note that ge(s · x) = se
g (x)
is clear on a generating set, and so ge is actually S−linear. Similarly, one can define an S−linear map in the other
direction, and they’ll clearly compose to the identity on a generating set, and hence will be inverses.
Example 281. Let M, N be R−modules, and I an ideal of R such that IM = IN = 0. Then note that for A = M
or A = N , we have A ⊗R R/I ∼
= A/IA ∼
= A/0 ∼
= A since IA = 0. Then,
M ⊗R N ∼
= M ⊗R/I N
= (M ⊗R/I RI ) ⊗R N ∼
= M ⊗R/I (R/I ⊗R N ) ∼
by using the properties in Proposition 254.
Example 282. The exercise on the homework which states “Let (R, m) be a quasi-local ring and M, N finitely
generated R−modules. Then M ⊗R N = 0 if and only if M = 0 or N = 0.” is false if R is not quasi local or if M
or N is not finitely generated. In particular, note that Z/2Z 6= 0 and Z/3Z 6= 0. However, Z is not quasi-local and
Z/2Z
Z/2Z ⊗Z Z/3Z ∼
= 0. Also, if R = Z(2) , then R 6= 0 is local; however, R/(2)(2) ⊗R Q ∼
=
= Z(2) /2Q(2) ∼
= Q/Q = 0
3(Z/2Z)
despite the fact that R/(2)(2) 6= 0 and Q 6= 0. This is okay because Q is not a finitely generated Z(2) −module.
Note. Anything we’ve said about the functor M ⊗R − also holds for the functor − ⊗R M since they’re naturally
equivalent functors.
Definition 283. Let M be an R−module and define a functor HomR (M, −) : R − mod → R − mod on objects as N 7−→ HomR (M, N ) = {f : M → N | f is an R − module homomorphism} and on morphisms as
(f : N1 → N2 ) 7−→ (f∗ : HomR (M, N1 ) → HomR (M, N2 )) where f∗ (g) = f g.
Note. The functor HomR (M, −) is covariant, additive, and multiplicative.
6 That
result is due to Kunz in 1969.
55
4.6
4.6.1
Friday 20 April 2012
Tangent in Commutative Algebra
Lemma 284. Let k be an infinite field and f (x1 , . . . , xn ) ∈ k[x1 , . . . , xn ] \ {0}. then there exists some (a1 , . . . , an ) ∈ k n
such that f (a1 , . . . , an ) 6= 0.
Proof. We’ll prove this using induction on n. When n = 1, then every nonzero polynomial, f (x) ∈ k[x] has only finitely
many roots. Since k is an infinite field, then there must be some a ∈ k such that f (a) = 0. For the inductive step,
let n > 1 and let f ∈ k[x1 , . . . , xn ] be such that f 6= 0. Note that we can write f (x1 , . . . , xn ) = fr (x1 , . . . , xn−1 )xrn +
fr−1 (x1 , . . . , xn−1 )xr−1
+ . . . + f0 (x1 , . . . , xn−1 ) where fr 6= 0. By induction, choose (a1 , . . . , an−1 ) ∈ k n−1 such that
n
fr (a1 , . . . , an−1 ) 6= 0. Then, let g(xn ) = f (a1 , . . . , an−1 , xn ) = fr (a1 , . . . , an−1 )xrn + . . . + f0 (a1 , . . . , an−1 ). By the
n = 1 case, we have that there is some an ∈ k such that f (a1 , . . . , an−1 , an ) = g(an ) 6= 0.
Note. Lemma 284 is false if k is a finite field. This is easily seen by considering xp − x.
Proposition 285. Let k be an infinite field and V a finite dimensional vector space. Then V is not the union of
finitely many proper subspaces.
Proof. Let v1 , . . . , vn be a basis for V . Suppose that V = W1 ∪ . . . ∪ Wt where Wi are proper subspaces of V . Without
loss of generality, we can assumePthat dim Wi = n − 1 for all i. Let W ⊆ V be any n − 1 dimensional subspace with
n
basis {u1 , . . . , un−1 }, then ui = j=1 ai,j vj for some ai,j ∈ k. Thus, [aij ] is an n − 1 by n matrix of rank n − 1. Let
Pn
v ∈ V . Then v = i=1 ci vi for some ci ∈ k. Then v ∈ W if and only if

a1,1
a2,1
..
.
···
···
..
.
...
...
a1,2
a2,2
..
.


rank 

 a
n−1,1
c1
an−1,2
c2
a1,1
a2,1
..
.
a1,2
a2,2
..
.
a1,n
a2,n
..
.

an−1,n
cn


=n−1


which is true if and only if



det 

 a
n−1,1
c1
Let

an−1,2
c2
a1,1
a2,1
..
.


fW (x1 , . . . , xn ) = det 

 a
n−1,1
x1
···
···
..
.
...
...
a1,2
a2,2
..
.
an−1,2
x2
a1,n
a2,n
..
.
an−1,n
cn
···
···
..
.
...
...



 = 0.


a1,n
a2,n
..
.
an−1,n
xn



 ∈ k[x1 , . . . , xn ]


