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MATH 8240: COMMUTATIVE ALGEBRA & GEOMETRY
HOMEWORK SETS AND EXAMS
Yongwei Yao
2016 FALL SEMESTER
GEORGIA STATE UNIVERSITY
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Contents
HW Set #01, Solutions
HW Set #02, Solutions
HW Set #03, Solutions
HW Set #04, Solutions
Midterm I, Solutions
HW Set #05, Solutions
HW Set #06, Solutions
HW Set #07, Solutions
HW Set #08, Solutions
Midterm II, Solutions
HW Set #09, Solutions
HW Set #10, Solutions
HW Set #11, Solutions
Final Exam, Solutions
Extra Credit Set, Solutions—not really
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
Note. Throughout the semester, all rings are assumed to be commutative with one, unless
they are stated otherwise explicitly. In other words, if R is a ring, then xy = yx and there
exists 1R ∈ R (or simply 1 ∈ R) such that 1R x = x (or simply 1x = x) for all x, y ∈ R.
Moreover, if f : R → S is a ring homomorphism, it is always assumed that f (1R ) = 1S .
These assumptions will not be repeated.
Problems for extra credits are available; see the last page of this file.
There are three (3) PDF files for the homework sets and exams, one with the problems
only, one with hints, and one with solutions. Links are provided below.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
√
I
Math 8240 (Fall 2016)
Homework Set #01 (Due 08/31)
Solutions
Problem
1.1. Let R beTa ring and I1 , I2 , . . . , In be ideals of R, with n ∈ N. Prove that
Tn
n
if
I
is
Ii = Ic for some c ∈ {1, . . . , n}. (It follows immediately that
Tn i=1 i prime then Ti=1
n
i=1 Ii ∈ Spec(R) ⇐⇒
i=1 Ii ∈ {I1 , . . . , In } ∩ Spec(R).)
T
Proof. To prove the claim by contraposition, suppose ni=1 Ii 6= Ij for all j ∈ {1, . . . , n}. By
set theory, we see that
n
\
Ii ( Ij for all j ∈ {1, . . . , n}.
i=1
Thus, for each j ∈ {1, . . . , n}, there exists rj ∈ Ij \
n
Y
j=1
rj ∈
n
\
j=1
Ij =
n
\
Ii
while rj ∈
/
i=1
n
\
Tn
I
. It follows that
i
i=1
Ii for all j ∈ {1, . . . , n}.
i=1
T
This implies thatT ni=1 Ii is not a prime Tideal. Now the proof is complete. (It follows
immediately that ni=1 Ii ∈ Spec(R) ⇐⇒ ni=1 Ii ∈ {I1 , . . . , In } ∩ Spec(R).)
Problem 1.2. Let R be a ring and I an ideal of R. For each r ∈ R, write r = r + I ∈ R/I.
For any x ∈ R, recall that (I :R x) = {r ∈ R | rx ∈ I}. Prove or disprove each of the
following statements (with a proof or a counterexample).
(1) For every element x ∈ R, (I :R x)/I = {r ∈ R/I | rx = 0 ∈ R/I}.
(2) For every element x ∈ R, x is a ZD (zero-divisor) of R/I if and only if (I :R x) 6= I.
Proof. We begin by observing the containment (I :R x) ⊇ I, which is very easy to prove.
(1) The statement is true. Indeed, for r ∈ R/I with r ∈ R, we have
r ∈ (I :R x)/I ⇐⇒ r ∈ (I :R x) ⇐⇒ rx ∈ I ⇐⇒ rx = 0,
which proves the claim.
(2) The statement is true. Indeed,
x is a ZD ⇐⇒ {r ∈ R/I | rx = 0 ∈ R/I} =
6 {0}
(1)
⇐⇒ (I :R x)/I 6= {0} ⇐⇒ (I :R x) 6= I.
√
Problem
1.3. Let R be a ring and I be an ideal of R. We say √
R is reduced iff 0 = 0 (or
p
{0} = {0}, to be precise). Also, we say I is a radical ideal iff I = I.
√
(1) In Z (the ring of integers), consider the ideal I = (12). Compute I and determine
whether I is radical.
(2) Consider Z12 = Z/(12), which is a ring. Determine whether it is reduced.
(3) Let n ∈ Z such thatp|n| = pr11 · · · prss where p1 , . . . , ps are distinct prime numbers and
ri > 1. Determine (n). No proof is necessary.
p
Solution/Proof. (1) We claim that (12) = (6) ) (12) and, hence, I = (12) is not radical.
(2) We claim that Z12 p
= Z/(12) is not reduced. Indeed, [6] 6= [0] but [6]2 = [36] = [0] in
Z12 . (In fact, Nil(Z12 )p
= {[0]} = {[0], [6]}.)
(3) We claim that (n) = (p1 · · · ps ). Here is a proof: For any x ∈ (p1 · · · ps ) so that
x = p1 · · · ps kpwith k ∈ Z,pit is clear that xr = pr1 · · · prs k r ∈ (n) for r = max{r1 , . . . , rs },
showing x ∈ (n). Thus (n) ⊇ (p1 · · · ps ).
1
p
Conversely, let y ∈ (n) ⊆ Z. Then there exists an integer t > 0 such that y t ∈ (n), that
is, n | y t . This implies pi | y t for all i = 1, . . . , s. As p1 , . . . , ps are distinct prime numbers,
we see
pi | y for i = 1, . . . , s, hence p1 · · · ps | y, that is, y ∈ (p1 · · · ps ).
p
p
This shows (n) ⊆ (p1 · · · ps ). Consequently, (n) = (p1 · · · ps ) as claimed.
√
Problem 1.4. Let R be a ring and I be an ideal of R. Prove that I is radical (i.e., I = I)
if and only if R/I is reduced.
√
Proof. First, we provide a conceptual proof by using the fact that I/I = Nil(R/I).
Indeed,
√
√
I = I ⇐⇒ I/I = 0 ⇐⇒ Nil(R/I) = 0 ⇐⇒ R/I is reduced.
Alternatively, we √
give a elementary√proof as follows.
√ “If”: Assume R/I is reduced. It
is always true that I ⊇ I. To show I ⊆ I, let x ∈ I. Then x + I is nilpotent in R/I.
As R/I is reduced, it is forced that
x + I = 0R/I = 0 + I,
√
√
which implies x ∈ I. This
proves
I
⊆
I.
Hence
I = I, i.e., I is a radical ideal.
√
“Only if”: Assume I = I. To show R/I is reduced, let x + I be any nilpotent element in
n
n
R/I, with x ∈ R.
√ Then there exists n ∈ N such that (x + I) = 0 + I, which implies x ∈ I
and hence x ∈ I = I, which implies
x + I = 0 + I = 0R/I .
This shows that 0 + I, the zero element of R/I, is the only nilpotent element of R/I. In
other words, R/I is reduced. This completes the proof.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
2
√
I
Math 8240 (Fall 2016)
Homework Set #02 (Due 09/07)
Solutions
Problem 2.1. Let f : R → S be a ring homomorphism (so that f (1R ) = 1S , as this is always
assumed). Suppose that Q be an ideal of S, and let P = f −1 (Q).
(1) True or false: P is an ideal of R.
(2) Prove that if Q ∈ Spec(S) (i.e., Q is a prime ideal of S) then P ∈ Spec(R).
(3) Prove that if S is a domain then Ker(f ) ∈ Spec(R).
Proof. (1) True. First, we see 0R ∈ f −1 (Q) = P as f (0R ) = 0S ∈ Q. Next, let a, b ∈ P and
r ∈ R, so that f (a) ∈ Q and f (b) ∈ Q. Then
f (a − b) = f (a) − f (b) ∈ Q and f (ra) = f (r)f (a) ∈ Q
since Q is an ideal of S. So a − b ∈ P and ra ∈ P . Thus P is an ideal of R.
(2) As f (1R ) = 1S ∈
/ Q, we see 1R ∈
/ f −1 (Q) and hence P = f −1 (Q) is a proper ideal
of R. Suppose, for a, b ∈ R, we have ab ∈ P , which simply means f (a)f (b) = f (ab) ∈ Q.
Then, as Q is prime, either f (a) ∈ Q or f (b) ∈ Q; that is, either a ∈ f −1 (Q) or b ∈ f −1 (Q).
Consequently, P is a prime ideal of R, i.e., P ∈ Spec(R).
(3) Observe that Ker(f ) is no other than f −1 ({0S }). Also, we see that {0S } is prime
in S, since S is a domain. Thus, by part (2) above, we see Ker(f ) ∈ Spec(R). (To
prove this from scratch, let a, b ∈ R be such that ab ∈ Ker(f ), which simply means
f (a)f (b) = f (ab) = 0S . As S is a domain, either f (a) = 0S or f (b) = 0S ; that is, either
a ∈ Ker(f ) or b ∈ Ker(f ). Finally, as f (1R ) = 1S 6= 0S , we see Ker(f ) is a proper ideal.
Consequently, Ker(f ) is a prime ideal of R.)
Problem 2.2. Let R be a ring and X a non-empty set/collection of (certain) ideals of R.
Assume that (X, ⊆) is a totally ordered set (i.e., for all I1 , I2 ∈ X, either I1 ⊆ I2 or I2 ⊆ I1 ).
S
(1) True or false: I∈X I, the union of all ideals
T in X, is an ideal of R.
(2) Further assume X ⊆ Spec(R). Prove that P ∈X P ∈ Spec(R).
S
Proof. (1) True. S
Since X 6= ∅, we may pick an ideal I0 ∈ X. Hence 0R ∈ I0 ⊆ I∈X I.
R. There exist I1 , I2 ∈ X such that a ∈ I1 and b ∈ I2 .
Next, let a, b ∈ I∈X I and r ∈ S
Immediately, we see ra ∈ I1 ⊆ I∈X I. As (X, ⊆) is totally ordered, either I1 ⊆ I2 or
I2 ⊆ I1 , which implies either a, b ∈ I2 or a, b ∈ I1 , which further implies
a − b ∈ I2
or a − b ∈ I1 .
S
In either case, a T
− b ∈ I∈X I. This proves that I∈X I is an ideal of R.
(2) Note that P ∈X P is an ideal of R, since the intersection of any collection of ideals of
R remains an ideal of R. Since X 6= ∅, we see that
\
P ⊆ P0 ( R for any P0 ∈ X ⊆ Spec(R).
S
P ∈X
T
This shows that P ∈X P is a proper
ideal.
T
Next, let r, s ∈ R \
/ P1 and s ∈
/ P2 . As
P ∈X P . There exist P1 , P2 ∈ X such that r ∈
(X, ⊆) is totally ordered, either P1 ⊆ P2 or P2 ⊆ P1 , which implies that either r, s ∈
/ P1 or
r, s ∈
/ P2 , which further implies
rs ∈
/ P1
or rs ∈
/ P2
T
since
P
,
P
∈
X
⊆
Spec(R).
In
either
case,
rs
∈
/
1
2
P ∈X P . This establishes the claim
T
P ∈X P ∈ Spec(R), as required.
3
Problem 2.3. Let R be a ring and S ⊆ R \ 0 (that is, S ⊆ R \ {0R }). Consider
Ω = {I | I is an ideal of R such that I ∩ S = ∅}.
(1) Prove that the partially ordered set (Ω, ⊆) admits a maximal element.
(2) In the case of S = {1R } ⊆ R \ 0, characterize the maximal element(s) of (Ω, ⊆).
(3) In the case of R = Z and S = {60}, find/list all the maximal element(s) of (Ω, ⊆).
Proof. (1) The zero ideal satisfies {0R } ∩ S = ∅. So {0R } ∈ Ω and hence Ω 6=S∅. Let X
be any non-empty subset of Ω such that (X, ⊆)) is totally ordered. Let J = I∈X I. By
Problem 2.2(1), J is an ideal of R. Moreover, by set theory, we have
[ [
[
J ∩S =
I ∩S =
(I ∩ S) =
∅ = ∅, which shows J ∈ Ω.
I∈X
I∈X
I∈X
By the construction of J, it is clear that I ⊆ J for all I ∈ X. Therefore J is an upper bound
of X in (Ω, ⊆). By Zorn’s lemma, the poset (Ω, ⊆) admits a maximal element.
(2) In case S = {1R } ⊆ R \ 0, the maximal element(s) of (Ω, ⊆) are precisely the maximal
ideal(s) of R.
(3) In case R = Z and S = {60}, the maximal element(s) of (Ω, ⊆) are precisely the
following: (23 ), (32 ), (52 ), and (p) for all prime numbers p > 7.
Problem 2.4. Let R be a ring and ∅ 6= S ⊆ R \ 0 such that S is multiplicatively closed
(that is, st ∈ S for all s, t ∈ S). Let P be any maximal element of the poset (Ω, ⊆) where
Ω = {I | I is an ideal of R such that I ∩ S = ∅}.
