Math 902
Solutions to Exam I
Except where specified, R denotes a commutative ring with identity.
Note: Thanks to Katie Tucker for letting me use her tex file for several of these solutions.
1. Assume R is Noetherian and let M be a nonzero R-module. Let Λ be the set of all ideals
of the form (0 :R x) for some nonzero x ∈ M . Note that Λ is nonempty as M 6= 0. As
R is Noetherian (not by Zorn’s lemma!), every ideal in Λ is contained in some maximal
element of Λ. Prove that each maximal element of Λ is a prime ideal.
Solution: Let p be maximal in Λ, so p = (0 :R x) for some nonzero x ∈ M . Suppose
a, b ∈ R such that ab ∈ p and without loss of generality suppose a 6∈ p. Then we have
abx = 0 and ax 6= 0, and as R is commutative, this gives b(ax) = 0 and so b ∈ (0 :R ax),
which is an element of Λ. For all r ∈ p, we further have rx = 0 and so r(ax) = a(rx) = 0
and so p ⊆ (0 :R ax). As p is maximal in Λ we thus have p = (0 :R ax) and thus b ∈ p.
2. Let R be Noetherian and M an R-module. A prime ideal of R of the form (0 :R x) for
some x ∈ M is called an associated prime of M . The set of all associated primes of M
is denoted AssR M . By Problem # 1, M 6= 0 if and only if AssR M 6= ∅. Prove that the
following are equivalent for a prime ideal p:
(a) p ∈ AssR M .
(b) There exists an injective R-homomorphism R/p → M .
(c) HomRp (k(p), Mp ) 6= 0, where k(p) = Rp /pRp .
Solution: (a) implies (b): Let p ∈ AssR M . Then there exists a nonzero x ∈ M such that
p = (0 :R x) = {r ∈ R : rx = 0}. Define φ : R → M by φ(r) = rx. Then p = ker φ. Thus,
the induced homomorphism φ : R/p → M defined by φ(r + p) = rx is well-defined and
injective.
(b) implies (c): For p a prime ideal of R, suppose φ : R/p → M is an injective R-module
homomorphism. Localizing at p and using that localization is exact, we obtain an injective
Rp -homomorphism (R/p)p → Mp . As k(p) ∼
= (R/p)p , we obtain HomRp (k(p), Mp ) 6= 0.
(c) implies (a): Suppose HomRp (k(p), Mp ) 6= 0 and let f ∈ HomRp (k(p), Mp ) be a nonzero
element. As k(p) is a field we thus have f is injective. Let f (1) = xs ∈ Mp . As f is injective,
we obtain that pRp = (0 :Rp xs ) = (0 :Rp x1 ). Now, if r ∈ (0 :R x) then 1r ∈ (0 :Rp x1 ) = pRp
and hence r ∈ p. Thus, (0 :R x) ⊆ p. As R is Noetherian, p is finitely generated, say
p = (y1 , . . . , yn ) for some yi ∈ p. Then for each i we have y1i ∈ pRp , so y1i x = 0 and thus
there exists an si ∈ R\p such that si yi x = 0. Set s = s1 · · · sn and note s ∈ R\p as p is
prime. We then have that yi sx = (s1 · · · si−1 si+1 · · · sn )si yi x = 0 since R is commutative,
so yi ∈ (0 :R sx) for all i. Thus p ⊆ (0 :R sx). Furthermore, for any z ∈ (0 :R sx) we have
zsx = 0 and so zs ∈ (0 :R x) ⊆ p. Since p is prime and s 6∈ p we thus have z ∈ p and so
p = (0 :R sx). Thus p ∈ AssR M .
3. Let R be Noetherian and M a nonzero R-module. Let
ZDR (M ) = {r ∈ R | ru = 0 for some u ∈ M, u 6= 0}.