and note that fW (c1 , . . . , cn ) = 0 if and only if v ∈ W . Also note that fW 6= 0 since W is a proper subspace of V , so
Qt
there is some v 0 ∈ V \ W . Now, let fi = fWi as constructed. Then set f = i=1 fi ∈ k[x1 , . . . , xn ] and f 6= 0 since
each fi =
6 0. As |k| = ∞, there is some (c1 , . . . , cn ) ∈ k n such that f (c1 , . . . , cn ) 6= 0. Thus, fi (c1 , . . . , cn ) 6= 0 for each i
and so v ∈
/ Wi for all i. This is a contradiction since V = W1 ∪ . . . ∪ Wt and so we must have that V cannot be written
as the union of finitely many proper subspaces.
Proposition 286. Let (R, m) be a quasi-local ring and let M be a finitely generated R−module such that R/m is an
infinite field. Then M is not the union of finitely many proper subspaces.
N1 + mM
Proof. Suppose that M = N1 ∪ . . . ∪ Nt where N1 , . . . , Nt are submodules of M . Then M/mM =
∪ ... ∪
mM
Nt + mM
Ni + mM
. Note that M/mM is a finitely generated R/m−vector space. By the field case, M/mM =
for
mM
mM
some i. Then we have that M = Ni + mM and so by Nakayama’s Lemma (Lemma 231), we have that M = Ni .
Remark 287. The following is a useful way to force the quotient field to be infinite: Suppose (R, m) is a quasi-local
ring, and R/m is a finite field. Consider the ring S = R[x]mR[x] . Note that mR[x] = m[x] is prime in R[x] since
56
R[x]/mR[x] ∼
= (R/m)[x] is a domain since R/m is a field. So S is a quasi-local ring. Let n = mR[x]mR[x] and note that
n is the unique maximal ideal of S. Then,
R[x]mR[x] ∼
R[x]
S
∼
=
=
= (R/m)[x]0
n
mR[x]mR[x]
mR[x] mR[x]
which is the field of fractions of (R/m)[x] and hence equal to (R/m)(x) which is an infinite field!
Proposition 288. Let φ : R → S be a ring homomorphism where S is a flat R−module (or in equivalent language,
φ is a flat ring map). Let M be a flat S−module. Then M is a flat R−module.
Proof. Let 0 → A → B be an injection of R−modules (that is, an exact sequence). Since S is flat as an R−module,
then 0 → A ⊗R S → B ⊗R S is exact and now these are S−module homomorphisms. Since M is a flat S−module, then
0 → (A ⊗R S) ⊗S M → (B ⊗R S) ⊗S M is also exact. Using the associative property we proved for tensor products,
then 0 → A ⊗R (S ⊗S M ) → B ⊗R (S ⊗S M ) is exact. Since S ⊗S M ∼
= M , then we have that 0 → A ⊗R M → B ⊗R M
is exact using Lemma 273 and hence M is a flat R−module since these are now R−module homomorphisms.
Corollary 289. If φ : R → S and ψ : S → T are flat ring maps, then ψφ : R → T is also a flat ring map.
Proof. This is done by taking M = T from the previous proposition.
Remark 290. The natural ring homomorphism f : R → S where R and S are as in Remark 287 is flat. Also, in the
same context the maximal ideal of R extended to S is mS = n.
Proof. The map φ : R → R[x] is flat since R[x] is a free R−module. Also, ψ : R[x] → R[x]mR[x] is flat because localization
is always flat. Thus, ψφ = f : R → R[x]mR[x] (which is the natural ring homomorphism) is a flat map by Corollary 289.
Remark 291. Let (R, m) be quasi-local and M a finitely generated R−module. Consider N = M ⊗R S where S =
R[x]mR[x] . Then N is a finitely generated S−module and |S/n| = ∞ where n is the maximal ideal of the quasi-local ring S.
Proposition 292. In particular, in the case of the previous remark, we have λR (M ) = λS (N ).
Proof. We proceed by induction on λR (M ). When λR (M ) = 1, then M is a simple module and so M ∼
= R/m. Thus,
N = M ⊗R S ∼
= R/m ⊗R S ∼
= S/mS ∼
= S/n which is a simple S−module, and so λS (N ) = 1 as well. For the
inductive step, assume that λR (M ) > 1 and let A 6= 0 be a proper submodule of M . Then 0 → A → M → M/A → 0
is an exact sequence of R−modules. By additivity of length on exact sequences (see Proposition 44), we have
λR (M ) = λR (A) + λR (M/A). Note that S is flat by Remark 290. Thus, applying the functor − ⊗R S to the short
exact sequence 0 → A → M → M/A → 0 gives the short exact sequence
0 → A ⊗R S → M ⊗R S → M/A ⊗R S → 0.
Since λR (A), λR (M/A) < λR (M ), then by our inductive hypothesis λR (A) = λS (A ⊗R S) and λR (M/A) =
λS (M/A ⊗R S). Again using Proposition 44, we have that
λS (N ) = λS (M ⊗R S) = λS (A ⊗R S) + λS (M/A ⊗R S) = λR (A) + λR (M/A) = λR (M )
as desired.
4.7
4.7.1
Monday 23 April 2012
Covariant Hom Functor
Recall. If M, N are R−modules, then HomR (M, N ) := {f : M → N | f is an R − module homomorphism} is an
R−module via (rf )(m) = rf (m) and (f1 + f2 )(m) = f1 (m) + f2 (m).
Proposition 293. Here are some useful isomorphisms:
1. For any R−module N , HomR (R, N ) ∼
= N via the map f 7−→ f (1).
2. If I is an ideal of R and N is an R−module, then HomR (R/I, N ) ∼
= (0 :N I) via the map f 7−→ f (1). Note that
(0 :N I) := {u ∈ N | Iu = 0}.
3. If A, B, N are R−modules, then HomR (A ⊕ B, N ) ∼
= HomR (A, N ) ⊕ HomR (B, N ) via the map f 7−→ (f |A , f |B ).
Proof. 1. Let φ : HomR (R, N ) → N be defined by φ(f ) = f (1). Then for any f, g ∈ HomR (R, N ), we have
φ(f + g) = (f + g)(1) = f (1) + g(1) = φ(f ) + φ(g). Also, if r ∈ R, then φ(rf ) = (rf )(1) = rf (1) = rφ(f )
and so φ is an R−module homomorphism. Next, suppose φ(f ) = φ(g), then f (1) = g(1) and for any r ∈ R,
f (r) = f (r · 1) = rf (1) = rg(1) = g(r · 1) = g(r) and so f = g, which shows that φ is injective. Now, let n ∈ N .