(Such P exists, by Problem 2.3(1).) Prove that P ∈ Spec(R). (Consequently, there exists
P ∈ Spec(R) such that P ∩ S = ∅.)
Proof. Note that P ∩S = ∅ and P ( R. By way of contradiction, suppose that P ∈
/ Spec(R).
Then there must exist a1 , a2 ∈ R \ P such that a1 a2 ∈ P . Consider ideals
I1 = (a1 ) + P
and I2 = (a2 ) + P.
Clearly, P ( Ii for i = 1, 2. Since P is a maximal element in (Ω, ⊆), we see Ii ∈
/ Ω, hence
Ii ∩ S 6= ∅ for i = 1, 2. Choose (and fix) any si ∈ Ii ∩ S for i = 1, 2.
On one hand, as si ∈ Ii = (ai ) + P and a1 a2 ∈ P , we see (details omitted)
s1 s2 ∈ ((a1 ) + P )((a2 ) + P ) ⊆ P.
(†)
On the other hand, as si ∈ S for i = 1, 2 and S is multiplicatively closed, we see
s1 s2 ∈ S.
(‡)
Now, putting (†) and (‡) together, we see P ∩ S 6= ∅, which is a contradiction. Consequently, P must be a prime ideal of R.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
4
√
I
Math 8240 (Fall 2016)
Homework Set #03 (Due 09/14)
Solutions
Problem 3.1. Prove that if R is an Artinian ring then |Max(R)| < ∞ (i.e., there are only
finitely many maximal ideals of R).
Proof. Let R be an Artinian ring. To prove by contradiction, suppose |Max(R)| = ∞. Thus
we may choose (and fix) mutually distinct maximal ideals
m1 , m2 , . . . , mi , mi+1 , . . . ∈ Max(R).
Consider the following descending chain of ideals:
m1 ⊇ m1 ∩ m2 ⊇ · · · ⊇
j
\
j+1
mi ⊇
i=1
\
mi ⊇ · · ·
i=1
Fix any k ∈ N. For each i = 1, . . . , k, choose ai ∈ mi \mk+1 as mi and mk+1 are incomparable.
Q
T
Q
Q
T
Then ki=1 ai ∈ ki=1 mi but ki=1 ai ∈
/ mk+1 since mk+1 ∈ Spec(R). Thus ki=1 ai ∈
/ k+1
i=1 mi ,
Tk
Tk+1
which implies i=1 mi 6= i=1 mi . Therefore the descending chain of ideals
m1 )
2
\
mi ) · · · )
i=1
j
\
i=1
j+1
mi )
\
mi ) · · ·
i=1
does not stabilize. This is a contradiction (to the assumption that R is Artinian). Hence
there are only finitely many maximal ideals, i.e., |Max(R)| < ∞.
Problem 3.2. Let R be an Artinian ring (with 1R ∈ R, as always), and J = ∩m∈Max(R) m
(that is, J = Jac(R), the Jacobson radical of R). For each k ∈ N, prove that the following
are equivalent:
(1) (0 :R J k ) 6= R.
(2) J k 6= J k+1 .
(3) J k 6= 0.
Proof. (1) ⇒ (2): Assume (0 :R J k ) 6= R; and denote I := (0 :R J k ). Since R is Artinian,
we see that R/I is a non-zero Artinian R-module. Therefore it contains a (non-zero) simple
R-submodule, which must be of the form K/I with I K ≤ R. Moreover, K/I ∼
= R/m for
some m ∈ Max(R) by the structure of simple modules. In particular, we have
m(K/I) = 0,
which means mK ⊆ I.
Now we see
J k+1 K = J k (JK) ⊆ J k I = 0,
which implies
(0 :R J k+1 ) ⊇ K ) I = (0 :R J k ).
Consequently we see J k 6= J k+1 .
The implication (2) ⇒ (3) and the implication (3) ⇒ (1) are straightforward. Indeed, just
consider their contrapositives.
Problem 3.3. Let R be an Artinian ring (with 1R ∈ R, as always), and J = Jac(R).
(1) Prove that there exists k ∈ N such that J k = J k+i for all i ∈ N.
(2) Prove that there exists k ∈ N such that J k = 0.
(3) Prove that Jac(R) = Nil(R).
5
Proof. (1) As R is Artinian, the claim follows from the descending chain of ideals
J 1 ⊇ J 2 ⊇ · · · ⊇ J i ⊇ J i+1 ⊇ · · · .
(2) This follows from (1) above and Problem 3.2. Indeed, we have J k = J k+i by (1), hence
(0 :R J k ) = R by Problem 3.2, which implies 1R ∈ (0 :R J k ), which forces J k = 0.
(3) First, note that
Jac(R) = ∩m∈Max(R) m ⊇ ∩P ∈Spec(R) P = Nil(R),
which is independent of R being Artinian. Second, as J k = 0 from (2) above, we see
Jac(R) ⊆ Nil(R).
Therefore Jac(R) = Nil(R).
Problem 3.4. Prove Spec(R) = Max(R) = Min(R) for every Artinian ring R. (Equivalently,
for every Artinian ring R, there exist no P, Q ∈ Spec(R) such that P ( Q.)
Proof. By Problem 3.1 and Problem 3.3, |Max(R)| < ∞ and Nil(R) = Jac(R); say Max(R) =
{m1 , . . . , mn }. Let P ∈ Spec(R) be an arbitrary prime ideal of R. Then
n
\
mi = Jac(R) = Nil(R) ⊆ P.
i=1
This necessarily implies mc ⊆ P for some c ∈ {1, . . . , n}. (Otherwise there would exist
/ P, a
xi ∈ mi \ P for all i ∈ {1, . . . , n}, which would imply Nil(R) = ∩ni=1 mi 3 x1 · · · xn ∈
contradiction.) This forces P = mc ∈ Max(R). So Spec(R) ⊆ Max(R). Since Spec(R) ⊇
Max(R) holds trivially, we see Spec(R) = Max(R). Therefore Spec(R) = Max(R) = Min(R).
Without using Problem 3.1 or Problem 3.3, we give another proof : Suppose the claim
fails. Then there must exist P ∈ Spec(R) such that P is not maximal. Let S := R/P .
(From now on, we forget R and just study S.) Note that S is an Artinian domain with the
zero ideal not maximal, i.e., 0 ∈
/ Max(S). Fix any m ∈ Max(S); so we have 0 ( m. Choose
any 0 6= x ∈ m. Since S is Artinian, the descending chain of ideals
(x) ⊇ (x2 ) ⊇ · · · ⊇ (xi ) ⊇ (xi+1 ) ⊇ · · ·
must stabilize; so there exists n ∈ N such that (xn ) = (xn+1 ). Thus there exists s ∈ S such
that xn = sxn+1 , which implies xn (1 − sx) = 0. Since S is a domain and x 6= 0, we see
1 − sx = 0 hence sx = 1. This implies x ∈ U(S), contradicting x ∈ m ( S.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
6
√
I
Math 8240 (Fall 2016)
Homework Set #04 (Due 09/21)
Solutions
Problem 4.1. Let I and J be proper ideals of a ring R.
√
√
(1) True or false: √I ⊆ √J if and only if V(I) ⊇ V(J).
(2) True or false: I = √
J if and
√ only if V(I) = V(J).
(3) Prove or disprove: √I ⊆ √J if and only if Min(I) ⊇ Min(J).
(4) Prove or disprove: I = J if and only if Min(I) = Min(J).
Solution/Disproof/Proof. (1) True. (2) True.
(3) We disprove the (false) statement with a counterexample: Let
R = Z,
I=0
and
J = (2).
√
In this case, I = I ⊆ J = J but Min(I) = {I} + {J} = Min(J). (As a matter of fact,
the ‘if ’ part of the statement holds and can be proved very easily.)
(4) We prove the (true) statement as follows: If Min(I) = Min(J), then
\
\
√
√
I=
P =
P = J.
√
P ∈Min(I)
√
P ∈Min(J)
√
Conversely, if I = J then V(I) = V(J) by part (2). Hence Min(I)
= Min(J) because
Min(K) consisting of the minimal elements of the poset V(K), ⊆ for any ideal K.
Problem 4.2. Let R be a ring, and Ii , Iλ be ideals of R with i ∈ {1, . . . , m} and λ ∈ Λ, in
which m ∈ N and Λ is an arbitrary set. Also let {P1 , . . . , Pn } ⊆ Spec(R), with n ∈ N.
Q
(1) True or false: For P ∈ Spec(R), m
i=1 Ii ⊆ P ⇐⇒ Ia ⊆ P for some a ∈ {1, . . . , m}.
m
n
(2) True or false: ∩i=1 Ii ⊆ ∪j=1 Pj ⇐⇒ Ia ⊆ Pb for some 1 6 a 6 m and 1 6 b 6 n.
p Qm
p m
Q
m
I
=
∩i=1 Ii , and hence V( m
(3) True or false:
i
i=1 Ii ) = V(∩i=1 Ii ).
i=1
m
m
(4) Prove or disprove: V(∩
Pi=1 Ii ) = ∪i=1 V(Ii ).
(5) Prove or disprove: V( λ∈Λ Iλ ) = ∩λ∈Λ V(Iλ ).
Q
Proof. (1) True. If Ii * P for all i ∈ {1, . . . , m} then one can construct x ∈ ( m
i=1 Ii ) \ P .
(2) True. This follows
from
Prime
Avoidance
and
part
(1)
above.
Qm
Q
m
m
m
(3) True. We have m
i=1 Ii .
i=1 Ii ⊆ ∩i=1 Ii and (∩i=1 Ii ) ⊆
(4) We prove it as follows: For all P ∈ Spec(R),
m
P ∈ V(∩m
i=1 Ii ) ⇐⇒ P ∈ V(∩i=1 Ii )
⇐⇒ ∩m
i=1 Ii ⊆ P
⇐⇒ Ii ⊆ P
for some i ∈ {1, . . . , m}
⇐⇒ P ∈ V(Ii ) for some i ∈ {1, . . . , m}
⇐⇒ P ∈ ∪m
i=1 V(Ii ).
(5) We prove it as follows: For all P ∈ Spec(R),
X X
P ∈V
Iλ ⇐⇒
Iλ ⊆ P
λ∈Λ
λ∈Λ
⇐⇒ Iλ ⊆ P
for all λ ∈ Λ
⇐⇒ P ∈ V(Iλ ) for all λ ∈ Λ
⇐⇒ P ∈ ∩λ∈Λ V(Iλ ).
7
Problem 4.3. Let R be a ring and consider the power series ring R[[x, y]]. Let
X
f=
aij xi y j = a00 + a10 x + a01 y + a20 x2 + a11 xy + a02 y 2 + · · · ∈ R[[x, y]]
i, j>0
in which aij ∈ R. Prove that f ∈ U(R[[x, y]]) ⇐⇒ a00 ∈ U(R), via the following steps:
(1) If f ∈ U(R[[x, y]]), verify that a00 ∈ U(R).P
(2) If a00 ∈ U(R), prove that there exists g = i, j>0 bij xi y j ∈ R[[x, y]] such that f g = 1.
P
Proof. (1) Assume f ∈ U(R[[x, y]]); say f −1 = i, j>0 bij xi y j ∈ R[[x, y]] with bij ∈ R. Since
the constant term of f f −1 must be 1, we see a00 b00 = 1, which proves a00 ∈ U(R).
(2) Assume a00 ∈ U(R). Define bmn ∈ R, with m, n ∈ N ∪ {0}, inductively as follows
X
am−i,n−j bij ∈ R when m + n > 0.
and
bmn = −a−1
b00 = a−1
00
00 ∈ R
06 i6m
06 j 6n
i+j < m+n
P
Then g := i, j>0 bij xi y j ∈ R[[x, y]] satisfies f g = 1, which proves f ∈ U(R[[x, y]]).
Combining (1) and (2), we establish f ∈ U(R[[x, y]]) ⇐⇒ a00 ∈ U(R). (It is not hard to
see that f ∈ R[[x1 , . . . , xn ]] is invertible ⇐⇒ the constant term of f is invertible.)
Problem 4.4. Let R be a ring and consider the power series ring R[[x, y]]. Prove that
R[[x, y]] is quasi-local if and only if R is quasi-local.
Proof.
To prove ‘if’, assume that (R, m) is quasi-local. By Problem 4.3, a power series
P
i j
i, j>0 aij x y is not invertible if and only if a00 ∈ R \ U(R) = m. In other words,
(
)
X
R[[x, y]] \ U(R[[x, y]]) =
aij xi y j a00 ∈ m ,
i, j>0
which is closed under addition. (Also, R[[x, y]] is non-zero.) Thus R[[x, y]] is quasi-local.