That is, ZDR (M ) is the set of elements of R which are zero-divisors on M . Prove that
[
ZDR (M ) =
p.
p∈AssR M
Solution: Note if r ∈ p ∈ AssR M , then as p = (0 :R x) for some nonzero x ∈ M we
have rx = 0 so r ∈ ZDR (M ). Now let r ∈ ZDR (M ) be given. Then there exists nonzero
x ∈ M such that rx = 0. By problem #1, AssR (Rx) is nonzero. Thus there exists a
p = (0 :R bx) ∈ AssR (Rx) for some b ∈ R. We then have rbx = brx = 0, so r ∈ p. As
x ∈ M we also have bx ∈ M and thus p ∈ AssR (M ). Thus r ∈ p ⊆ ∪q∈AssR M q and hence
ZDR (M ) = ∪q∈AssR (M ) q.
4. Let R be Noetherian and M and N R-modules. Assume M is finitely generated. Prove
that
AssR HomR (M, N ) = SuppR M ∩ AssR N.
(Hint: Use Problem # 2 and Hom-tensor adjunction. Also note that for a finitely generated
R-module L, L ⊗R k(p) ∼
= k(p)` , where ` = µRp (Lp ) by Nakayama.)
Solution: Let p ∈ Spec R. Since M is finitely generated and R is Noetherian, HomR (M, N )p ∼
=
HomRp (Mp , Np ). Let r = µR (Mp ) = dimk(p) Mp /pp Mp = dimk(p) k(p) ⊗Rp Mp . Using Homtensor adjunction, we have
HomRp (k(p), HomR (M, N )p ) ∼
= HomRp (k(p), HomRp (Mp , Np ))
∼
= HomRp (k(p) ⊗Rp Mp , Np )
∼
= HomRp (k(p)r , Np )
∼
= HomRp (k(p), Np )r .
Hence, HomRp (k(p), HomR (M, N )p ) 6= 0 if and only if r > 0 and HomRp (k(p), Np ) 6= 0,
which holds if and only if Mp 6= 0 and HomRp (k(p), Np ) 6= 0. Thus, by Problem #2, we
obtain that p ∈ AssR HomR (M, N ) if and only if p ∈ SuppR M and p ∈ AssR N .
5. Let φ : R → S be a homomorphism of commutative rings. Let F be a flat S-module and
E an injective R-module. Prove that HomR (F, E) is an injective S-module.
Solution: Note HomR (F, E) is an S-module as F is an R−S bimodule under the R-action
given by φ. By an exercise from class, it is enough to show that HomS (−, HomR (F, E)) is
exact. This functor is left exact by homework set 4 problem 2, so we need only show right
exactness. Thus let 0 → M → N be an exact sequence of left S-modules. As F is flat,
the sequence 0 → F ⊗S M → F ⊗S N is also exact as left S-modules (and thus as left Rmodules). Since E is injective, the sequence HomR (F ⊗S N, E) → HomR (F ⊗S M, E) → 0
is also exact. By Hom −⊗ adjointness, we get the commutative diagram
HomR (F ⊗S N, E)

∼
=
y
−−−→
HomR (F ⊗S M, E)

∼
=
y
−−−→ 0
HomS (N, HomR (F, E)) −−−→ HomS (M, HomR (F, E)) −−−→ 0
Since the top sequence is exact, so is the bottom sequence (by the Five Lemma) and thus
HomR (F, E) is an injective S-module.
6. Let M be a finitely presented R-module and {Ai | i ∈ I} a set of R-modules. Prove that
there is an isomorphism
Y
Y
∼
=
ψ : M ⊗R
Ai −
→
(M ⊗R Ai )
i
i
such that ψ(m ⊗ (ai )) = (m ⊗ ai ) for m ∈ M, ai ∈ Ai . (Hint: Use the right exactness of
the tensor product and a finite presentation for M .)
Solution: First, we make an observation. Let I be an index set and consider a family of
fi
gi
sequences of R-modules Li −
→ Mi −
→ Ni for each i ∈ I. Denote this sequence by (#)i . By
taking products, we have a sequence of R-modules (#):
Q
Q
Y
fi Y
gi Y
Mi −−→
Ni −−→
Ni .
i
i
i
It is easily verified that (#) is exact if and only if (#)i is exact for all i.