Then define an R−linear map f : R → N by f (1) = n. Then φ(f ) = f (1) = n so that φ is surjective.
57
2. Let φ : HomR (R/I, N ) → (0 :N I) be defined by φ(f ) = f (1). First we should check that the image of φ is contained in
(0 :N I). So let f ∈ HomR (R/I, N ). Then since f is an R−module homomorphism, If (1) = f (I1) = f (0) = 0 so that
f (1) ∈ (0 :N I) as claimed. Next, we show that φ is an R−module homomorphism. Let f, g ∈ HomR (R/I, N ). Then
φ(f +g) = (f +g)(1) = f (1)+g(1) = φ(f )+φ(g). If additionally, r ∈ R, then φ(rf ) = (rf )(1) = rf (1) = rφ(f ). Now,
if φ(f ) = φ(g), then f (1) = g(1). The for any r ∈ R, we have that since f, g are R−module homomorphisms, f (r) =
f (r1) = rf (1) = rg(1) = g(r1) = g(r) so that f = g and φ is injective. Finally, if n ∈ (0 :N I), then In = 0. This
means that we can define an R−module homomorphism f : R → N by f (1) = n and that I ⊆ ker f and hence we have
an induced R−module homomorphism fe : R/I → N where fe(1) = n. Hence, φ(fe) = fe(1) = n and so φ is surjective.
3. Let φ : HomR (A ⊕ B, N ) → HomR (A, N ) ⊕ HomR (B, N ) be defined by φ(f ) = (f |A , f |B ). First, if f, g ∈
HomR (A⊕B, N ), then φ(f +g) = ((f +g)|A , (f +g)|B ) = (f |A +g|A , f |B +g|B ) = (f |A , f |B )+(g|A , g|B ) = φ(f )+φ(g).
If r ∈ R, then φ(rf ) = ((rf )|A , (rf )|B ) = (rf |A , rf |B ) = r(f |A , f |B ) = rφ(f )and so φ is an R−module homomorphism. Next, if φ(f ) = φ(g), then f |A = g|A and f |B = g|B so that for any a + b ∈ A ⊕ B, we have f (a + b) =
f |A (a) + f |B (b) = g|A (a) + g|B (b) = g(a + b) so that φ is injective. Finally, if (f, g) ∈ HomR (A, N ) ⊕ HomR (B, N ),
then define f ⊕ g : A ⊕ B → N by (f ⊕ g)(a ⊕ b) = f (a) + g(b). Then f ⊕ g is an R−module homomorphism since
f and g are, and hence φ(f ⊕ g) = ((f ⊕ g)|A , (f ⊕ g)|B ) = (f, g) and so φ is surjective.
Recall. Recall from Definition 283 that HomR (M, −) as a covariant, additive, multiplicative functor.
Remark 294. If S is an R−algebra and M is an S−module, then for any R−module, N , then HomR (M, N ) has the
structure of an S−module via (sf )(m) := f (sm). Furthermore, if g : N1 → N2 is an R−module homomorphism, then
g∗ as defined in Definition 283 is an S−module homomorphism. Thus, in this case, we have that HomR (M, −) is a
functor from R − mod to S − mod.
Proposition 295. Let R be a commutative ring and M an R−module. Then the functor HomR (M, −) is left exact.
Proof. Let
f
0
g
A
B
C
be an exact sequence of R−modules. Then we must show that
0
HomR (M, A)
f∗
HomR (M, B)
g∗
HomR (M, C)
is exact. We’ll start by showing that f∗ is injective. So suppose that f∗ (α) = 0 for some α ∈ HomR (M, A). Then we
have that f α = 0. Since f is injective, then we must have that α = 0 and hence f∗ is injective as well. Next, since
HomR (M, −) is a covariant functor and gf = 0 since the original sequence is exact, then g∗ ◦ f∗ = (gf )∗ = 0∗ = 0.
Thus, it only remains to show that ker(g∗ ) ⊆ image(f∗ ). Let β : M → B be an R−module homomorphism such that
g∗ (β) = 0. Then by definition of g∗ , we have gβ = 0. This gives that image(β) ⊆ ker(g) = image(f ) since the original
sequence is exact. Define h : M → A by h(m) := f −1 (β(m)). This is well defined since β(m) ∈ image(f ) for all m ∈ M
and f is injective. Then h is an R−module homomorphism since both f and β are. Also, f∗ (h) = f h = f f −1 β = β
and so β ∈ image(f∗ ). Hence ker(g∗ ) = image(f∗ ), which completes the proof.
Example 296. This example shows that the functor HomR (M, −) need not preserve surjections. Consider the map
π : Z → Z/2Z which is surjective. Applying HomZ (Z/2Z, −) gives the map π∗ : HomZ (Z/2Z, Z) → HomZ (Z/2Z, Z/2Z).
Since HomZ (Z/2Z, Z) ∼
= (0 :Z 2Z) = 0 and HomZ (Z/2Z, Z/2Z) ∼
= Z/2Z by Proposition 293, then π∗ cannot be surjective.
Proposition 297. Let P be an R−module. Then P is projective if and only if HomR (P, −) is exact (which is equivalent
by Proposition 295 to HomR (P, −) preserving surjections).
Proof. (⇒) Let π : A → B be a surjection. We need to show that π∗ : HomR (P, A) → HomR (P, B) is also surjective.
Note that π∗ (α) = πα. Let g : P → B be an R−module homomorphism, that is, g ∈ HomR (P, B). Then, as P is
projective, there is a map h : P → A such that the following diagram commutes:
P
h
A
g
π
B
58
0.
Equivalently, π ◦ h = π∗ (h) = g and so πa st is surjective.
(⇐) Let HomR (P, −) be exact, and suppose that π : A → B is a surjection. Then π∗ : HomR (P, A) → HomR (P, B)
is surjective as well. Then for any g ∈ HomR (P, B) there is a map h ∈ HomR (P, A) such that π ◦ h = π∗ (h) = g.
Equivalently, given any diagram as below with exact row, there is a map h : P → A such that the diagram commutes.
P
h
A
4.7.2
g
π
0.
B
5 Lemma
Lemma 298 (5-Lemma). Consider the following commutative diagram of R−modules, with exact rows.
A1
α1
h1
B1
A2
α2
A3
h2
β1
B2
α3
A4
h3
β2
B3
α4
A5
h4
β3
B4
h5
β4
B5
If h1 , h2 , h4 , h5 are all isomorphisms, then h3 is also an isomorphism.
Proof. We’ll first show that h3 is surjective. Let b3 ∈ B3 . Then there is some a4 ∈ A4 such that h4 (a4 ) = β3 (b3 ).
Notice that h5 (α4 (a4 )) = β4 (h4 (a4 )) = β4 (β3 (b3 )) = 0 by commutativity of the diagram and since the bottom
row is exact. Since h5 is an isomorphism, then α4 (a4 ) = 0. Hence a4 ∈ ker(α4 ) = image(α3 ). Thus, there is