Conversely, to prove ‘only if’, assume that R[[x, y]] is quasi-local. Let a, b ∈ R \ U (R).
Viewing a, b as power series, we see a, b ∈
/ U(R[[x, y]]) by Problem 4.3. Thus a + b remains
non-invertible in R[[x, y]] as R[[x, y]] is quasi-local. Now, by Problem 4.3 again, we see
a + b ∈ R \ U(R). This shows that R \ U(R) is closed under addition. (Also note that R is
non-zero.) Therefore R is quasi-local as required. (For a general n ∈ N, it is not hard to see
that R[[x1 , . . . , xn ]] is quasi-local if and only if R is quasi-local.)
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
8
√
I
Math 8240 (Fall 2016)
Midterm Exam I (09/28)
Solutions
Problem I.1. Let R be a ring and I a proper ideal of R (i.e., I R). (Also, a ring S is
said to be quasi-semi-local iff 0 < |Max(S)| < ∞.)
(1) Prove or disprove: If R is quasi-semi-local then so is R/I.
(2) Prove or disprove: If R/I Q
is quasi-semi-local then so is R.
(3) Prove or disprove: If I = ri=1 mni i with r, ni positive integers and mi ∈ Max(R)
for all i = 1, . . . , r then R/I is quasi-semi-local.
Solution/Proof. (1) True, with Max(R/I) = Max(R) ∩ V(I). First, we see R/I 6= 0 and
hence 0 < |Max(R/I)|. Second, we have (quite generally, over any ring R)
Max(R/I) = {m/I : m ∈ Max(R) and I ⊆ m},
which implies |Max(R/I)| 6 |Max(R)|. Therefore 0 < |Max(R/I)| 6 |Max(R)| < ∞. This
proves that R/I is quasi-semi-local, with Max(R/I) = Max(R) ∩ V(I).
(2) False. Here is a counterexample: Let R be any non-zero ring that is not quasi-semilocal (e.g., R = Z) and let I be any maximal ideal of R (say I = (2) ∈ Max(Z)). Then R/I
is a field and hence quasi-semi-local, but R is not quasi-semi-local (by choice).
(3) True, with Max(R/I) = {m1 /I, . . . , mr /I}. Indeed, we have
r
Y
m/I ∈ Max(R/I) ⇐⇒ m ∈ Max(R) and I ⊆ m, i.e.,
mni i ⊆ m
i=1
⇐⇒ m ∈ Max(R) and mi ⊆ m for some i ∈ {1, . . . , r}
⇐⇒ m = mi for some i ∈ {1, . . . , r}.
This shows Max(R/I) = {m1 /I, . . . , mr /I}. Hence R/I is quasi-semi-local.
Problem I.2. Let R be a ring, and I, P ideals of R. Consider the polynomial ring R[x].
For every ideal J of R, denote J[x] := {a0 + · · · + an xn | 0 6 n ∈ Z and ai ∈ J}, which is an
ideal of R[x]. Prove the following:
(1) Prove R[x] ∼
= R [x] (i.e., R[x]/I[x] ∼
= (R/I)[x]) as rings.
I[x]
I
(2) Prove that P ∈ Spec(R) if and only if P [x] ∈ Spec(R[x]).
P
P
Proof.
(1) First, define a map h : R[x] → RI [x] by h( ni=0 ai xi ) = ni=0 ai xi ∈ RI [x] for all
Pn
i
i=0 ai x ∈ R[x], with ai := ai + I ∈ R/I. It is routine to verify that h is an onto ring
homomorphism from R[x] to RI [x] with Ker(h) = I[x]. By the Fundamental Theorem For
Ring Homomorphisms, we see
R[x]/I[x] = R[x]/ Ker(h) ∼
= Im(h) = R [x].
I
(2) Given (1), we see
P ∈ Spec(R) ⇐⇒ R/P is a domain
⇐⇒
R
[x]
P
is a domain
(1)
⇐⇒ R[x]/P [x] is a domain ⇐⇒ P [x] ∈ Spec(R[x]).
Problem I.3. Prove that, for any ring R, the polynomial ring R[x] is never quasi-local.
Proof. This follows from the fact x, x + 1 ∈ R[x] \ U(R[x]) but (x + 1) − x = 1 ∈ U(R[x]).
(Since the case of R = 0 is trivial, the above is under the assumption that R 6= 0.)
9
Problem I.4. Let R be a domain and consider the polynomial ring R[x].
(1) True or false: U(R[x]) = U(R). No justification is necessary.
(2) Prove that Jac(R[x]) = 0.
Solution/Proof. (1) True. (It is clear that U(R[x]) ⊇ U(R). For every f = f (x) ∈ U(R[x]),
there exists g = g(x) ∈ R[x] such that f g = 1, which implies deg(f ) + deg(g) = deg(f g) = 0
since R is a domain, which forces f, g ∈ R, hence f ∈ U(R). So U(R[x]) ⊆ U(R).)
(2) It is clear that {0} ⊆ Jac(R[x]). Next, let f = f (x) be an arbitrary non-zero polynomial
in R[x]. It is easy to see deg(1 + xf (x)) > 1. By (1), we see
1 + xf (x) ∈
/ U(R[x]).
Thus f (x) ∈
/ Jac(R[x]). This proves Jac(R[x]) ⊆ {0}. In conclusion, Jac(R[x]) = {0}.
Problem I.5. Let R be a ring, and P1 , . . . , Pn be pairwise incomparable prime ideals of R
(i.e., Pi * Pj and Pj * Pi for all i 6= j), with 0 < n ∈ Z. Let S = R \ ∪ni=1 Pi . Also consider
Ω = {I | I ≤ R such that I ∩ S = ∅}.
Let Max(Ω, ⊆) denote the set of all maximal elements of the poset (Ω, ⊆).
(1) True or false: S 6= ∅ and S is multiplicatively closed. No justification is necessary.
(2) True or false: ∅ 6= Max(Ω, ⊆) ⊆ Spec(R). No justification is necessary.
(3) Prove or disprove: Max(Ω, ⊆) = {P1 , . . . , Pn }.
Solution/Proof. (1) True. Indeed, x, y ∈ S =⇒ x, y ∈
/ Pi , ∀i =⇒ xy ∈
/ Pi , ∀i =⇒ xy ∈ S.
(2) True. This follows from Problem 2.4.
(3) We prove the claim as follows: Let j ∈ {1, . . . , n}. Clearly Pj ∈ Ω. Let Q 6 R
such that Pj ( Q. By assumption, we must have Q * Pi for all i ∈ {1, . . . , n}. By Prime
Avoidance, we see
Q * ∪ni=1 Pi ,
which simply means Q ∩ S 6= ∅ and hence Q ∈
/ Ω. Thus Pj ∈ Max(Ω, ⊆).
Conversely, let P ∈ Max(Ω, ⊆). In particular, P ∩ S = ∅ which is equivalent to P ⊆
∪ni=1 Pi . By Prime Avoidance, we see
P ⊆ Pk ∈ Ω for some k ∈ {1, . . . , n},
which forces P = Pk (in light of P ∈ Max(Ω, ⊆)). Now the proof is complete.
Extra Credit Problem I.6 (1 point, no partial credit). Prove or disprove: For any ring
R, the polynomial ring R[x] is never quasi-semi-local.
Proof. We prove the statement: If R is zero then clearly R[x] is not quasi-semi-local. Assume
that R is non-zero, so there exists m ∈ Max(R). Denote k := R/m, which is a field. By
Problem I.2, we see k[x] ∼
= R[x]/m[x]. By Problem I.1, if R[x] is quasi-semi-local then k[x]
must be quasi-semi-local. However, direct reasoning shows that k[x] is not quasi-semi-local.
(Similar to showing that there are infinitely many prime numbers, one can show that there
are infinitely many monic irreducible polynomials in k[x], which give rise to infinitely many
maximal ideals of k[x].) So R[x] is never quasi-semi-local.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
10
√
I
Math 8240 (Fall 2016)
Homework Set #05 (Due 10/05)
Solutions
Problem 5.1. Consider the commutative ring R = Z12 = Z/(12) = {[0], [1], . . . , [11]}. Let
S be the multiplicative subset of R generated by [3] ∈ R; that is, S = {[3]n | n > 0}.
(1) Partition all formal fractions rs (with r ∈ R and s ∈ S) such that two formal fractions
are in the same partition if and only if they determine the same element in S −1 R.
(2) Determine the ring structure of S −1 R. Is it isomorphic to a ring familiar to us?
Solution. (1) Since [3]2 = [9], [3]3 = [27] = [3], . . . , we see S = {[1], [3], [9]} ⊂ Z12 . Now
that |R| = 12 and |S| = 3, there are altogether 12 · 3 = 36 possible formal fractions of the
[0] [1]
form rs with r ∈ R and s ∈ S, namely, [1]
, [1] , . . . , [11]
; [0] , [1] , . . . , [11]
; [0] , [1] , . . . , [11]
. However,
[1] [3] [3]
[3] [9] [9]
[9]
−1
−1
there are exactly four (4) elements in S R (according to the construction of S R):
[0]
[1]
=
[4]
[1]
=
[8]
[1]
=
[0]
[3]
=
[4]
[3]
=
[8]
[3]
[1]
[1]
=
[5]
[1]
=
[9]
[1]
=
[3]
[3]
=
[7]
[3]
=
[11]
[3]
=
[0]
[9]
=
=
[1]
[9]
[4]
[9]
=
=
[5]
[9]
[8]
[9] ,
=
[2]
[1]
[9]
[9] ,
=
[3]
[1]
[6]
[1]
=
=
[7]
[1]
[10]
[1]
=
=
[11]
[1]
[2]
[3]
=
=
[1]
[3]
[6]
[3]
=
=
[5]
[3]
[10]
[3]
=
[2]
[9]
=
[6]
[9]
=
[10]
[9] ,
[9]
[3]
=
[3]
[9]
=
[7]
[9]
=
[11]
[9] .
=
(For example, [7]
= [5]
follows from [3]([7][3] − [1][5]) = [0]. However, we have [2]
6= [7]
as
[1]
[3]
[3]
[9]
[1]([2][9] − [3][7]) 6= [0], [3]([2][9] − [3][7]) 6= [0] and [9]([2][9] − [3][7]) 6= [0].)
[0] [1] [2]
(2) The four (4) distinct elements of S −1 R could be represented by [1]
, [1] , [1] and [3]
, for
[1]
[i]
for i ∈ {0, 1, 2, 3}. It is
example. Define a map h from Z4 = {0, 1, 2, 3} to S −1 R by i 7→ [1]
−1
routine to verify that h is a ring isomorphism. Consequently, S R ∼
= Z4 .
Problem 5.2. Let S be a non-empty multiplicative subset of a ring R (with 1R ∈ S). Let
ϕ : R → S −1 R be the natural ring homomorphism defined by ϕ(r) = 1r for all r ∈ R.
(1) Prove Ker(ϕ) = {r ∈ R | there exists s ∈ S such that sr = 0R } = ∪s∈S AnnR (s).
(2) Prove or disprove: ϕ is injective if and only if S contains no zero-divisor.
Proof. (1) Indeed, for r ∈ R,
r
0
=
1
1
⇐⇒ ∃s ∈ S such that sr = 0R
r ∈ Ker(ϕ) ⇐⇒ ϕ(r) = 0S −1 R ⇐⇒
⇐⇒ ∃s ∈ S such that r ∈ AnnR (s).
(2) We prove the statement as follows:
ϕ is injective ⇐⇒ Ker(ϕ) = {0} ⇐⇒ ∪s∈S AnnR (s) = {0}
⇐⇒ AnnR (s) = {0} for all s ∈ S ⇐⇒ S contains no zero-divisor.
Problem 5.3. Let S be a non-empty multiplicative subset of a ring R (with 1R ∈ S). Let
ϕ : R → S −1 R be the natural ring homomorphism defined by ϕ(r) = 1r for all r ∈ R. Prove
that the following are equivalent:
(1) The natural ring homomorphism ϕ is onto.
(2) For every s ∈ S, there exists x ∈ R such that xs − 1 ∈ Ker(ϕ).
(3) For every s ∈ S, there exist x ∈ R and s0 ∈ S such that s0 xs = s0 .
Proof. (1) ⇒ (2): Let s ∈ S, so that 1s ∈ S −1 R. Since ϕ is onto, there exists x ∈ R such
that ϕ(x) = 1s , that is, x1 = 1s . Therefore
ϕ(xs − 1) =
xs − 1
x s 1
1 s 1
s 1
= · − = · − = − = 0S −1 R ,
1
1 1 1
s 1 1
s 1
11
which implies xs − 1 ∈ Ker(ϕ).
