Download Math 905 Solutions to Homework # 3 Throughout R

Math 905
Solutions to Homework # 3
Throughout R denotes a commutative ring with identity.
1. Prove the following statements are equivalent:
(a) R is reduced
(b) Rp is reduced for all prime ideals p.
Also, prove that if R is reduced then Rp is a field for all minimal primes p.
Solution: Suppose R is reduced. Let p be a prime ideal and rs a nilpotent element of
n
Rp . Then rsn = 0 for some n ≥ 1. Then there exists t 6∈ p such that trn = 0. Hence,
(tr)n = 0. Since R is reduced, this implies that tr = 0. Therefore, rs = 0, and Rp is
reduced. Conversely, suppose Rp is reduced for all prime ideals p of R. Let r ∈ R and
suppose rn = 0. Let I = (0 :R r) = {t ∈ R | tr = 0}, which is an ideal of R. Let p be
an arbitrary prime ideal of R. Then ( 1r )n = 0 in Rp . Since Rp is reduced, we must have
r
= 0. This implies that there is an element t ∈ R, t 6∈ p such that tr = 0; i.e., I 6⊂ p.
1
Since p is arbitrary, we have I = R. In particular, 1 ∈ I = (0 :R r), so r = 0. Hence, R is
reduced.
Now, suppose R is reduced and p is a minimal prime of R. By above, we have that Rp is
reduced. Now, as p is minimal, the only prime ideal of Rp is pp . We also know that, in
any ring, the set of nilpotents is the intersection of all prime ideals of the ring. Thus, pp
is the set of nilpotents of Rp . But since Rp is reduced, this means that pp = 0; i.e., 0 is a
maximal ideal of Rp . Thus, Rp is a field.
2. Give an example of a ring which is not a domain but with the property that Rp is a domain
for all primes p of R.
Solution: Let R = Z6 . The prime ideals of R are p = (2) and q = (3). Since 3 · 2 = 0 and
3 6∈ (2), we see that 12 = 0 in Rp . Hence, pp = 0 and Rp is a field. Similarly, Rq is a field.
3. Let M be a finitely generated R-module and p a prime ideal of R. Prove that Mp = 0 if
and only if p 6⊃ AnnR M . (Here, AnnR M = {r ∈ R | rM = 0}.) Also, give of an example
of a ring R, an R-module M , and a prime ideal p such that p ⊇ AnnR M and Mp = 0.
Solution: First, suppose p 6⊃ AnnR M . Let t ∈ AnnR M , t 6∈ p. For any m ∈ M , tm = 0
which implies m1 = 0 in Mp . Thus, Mp = 0. Conversely, suppose Mp = 0. As M is finitely
generated, M = Rx1 + Rx2 + · · · + Rxn for some x1 , . . . , xn ∈ M . Since x1i = 0 in Mp , for
each i there exists ti ∈ R, ti 6∈ p such that ti xi = 0. Let t = t1 t2 · · · tn . Then t 6∈ p and
txi = 0 for all i. Since x1 , . . . , xn generate M , tM = 0. Hence, AnnR M 6⊂ p.
Consider the Z-module M = Q/Z. Now, AnnZ M = (0). (For, if tM = 0 then t · p1 ∈ Z
for any prime p. That is, every prime p divides t. Hence, t = 0.) On the other hand, if
p = (0) then Zp = Qp = Q. Hence, Mp = 0.
4. Let I be a finitely generated ideal and suppose for every maximal ideal of R, either
Im = 0 or Im = Rm . Prove that I is generated by an idempotent. (Hint: Prove that
I + (0 :R I) = R.)
Solution: Let m be a maximal ideal. If Im = 0 then by exercise 3 above, there exists
t ∈ AnnR I with t 6∈ m; i.e., AnnR I 6⊂ m. On the other hand, Im = Rm implies I 6⊂ m. So
the hypothesis implies that for every maximal ideal m of R, either I 6⊂ m or AnnR I 6⊂ m.
Thus, I +AnnR I is not contained in any maximal ideal, so I +AnnR I = R. Then 1 = i+j
for some i ∈ I and j ∈ AnnR I. Note that i = i(i + j) = i2 + ij = i2 , so i is idempotent.
For any x ∈ I, x = x(i + j) = xi + xj = xi ∈ (i), which implies that I = (i).
5. Let R be a quasi-local ring and M and N finitely generated R-modules. Prove that
M ⊗R N = 0 if and only if M = 0 or N = 0. Give examples to show that the statement
is false if R is not quasi-local, or if either module is not finitely generated.