some a3 ∈ A3 such that α3 (a3 ) = a4 . Notice that β3 (b3 − h3 (a3 )) = β3 (b3 ) − β3 (h3 (a3 )) = β3 (b3 ) − h4 (α3 (a3 )) =
β3 (b3 ) − h4 (a4 ) = β3 (b3 ) − β3 (b3 ) = 0. Hence, b3 − h3 (a3 ) ∈ ker(β3 ) = image(β2 ) and so there exists some b2 ∈ B2 such
that β2 (b2 ) = b3 − h3 (a3 ). Since h2 is an isomorphism, then there exists some a2 ∈ A2 such that h2 (a2 ) = b2 . By commutativity of the diagram, we then have h3 (α2 (a2 )) = β2 (h2 (a2 )) = β2 (b2 ) = b3 − h3 (a3 ). Hence, b3 = h3 (α2 (a2 )) + h3 (a3 )
and thus, b3 = h3 (α2 (a2 ) + a3 ) ∈ image(h3 ) so that h3 is surjective.
Next, we’ll show that h3 is injective. Let h3 (a3 ) = 0. We wish to show that a3 = 0. Note first that 0 = β3 (0) =
β3 (h3 (a3 )) = h4 (α3 (a3 )) since the diagram commutes and β3 is a homomorphism. As h4 is an isomorphism, then
α3 (a3 ) = 0. Thus, a3 ∈ ker(α3 ) = image(α2 ) since the top row is exact. So let a2 ∈ A2 be such that α2 (a2 ) = a3 .
Then β2 (h2 (a2 )) = h3 (α2 (a2 )) = h3 (a3 ) = 0 since the diagram commutes. Hence, h2 (a2 ) ∈ ker(β2 ) = image(β1 ). So
let b1 ∈ B1 be such that β1 (b1 ) = h2 (a2 ). Since h1 is an isomorphism, then there is some a1 ∈ A1 such that h1 (a1 ) = b1 .
By commutativity of the diagram we get that h2 (α1 (a1 )) = β1 (h1 (a1 )) = β1 (b1 ) = h2 (a2 ). Since h2 is an isomorphism,
then α1 (a1 ) = a2 . Also, since the top row is exact, then 0 = α2 (α1 (a1 )) = α2 (a2 ) = a3 and hence h3 is injective.
4.8
4.8.1
Wednesday 25 April 2012
5 Lemma Consequences
Proposition 299. This is a special case of the 5-Lemma (Lemma 298). Consider the following commutative diagram
with exact rows where h2 , h3 are isomorphisms.
0
A1
α1
h1
0
B1
A2
α2
h2
β1
B2
A3
h3
β2
B3
Then there exists an isomorphism h1 : A1 → B1 such that the diagram commutes.
Proof. If there is a homomorphism h1 : A1 → B1 such that the diagram commutes, then using the 5-Lemma, since
the larger diagram
59
0
A1
0
h4
h5
0
α1
A2
h1
B1
0
α2
A3
h2
β1
B2
h3
β2
B4
commutes, both rows are exact, and h2 , h3 , h4 , h5 are isomorphisms, then h1 is an isomorphism. So let a1 ∈ A1 and note
that β2 (h2 (α1 (a1 ))) = h3 (α2 (α1 (a1 ))) = h3 (0) = 0. Thus, image(h2 α1 ) ⊆ ker(β2 ) = image(β1 ). Since β1 is injective,
then we can thus define h1 = β1−1 h2 α1 and so by definition, the diagram commutes.
Proposition 300. Similarly, we have the following special case of the 5-Lemma (Lemma 298).7 Consider the following
commutative diagram with exact rows, where h1 , h2 are isomorphisms.
A1
α1
h1
B1
α2
A2
A3
h2
β1
0
h3
β2
B2
B3
0
Then there exists an isomorphism h3 : A3 → B3 such that the diagram commutes.
Proof. Again, if there is a homomorphism h3 : A3 → B3 such that the diagram commutes, then using the 5-Lemma,
since the larger diagram
A1
α1
h1
B1
A2
α2
A3
h2
β1
B2
0
h3
β2
0
h4
B3
h5
0
0
commutes, both rows are exact, and h1 , h2 , h4 , h5 are isomorphisms, then h3 is an isomorphism. So choose any a3 ∈ A3 .
Since α2 is surjective, there is some a2 ∈ A2 such that α2 (a2 ) = a3 . Suppose that α2 (a2 ) = α2 (a02 ) = a3 . Then,
α2 (a2 − a02 ) = 0 and so a2 − a02 ∈ ker(α2 ) = image(α1 ) and there exists some a1 ∈ A1 such that α1 (a1 ) = a2 − a02 . Since
the bottom row is exact, and since the original diagram commutes, we have 0 = β2 (β1 (h1 (a1 ))) = β2 (h2 (α1 (a1 ))) =
β2 (h2 (a2 − a02 )) = β2 (h2 (a2 )) − β2 (h2 (a02 )) and so β2 h2 is constant on α2−1 (a3 ). Hence, the map h3 : A3 → B3 defined
by h3 (a3 ) = β2 (h2 (a2 )) for any a2 ∈ A2 such that α2 (a2 ) = a3 is well defined, and a homomorphism. The diagram
commutes since h3 (α2 (a2 )) = β2 (h2 (a2 )) by definition of h3 .
4.8.2
Contravariant Hom Functor
Definition 301. Let N be an R−module, and g : M1 → M2 an R−module homomorphism. Then, define g ∗ :
HomR (M2 , N ) → HomR (M1 , N ) via α 7−→ αg. Then g ∗ is an R−module homomorphism. We can thus define a functor
HomR (−, N ) : R − mod → R − mod on objects by sending M to HomR (M, N ) and on morphisms g : M1 → M2 to g ∗ .
Note. The functor HomR (−, N ) is contravariant, additive, and multiplicative. Also, as before, if N is also an S−module,
for some R−algebra, S, then HomR (−, N ) can be thought of as a functor from R − mod to S − mod since for
f ∈ HomR (M, N ) and s ∈ S, we have (sf )(m) = sf (m).
Proposition 302. If N is an R−module, then HomR (−, N ) is left exact. That is, given an exact sequence A → B →
C → 0, the sequence 0 → HomR (C, N ) → HomR (B, N ) → HomR (A, N ) is also exact with the appropriate maps.
Proof. Let the sequence
f
A
g
B
C
be exact. Then we must show that the sequence
7 He
alluded to this in class, but didn’t even bother to state it.
60
0
HomR (C, N )
0
g∗
HomR (B, N )
f∗
HomR (A, N )
is also exact. We’ll first show that g ∗ is injective. So let γ ∈ HomR (C, N ) and suppose that g ∗ (γ) = 0. Then α ◦ g = 0.
Since g is surjective, this then gives that α = 0. Next, since the original sequence is exact, then gf = 0 and so
applying the contravariant functor HomR (−, N ) gives that f ∗ ◦ g ∗ = (gf )∗ = 0∗ = 0. It thus only remains to show
that ker(g ∗ ) ⊆ image(f ∗ ). So let β ∈ HomR (B, N ) such that βf = f ∗ (β) = 0. Suppose that g(b) = g(b0 ), then b − b0 ∈
ker(g) = image(f ) by the exactness of the original sequence and so there exists some a ∈ A such that f (a) = b − b0 . Thus,