(2) ⇒ (3): By (2), for every s ∈ S, there exists x ∈ R such that xs − 1 ∈ Ker(ϕ). Then
by Problem 5.2(1), there exists s0 ∈ S such that s0 (xs − 1) = 0, which yields s0 xs = s0 .
(3) ⇒ (1): For every rs ∈ S −1 R with r ∈ R and s ∈ S, there exist x ∈ R and s0 ∈ S such
that s0 xs = s0 , hence we have
r
s0 xr
s0 xr
xr
= 0 = 0 =
= ϕ(xr).
s
s xs
s
1
This proves that ϕ is onto.
Problem 5.4. Let S be a non-empty multiplicative subset of a ring R (with 1R ∈ S), and
I an ideal of R.
(1) Prove S −1 I = S −1 R if and only if I ∩ S 6= ∅.
(2) For any x ∈ R, show S −1 (I :R x) = (S −1 I :S −1 R x1 ).
Proof. (1) Indeed, we have
S −1 I = S −1 R ⇐⇒
1
∈ S −1 R
1
1
i
=
1
s
0
⇐⇒ ∃i ∈ I, s ∈ S, s ∈ S such that s0 (i − s) = 0R
⇐⇒ ∃i ∈ I, s ∈ S such that
⇐⇒ ∃i ∈ I, s ∈ S, s0 ∈ S such that I 3 s0 i = s0 s ∈ S
†
⇐⇒ I ∩ S 6= ∅,
†
in which ⇐= follows from the choice I 3 i = s ∈ S and any s0 ∈ S.
(2) Let f ∈ S −1 (I :R x), so that f = rs for some r ∈ (I :R x) and s ∈ S. Therefore
rx
rx
r −1
x
x
−1
=
∈ S I, which shows f = ∈ S I :S −1 R
.
f· =
1
s1
s
s
1
This establishes the inclusion S −1 (I :R x) ⊆ (S −1 I :S −1 R x1 ).
To show the other inclusion, let f = rs ∈ (S −1 I :S −1 R x1 ) with r ∈ R and s ∈ S. This
simply says rs · x1 ∈ S −1 I, i.e., rx
∈ S −1 I. That is, there exist a ∈ I and t ∈ S such
s
a
rx
−1
that s = t ∈ S R. In light of the construction of S −1 R, there exists u ∈ S such that
u(rxt − sa) = 0. Thus utrx = usa ∈ I and, hence, utr ∈ (I :R x). Consequently, we see
r
utr
f= =
∈ S −1 (I :R x) since utr ∈ (I :R x) and uts ∈ S.
s
uts
This establishes the inclusion S −1 (I :R x) ⊇ (S −1 I :S −1 R x1 ). The proof is complete. (More
generally, S −1 (K :R N ) = (S −1 K :S −1 R S −1 N ) for all R-modules K, N ⊆ M with N finitely
generated.)
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
12
√
I
Math 8240 (Fall 2016)
Homework Set #06 (Due 10/12)
Solutions
Problem
√ 6.1. Let R
√ be a ring, S a multiplicative subset (with 1 ∈ S) and I an ideal of R.
−1
−1
Prove S I = S
I.
√
√
Proof. Let f ∈ S −1 I, so that f = rs with r ∈ I and s ∈ S. Say rn ∈ I with n ∈ N. Then
√
√
√
n
n
f n = rs = rsn ∈ S −1 I, which shows f ∈ S −1 I. Therefore S −1 I ⊇ S −1 I.
√
Conversely, let g = st1 ∈ S −1 I with t ∈ R and s1 ∈ S. Then there exists an integer
m
m
m > 0 such that g m = st m ∈ S −1 I, that is, st m = si2 for some i ∈ I and s2 ∈ S. From the
1
1
construction of S −1 R, there exists s3 ∈ S such that (tm s2 − ism
1 )s3 = 0, which gives
√
m
tm s2 s3 = ism
s
∈
I,
which
implies
(ts
s
)
∈
I
and
hence
ts
s
∈
I.
3
2
3
2
3
1
√
√
√
Thus g = st1 = sts1 s22ss33 ∈ S −1 I. This shows S −1 I ⊆ S −1 I. Now the proof is complete. Problem 6.2. Let R be a ring, I a proper ideal of R, and S = {1 + r | r ∈ I}.
(1) Prove that S is a non-empty multiplicative subset of R and I ∩ S = ∅.
(2) Prove that S −1 R is a non-zero ring.
(3) Prove S −1 I ⊆ Jac(S −1 R).
Proof. (1) Let si = 1 + xi ∈ S with xi ∈ I for i = 1, 2. Then
s1 s2 = (1 + x1 )(1 + x2 ) = 1 + x1 + x2 + x1 x2 ∈ S
as x1 + x2 + x1 x2 ∈ I,
which shows S is multiplicatively closed. (And, clearly, 1 = 1 + 0 ∈ S.) Suppose there is
r ∈ I ∩ S. Then I 3 r = 1 + x for some x ∈ I. Hence 1 ∈ 1 + x − x = r − x ∈ I, which
contradicts the assumption that I is a proper ideal. This shows I ∩ S = ∅.
(2) Since I ∩ S = ∅, we see 0R ∈
/ S. Thus S −1 R is a non-zero ring. (Also, from I ∩ S = ∅,
we see S −1 I is a proper ideal of S −1 R. Thus S −1 R must be a non-zero ring.)
(3) As S −1 I is an ideal of S −1 R, it suffices to show 1S −1 R + f ∈ U(S −1 R) for all f ∈ S −1 R.
Let f ∈ S −1 I, so that f = xs for some x ∈ I and s ∈ S. We have
1S −1 R + f =
1
1
+
x
s
=
1
1
+
x
s
=
s+x
.
s
Now we claim s + x ∈ S. (Indeed, as s ∈ S, we see s = 1 + y for some y ∈ I. Hence
s+x = (1+y)+x = 1+(y +x) ∈ S as y +x ∈ I.) Consequently, 1S −1 R +f = s+x
∈ U(S −1 R)
s
s
−1
−1
with (1S −1 R + f ) = s+x ∈ S R. This completes the proof.
Problem 6.3. Let I R and S = 1R + I ⊂ R be as in Problem 6.2 above.
(1) Prove or disprove: For every ideal J of R, S −1 J = S −1 R ⇐⇒ I + J = R.
(2) Prove or disprove: Spec(S −1 R) = {S −1 P | P ∈ V(I)}
(3) Prove or disprove: Max(S −1 R) = {S −1 M | M ∈ Max(R) ∩ V(I)}
(4) True or false: S −1 R is quasi-local ⇐⇒ R/I is quasi-local.
Proof/Solution. (1) We prove it as follows:
5.4(1)
S −1 J = S −1 R ⇐⇒
⇐⇒
⇐⇒
⇐⇒
J ∩ S 6= ∅
∃j ∈ J, ∃s ∈ S such that j = s
∃j ∈ J, ∃i ∈ I such that j = 1 + i
∃j ∈ J, ∃i ∈ I such that − i + j = 1 ⇐⇒ I + J = R.
13
(2) We disprove it as follows: Let R = Z and I = (2), hence S = 1 + (2) consists of all
odd integers. Then 0 = (0) ∈
/ V(I) but S −1 0 ∈ Spec(S −1 R). (In fact, it is straightforward
−1
to see that Spec(S R) = {S −1 P | P ∈ Spec(R) such that I + P 6= R}.)
(3) We prove it as follows: Let S −1 P ∈ Max(S −1 R) (and we may assume P ∈ Spec(R)).
By (1) above, I + P 6= R, hence I + P 6 M for some M ∈ Max(R), which implies M ∈ V(I).
By (1) and Problem 5.4, I + M 6= R ⇐⇒ S −1 M 6= S −1 R ⇐⇒ M ∩ S = ∅. Therefore
S −1 M ∈ Spec(S −1 R), hence S −1 P = S −1 M .
Conversely, let M ∈ Max(R) ∩ V(I). As above, I + M = M 6= R yields M ∩ S = ∅, hence
S −1 M ∈ Spec(S −1 R). Now it is clear that S −1 M is a maximal ideal in S −1 R.
(4) True. By (3), we have |Max(S −1 R)| = |Max(R) ∩ V(I)| = |Max(R/I)|.
Problem 6.4. Let R be a non-zero ring. Prove that the following statements are equivalent:
(1) R is reduced.
(2) S −1 R is reduced for every non-empty multiplicative subset S of R.
(3) RP is reduced for every P ∈ Spec(R).
(4) Rm is reduced for every m ∈ Max(R).
p
Proof. (1) ⇒ (2): Assume that R is reduced, i.e.,
(0R ) = (0R ). For any non-empty
multiplicative subset S of R, we have
p
p
p
6.1
(0S −1 R ) = S −1 (0R ) = S −1 (0R ) = S −1 (0R ) = (0S −1 R )
by applying Problem 6.1 to the ideal I = (0R ). This proves that S −1 R is reduced.
(2) ⇒ (3) ⇒ (4): This is obvious.
(4) ⇒ (1): Statement (4) and Problem 6.1 imply
p
p
6.1 p
(2)
(0R ) m = (0R )m = (0Rm ) = (0Rm ) = (0R )m .
p
Simply put,
(0R ) m = (0R )m for all m ∈ Max(R). By the result that “Im = Jm for all
p
m ∈ Max(R) implies I = J”, we see that (0R ) = (0R ) and hence R is reduced.
(Here is
p
x
n
another argument: ∀x ∈ Nil(R), ∃n > 1 such that x = 0, which shows 1 ∈ (0Rm ) = (0Rm ),
i.e., x1 = 0Rm for all m ∈ Max(R). This implies x = 0R . Thus Nil(R) = (0R ).)
Problem 6.5 (Extra Credit, 1 point). Let R be a ring, S a non-empty multiplicative subset
of R. Spell out a necessary-and-sufficient condition (on S) for the natural map R → S −1 R
to be an isomorphism. No need to justify.
Solution. The natural map R → S −1 R is an isomorphism ⇐⇒ S ⊆ U(R). This is an easy
consequence of Problem 5.2 and Problem 5.3.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
14
√
I
Math 8240 (Fall 2016)
Homework Set #07 (Due 10/19)
Solutions
Problem 7.1. Let R be a ring, P ∈ Spec(R), M an R-module, and x ∈ M .
(1) Prove AssR (R/P ) = {P }.
(2) Prove that Rx ∼
= R/ AnnR (x) as R-modules.
(3) True or false: If M1 ∼
= M2 as R-modules then AssR (M1 ) = AssR (M2 ).
Proof/Solution. (1) We denote x = x + P for all x ∈ R. If x = 0, then clearly AnnR (x) = R.
If x 6= 0, then x ∈
/ P and for all r ∈ R,
r ∈ AnnR (x) ⇐⇒ rx = 0 ⇐⇒ rx = 0 ⇐⇒ rx ∈ P ⇐⇒ r ∈ P.
This shows that AnnR (x) = P for every x 6= 0 ∈ R/P . Consequently, P is in AssR (R/P and
P is the only prime ideal in AssR (R/P . This completes the proof that AssR (R/P ) = {P }.
(2) Define h : R → M by h(r) = rx. It is routine to verify h ∈ HomR (R, M ); and it is
clear that h(1R ) = x. It is also routine/easy to verify Ker(h) = AnnR (x) and Im(h) = Rx.
Therefore
Rx = Im(h) ∼
= R/ Ker(h) = R/ AnnR (x).
(3) True. (The proof is easy and elementary.)
Problem 7.2. For any ring R, any R-module M and any P ∈ Spec(R), prove that the
following statements are equivalent:
(1) P ∈ AssR (M ).
(2) There exists an R-submodule J 6 M such that J ∼
= R/P as R-modules.
(3) There exists a finitely generated R-submodule K 6 M such that AssR (K) = {P }.
(4) There exists an R-submodule L 6 M such that P ∈ AssR (L).
Proof. (1) ⇒ (2): Assume P ∈ AssR (M ). Then there exists x ∈ M such that AnnR (x) = P .
Letting J := Rx 6 M , we see
J = Rx ∼
= R/ AnnR (x) = R/P.
(2) ⇒ (3): Let K := J 6 M as in (2). Since K ∼
= R/P , we see that K is finitely generated
over R (since R/P is so) and AssR (K) = AssR (R/P ) = {P } by Problem 7.1(1)&(3).
(3) ⇒ (4): This is obvious, via letting L := K.
(4) ⇒ (1): Since L 6 M and P ∈ AssR (L), we see P ∈ AssR (L) ⊆ AssR (M ). Now the
proof is complete.
Problem 7.3. Let R be a ring, S a non-empty multiplicative subset of R, P ∈ Spec(R),
and M a finitely generated R-module. Also denote SuppR (M ) := {P ∈ Spec(R) | MP 6= 0},
called the support of M .
