Math 902
Solutions to Homework # 3
Throughout R denotes a commutative ring with identity.
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1. A ring R is called reduced if (0) = (0). Prove that the following are equivalent:
(a) R is reduced.
(b) RS is reduced for every multiplicatively closed set S of R.
(c) Rm is reduced for every maximal ideal m.
Solution: (a) =⇒ (b) : Suppose rs ∈ RS and ( rs )n = 0 for some n. Then there exists
t ∈ S such that trn = 0. Hence, (tr)n = 0. As R is reduced, tr = 0. Thus, rs = 0.
(b) =⇒ (c) : This is clear.
(c) =⇒ (a) : Let r ∈ R and suppose rn = 0. Then ( 1r )n = 0 in Rm for every maximal
ideal m of R. As Rm is reduced, we have 1r = 0 in Rm . Thus, for every m, there exists
tm ∈ R \ m such that tm r = 0. Consider the ideal (0 :R r) = {a ∈ R | ar = 0}. We’ve
shown that (0 :R r) 6⊂ m for every maximal ideal m. Thus, (0 :R r) = R; i.e., r = 0.
Hence, R is reduced.
2. Prove that if R is reduced then Rp is a field for every minimal prime p. Also, give an
example to show that the converse is false.
Solution: Let p be a minimal prime of R. Then Spec Rp = {pRp }. By Problem
#1, Rp is reduced. Hence, the nilradical of Rp is 0. But the nilradical of Rp is the
intersection of all prime ideals of Rp . Since pRp is the only prime ideal of Rp , we must
have pRp = 0. But pRp is the maximal ideal of Rp . Hence, Rp is a field.
Let k be a field and S = k[x, y], a polynomial ring in two variables over k. Let
I = (x2 , xy)S and R = S/I. Clearly, R is not reduced, as the image of x in R is a
nonzero nilpotent. The prime ideals of R correspond to the prime ideals of S which
contain I. First note that p := (x)S is a prime ideal containing I. If Q is a prime
ideal S containing I then x2 ∈ Q and hence x ∈ Q. Thus, Q ⊇ p and p is the
unique prime minimal over I. Consequently, q := pR = p/I is the unique minimal
prime ideal of R. Note that Ip = (x)Sp = pSp since y is a unit in k[x, y]p . Thus
Rq ∼
= (S/I)q ∼
= Sp /Ip ∼
= Sp /pSp is a field, since pSp is the maximal ideal of Sp .
3. Define the support of an R-module M by SuppR M := {p ∈ Spec R | Mp 6= 0}. Prove
that if M is finitely generated then SuppR M = V (AnnR M ). Also, give an example of
a Z-module M such that SuppR M is not closed.
Solution: Let p ∈ Spec R. We’ll prove that Mp = 0 if and only if p 6⊃ AnnR M . Let
M = Rx1 +· · ·+Rxn and suppose Mp = 0. Then x1i = 0 in Mp for all i. Hence, for each
i, there exists ti ∈ R \ p such that ti xi = 0. Let t = t1 t2 · · · tn . Then t 6∈ p and txi = 0
for all i, i.e., t ∈ AnnR M . Thus, p 6⊃ AnnR M . Conversely, suppose p 6⊃ AnnR M . Let
t ∈ AnnR M , t 6∈ p. Then for all x ∈ M , x1 = 0 in Mp as tx = 0. Thus, Mp = 0.
Let R = Z and M = ⊕p Z/pZ, where the sum is over all (positive) prime integers p.
Then SuppR M = {pZ | p a prime}. Suppose SuppR M is closed. Then SuppR M =
V(I) for some ideal I of Z. As Z is a PID, I = (d) for some d ∈ Z. As every nonzero
prime ideal (p) contains I, we see that p divides d for every prime integer p. Clearly,
this implies d = 0. But then (0) ∈ V(I), but (0) 6∈ SuppR M .
4. Let I be a fiinitely generated ideal and suppose that for every prime ideal p ⊇ I one
has Ip = 0. Prove that I is generated by an idempotent. (Hint: Use the previous
problem to show that I + AnnR I = R.)
Solution: Let p be a prime ideal of R and suppose p ⊇ I. Then Ip = 0. As I is finitely
generated, we have p 6⊃ AnnR I by Problem #3. Hence, no prime ideal contains both
I and AnnR I. Therefore, I + AnnR I = R. Thus, 1 = a + b for some a ∈ I and
b ∈ AnnR I. Clearly, (a) ⊆ I. Let r ∈ I. Then r = r(a + b) = ra + rb = ra ∈ (a) as
rb = 0. Hence, I = (a). Also, observe that a2 = a(1 − b) = a − ab = a, so a is an
idempotent.
5. Suppose R is Noetherian and Spec R is connected. Prove that R is a domain if and
only if Rp is a domain for every p ∈ Spec R.