0 = β(f (a)) = β(b − b0 ) = β(b) − β(b0 ) and so β is constant on g −1 (c). Hence, there is some γ ∈ HomR (C, N ) such that
γ = βg −1 . Since g ∗ (γ) = g ∗ (βg −1 ) = βg −1 g = β, then β ∈ image(g ∗ ) and hence ker(f ∗ ) ⊆ image(g ∗ ) as required.
Proposition 303. The functor HomR (−, N ) is exact if and only if N is injective. Note here that we know by
Proposition 302 HomR (−, N ) is exact if and only if it takes injections to surjections.
Proof. (⇒) Let f : A → B be an injection. Since f ∗ : HomR (B, N ) → HomR (A, N ) is surjective, then for any
α ∈ HomR (A, N ) there is some β ∈ HomR (B, N ) such that βf = f ∗ (β) = α. That is, in the diagram
N
β
α
0
A
B
f
with exact row there exists an R−module homomorphism β : B → N such that the diagram commutes. Hence, by
definition, N is injective.
(⇐) Now suppose that N is an injective R−module. Then suppose f : A → B is an injective R−module homomorphism,
and let α ∈ HomR (A, N ). Then in the commutative diagram
N
β
α
0
A
f
B
with exact row, there exists an R−module homomorphism β such that the diagram commutes. Hence, f ∗ (β) = βf = α
and so f ∗ is surjective as desired.
Definition 304. An R−module is called finitely presented if there exists an exact sequence of the form
Rm → Rn → M → 0.
That is, both M and ker(Rn → M ) are finitely generated.
Definition 305. Let f : A → B be an R−module homomorphism. Then coker(f ) = B/f (A) is called the cokernel of f .
Lemma 306 (Snake Lemma8 ). Let
A
α
β
B
g
f
0
A
0
α0
B
0
C
0
h
β0
C0
be a commutative diagram of R−modules with exact rows. Then there is a map δ : ker(h) → coker(f ) such that the sequence
8 This
wasn’t even stated in class, but is super useful in commutative algebra and is used to prove things that also weren’t shown in class.
61
ker(f )
α
b
ker(g)
βb
ker(h)
δ
coker(f )
αb0
coker(g)
βb0
coker(h)
is exact.
Proof. First, note that ker(f ) ⊆ A, ker(g) ⊆ B, and ker(h) ⊆ C. Also, if a ∈ ker(f ), then gα(a) = α0 f (a) = α0 (0) = 0
by commutativity, and so α(a) ∈ ker(g). Thus, we define α
b to be α restricted to ker(f ). Similarly, for any b ∈ ker(g), then
b
β(b) ∈ ker(h) and so we define β to be β restricted to ker(g). Next, if a0 +f (A) ∈ coker(f ), then α0 (a0 )+g(B) ∈ coker(g),
and for b0 + g(B) ∈ coker(g), then β 0 (b0 ) + h(C) ∈ coker(h), and we define αb0 (a0 + f (A)) = α0 (a0 ) + g(B) and
βb0 (b0 + g(B)) = β 0 (b0 ) + h(C). We must show these are well defined however, so let a01 + f (A) = a02 + f (A). Then there
is some a ∈ A such that a01 = a02 + f (a). Then,
αb0 (a01 + f (A))
=
=
=
=
=
=
α0 (a01 ) + g(B)
α0 (a02 + f (a)) + g(B)
α0 (a02 ) + α0 (f (a)) + g(B)
α0 (a02 ) + g(α(a)) + g(B)
α0 (a02 ) + g(B)
αb0 (a2 + f (A))
due to the commutativity of the diagram, the fact that α0 is an R−module homomorphism and since g(α(a)) ∈ g(B).
The proof that βb0 is well defined is the same as that for αb0 , mutatis mutandis.
Next, we’ll construct the map δ : ker(h) → coker(f ), so let c ∈ ker(h). Then there is some b ∈ B such that β(b) = c
since β is surjective. By commutativity of the diagram and since c ∈ ker(h), we have that β 0 g(b) = hβ(b) = h(c) = 0.
Thus, g(b) ∈ ker(β 0 ) = image(α0 ) since the bottom row is exact. We hence have that there is some a0 ∈ A0 such that
α0 (a0 ) = g(b). Set δ(c) = a0 + f (A) so that δ(c) ∈ coker(f ) as desired.
We need to show that using this definition, δ is a well defined R−module homomorphism. Since α0 is injective,
then the choice of a0 given the element b is unique, so to show the map is well defined, we only need to show it
is independent of the choice of b. So suppose that β(b) = β(eb) = c. Then let a0 , ae0 ∈ A0 be the unique elements
0 e0
0
e
e0
such that α0 (a0 ) = g(b) and
α (a ) = g(b). We need to show that a + image(f ) = a + image(f ). Since β is an
R−module, then β b − eb = β(b) − β eb = c − c = 0. Since the top row is exact, then there is some a ∈ A
0
such that α(a) = b − eb. Since the diagram commutes,
and since
homomorphisms, we then have
g, α are
R−module
0
0 0
0 e0
0
0
0
e
e
e
α (f (a)) = g(α(a)) = g b − b = g(b) − g b = α (a ) − α a = α a − a . Since α0 is injective, then we have
that f (a) = a0 − ae0 and so a0 = ae0 + f (a). Thus a0 + image(f ) = ae0 + image(f ) and so δ is well defined.
To show δ is an R−module homomorphism, let c1 , c2 ∈ ker(h) and r ∈ R. Then, we have a01 , a02 ∈ A0 such that
α0 (a01 ) = g(b1 ) and α0 (a02 ) = b2 where β(b1 ) = c1 and β(b2 ) = c2 . We thus have that rc1 + c2 ∈ ker(h) since
ker(h) is an R−module and that β(rb1 + b2 ) = rβ(b1 ) + β(b2 ) = rc1 + c2 since β is an R−module homomorphism.
Also, g(rb1 + b2 ) = rg(b1 ) + g(b2 ) = rα0 (a01 ) + α0 (a02 ) = α0 (ra01 + a02 ). Thus, δ(rc1 + c2 ) = ra01 + a02 + f (A) =
r(a01 + f (A)) + (a02 + f (A)) = rδ(c1 ) + δ(c2 ) so that δ is an R−module homomorphism.
b α(a)) = β(α(a)) = 0 since the original
It remains to show that the sequence is exact. So let a ∈ ker(f ). Then, β(b
b then β(b) = 0 and so by exactness of the original diagram, there is some
diagram has exact rows. Now, if b ∈ ker(β),
a ∈ A such that α(a) = b. If a ∈ ker(f ), then we’ll have that ker βb = image α
b. By commutativity of the diagram, we
have that α0 (f (a)) = g(α(a)) = g(b) = 0. Since α0 is injective, then f (a) = 0 and so a ∈ ker(f ) as desired.