Q
Q
Now, there is a natural isomorphism φ : R ⊗R i Ai → i (R ⊗R Ai ) given by φ(r ⊗
Q(ai )) =
n
(r ⊗ ai ). By taking (finite) direct sums, we get a natural isomorphism ψ : R ⊗R i Ai →
Q
n
finitely presented, there
i (R ⊗R Ai ) where ψ((rj ) ⊗ (ai )) = ((rj ) ⊗ ai ). Now, as M isQ
exists an exact sequence Rm → Rn → M → 0. Tensoring with Ai and using the right
exactness of the tensor product, we have
Y
Y
Y
R m ⊗R
A i → R n ⊗R
Ai → M ⊗R
Ai → 0.
i
i
i
On the other hand, tensoring with each Ai first, we have for each i an exact sequence
Rm ⊗R Ai → Rn ⊗R Ai → M ⊗R Ai → 0.
Taking the product of these exact sequences (using the observation above) gives an exact
sequence:
Y
Y
Y
(Rm ⊗R Ai ) →
(Rn ⊗R Ai ) →
(M ⊗R Ai ) → 0.
i
i
i
Consider the commutative diagram
Q
Q
Q
Rm ⊗R i Ai −−−→ Rn ⊗R i Ai −−−→ M ⊗R i Ai −−−→ 0


∼
∼
y=
y=
Q m
Q n
Q
i (R ⊗R Ai ) −−−→
i (R ⊗R Ai ) −−−→
i (M ⊗R Ai ) −−−→ 0.
The vertical arrows are isomorphismsQ
by the discussion
above. Hence, by the Five Lemma,
Q
there exists an isomorphism M ⊗i Ai → i (M ⊗R Ai ) making the above diagram
commute.
7. Let C[0, 1] denote the set of continuous functions f : [0, 1] → R, where [0, 1] denotes the
closed interval of real numbers from 0 to 1. It is easily seen that C[0, 1] is a commutative
ring under addition and multiplication (not composition!) of real-valued functions. (You
do not need to show this.)
(a) Describe the units of C[0, 1].
(b) Prove that Spec C[0, 1] is connected.
(c) For a ∈ [0, 1], show that ma := {f ∈ C[0, 1] | f (a) = 0} is a maximal ideal of C[0, 1].
Solution:
Part (a): If f ∈ C[0, 1] is a unit, then for some g ∈ C[0, 1] and all x ∈ [0, 1] we have
f (x)g(x) = 1 since the multiplicative identity in C[0, 1] is the function i(x) = 1. Thus for
f to be a unit, we must have f (x) 6= 0 for all x ∈ [0, 1]. When this holds, we can define
1
, which is continuous by an 825-826 result. Hence f is a unit if
g ∈ [0, 1] via g(x) = f (x)
and only if f (x) 6= 0 for all x ∈ [0, 1].
Part (b): By a theorem from class, it is enough to show that C[0, 1] has no nontrivial
idempotents. Thus let f ∈ C such that f 2 = f . Let x ∈ [0, 1]. Then [f (x)]2 = f (x) so
we have f (x)[f (x) − 1] = 0. As R is a domain, f (x) = 0 or f (x) = 1. Note that this
holds for all x ∈ [0, 1], so the only values f takes on are 0 and 1. Now suppose there exist
x, y ∈ [0, 1] such that f (x) = 0 and f (y) = 1. As f is continuous, the Intermediate Value
Theorem yields that there is some z between x and y such that, in particular f (z) = 21 .
However this is a contradiction and so no such x and y exist. Hence f is either the zero
function or the identity function, and so C[0, 1] has no nontrivial idempotents.
Part (c): To see ma is maximal, it is enough to show that C[0, 1]/ma is a field. Thus
consider the map εa : C[0, 1] → R given by εa (f ) = f (a). Then ker εa = ma by definition.
Furthermore εa is surjective since for any r ∈ R, the map gr : [0, 1] → R given by gr (x) = r
is continuous and gr (a) = r in particular. Thus C[0, 1]/ma ∼
= R by the first isomorphism
theorem and so C[0, 1]/ma is a field. Therefore ma is maximal.
8. Prove that every maximal ideal of C[0, 1] has the form ma for some a ∈ [0, 1], as defined
in the previous problem. (Hint: Suppose I is an ideal not contained in any ma . Use the
compactness of [0, 1] to show I contains a unit. Let a ∈ [0, 1]. Then there exists fa ∈ I
such that fa (a) 6= 0. Since fa is continuous, there exists an open set Oa containing a such
that fa (u) 6= 0 for all u ∈ Oa .)