(1) Prove that S −1 M = 0 ⇐⇒ AnnR (M ) ∩ S 6= ∅.
(2) Prove or disprove: MP 6= 0 ⇐⇒ AnnR (M ) ⊆ P .
(3) Prove or disprove: SuppR (M ) is a closed subset of Spec(R) in Zariski topology.
Solution. (1) For one direction, assume t ∈ AnnR (M ) ∩ S 6= ∅. Then, for all ms ∈ S −1 M
with s ∈ S and m ∈ M , we have
m
tm
0
=
=
= 0S −1 M ,
s
ts
ts
showing that S −1 M = 0. (This direction does not rely on M being finitely generated.)
15
Conversely, assume S −1 M = 0. Since M is a finitely generated R-module, say that M is
generated by m1 , . . . , mk ∈ M . For each i = 1, . . . , k, as m1i ∈ S −1 M = 0, there is si ∈ S
Q
such that si mi = 0M . Letting s := ki=1 si ∈ S, we see
smi = 0M
for all i = 1, . . . , k,
which implies s ∈ AnnR (M ) ∩ S.
(2) We prove it as follows: By (1) above, we see
MP 6= 0 ⇐⇒ AnnR (M ) ∩ (R \ P ) = ∅ ⇐⇒ AnnR (M ) ⊆ P.
(3) We prove it as follows: By (2) above, we see
SuppR (M ) = V(AnnR (M )),
which is a closed subset of Spec(R) in Zariski topology.
Problem 7.4. Let Z be the ring of integers and Q be the field of rational numbers.
| m, n ∈ Z and gcd(21, n) = 1}. Realize R as a ring of fractions of Z.
(1) Let R = { m
n
That is, find a non-empty multiplicative subset S of Z such that R = S −1 Z.
(2) Identify each of the prime ideals of R explicitly. Which of them are maximal?
(3) By choosing a suitable multiplicative subset U of Z, construct a ring of fractions
T = U −1 Z such that T has precisely five (5) prime ideals.
Solution/Proof. (1) Let S = {n ∈ Z | gcd(21, n) = 1}, which is clearly multiplicatively closed
in Z. As 21 = 3 · 7, we may equivalently write
S = {n ∈ Z | 3 - n and 7 - n} = Z \ (3Z ∪ 7Z).
Then it is clear that R = S −1 Z. (To be rigorous, you might want to write R ∼
= S −1 Z.)
(2) The prime ideals of Z are precisely {0} and (p) for all prime numbers p. It is clear
that {0} ∩ S = ∅. Also, given any prime number p, we see
(p) ∩ S = ∅ ⇐⇒ p = 3 or p = 7.
Thus, {P ∈ Spec(Z) | P ∩ S = ∅} consists of P0 = {0}, P3 = (3) and P7 = (7) precisely.
Correspondingly, we see
Spec(S −1 Z) = {S −1 P0 , S −1 P3 , S −1 P7 } and
Max(S −1 Z) = {S −1 P3 , S −1 P7 }.
(3) Let U = {n ∈ Z | gcd(210, n) = 1}, which is a multiplicatively closed subset of Z. Also
note that U = Z \ ((2) ∪ (3) ∪ (5) ∪ (7)) as 210 = 2 · 3 · 5 · 7. By reasoning similar to that in
part (2) above, we see that the prime ideals in Spec(Z) not intersecting with U are P0 = {0},
P2 = (2), P3 = (3), P5 = (5) and P7 = (7) precisely. Correspondingly, there are precisely
five (5) prime ideals of T = U −1 Z, namely, U −1 P0 = {0}, U −1 P2 , U −1 P3 , U −1 P5 and U −1 P7 .
(Note that Max(U −1 Z) = {U −1 P2 , U −1 P3 , U −1 P5 , U −1 P7 }.) Also see Problem I.5.
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
16
√
I
Math 8240 (Fall 2016)
Homework Set #08 (Due 10/26)
Solutions
Problem 8.1. Let R be a ring, M be an R-module, Q be a P -primary R-submodule of M ,
and x ∈ M such that x ∈
/ Q. Prove that (Q :R x) is a P -primary ideal of R.
Proof. Consider 0 6= (Q + Rx)/Q 6 M/Q. As M/Q is P -coprimary, so is (Q + Rx)/Q. Also,
it is straightforward to see that (Q + Rx)/Q ∼
= R/(Q :R x) (cf. Problem 7.1(2)). Hence
R/(Q :R x) is P -coprimary, which proves that (Q :R x) is P -primary in R. (In fact, we could
have assumed Q = 0 without loss of generality.)
Alternatively, here is an elementary approach: First of all, (Q :R x) is a proper
ideal of R since x ∈
/p
Q.
p
p
Next, we prove (Q :R x) = P . It is clear that P =
Ann
(M/Q)
⊆
(Q :R x).
R
p
Conversely, for any r ∈ (Q :R x), we have rx ∈ Q and hence r ∈ Ann
pR (M/Q) = P√since Q
is P -primary
/ Q. This shows (Q :R x) ⊆ P , which yields (Q :R x) ⊆ P = P .
p in M and x ∈
Therefore (Q :R x) = P .
It remains to prove that (Q :R x) is primary (in R). To this end, let a, b ∈ R such that
ab ∈ (Q :R x) and b ∈
/ (Q :Rpx). This simply says p
abx ∈ Q and bx ∈
/ Q. Thus, as Q is
P -primary in M , we see a ∈ AnnR (M/Q) = P = (Q :R x). This proves that the ideal
(Q :R x) is primary, and hence P -primary, in R.
Problem 8.2. Let R be a ring, M be an R-module, Q be a P -primary R-submodule of M ,
and r ∈ R such that r ∈
/ AnnR (M/Q). Prove that (Q :M r) is a P -primary submodule of M .
r
Proof. Consider the R-linear map h : M/Q → M/Q defined by multiplication by r, that is,
h(m + Q) = rm + Q for all m + Q ∈ M/Q. Observe that Ker(h) = (Q :M r)/Q. Therefore
M/(Q :M r) ∼
=
M/Q
M/Q ∼
=
= Im(h) 6 M/Q.
(Q :M r)/Q
Ker(h)
Moreover, 0 6= Im(h) since r ∈
/ AnnR (M/Q). Finally, note that M/Q is P -coprimary, which
then implies that M/(Q :M r) ∼
= Im(h) is P -coprimary, which eventually proves that (Q :M r)
a P -primary submodule of M . (Once again, we could have assumed Q = 0 WLOG. Also,
an elementary approach is available, which is omitted here.)
Problem 8.3. Let R be a ring, M be an R-module, Q be a P -primary R-submodule of M ,
and r ∈ R. Consider the following ascending chain of R-submodules of M :
(∗)
(Q :M r0 ) ⊆ (Q :M r1 ) ⊆ · · · ⊆ (Q :M ri ) ⊆ (Q :M ri+1 ) ⊆ · · ·
(1) In the case of r ∈
/ P , prove that the ascending chain (∗) stabilizes eventually.
(2) In the case of r ∈ P , prove that the ascending chain (∗) stabilizes eventually.
(3) In each case, determine the (stabilized) value of (Q :M ri ) for i 0.
Proof/Solution. (1) We claim that Q = (Q :M ri ) for all i > 0 if r ∈
/ P . As r ∈
/ P and
P ∈ Spec(R), we see ri ∈
/ P for all i > 0. Fix any i > 0 and let x ∈ (Q :M ri ), so
that ri x ∈ Q. As Q is P -primary in M and ri ∈
/ P , it is forced that x ∈ Q. This shows
(Q :M ri ) ⊆ Q. Moreover, the inclusion Q ⊆ (Q :R ri ) is automatic. Therefore (Q :M ri ) = Q
for all i > 0. In particular,
the ascending chain (∗) stabilizes.
p
(2) Since r ∈ P = AnnR (M/Q), there exists n ∈ N such that rn ∈ AnnR (M/Q), hence
i
r ∈ AnnR (M/Q) for all i > n. Thus (Q :M ri ) = M for all i > n. Therefore the ascending
chain (∗) stabilizes eventually.
17
(
Q for all i > 0 if r ∈
/P
ri ) =
by the work in (1) and (2). M for all i 0 if r ∈ P
(3) We see that (Q :M
Problem 8.4. Let R be a ring, K ( M be R-modules, and r ∈ R. Assume that K admits
a primary decomposition in M ; say K = Q1 ∩ · · · ∩ Qs in which Qi is Pi -primary in M for
i = 1, . . . , s with s ∈ N. Consider the following ascending chain of R-submodules of M :
(K :M r0 ) ⊆ (K :M r1 ) ⊆ · · · ⊆ (K :M ri ) ⊆ (K :M ri+1 ) ⊆ · · ·
(∗∗)
(1) Prove that the ascending chain (∗∗) stabilizes eventually.
(2) Determine the (stabilized) value of (K :M ri ) for i 0.
Proof/Solution. (1)&(2) By Problem 8.3, we see (for all i 0)
!
s
s
\
\
(K :M ri ) =
Qj :M ri =
(Qj :M ri )
j=1
j=1

=

\
(Qj :M ri )
!
\
\
r∈Pk
r∈P
/ j

=

\
(Qk :M ri )
Qj 

!
\
\
M
=
r∈Pk
r∈P
/ j

\
\
Qj 
\
M=
r∈P
/ j
Qj
r∈P
/ j
for all i 0. (More concretely, we may partition {1, . . . , s} as A ∪ B such that r ∈
/ Pj if
and only if j ∈ A while r ∈ Pk if and only if k ∈ B. By Problem 8.3(1), we see
(Qj :M ri ) = Qj
for all j ∈ A and i > 0.
For each k ∈ B, there exists nk > 0 such that rnk ∈ AnnR (M/Qk ) and hence (Qk :M ri ) = M
for all i > nk ; see Problem 8.3(2). Let n = max{nk | k ∈ B}. Then
(Qk :M ri ) = M
for all k ∈ B and i > n.
Consequently (for all i > n)
!
s
s
\
\
(K :M ri ) =
Qj :M ri =
(Qj :M ri ) =
j=1
j=1
\
j∈A
Qj
(Qj :M ri )
!
\
j∈A
!
=
!
\
\
M
k∈B
(Qk :M ri )
k∈B
!
\
\
!
=
\
j∈A
Qj
\
M=
\
Qj =
j∈A
\
Qj
r∈P
/ j
for all i > n.)
In conclusion, (∗∗) stabilizes with (K :M ri ) = ∩r∈P
/ j Qj for all i 0. (In case r ∈ Pj for
all j = 1, . . . , s, the above expression ∩r∈P
Q
yields
the
intersection of none of the Qj , which
/ j j
is M by convention, which still agrees with the stabilized value of (K :M ri ) when i 0.) PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
18
√
I
Math 8240 (Fall 2016)
Midterm Exam II (11/02)
Solutions
Problem II.1. Let R be a ring and P ∈ Spec(R). Among the following 6 statements
concerning P , determine which implies (or does not imply) which.
(1) P ∈ Min(R).
(4) dim(RP ) = 0.
(2) | Spec(RP )| = 1.
(5) RP is Artinian.
(3) PP = Nil(RP ).
(6) RP is a field.
Solution. We have (6) ⇒ (5) ⇒ (1) ⇔ (2) ⇔ (3) ⇔ (4) 6⇒ (5) 6⇒ (6). All the implications
should be straightforward (details skipped). For (4) 6⇒ (5) (hence for (4) 6⇒ (6) as well), let
K be any field and R = K[xi | i ∈ N]/m2 where m = (xi | i ∈ N). Then Spec(R) = {P :=
m/m2 }, hence dim(R) = 0. But RP ∼
= R is not Artinian. For (4) 6⇒ (6), see R = Z4 and
P = 2Z4 for example. (If RP is Noetherian, then (4) ⇒ (5) holds, but still (4) 6⇒ (6).)
Problem II.2. Let f : R → T be a ring homomorphism (with R and T rings) and S be a
(non-empty) multiplicative subset of R. Denote U = f (S), a multiplicative subset of T .
(1) Construct a ring homomorphism g : S −1 R → U −1 T explicitly.
(2) Denote K = Ker(f ). Prove or disprove: Ker(g) = S −1 K.
(3) Denote A = Im(f ), a subring of T . Prove or disprove: Im(g) = U −1 A.
Solution/Proof. (1) Let hR : R → S −1 R and hT : T → U −1 T be the ring homomorphisms
defined by hR (r) = 1rR and hT (t) = 1tT for all r ∈ R and t ∈ T . For every s ∈ S,
hT f (s) = hT (f (s)) = f1(s)
, which is invertible in U −1 T . By the universal property of S −1 R,
T
there is a (unique) ring homomorphism g : S −1 R → U −1 T (such that ghR = hT f ) which is
−1 f (r)
defined by g( rs ) = (hT f (s))−1 hT f (r) = f1(s)
= ff (r)
, where r ∈ R and s ∈ S.