b
Let b ∈ ker(g). Then, δ(β(b))
= δ(β(b)) = a0 + f (A) where a0 ∈ A is such that α0 (a0 ) = g(b). Also, g(b) = 0
0 0
b
since b ∈ ker(g), and so α (a ) = 0. Since α0 is injective, then a0 = 0, and so δ(β(b))
= 0 + f (A) = 0 ∈ coker(f ).
Next, let c ∈ ker(δ). Then there is some b ∈ B such that β(b) = c and there is some a0 ∈ A0 such that α0 (a0 ) = g(b).
Since δ(c) = 0, then a0 + f (A) = f (A) and so a0 ∈ f (A). Let a ∈ A be chosen so that f (a) = a0 . Then consider b − α(a). We have β(b − α(a)) = β(b) − βα(a) = β(b) by the exactness of the original diagram. Also,
b as desired.
g(b − α(a)) = g(b) − g(α(a)) = g(b) − α0 f (a) = g(b) − α0 (a0 ) = 0 and so c ∈ image(β)
0
0 0
Let c ∈ ker(h). Then, δ(c) = a + f (A) where α (a ) = g(b)
and β(b) = c. So α0 (δ(c))= αb0 (a0 + f (A)) =
0 0
α (a ) + g(B) = g(b) + g(B) = 0 + g(B) and so image(δ) ⊆ ker αb0 . Net, let a0 + f (A) ∈ ker αb0 . Then we have
0 0
0 0
α0 (a0 ) + g(B) = 0 ∈ coker(g) and
so
α (a ) ∈ g(B). Then let b ∈ B be such that g(b) = α (a ). Then by definition,
δ(β(b)) = a0 + f (A) and so ker αb0 = image(δ).
Finally, let a0 + f (A) ∈ coker(f ). Then, βb0 (αb0 (a0 + f (A))) = βb0 (α0 (a0) +g(B)) = β 0 (α0 (a0 )) + h(C) = 0 + h(C) since
the original diagram had exact rows. Conversely, let b0 + g(B) ∈ ker βb0 . Then 0 = βb0 (b0 + g(B)) = β 0 (b0 ) + h(C),
62
and so β 0 (b0 ) ∈ h(C). Thus, there is some c ∈ C such that h(c) = β 0 (b0 ). By exactness of the original top row, we
have that there is some b ∈ B such that β(b) = c. Then by commutativity of the original diagram, we have that
β 0 (g(b)) = h(β(b)) = h(c) = β 0 (b0 ), and so β 0 (b0 − g(b)) = 0. Hence, by exactness of the original bottom row, there is some
a0 ∈ A0 such that α0 (a0 ) = b − g(b). Hence, αb0 (a0 + f (A)) = α0 (a0 ) + g(B) = b0 − g(b) + g(B) = b0 + g(B) as desired.
Lemma 307. Suppose M is a finitely presented R−module and let
0→K→N →M →0
be a short exact sequence where N is finitely generated. Then K is also finitely generated.9
Proof. Let
f
0
K
g
N
0
M
be an exact sequence of R−modules where M is finitely presented and N is finitely generated. Then there is an exact
sequence of the form
Rs
φ
Rt
ψ
0
M
for some m, n ∈ N1 . Let e1 , . . . , et be an R−module basis for Rt . For each i, then ψ(ei ) ∈ M . Since g is surjective,
then there exist elements n1 , . . . , nt ∈ N such that g(ni ) = ψ(ei ) for each i. Define an R−module homomorphism
α : Rt → N by α(ei ) = ni . This gives the commutative diagram
φ
Rs
Rt
ψ
M
α
0
f
K
1M
g
N
0
M
0
with exact rows. The left square commutes since if (r1 , . . . , rt ) ∈ Rt , then
ψ(r1 , . . . , rt )
=
=
=
=
r1 ψ(e1 ) + . . . + rt ψ(et )
r1 g(n1 ) + . . . + rt g(nt )
r1 g(α(e1 )) + . . . + rt g(α(et ))
gα((r1 , . . . , rt ))
as desired. If we can define a homomorphism β : Rs → K such that the left square commutes, then we’ll be able to apply
the Snake Lemma (Lemma 306). Let 1 , . . . , s be a basis for Rs . For each i, we have that gαψ(i ) = φψ(i ) = 0 since the
top row is exact and by commutativity of the diagram. Hence, αφ(i ) ∈ ker(g) = image(f ) for all i. Since f is injective
there is thus a unique ki ∈ K such that f (ki ) = αφ(i ). We then define the R−module homomorphism β : Rs → K
by β(ei ) = ki . Then the diagram commutes by definition, and so by the Snake Lemma, the following sequence is exact:
ker(β) → ker(α) → 0 → coker(β) → coker(α) → 0
since the map 1M is an isomorphism on M and so ker(1M ) = 0 and coker(1M ) = M/M = 0. Hence,
K/β(Rs ) = coker(β) ∼
= coker(α) = N/α(Rt ).
Since N is finitely generated, then N/α(Rt ) is finitely generated, and so K/β(Rs ) is also finitely generated. Since Rs
is also finitely generated, then β(Rs ) is finitely generated. Thus, K is finitely generated.
Remark 308. If R is Noetherian, then M is finitely presented if and only if M is finitely generated.
Proof. (⇒) If M is finitely presented, then M is automatically finitely generated.
(⇐) Let f : Rn → M be surjective. Such a surjection exists since M is finitely generated. Let K = ker(f ). Then
K ⊆ Rn is a submodule of Rn which is a Noetherian R−module, and so K is also finitely generated. Thus, M is finitely
presented.
9 This
was not proved in class, it was only indicated that the Snake Lemma would be useful.
63
Remark 309. Let I be an ideal of R. Then R/I is finitely presented if and only if I is finitely generated.
Proof. Since I is an ideal of R, then the sequence 0 → I → R → R/I → 0 is exact.
(⇒) If R/I is finitely presented, then since R is clearly finitely generated (as an R−module), we have that I is finitely
generated by Lemma 307.
(⇐) If I is finitely generated, then since I = ker(π) where π is the canonical quotient map, we have that R/I is
finitely presented by definition.
Proposition 310. Let S be a flat R−algebra, let M, N be R−modules, and assume that M is finitely presented. Then
HomR (M, N ) ⊗R S ∼
= HomS (M ⊗R S, N ⊗R S) via the map f ⊗ s 7−→ f ⊗ s where f ⊗ s on the right is the map
f ⊗ s : M ⊗R S → N ⊗R S given by m ⊗ s0 7−→ f (m) ⊗ ss0 .
Proof. We first do the case when M = Rn . Then, using parts 1 and 3 of Proposition 293, we have HomR (M, N ) =
n
n
M
M
HomR (Rn , N ) ∼
HomR (R, N ) ∼
N = N n . Hence,
=
=
i=1
i=1
HomR (M, N ) ⊗R S
=
∼
=
∼
=
∼
=
∼
=