Solution: Suppose M ⊗R N = 0. Let k = R/m, where m is the maximal ideal of R.
Let M denote M/mM ∼
= M ⊗k k; similarly for N . Then 0 = k ⊗R (M ⊗R N ) ⊗R k ∼
=
∼
(k ⊗R M ) ⊗R (N ⊗R k) = M ⊗R N . By the associativity of tensor products, we have
0 = M ⊗R N ∼
= (M ⊗k k) ⊗R N ∼
= M ⊗k (k ⊗R N ) ∼
= M ⊗k N .
Since k is a field and M and N are finitely generated k-modules, M ∼
= k r and N ∼
= k s for
some integers r, s. Thus,
0 = M ⊗k N ∼
= k r ⊗k k s ∼
= (k r )s ∼
= k rs .
Thus, either r = 0 or s = 0. That is, either M = M/mM = 0 or N = N/mN = 0. By
Nakayama’s lemma, this implies either M = 0 or N = 0.
6. Let M and N be finitely generated R-modules. Prove that
p
p
Ann M ⊗R N = AnnR M + AnnR N .
Solution: Since the radical of any ideal is the intersection of all the primes containing it,
it suffices to show that for any prime p, p 6⊃ AnnR M ⊗R N if and only if p 6⊃ AnnR M +
AnnR N . Suppose p 6⊃ AnnR M ⊗R N . Then by exercise 3, (M ⊗R N )p = 0. Using the
associativity of tensor products,
0 = (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 ⊗Rp Np .
Since Mp and Np are finitely generated Rp -modules, exercise 5 implies that Mp = 0 or
Np = 0. By exercise 3, this implies p 6⊃ AnnR M or p 6⊃ AnnR N ; that is, p 6⊃ AnnR M +
AnnR N . Conversely, suppose p 6⊃ AnnR M + AnnR N . Without loss of generality, suppose
p 6⊃ AnnR M . Let t ∈ AnnR M with t 6∈ p. Then tM = 0, so t(m ⊗ n) = (tm) ⊗ n = 0 for
all m ∈ M , n ∈ N . Since the elements m ⊗ n generate M ⊗R N , we see that t annihilates
M ⊗R N . Thus, p 6⊃ AnnR M ⊗R N .
7. Let A, B and C be R-modules. Prove that A ⊗R (B ⊕ C) ∼
= (A ⊗R B) ⊕ (A ⊗R C).
Solution: Define a function f : A × (B ⊕ C) → (A ⊗R B) ⊕ (A ⊗R C) by f (a, (b, c)) =
(a⊗b, a⊗c). It is easily seen that f is R-bilinear. This induces an R-module homomorphism
fˆ : A ⊗R (B ⊕ C) → (A ⊗R B) ⊕ (A ⊗R C) where fˆ(a ⊗ (b, c)) = (a ⊗ b, a ⊗ c). Now
define a function g1 : A × B → A ⊗R (B ⊕ C) by g1 (a, b) = a ⊗ (b, 0). Again, it is easily
seen that g1 is R-bilinear. Thus, there exists an R-module homomorphism gˆ1 : A ⊗R B →
A ⊗R (B ⊕ C) with gˆ1 (a ⊗ b) = a ⊗ (b, 0). In the same way, we can construct an Rmodule homomorphism g2 : A ⊗R C → A ⊗R (B ⊕ C) where g2 (a ⊗ c) = a ⊗ (0, c). Then
g = g1 ⊕ g2 : (A ⊗R B) ⊕ (A ⊗R C) → A ⊗R (B ⊕ C) is an R-module homomorphism such
that g(a1 ⊗ b, a2 ⊗ c) = a1 ⊗ (b, 0) + a2 ⊗ (0, c) for all a ∈ A, b ∈ B, c ∈ C. Now, for all
a1 , a2 ∈ A, b ∈ B, c ∈ C, we have
fˆĝ(a1 ⊗ b, a2 ⊗ c) = fˆ(a1 ⊗ (b, 0) + a2 ⊗ (0, c))
= fˆ(a1 ⊗ (b, 0)) + fˆ(a2 ⊗ (0, c))
= (a1 ⊗ b, 0) + (0, a2 ⊗ c)
= (a1 ⊗ b, a2 ⊗ c).
Since (A⊗R B)⊕(A⊗R C) is generated by elements of the form (a1 ⊗b, a2 ⊗c), we conclude
that fˆĝ is the identity map. In a similar way, one can show that ĝ fˆ is the identity map.
Hence, fˆ is an isomorphism.