1T
(s)
T
=
(2) For every ks ∈ S −1 K with k ∈ K = Ker(f ) and s ∈ S, it is clear that g rs = ff (k)
(s)
0
k
r
= 0U −1 T and hence s ∈ Ker(g). Conversely, let α = s ∈ Ker(g) with r ∈ R and s ∈ S so
f (s)
that ff (r)
= g rs = 0U −1 T = 01TT . There exists u ∈ U = f (S) such that uf (r) = 0T . Writing
(s)
u = f (s0 ) with s0 ∈ S, we see f (s0 r) = f (s0 )f (r) = 0. Therefore s0 r ∈ Ker(f ) = K and hence
0
α = rs = ss0 rs ∈ S −1 K. This proves Ker(g) = S −1 K.
r ∈ R and s ∈ S =
(3) It is straightforward to see Im(g) = g rs rs ∈ S −1 R = ff (r)
(s)
a a ∈ Im(f ) and u ∈ f (S) = a a ∈ A and u ∈ U = U −1 A.
u
u
Problem II.3. Let R be a ring, M a finitely generated R-module, and P ∈ Spec(R). Denote
n := µRP (MP ), the minimal number of generators of MP over RP . Prove that there exists
an open subset D of Spec(R) such that P ∈ D and µRQ (MQ ) 6 n for all Q ∈ D.
Proof. Say MP is generated by xsii ∈ MP , 1 6 i 6 n, with xi ∈ M and si ∈ R \ P . Let
P
N := ni=1 Rxi 6 M . Then NP = MP , or equivalently, (M/N )P = 0. Observe that M/N is
finitely generated (as M is finitely generated). By Problem 7.3(3), we see that
D := {Q ∈ Spec(R) | (M/N )Q = 0}
is an open subset of Spec(R) in Zariski topology; clearly P ∈ D.
Finally, for all Q ∈ D, since (M/N )Q = 0, we have MQ = NQ , which then implies
µRQ (MQ ) = µRQ (NQ ) 6 n, which completes the proof. (This shows that, for any m ∈ Z,
Dm := {p ∈ Spec(R) | µRp (Mp ) 6 m} is open in Spec(R), which proves that the function
defined by p 7→ µRp (Mp ), from Spec(R) to Z, is upper semi-continuous.)
19
Problem II.4. Let R be a ring and let {Mλ }λ∈Λ be a family of R-modules (with Λ 6= ∅).
Prove AssR (⊕λ∈Λ Mλ ) = ∪λ∈Λ AssR (Mλ ).
Proof. For each λ ∈ Λ, there is an injective R-linear map iλ : Mλ → ⊕λ∈Λ Mλ , which implies
AssR (Mλ ) ⊆ AssR (⊕λ∈Λ Mλ ); hence ∪λ∈Λ AssR (Mλ ) ⊆ AssR (⊕λ∈Λ Mλ ).
Conversely, let P ∈ AssR (⊕λ∈Λ Mλ ), so P = AnnR (x) for some x ∈ ⊕λ∈Λ Mλ . By the
construction of ⊕λ∈Λ Mλ , x exists in a direct sum of finitely many summands. More precisely,
up to isomorphism, we have x ∈ ⊕ri=1 Mλi 6 ⊕λ∈Λ Mλ for some λ1 , . . . , λr ∈ Λ with r ∈ N.
Hence P ∈ AssR (⊕ri=1 Mλi ) = ∪ri=1 AssR (Mλi ) ⊆ ∪λ∈Λ AssR (Mλ ).
Problem II.5. Let R be a ring and S a (non-empty) multiplicative subset of R. Let M be
an R-module. Prove each of the following:
(1) AssS −1 R (S −1 M ) ⊇ {S −1 P | P ∈ AssR (M ) and P ∩ S = ∅}.
(2) AssS −1 R (S −1 M ) = {S −1 P | P ∈ AssR (M ) and P ∩ S = ∅} if R is Noetherian.
Proof. (1) Let P ∈ AssR (M ) such that P ∩ S = ∅. Then there exists x ∈ M such that
AnnR (x) = P . By an argument similar to Problem 5.4(2), we see that (details omitted)
AnnS −1 R x1 = S −1 AnnR (x) = S −1 P ∈ Spec(S −1 R).
Thus P ∈ AssS −1 R (S −1 M ). So AssS −1 R (S −1 M ) ⊇ {S −1 P | P ∈ AssR (M ) and P ∩ S = ∅}.
(2) Let Q ∈ AssS −1 R (S −1 M ). Then there exist x ∈ M , s ∈ S, and P ∈ Spec(R) with
P ∩ S = ∅ such that AnnS −1 R ( xs ) = Q = S −1 P . Hence
∗
S −1 P = AnnS −1 R xs = AnnS −1 R x1 = S −1 AnnR (x).
∗
(Again, = is similar to Problem 5.4(2).) Since S −1 AnnR (x) = S −1 P , we must have
AnnR (x) ⊆ (S −1 AnnR (x))c = (S −1 P )c = P .
Next, we prove that P is minimal over AnnR (x) as follows: If there exists P 0 ∈ Spec(R)
such that AnnR (x) ⊆ P 0 ( P , then we would get S −1 P = S −1 AnnR (x) ⊆ S −1 P 0 ( S −1 P ,
which is a contradiction. Hence P is minimal over AnnR (x).
Now that P is minimal over AnnR (x), we see
P ∈ Min(R/(AnnR (x)) = Min(Rx) ⊆ AssR (Rx) ⊆ AssR (M )
since R is Noetherian. Thus AssS −1 R (S −1 M ) ⊆ {S −1 P | P ∈ AssR (M ) and P ∩ S = ∅}. Extra Credit Problem II.6 (1 point, no partial credit). Let R be a ring. Prove or
disprove: R is a domain if and only if RP is a domain for all P ∈ Spec(R).
Counterexample. Let R = F1 × F2 , with Fi any fields. The ideals of R are precisely
P1 := (0F1 ) × F2 , P2 := F1 × (0F2 ) and R,
with Spec(R) = {P1 , P2 }. Moreover RP1 ∼
= F1 and RP2 ∼
= F2 (details omitted). Therefore
RP is a domain (actually a field) for every P ∈ Spec(R), but R is not a domain.
(0R ),
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
20
√
I
Math 8240 (Fall 2016)
Homework Set #09 (Due 11/09)
Solutions
Problem 9.1. Let R be a ring (not necessarily Noetherian), S a multiplicative subset of R,
M an R-module, and r ∈ R. Recall that ZDR (M ) = {a ∈ R | (0 :M a) 6= 0}.
(1) Prove that if r ∈
/ ZDR (M ) then 1r ∈
/ ZDS −1 R (S −1 M ).
(2) Disprove: if r ∈ ZDR (M ) then 1r ∈ ZDS −1 R (S −1 M ).
(3) Prove ∪P ∈Min(R) P ⊆ ZD(R). (Assume R is non-zero, as otherwise this is trivial.)
Proof/Solution. (1) Let r ∈
/ ZDR (M ), so that (0 :M r) = {0M }. By an argument similar
to Problem 5.4(2), we see (0 :S −1 M 1r ) = S −1 (0 :M r) = S −1 {0M } = {0S −1 M } and hence
r
∈
/ ZDS −1 R (S −1 M ). (In fact, being a non-zero-divisor is a local property.)
1
(2) Let M = R = Z12 and S = {[1], [3], [9]} be as in Problem 5.1. Consider r = [3] ∈ R.
It is straightforward to see that r ∈ ZD(R) but 1r ∈
/ ZD(S −1 R).
(3) Suppose r ∈
/ ZD(R) for some r ∈ P ∈ Min(R). Let W = {rn s | n > 0 and s ∈ R \ P }.
It is easy to see that W is multiplicatively closed and
∅ 6= R \ P ( {r} ∪ (R \ P ) ⊆ W.
Moreover, the assumption r ∈
/ ZD(R) implies 0R ∈
/ W . So there exists Q ∈ Spec(R) such
that Q ∩ W = ∅. This implies Q ( P , contradicting the assumption that P ∈ Min(R).
Therefore ∪P ∈Min(R) P ⊆ ZD(R).
(Here we give an alternative proof of (3) via localization: If r ∈ P ∈ Min(R), then
r
∈
PP = Nil(RP ) (cf. Problem II.1), hence 1r ∈ ZD(RP ), therefore r ∈ ZD(R) by (1).) 1
Problem 9.2. Let R be a ring, I a finitely generated ideal of R, and M an R-module.
n
Consider K = ∩∞
n=1 (I M ), which is an R-submodule of M . Assume that
IK = Q1 ∩ · · · ∩ Qs
with Qi being Pi -primary in M
for i = 1, . . . , s.
(This is the case when R is Noetherian and M is finitely generated over R.)
(1) In the case of I ⊆ Pi , prove K ⊆ Qi .
(2) In the case of I * Pi , prove K ⊆ Qi .
n
∞
n
(3) Prove IK = K, i.e., I ∩∞
n=1 (I M ) = ∩n=1 (I M ).
p
Proof. (1) Since I ⊆ Pi = AnnR (M/Qi ) and I is finitely generated, there exists n ∈ N
such that I n ⊆ AnnR (M/Qi ) and hence I n M ⊆ Qi . Thus K ⊆ I n M ⊆ Qi .
(2) Choose any r ∈ I \ Pi . Clearly rK ⊆ IK ⊆ Qi . Now we see K ⊆ (Qi :M r) = Qi ,
since Qi is Pi -primary in M (plus r ∈
/ Pi ); see the proof of Problem 8.3(1).
(3) By (1) and (2) above, we see K ⊆ Qi for all i = 1, . . . , s. Therefore
IK ⊆ K ⊆ Q1 ∩ Q2 ∩ · · · ∩ Qs = IK.
(When M is Noetherian over R, IK = K can also be proved via Artin-Rees lemma.)
Problem 9.3. Let R be a ring, K M be R-modules such that K admits a primary
decomposition K = Q1 ∩ · · · ∩ Qs , in which Qi is Pi -primary in M for i = 1, . . . , s. For any
submodule N 6 M , prove the following:
(1) For each i = 1, . . . , s, if N ∩ Qi 6= N then N ∩ Qi is Pi -primary in N .
(2) If N * K then N ∩K admits a primary decomposition in N . (Also see Problem 10.5.)
21
Proof. (1) For every i = 1, . . . , s, we have
N
N + Qi
M
∼
6
.
=
N ∩ Qi
Qi
Qi
Note that M/Qi is Pi -coprimary. Thus, either N ∩ Qi = N or N/(N ∩ Qi ) is Pi -coprimary.
(2) If N ∩ K N , then
s
\
\
\
N ∩ K = (N ∩ Qi ) =
(N ∩ Qi ) =
(N ∩ Qi ),
i=1
N ∩Qi 6=N
N *Qi
which is a primary decomposition of N ∩ K in N , by (1) above. (A priori, we don’t know
(or care) whether the above intersection is minimal. If you care, see Problem 10.5.)
Problem 9.4. Let I be a finitely generated ideal of a ring R, and K M be R-modules
such that K admits a primary decomposition K = Q1 ∩ · · · ∩ Qs , in which Qi is Pi -primary
i
in M for i = 1, . . . , s. Also for any L 6 M , let (L :M I ∞ ) := ∪∞
i=1 (L :M I ) 6 M .
(1) In the case of I Pi , determine (Qi :M I ∞ ) explicitly.
(2) In the case of I 6 Pi , determine (Qi :M I ∞ ) explicitly.
(3) True or false: There exists n ∈ N such that (Qi :M I ∞ ) = (Qi :M I n ).
(4) Determine (K :M I ∞ ) explicitly.
(5) True or false: There exists n ∈ N such that (K :M I ∞ ) = (K :M I n ).
Solution. We provide answers as follows. (Proofs are omitted here, as they are similar to
Problem 8.3 and Problem 8.4.)
(1) If I Pi , then (Qi :M I ∞ ) = (Qi :M I n ) = Qi for all n > 0.
(2) If I 6 Pi , then (Qi :M I ∞ ) = (Qi :M I n ) = M for all n such that I n ⊆ AnnR (M/Qi ).
(3) True. This follows from (1) and (2). T
(4) By (1) and (2), we have (K :M I ∞ ) = I*Pi Qi , see (5) below.
(5) True. By (1) and (2), (K :M I ∞ ) = (K :M I n ) for all n such that
\
In ⊆
AnnR (M/Qi ).
I⊆Pi
Such n exists, as I is finitely generated. (This also explains the existence of n in (2).)