∼
=
∼
=
HomR (Rn , N ) ⊗R S
N n ⊗R S
(N ⊗R S)n
(HomS (S, N ⊗R S))n
HomS (S n , N ⊗R S)
HomS (Rn ⊗R S, N ⊗R S)
HomS (M ⊗R S, N ⊗R S).
Chasing through the isomorphisms, we have the map as given in the statement.10
Now, we’ll do the general case. As M is finitely presented, then there is an exact sequence
Rm → Rn → M → 0.
(4.1)
Applying − ⊗R S to sequence (4.1) gives that the sequence
R m ⊗R S → R n ⊗R S → M ⊗R S → 0
(4.2)
is also exact. Then applying the functor HomS (−, N ⊗R S) to sequence (4.2) gives that the sequence
0 → HomS (M ⊗R S, N ⊗R S) → HomS (Rn ⊗R S, N ⊗R S) → HomS (Rm ⊗R S, N ⊗R S)
(4.3)
is exact. Starting again with sequence (4.1), and applying the functor HomR (−, N ) gives that the sequence
0 → HomR (M, N ) → HomR (Rn , N ) → HomR (Rm , N )
(4.4)
is exact. Since S is flat, then the functor − ⊗R S is exact, and so applying it to sequence (4.4) gives that the sequence
0 → HomR (M, N ) ⊗R S → HomR (Rn , N ) ⊗R S → HomR (Rm , N ) ⊗R S
(4.5)
is also exact. We then have the following commutative diagram
0
HomR (M, N ) ⊗R S
HomR (Rn , N ) ⊗R S
h1
0
HomS (M ⊗R S, N ⊗R S)
h2
HomS (Rn ⊗R S, N ⊗R S)
HomR (Rm , N ) ⊗R S
h3
HomS (Rm ⊗R S, N ⊗R S)
with exact rows by sequences (4.5) and (4.3), and isomorphisms h2 , h3 by the first case. By Proposition 299, then there
exists an isomorphism h1 : HomR (M, N ) ⊗R S → HomS (M ⊗R S, N ⊗R S) such that the entire diagram commutes.
In particular, we have that HomR (M, N ) ⊗R S ∼
= HomS (M ⊗R S, N ⊗R S) as desired.
Corollary 311. If W is a multiplicatively closed subset of R, M is a finitely presented R−module, and N is any
R−module, then HomR (M, N )W ∼
= HomRW (MW , NW ).
10 I
got a bit lost in checking it, because of poorly named maps, but I’m sure it works.
64
Proof. Note that RW is a flat R−module by Proposition 276. Hence, by Proposition 310 and part 3 of Proposition
254, we have that HomR (M, N )W ∼
= HomR (M, N ) ⊗R RW ∼
= HomRW (M ⊗R RW , N ⊗R RW ) ∼
= HomRW (MW , NW )
as desired.
Theorem 312. Let M be a finitely presented R−module. Then the following are equivalent:
1. M is projective,
2. MP is a free RP −module for every P ∈ Spec(R), and
3. Mm is a free Rm −module for every maximal ideal m of R.
Proof. (1 ⇒ 2) By Proposition 243, then every finitely generated projective RP −module is free. Since M is finitely
presented, then M is also finitely generated, and by the last part of Remark 223, we know that MP is a finitely generated
RP −module. By Corollary 229, then MP is a projective RP −module as well. Hence, MP is a free RP −module for
every P ∈ Spec(R).
(2 ⇒ 3) This is trivial since every maximal ideal is prime.
(3 ⇒ 1) Note that it is enough by Proposition 297 to show that HomR (M, −) is exact, or equivalently, that it
preserves surjections. So let f : A → B be a surjection of R−modules. We then need to show that f∗ : HomR (M, A) →
HomR (M, B) is surjective as well. Let C = coker(f∗ ) = HomR (M, B)/ image(f∗ ). Then the sequence
HomR (M, A)
f∗
HomR (M, B)
0
C
is exact. We need to show that C = 0. Let m be an arbitrary maximal ideal of R, and localizing the sequence at m gives that
f∗
1
HomR (M, A)m
HomR (M, B)m
Cm
0
is exact as well. Starting again with f : A → B, localizing at m gives that
Am
f
1
Bm
0
is exact. Applying the functor HomRm (Mm , −) to this sequence gives that
HomRm (Mm , Am )
f
1
HomRm (Mm , Bm )
0
is exact as well. We thus have the following commutative diagram
HomR (M, A)m
α1
HomR (M, B)m
h1
HomRm (Mm , Am )
α2
Cm
h2
β1
HomRm (Mm , Bm )
0
h3
β2
0
0
with exact rows as shown above, and homomorphisms h1 , h2 by Corollary 311. Thus, by Proposition 300, we have that
Cm ∼
= 0. Then by Proposition 225, we have that C = 0 which completes the proof.
4.9
4.9.1
Friday 27 April 2012
Adjointness of Hom and Tensor
Theorem 313 (Adjointnes of Tensor Product and Hom). Let S be an R−algebra, A, B S−modules, and C an
R−module. Then there exists a natural isomorphism φ : HomR (A ⊗S B, C) → HomS (A, HomR (B, C)) given by
(f : A ⊗S B → C) 7−→ (φ(f ) : A → HomR (B, C)) where φ(f )(a) : B → C is given by b 7−→ f (a ⊗ b).11
Theorem 314. Another version of the Hom-Tensor Adjointness is: If A is an R−module, and B, C are S−modules,
then there is an isomorphism HomS (A ⊗R B, C) → HomR (A, HomS (B, C)) as abelian groups.12
11 Tom didn’t prove this in class, but he said it was an exercise. I’m going to prove the second version, but not this one. I’m pretty
sure this version has a similar proof.
12 Tom didn’t explicitly state it, but I’m pretty sure we need here that S is an R−algebra.
65
Proof. The map φ : HomS (A ⊗R B, C) → HomR (A, HomS (B, C)) is defined for each f : A ⊗R B → C by (φ(f (a)))(b) =
f (a ⊗ b). We first must check that φ(f (a)) : B → C is an S−module homomorphism. So let b1 , b2 ∈ B and s ∈ S. Then
φ(f (a))(b1 + sb2 )
=
=
=
=
=
f (a ⊗ (b1 + sb2 ))
f (a ⊗ b1 + a ⊗ sb2 )
f (a ⊗ b1 ) + f (a ⊗ sb2 )
f (a ⊗ b1 ) + sf (a ⊗ b2 )
φ(f (a))(b1 ) + sφ(f (a))(b2 )
since f is an S−module homomorphism and tensor products are linear. Next, we must check that φ(f ) : A → HomS (B, C)
is an R−module homomorphism. So let a1 , a2 ∈ A, r ∈ R, and b ∈ B. Then,
φ(f (a1 + ra2 ))(b)
=
=
=
=
f ((a1 + ra2 ) ⊗ b)
f (a1 ⊗ b + ra2 ⊗ b)
f (a1 ⊗ b) + rf (a2 ⊗ b)
φ(f (a1 ))(b) + rφ(f (a2 ))(b)