Solution: Clearly, if R is a domain, so is Rp for all p ∈ Spec R. Conversely, suppose
Rp is a domain for all p ∈ Spec R. Let a ∈ R, a 6= 0 and I = (0 :R a). Suppose that
p ⊇ I. We claim that Ip = 0. First, since p ⊇ (0 :R a), a1 6= 0 in Rp . Let i ∈ I. Then
ia = 0. Thus, 1i a1 = 0 in Rp . As Rp is a domain and a1 6= 0, we must have 1i = 0. Thus,
i
= 0 for all s 6∈ p. Hence, Ip = 0. Now, as R is Noetherian, I is finitely generated. By
s
Problem #4, we have that I is generated by an idempotent. Since Spec R is connected,
the only idempotents are 0 and 1. Hence, I = 0 or I = R. Since I 6= R (as a 6= 0),
we have that I = (0 :R a) = (0). Thus, a is a non-zero-divisor. As a was an arbitrary
nonzero element of R, we conclude that R is a domain.
6. Prove that M ⊗R RS ∼
= MS for every R-module M and multiplicatively closed set S.
. This is well-defined
Solution: Define a function f : M × RS → MS by f (m, rs ) = rm
s
and R-biadditive. Hence, we get an induced map of abelian groups φ : M ⊗R RS → MS
such that φ(m ⊗ rs ) = rm
. It is easily seen that that φ is in fact RS -linear (acting on
s
the second component of the tensor product). Now define ψ : MS → M ⊗R RS by
ψ( ms ) = m ⊗ 1s . We claim that ψ is well-defined. Suppose ms = nt . Then there
exists u ∈ S such that utm = usn. Then utm ⊗ 1 = usn ⊗ 1 in M ⊗R RS . Thus,
1
m ⊗ ut = n ⊗ us. Multiplying both sides by ust
(using the RS -module structure on
1
1
M ⊗R RS ), we obtain that m ⊗ s = n ⊗ t . As ψ is well-defined, it is easy to see that
ψ is an RS -module homomorphism. Checking on a generating set, one obtains that φ
and ψ are inverse maps, and thus are isomorphisms.
7. Suppose (R, m) is a quasi-local ring and M, N are finitely generated R-modules. Prove
that M ⊗R N = 0 if and only if either M = 0 or N = 0.
Solution: Suppose M ⊗R N = 0. Then R/m ⊗R (M ⊗R N ) = 0. Using associativity
and that R/m ⊗R M ∼
= M/mM , we obtain that M/mM ⊗R N = 0. Then
0 = M/mM ⊗R N ∼
= (M/mM ⊗R/m R/m) ⊗R N
∼
= M/mM ⊗R/m (R/m ⊗R N )
∼
= M/mM ⊗R/m N/mM.
Since R/m is a field, we must have M/mM = 0 or N/mN = 0. (Recall this is because
the tensor product of two free modules of ranks r and s is a free module of rank rs,
and that every module over a field is free.) Hence, M = mM or N = mN . As M and
N are finitely generated, we conclude by Nakayama’s lemma that M = 0 or N = 0.
8. Give examples of nonzero R-modules M and N such that M ⊗R N = 0 in the following
cases:
(a) M and N are finitely generated but R is not quasi-local.
(b) R is quasi-local but one of M or N is not finitely generated.
Solution: For (a), consider R = Z, M = Z/2Z and N = Z/3Z. Then M and N
are nonzero finitely generated Z-modules. Furthemore, M ⊗Z N ∼
= Z/(2, 3)Z = 0, as
(2, 3)Z = Z.
For (b), consider R = Z(2) , the localization of Z localized at the prime ideal (2). Let
M = Q (which is the field of fractions of R) and N = R/(2)R (the residue field of R).
Then R is quasi-local and N is finitely generated. Moreover, M ⊗R N ∼
= Q/2Q = 0 as
2Q = Q.
9. Let M and N be finitely generated R-modules. Prove that
p
p
AnnR M ⊗R N = AnnR M + AnnR N .
Solution: First observe that as M and N are finitely generated, so is M ⊗R N .
(If {x1 , . . . , xm } and {y1 , . . . , yn } are generating sets for M and N , respectively, then
{xi ⊗ yj } generates M ⊗R N .) Since the radical of an ideal is the intersection of all
prime ideals containing it, it suffices to prove that for any p ∈ Spec R, p ⊇ AnnR M +
AnnR N if and only if p ⊇ AnnR M ⊗R N . Suppose first that p ⊇ AnnR M + AnnR N .
By Problem #3, we have Mp 6= 0 and Np 6= 0. Hence, by Problem #7, we obtain
Mp ⊗Rp Np 6= 0. Now, using Problem #6 and associativity of tensor products,
(M ⊗R N )p ∼
= (M ⊗R N ) ⊗R Rp
∼
= M ⊗R (N ⊗R Rp )
∼
= M ⊗R Np
∼
= M ⊗R (Rp ⊗Rp Np )
∼
= (M ⊗R Rp ) ⊗Rp Np
∼
= Mp ⊗R Np .
p
In particular, Mp ⊗Rp Np 6= 0 implies (M ⊗R N )p 6= 0. As M ⊗R N is finitely generated,
we have that p ⊇ AnnR M ⊗R N (by Problem #3 again).
Conversely, if p ⊇ AnnR M ⊗R N , we have (M ⊗R N )p 6= 0. Using the isomorphism
above, this gives that Mp ⊗Rp Np 6= 0. Hence, Mp 6= 0 and Np 6= 0. Thus, p ⊇ AnnR M
and p ⊇ AnnR N , and so p ⊇ AnnR M + AnnR N .