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
22
√
I
Math 8240 (Fall 2016)
Homework Set #10 (Due 11/16)
Solutions
Problem 10.1. Let R ⊆ T be an extension of rings (hence 1R = 1T ) and t ∈ T . Prove that
the following statements are equivalent (i.e., being integral is a local property):
(1) The element t is integral over R.
(2) The element 1t ∈ S −1 T is integral over S −1 R for every multiplicative subset S of R.
(3) The element 1t ∈ TP is integral over RP for every P ∈ Spec(R).
(4) The element 1t ∈ Tm is integral over Rm for every m ∈ Max(R).
Note that, for P ∈ Spec(R), the notation TP stands for (R \ P )−1 T .
T
Proof. Let R := R be the integral closure of R in T . (So R is naturally an R-module.)
(1) ⇒ (2): Assume that t is integral over R, i.e., t ∈ R. Then
T
t
S −1 T
∈ S −1 R = S −1 R = S −1 R
,
1
that is, 1t ∈ S −1 T is integral over S −1 R, for every multiplicative subset S of R.
(2) ⇒ (3) ⇒ (4): This is obvious.
(4) ⇒ (1): The assumption that 1t ∈ Tm is integral over Rm for every m
simply
∈TMax(R)
T
m
says that 1t ∈ Rm for every m ∈ Max(R), which is equivalent to 1t ∈ R
= R m for
m
every m ∈ Max(R), which yields t ∈ R, as required.
(In short, since integral closure localizes, this problem is reduced to the local property of
membership (the membership of t in R, to be precise), which has been established already.) Problem 10.2. Let R = K[x] be the polynomial ring over a field K on x. Define A =
{k0 + k1 x + · · · + km xm | 0 6 m ∈ Z, ki ∈ K, k1 = 0}, which is (clearly) a subring of R.
(1) Prove that R is module-finite over A (and hence R is integral over A).
(2) Find a monic polynomial f (y) ∈ A[y] such that f (x + 1) = 0.
(3) Prove that A is not integrally closed.
Proof. (1) We claim that R = A1 + Ax. It is clear that R ⊇ A1 + Ax. Conversely, for any
f (x) = k0 + k1 x + · · · + km xm ∈ K[x] = R with ki ∈ K, we have
f (x) = k0 + k1 x + · · · + km xm = (k0 + k2 x2 · · · + km xm ) + k1 x ∈ A1 + Ax
as k0 + k2 x2 · · · + km xm ∈ A and k1 ∈ A. This proves R = A1 + Ax as A-modules, showing
that R is module-finite (and hence integral) over A.
(2) Let f (y) = y 2 − 2y − (x2 − 1)y 0 ∈ A[y] (i.e., f (y) = y 2 + a1 y + a2 with a1 = −2 ∈ A
and a2 = −(x2 − 1) ∈ A). It is routine to see f (x + 1) = (x + 1)2 − 2(x + 1) − (x2 − 1) = 0.
3
(3) Note that x ∈
/ A, x = xx2 is in the fraction field of A, and x is integral over A as x ∈ R.
Therefore A is not integrally closed (in its fraction field).
Problem 10.3. Let R ⊆ T be an integral extension of (non-zero) rings (hence 1R = 1T ).
(1) Prove U(R) = U(T ) ∩ R.
(2) For every t ∈ T \ ZD(T ), prove {0} ( T t ∩ R (i.e., ∃s ∈ T such that 0 6= st ∈ R).
Proof. (1) It is clear that U(R) ⊆ U(T ) ∩ R. To show U(R) ⊇ U(T ) ∩ R, let u ∈ U(T ) ∩ R,
which simply means u ∈ R and u−1 ∈ T . In particular, u−1 is integral over R. Thus
(u−1 )n + r1 (u−1 )n−1 + · · · + rn−1 (u−1 ) + rn = 0
23
with n > 1 and ri ∈ R for i = 1, . . . , n. This yields
u−1 = −(r1 + · · · + rn−1 un−2 + rn un−1 ) ∈ R,
which implies u ∈ U(R).
(2) Let t ∈ T \ ZD(T ). As t is integral over R, there are monic polynomials in R[x]
vanishing at x = t. Let f (x) = xm + s1 xm−1 + · · · + sm−1 x + sm ∈ R[x] (with m > 1 and
si ∈ R) be such that it has the least degree among all monic polynomials in R[x] vanishing
at x = t. In particular,
tm + s1 tm−1 + · · · + sm−1 t + sm = 0,
which gives T t 3 −(tm−1 + s1 tm−2 + · · · sm−1 )t = sm ∈ R, that is, sm ∈ T t ∩ R.
Suppose sm = 0, and thus (tm−1 + s1 tm−2 + · · · sm−1 )t = 0. As t ∈
/ ZD(T ), we have
tm−1 + s1 tm−2 + · · · sm−1 = 0,
contradicting the minimality of deg(f (x)). (In the case of m = 1, we have f (t) = t − s1 = 0,
which gives t = s1 = 0, a contradiction.) This shows 0 6= sm ∈ T t ∩ R, as required.
Problem 10.4. Let R be a non-zero ring (not necessarily Noetherian). Prove that Spec(R),
with the Zariski topology, is quasi-compact; that is, if Spec(R) = ∪λ∈Λ Oλ with each Oλ open
in Spec(R), then there exist finitely many λ1 , . . . , λn ∈ Λ such that Spec(R) = ∪ni=1 Oλi .
Proof. Suppose Spec(R) = ∪λ∈Λ Oλ with OP
for each λ ∈ Λ. For each λ ∈ Λ,
λ open in
Spec(R)
S
write Oλ = D(Iλ ) with Iλ 6 R. Then D
λ∈Λ D(Iλ ) = Spec(R), which forces
λ∈Λ Iλ =
X
Iλ = R.
λ∈Λ
P
there exist finitely many λ1 , . . . , λn ∈ Λ and xi ∈ Iλi such
In particular, 1R ∈ λ∈Λ Iλ . SoP
that 1R = x1 + · · · + xn . Hence ni=1 Iλi = R, which yields
n
n
X
[
D(Iλi ) = D
Iλi = D(R) = Spec(R).
i=1
i=1
This completes the proof that Spec(R), with the Zariski topology, is quasi-compact.
Problem 10.5 (Extra Credit, 1 point). In Problem 9.3(2), further assume that N * K and
that K = ∩si=1 Qi is minimal. Prove or disprove: N ∩ K = ∩N *Qi (N ∩ Qi ) is necessarily
a minimal primary decomposition of N ∩ K in N .
Solution. Let k be any field and R = k[x, y]. Consider M = R, K = (x2 , xy) 6 M ,
Q1 = (x) 6 M , Q2 = (x2 , xy, y 2 ) 6 M , and N = (x2 , y) 6 R. Then K = Q1 ∩ Q2 is a
minimal primary decomposition of K in M , with N * Qi for i = 1, 2. But the following
induced primary decomposition (of N ∩ K in N )
N ∩ K = (N ∩ Q1 ) ∩ (N ∩ Q2 )
is not minimal, as N ∩ K = K = (N ∩ Q1 ), which is already primary in N .
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
24
√
I
Math 8240 (Fall 2016)
Homework Set #11 (Due 11/30)
Solutions
Problem 11.1. Let R be a non-zero ring, and x, x1 , . . . , xn be indeterminates/variables.
Also let 0 6= a ∈ Z and denote by Z[a−1 ] the fraction ring of Z via inverting all powers of a.
(1) Prove that dim(R[x]) > dim(R) + 1.
(2) Prove that dim(R[x1 , . . . , xn ]) > dim(R) + n, for all 0 < n ∈ N.
(3) Prove that dim(Z[a−1 ][x1 , . . . , xn ]) > n + 1. (In fact dim(Z[a−1 ][x1 , . . . , xn ]) = n + 1.)
Proof. (1) Quite generally, for any chain P0 P1 · · · Pm of length m (with m > 0) of
distinct prime ideals of R, it is straightforward to see that
P0 [x] P1 [x] · · · Pm [x] Pm [x] + (x)
is a chain of length m + 1 of distinct prime ideals of R[x]. Consequently, we see
dim(R[x]) > dim(R) + 1,
which applies even when dim(R) = ∞.
(2) This follows from (1) by induction, details skipped. (If R is Noetherian, then
dim(R[x1 , . . . , xn ]) = dim(R) + n.)
(3) By (2), it suffices to show dim(Z[a−1 ]) > 1. Pick any prime number number p such that
p - a, which exists. Hence (p)∩Sa = ∅, in which Sa := {ai | i ∈ N}. Therefore Sa−1 0 Sa−1 (p)
is a chain of length 1 of distinct prime ideals of Sa−1 Z = Z[a−1 ]. Thus dim(Z[a−1 ]) > 1,
which proves dim(Z[a−1 ][x1 , . . . , xn ]) > n + 1. (In fact, dim(Z[a−1 ][x1 , . . . , xn ]) = n + 1, as
Z[a−1 ] is Noetherian and dim(Z[a−1 ]) = 1. Quite generally, dim(S −1 R) 6 dim(R) for every
multiplicative subset S of R, as any chain of distinct prime ideals must come from a chain
of distinct prime ideals of R. In particular, dim(Z[a−1 ]) 6 dim(Z) = 1.)
Problem 11.2. Let Z ⊆ R be an extension of rings, with R finitely generated over Z.
(1) True or false: There exist 0 6= a ∈ Z and r1 , . . . , rd ∈ R such that R[a−1 ] is
module-finite over Z[a−1 ][r1 , . . . , rd ] and Z[a−1 ][r1 , . . . , rd ] ∼
= Z[a−1 ][x1 , . . . , xd ], with
x1 , . . . , xd being indeterminates/variables. (No need to justify.)
(2) Prove that m ∩ Z 6= 0, for every m ∈ Max(R).
(3) Prove that, for every m ∈ Max(R), the quotient field R/m has prime characteristic.
Proof. (1) True. In fact, Z can be replaced by any domain, as we have proved in class.
(2) Suppose m ∩ Z = 0 for some m ∈ Max(R). Then the induced ring homomorphism
Z → R/m is injective. Up to isomorphism, Z is a subring of R/m. By (1) applied to
Z ⊆ R/m, there exist 0 6= a ∈ Z and r1 , . . . , rd ∈ R/m such that (R/m)[a−1 ] is module-finite
over Z[a−1 ][r1 , . . . , rd ] and Z[a−1 ][r1 , . . . , rd ] ∼
= Z[a−1 ][x1 , . . . , xd ]. Consequently, we get
0 = dim(R/m) = dim(Z[a−1 ][r1 , . . . , rd ]) = dim(Z[a−1 ][x1 , . . . , xd ]) = d + 1,
which is a contradiction.
(3) Let m ∈ Max(R). Then m ∩ Z = pZ for some prime number p, by (2) above as well as
Problem 2.1(2). Considering the induced injective homomorphism (extension of fields)
Zp = Z/pZ = Z/(m ∩ Z) ,→ R/m,
we conclude that R/m has prime characteristic p.
Problem 11.3. Let R = K[x, y] be a polynomial ring on variables x and y over a field K.
Consider ideals I = (x3 , xy 3 ) and J = (x3 , x + y).
(1) Determine Z(I) in A2K = K × K explicitly. No need to justify.
25
(2) Determine Z(J) in A2K = K × K explicitly. No need to justify.
(3) Determine I(Z(I)) explicitly, under the assumption K = K. No need to justify.
(4) Determine I(Z(J)) explicitly. No need to justify.
Solution. (1) We claim that Z(I) = {(0, β) ∈ A2K | β ∈ K}. Indeed, for (α, β) ∈ A2K , we have
(α, β) ∈ Z(I) ⇐⇒ f (α, β) = 0 for all f (x, y) ∈ {x3 , xy 3 }
⇐⇒ α3 = 0 and αβ 3 = 0 ⇐⇒ α = 0 and αβ 3 = 0 ⇐⇒ α = 0.
(2) We claim that Z(J) = {(0, 0)}. Indeed, for (α, β) ∈ A2K , we have
(α, β) ∈ Z(J) ⇐⇒ f (α, β) = 0 for all f (x, y) ∈ {x3 , x + y}
⇐⇒ α3 = 0 and α + β = 0 ⇐⇒ α = 0 = β ⇐⇒ (α, β) = (0, 0).
(3) We claim that I(Z(I)) = (x). Since x3 ∈ (x3 , xy 3 ) ⊆ (x) ∈ Spec(R), we see
p
p
(x) ⊆ (x3 , xy 3 ) ⊆ (x) = (x).
p
√
So (x3 , xy 3 ) = (x). Finally, since K = K, we conclude I(Z(I)) = I = (x).