again since f is an S−module homomorphism (and r ∈ R ⊆ S) and tensor products are linear. Next, we must check
that φ is a group homomorphism. So let f, g ∈ HomS (A ⊗R B, C). We must show that φ(f + g) = φ(f ) + φ(g). So
let a ∈ A, and b ∈ B, then
φ(f + g)(a)(b)
= (f + g)(a ⊗ b)
= f (a ⊗ b) + g(a ⊗ b)
= φ(f )(a)(b) + φ(g)(a)(b)
by definition of φ and by definition of a sum of maps, and so φ is a group homomorphism.
Next, we define a map ψ : HomR (A, HomS (B, C)) → HomS (A⊗R B, C) by ψ(f ) : A⊗R B → C by ψ(f )(a⊗b) = f (a)(b).
]) : A × B → C as ψ(f
])(a, b) = f (a)(b). Then, it is easy to check that
In order to do this we should first define ψ(f
]) is an R−bilinear map, and so it induces the map ψ(f ) we want.13 We must however check that ψ is a group
ψ(f
homomorphism. Let f, g ∈ HomR (A, HomS (B, C)), a ∈ A, and b ∈ B. Then
ψ(f + g)(a)(b)
=
=
=
(f + g)(a ⊗ b)
f (a ⊗ b) + g(a ⊗ b)
ψ(f )(a)(b) + ψ(f )(a)(b)
by definition of ψ and by definition of a sum of maps, and so ψ is a group homomorphism. It only remains then to show
that ψ and φ are inverses. Let f ∈ HomS (A⊗B , C), a ∈ A and b ∈ B. Then, ψ(φ(f )) : A ⊗R B → C and is given by
ψ(φ(f ))(a⊗b) = φ(f )(a)(b) = f (a⊗b) by definition of ψ and then φ respectively. Similarly, if g ∈ HomR (A, HomS (B, C)),
a ∈ A and b ∈B, then φ(ψ(g)) : A → HomS (B, C) and is given by φ(ψ(g))(a)(b) = ψ(g)(a ⊗ b) = g(a)(b) by definition
of φ and then ψ respectively. Thus, φ and ψ are inverses and this gives the isomorphism as desired.
Proposition 315. Let S be an R−algebra, and I and injective R−module. Then HomR (S, I) is an injective S−module.
Proof. It is enough to show that HomS (−, HomR (S, I)) is exact. It is always left exact, so it is exact if and only if
it takes surjections to injections. So let A → B → 0 be an exact sequence of S−modules. Applying the functor in
question gives the map HomS (B, HomR (S, I)) → HomS (A, HomR (S, I)), and we want to show that it is injective. By
the adjointness of tensor products and Hom (Theorem 313), we have the following commutative diagram
0
HomS (B, HomR (S, I))
HomS (A, HomR (S, I))
∼
=
0
∼
=
HomR (B ⊗S S, I)
HomR (A ⊗S S, I)
∼
=
0
13 I’m
∼
=
HomR (B, I)
HomR (A, I)
almost entirely positive that these details are easy, and I don’t want to do them.
66
with all vertical maps being isomorphisms. Since I is an injective module, the bottom row is exact, and so by commutativity
and Proposition 299, all three rows are exact, and in particular, the map HomS (B, HomR (S, I)) → HomS (A, HomR (S, I))
is injective, which proves that the functor HomS (−, HomR (S, I)) is exact.
4.9.2
Projective Dimension, Regular Local Rings
Definition 316. Let M be an R−module. Then there exists a projective R−module, P0 such that 0 → K0 → P0 →
M → 0 is exact, where K0 is the kernel of the map P0 → M . That is, one first finds a surjection from a projective
R−module to M , say f0 : P0 → M and then set K0 = ker(f0 ). Similarly, there exists a projective R−module, P1 such
that 0 → K1 → P1 → K0 is exact, where K1 is the kernel of the map P1 → K0 . Repeating in this way, gives a long
exact sequence
···
P3
P2
P1
K1
0
P0
M
0
K0
0
0
This is called a projective resolution of M . The Ki ’s are called the syzygyies of M . The projective dimension of M
is pdR (M ) = inf{lengths of all projective resolutions of M } = inf{n | Kn is projective}. If no projective resolution
is finite, then pdR (M ) = ∞. That is, we set inf ∅ = ∞.
Theorem 317 (Hilbert’s Syzygy Theorem (1890)). If R = k[x1 , . . . , xn ] where k is a field, then pdR (M ) ≤ n for every
R−module M .
Remark 318. Note in the previous theorem, that n = dim R and m = (x1 , . . . , xn ) is a maximal ideal.
Theorem 319 (Krull). Let (R, m) be a local ring. Then µR (m) ≥ dim R.
Definition 320. If µR (m) = dim R, then R is called a regular local ring (RLR).
Example 321. Some examples of regular local rings:
B The ring k[x1 , . . . , xn ](x1 ,...,xn ) where k is a field is a regular local ring.
B Any field is a regular local ring.
B Local PID’s are regular local rings.
B In particular Z(p) is a regular local ring for any prime p.
Remark 322. Here are some properties of regular local rings:
B Every regular local ring is a domain.
B Every regular local ring is a UFD.
B If m = (x1 , . . . , xd ) is the unique maximal ideal with d = dim R, then (xi1 , . . . , xik ) is a prime ideal for all k ≤ d
with {i1 , . . . , ik } ⊆ {1, . . . , d}.
Question. If (R, m) is a regular local ring and P ∈ Spec R, is RP a regular local ring?14
Theorem 323 (Serre-Auslander-Buchsbaum, 1957). Let (R, m) be a local ring. The following are equivalent.
1. R is a regular local ring,
2. pdR (M ) < ∞ for all R−modules, M ,
3. pdR (M ) ≤ dim R for all R−modules, M , and
4. pdR (R/m) < ∞
Answer. We’re now able to answer the question using the Serre-Auslander-Buchsbaum Theorem (Theorem 323). To
that end, let q ∈ Spec(R) and choose a projective resolution
0 → Pn → . . . → P1 → P0 → R/q → 0.
There is a finite resolution due to Theroem 323 since R/q is an R−module and R is a regular local ring. Localizing
at q gives a sequence of projective Rq −modules by Corollary 229. Thus, pdRq Rq /qRq ≤ n < ∞ and so Rq is a regular
local ring by Theorem 323.
14 This
was an open question for a long time.
67