Solution: Let I be an ideal of C[0, 1] such that I 6⊆ ma for all a ∈ [0, 1]. Then for each
a ∈ [0, 1], there is some fa ∈ I such that fa (a) 6= 0. As each fa is continuous, there exist
open sets Oa ⊆ [0, 1] such that a ∈ Oa and for all u ∈ Oa we have fa (u) 6= 0. Then
{Oa } forms an open cover for [0, 1], which is compact, so we can extract a finite subcover
{Oai }ni=1 such that ∪ni=1 Oai = [0, 1]. Set f = fa21 + · · · + fa2n ∈ I. Then f (x) 6= 0 for all
x ∈ [0, 1] as fai (x) 6= 0 for some ai and fa2j ≥ 0 for all j so no cancellation can occur. By
7a, we thus have that f is a unit and in I so I = C[0, 1]. Thus if M is a maximal ideal of
C[0, 1] then M ⊆ ma for some a and so M = ma .
Q
9. Let F be field and R = ∞
i=1 F , the product of countably many copies of F . Recall that
R is von Neumann regular. Let I = ⊕∞
i=1 F . Observe that I is an ideal of R. Prove that Ip
is an injective Rp -module for every prime p of R, but that I is not an injective R-module.
(Hint: Note that every nonzero ideal of R has nontrivial intersection with I.)
Solution: Recall from class that the localization of von Neumann regular ring is von
Neumann regular and that a (commutative) quasi-local von Neumann regular ring is a
field. Thus, if p ∈ Spec R, Rp is a field and hence every Rp -module is injective (and
projective, for that matter). So it suffices to prove that I is not an injective R-module.
Suppose I is injective. Then the inclusion map I → R splits, so R = I ⊕ J for some ideal
J of R. But then I ∩ J = 0. However, as J 6= 0, we must have J ∩ I 6= 0. (Multiply
a nonzero element of J by an element of R which has 1 in exactly one (carefully chosen)
component and zeros elsewhere.) Thus, I is not a direct summand of R and so I is not
injective.
10. Let M be a finitely generated R-module and let µR (M ) denote the least integer ` such
that M has a generating set consisting of ` elements. Note that for any multiplicatively
closed set, µRS (MS ) ≤ µR (M ). For any integer r ≥ 0, prove that the set
Ur := {p ∈ Spec R | µRp (Mp ) ≤ r}
is an open set of Spec R. (Hint: Suppose p ∈ Ur . Prove there exists c ∈ R such that
p ∈ D(c) ⊆ Ur . Recall that D(c) = {q ∈ Spec R | c 6∈ q}.)
Solution: Let p ∈ Ur . Then Mp can be generated over Rp by r elements, say xs11 , . . . , xsrr .
Then Mp can also be generated by x11 , . . . , x1r . Define an R-module homomorphism φ :
Rr → M by φ(ei ) = xi , where {e1 , . . . , er } is a basis for Rr . Let C = coker φ. Then we
have an exact sequence
φ
Rr →
− M → C → 0.
If we localize this sequence at p, then as localization is an exact functor and the map φ1 is
surjective, we get that Cp = 0. As C is finitely generated, this implies that p 6⊃ AnnR C.
Choose t ∈ AnnR C \ p. As tC = 0 we have Ct = 0. Localizing the exact sequence at the
multiplicatively closed set {tn | n ≥ 0}, we have
(Rt )r → Mt → 0
since localization is exact and Ct = 0. Note that at t 6∈ p, p ∈ D(t). Suppose q ∈ D(t). As
t 6∈ q then qt ∈ Spec Rt . Localizing at qt gives:
(Rt )rqt → (Mt )qt → 0.
But (Rt )qt ∼
= Rq and (Mt )qt ∼
= Mq . Hence, we have an exact sequence
(Rq )r → Mq → 0.
This implies that Mq is generated by r elements, so q ∈ Ur . Thus, p ∈ D(t) ⊆ Ur and
consequently Ur is open.