(4) We claim that I(Z(J)) = (x, y). Indeed, for f (x, y) ∈ K[x, y], we have
f (x, y) ∈ I(Z(J)) ⇐⇒ f (0, 0) = 0
⇐⇒ the constant term of f (x, y) is 0 ⇐⇒ f (x, y) ∈ (x, y).
Problem 11.4. Let K = K be an algebraically closed field. Consider U = Z(xy − z) and
V = Z(xy − z 2 ), both algebraic sets/varieties in A3K = K × K × K. Prove that U and V
are not isomorphic as algebraic sets/varieties.
Proof. Note that A[U ] = K[x, y, z]/(xy − z) ∼
= K[x, y] as rings. As K = K, every maximal
ideal of A[x, y] is of the form
(x − α, y − β)
with (α, β) ∈ K × K (cf. Hilbert’s Nullstellensatz). Hence every maximal ideal of A[U ] can
be generated by 2 elements. (In fact, for any field L and any n ∈ N, every maximal ideal of
L[x1 , . . . , xn ] can be generated by n elements.)
However, the maximal ideal m := (x, y, z)/(xy − z 2 ) ∈ Max(A[V ]) can not be generated
by 2 elements. In fact, it is straightforward to see (with petty details omitted)
3 3
m ∼
(x, y, z)
(x, y, z) ∼ K[x, y, z]
A[V ]
∼
=
.
=
=
=
m2
(x, y, z)2 + (xy − z 2 )
(x, y, z)2
(x, y, z)
m
Thus even a quotient (module) of m needs (at least) 3 generators.
In conclusion, A[U ] and A[V ] are not isomorphic as rings (hence they are not isomorphic
as K-algebras). Consequently U and V are not isomorphic as algebraic sets/varieties. (Other
ways are available to prove this. For example, one can show that A[U ] is a UFD while A[V ] is
not. Invoking regularity (not covered in Math 8240), one can also show that A[U ] is regular
while A[V ] is not. In terms of algebraic geometry, one sees that U is smooth everywhere
while V is not.)
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
26
√
I
Math 8240 (Fall 2016)
Final Exam (12/07)
Solutions
Problem III.1. Let R be a ring and M a non-zero R-module such that 0 = {0M } has a
primary decomposition in M . Prove ZDR (M ) = ∪P ∈Ass
g R (M ) P .
p
g R (M ), so that r ∈ P = AnnR (x) for some 0 6= x ∈ M . Denote
Proof. Let r ∈ P ∈ Ass
n := min{i ∈ N | ri x = 0M }, which is well-defined. Then n > 1 and r(rn−1 x) = 0M with
rn−1 x 6= 0M , showing that r ∈ ZD(M ). Thus ZD(M ) ⊇ ∪P ∈Ass
g R (M ) P , quite generally.
Conversely, let s ∈ ZD(M ), so that s ∈ R and sy = 0M for some y ∈ M \ {0M }. Fix a
/ Qj
minimal primary decomposition 0 = ∩m
i=1 Qi , with Qi being Pi -primary in M . Then y ∈
for some j ∈ {1, . . . , m}. Consequently s ∈ Pj ⊆ ∪P ∈Ass
P
(since
sy
=
0
∈
Q
,
y
∈
/
Qj ,
M
j
g R (M )
and Qj is Pj -primary in M ). Therefore ZD(M ) ⊆ ∪P ∈Ass
g R (M ) P .
Problem III.2. Let R be a Noetherian ring, M an R-module, and x ∈ M . Prove that
x = 0M if and only if x1 = 0MP ∈ MP for all P ∈ AssR (M ).
Proof. If x = 0M , then it is clear that x1 = 0MP for all P ∈ AssR (M ).
Suppose x 6= 0M , so that Rx is a non-zero R-module. Then ∅ 6= AssR (Rx) since R is
Noetherian. Choose any P ∈ AssR (Rx) ⊆ AssR (M ); hence P = AnnR (rx) ⊇ AnnR (x) for
some r ∈ R. Then, by an argument very similar to Problem 5.4(2), we see that
AnnRP ( x1 ) = (AnnR (x))P ⊆ PP ( RP ,
which proves x1 6= 0MP for this P ∈ AssR (Rx) ⊆ AssR (M ). (One can show x1 6= 0MP
from scratch: if x1 = 0MP , then there exists s ∈ R \ P such that sx = 0M , which implies
P ⊇ AnnR (x) 3 s ∈ R \ P , a contradiction.) Now the proof is complete. (From the proof,
we see that x = 0M if and only if x1 = 0MP ∈ MP for all P ∈ AssR (Rx). Another version is
that M = 0 if and only if MP = 0 for all P ∈ AssR (M ).)
Problem III.3. Let I be a proper ideal of a ring R. Prove that V(I)
√ is irreducible (that is,
V(I) is not a union of two closed proper subsets) if and only if I ∈ Spec(R).
√
Proof. Assume that I ∈ Spec(R). Let V(J
V(I), for i = 1, 2, be any two closed proper
√ i) ( √
subsets of V(I). By Problem 4.1(1)&(2), Ji ) I for i = 1, 2. Therefore
√
V(I) 3 I ∈
/ V(J1 ) ∪ V(J2 ),
which implies V(J1 ) ∪ V(J2 ) ( V(I). Thus V(I) is irreducible. √
√
Conversely,
assume
that
V(I)
is
irreducible.
Let
r
I;
let
J
I + (ri ), so
1 , r2 ∈ R \
i =
√
√
that Ji ) I, for i = 1, 2. By Problem 4.1(1)&(2), we see V(Ji ) ( V(I) for i = 1, 2. Now
the assumption that V(I) is irreducible forces
V(J1 ) ∪ V(J2 ) ( V(I).
√
So there exists P ∈ V (I) = V( I) but√P ∈
/ V(J1 ) ∪ V(J√2 ). This necessarily
/P
√ implies r1 ∈
and r2 ∈
/ P , which implies r1 r2 ∈
/ P ⊇ I, hence r1 r2 ∈
/ I. Now we see I ∈ Spec(R). Problem III.4. Let R ⊆ T be an extension of rings and t ∈ U(T ) so that t−1 ∈ T . Prove
that the following statements are equivalent:
(1) t is integral over R.
(2) t is integral over R[t−1 ].
(3) t ∈ R[t−1 ].
27
Proof. (1) ⇒ (2): This is obvious from definition, since R ⊆ R[t−1 ].
(2) ⇒ (3): That t is integral over R[t−1 ] simply means tm + a1 tm−1 + · · · + am−1
=0
Pt + am m−i
for some m > 1 and ai ∈ R[t−1P
]. Thus tm = −(a1 tm−1P+ · · · + am−1 t + am ) = − m
a
t
,
i=1 i
m
m
m 1−m
m−i 1−m
1−i
−1
which yields t = t t
= − ( i=1 ai t ) t
= − i=1 ai t ∈ R[t ].
(3) ⇒ (1): Since t ∈ R[t−1 ], we have t = r0 + r1 t−1 + · · · + rn (t−1 )n with n > 0 and ri ∈ R
for i = 0, . . . , n. Multiplying both sides by tn , we see tn+1 − r0 tn − · · · − rn−1 t − rn = 0,
which shows that t is integral over R. This completes the proof.
Problem III.5. Let R ⊆ T be an integral extension of rings. Prove the following:
(1) For every P ∈ Spec(R), R ∩ (P
P ec = P ).
√T ) = P (i.e.,
√
(2) For every I 6 R, R ∩ (IT ) ⊆ I (i.e., I ec ⊆ I).
Proof. (1) By Lying-over Theorem, there exists a (prime) ideal Q 6 T such that R ∩ Q = P .
In particular, we have P ⊆ Q and hence P ⊆ P T ⊆ Q. Thus
P ⊆ R ∩ P T ⊆ R ∩ Q = P,
which yields R ∩ P T √
= P.
(2) Denoting J := I and V(I) := VR (I), we have (with all steps straightforward)
!
\
I ⊆ J ⊆ R ∩ JT = R ∩
P T
P ∈V(I)
!
⊆R∩
\
(P T )
P ∈V(I)
=
\
√
(1) \
R ∩ PT =
P = J := I.
P ∈V(I)
P ∈V(I)
(The above actually shows (∗) : R ∩ JTP= J for every radical ideal J 6 R.) (Alternatively,
(2) can be proved as follows: Let r = nk=1 ik tk ∈ R ∩ IT with ik ∈ I and tk ∈ T . Denote
S := R[t1 , . . . , tn ], a subring of T . Then S is module-finite over R and rS 6 IS. By
the ‘determinantal trick’ (details skipped), √
r is a root of a monic polynomial with (lower)
coefficients in I, which in turn implies r ∈ I. Also, (2) implies (∗) and hence (1).)
Extra Credit Problem III.6 (1 point, no partial credit). Prove or disprove: For any
extensions of rings R ⊆ S ⊆ T with T module-finite over R, S is module-finite over R.
Solution. The statement is false. Let R be any ring that is not Noetherian so that R has an
ideal I that is not finitely generated (for example, R = Q[x1 , x2 , . . . ] and I = (x1 , x2 , . . . )R).
Let T = R[y]/(y 2 ), which is (can be regarded as) an extension ring of R. And let S = R1+Iy,
which is a subring of T containing R. Now T = R1 + Ry is module-finite over R, but S is
not module-finite over R (details omitted). (The statement holds if R is Noetherian.)
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
28
√
I
Math 8240 (Fall 2016)
Extra Credit Set (By 11/30)
Solutions
Each problem carries 3 points. You must solve a problem completely and correctly in
order to claim the extra credit. You may attempt a problem for as many times as you wish
by 11/30. Feel free to make use of the results we have covered.
The points you get here will be added to the total score from homework assignments.
Each F represents a correct solution submitted.
Q
Problem E-1. Let R be the product of a non-empty family of fields; that is, R = λ∈Λ Fλ
where Λ 6= ∅ and Fλ is a field for each λ ∈ Λ. Prove that every prime ideal of R is maximal
(hence also minimal). (In case Λ is infinite, this provides a ring that is not Noetherian
(hence not Artinian) but with all prime ideals maximal/minimal.)
FFF
Problem E-2. Let R consist of all continuous real-valued functions on [0, 1]. Note that R
is a ring under addition and multiplication for functions as in calculus. For each x ∈ [0, 1],
let mx = {f ∈ R | f (x) = 0}. Prove Max(R) = {mx | x ∈ [0, 1]}.
FF
Problem E-3. Prove that Min(R[x]) = {P [x] | P ∈ Min(R)}.
F
Problem E-4. Let R be a non-zero ring and consider R[x]. Denote N = Nil(R).
(1) Prove or disprove: Nil(R[x]) = Nil(R)[x], i.e., Nil(R[x]) = N [x].
(2) Prove U(R[x]) = {a0 + a1 x + · · · + an xn | a0 ∈ U(R) and ai ∈ Nil(R) for 1 6 i 6 n}.
(3) Determine Jac(R[x]) with justification.
FF
Problem E-5. Prove that, for any non-empty multiplicative subset S of R and for any
Artinian R-module M , the natural map ϕ : M → S −1 M is onto.
FFF
Problem E-6. Let R be a Noetherian domain and x ∈ P ∈ Spec(R) with PP = (x)P
(i.e., PP = x1 RP ). Show that @ Q ∈ Spec(R) such that {0} ( Q ( P .
FFF
Problem E-7. Prove or disprove: An R-module M is finitely generated over R if and only
if MP is finitely generated over RP for all P ∈ Spec(R) if and only if Mm is finitely generated
over Rm for all m ∈ Max(R).
F
Problem E-8. Let R be a ring, I an ideal and S a multiplicatively closed subset of R. Show
that I ∩ S = ∅ if and only if there exists m ∈ Max(R) such that Im ∩ ϕ(S) = ∅, in which
ϕ : R → Rm is the natural ring homomorphism.
F
Problem E-9. Let R be a non-zero ring. Prove or disprove each of the following:
(1) If R is reduced, then RP is a field (and hence PP = (0RP )) for every P ∈ Min(R).
(2) If RP is a field for every P ∈ Min(R), then R is reduced.
FF
Problem E-10. Let R ⊆ T be an extension of rings (not necessarily Noetherian) and t ∈ T .
For every ideal I of T , denote by hI : T → T /I the natural ring homomorphism. Prove that
the following are equivalent:
(1) t is integral over R.
(2) hP (t) is integral over hP (R) for all P ∈ Min(T ).
F
PROBLEMS
√
I=
T
P ∈V(I)
HINTS
SOLUTIONS
P . . . . . S −1 R . . . . . M = 0 ⇐⇒ Mm = 0, ∀m ∈ Max(R) . . . . . N = Q1 ∩ · · · ∩ Qs . . . . . K = K =⇒ I(Z(I)) =
√
I