MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #7

JOHN VOIGHT

Problem XVII.12. Prove that an R-module E is a generator if and only if it is balanced and finitely generated projective over End_{R} E.

Solution. Lang proves the (⇒) direction as Theorem 7.1, so it suffices to show that if E is balanced and finitely generated projective over End_{R} E, then R is a homomorphic image of a direct sum of E with itself.

Since E is finitely generated projective over End_{R} E, we have an isomorphism (End_{R} E)^{n} ≅ E ⊕ F for some End_{R} E-module F. Therefore we have the following isomorphisms of End_{R} E-modules:

E^{n} ≅ Hom_{End_{R} E}((End_{R} E)^{n}, E) ≅ Hom_{End_{R} E}(E ⊕ F, E)
≅ Hom_{End_{R} E}(F, E) ⊕ End_{End_{R} E}(E).

If we define the operation of End_{R} E to be composition of mappings on the left, these become isomorphisms over R. Since E is balanced, End_{End_{R} E}(E) ≅ R, so E is a generator.

Problem X.9(a). Let A be an artinian commutative ring. Prove all prime ideals are maximal. [Hint: Given a prime ideal 𝔭, let x ∈ A, x ∉ 𝔭. Consider the descending chain (x) ⊃ (x^{2}) ⊃ . . ..]

Solution. We show that any artinian domain is a field. Let 𝔭 be a prime ideal, so that A/𝔭 is a domain. Any quotient ring of an artinian ring is artinian (Proposition 7.1), so A/𝔭 is artinian. Let x ∈ A/𝔭 be nonzero. Then the descending chain (x) ⊃ (x^{2}) ⊃ … must terminate, so (x^{k}) = (x^{k+1}) for some integer k; therefore there exists a y ∈ A/𝔭 such that x^{k+1}y = x^{k}, which is to say x^{k}(1 − xy) = 0, so xy = 1, and x ∈ (A/𝔭)^{∗}. Therefore A/𝔭 is a field, so 𝔭 is maximal.

Problem X.9(b). There is only a finite number of prime, or maximal, ideals. [Hint: Among all finite intersections of maximal ideals, pick a minimal one.]

Solution. Let S be the set of finite intersections of maximal ideals in A. This set is nonempty, so by Exercise XVII.2(c), there exists a minimal such intersection 𝔪_{1} ∩ ⋯ ∩ 𝔪_{r}. If 𝔪 is any maximal ideal of A, then, 𝔪 ∩ ⋂_{i} 𝔪_{i} = ⋂_{i} 𝔪_{i} so 𝔪 ⊃ ⋂_{i} 𝔪_{i} ⊃ 𝔪_{1}𝔪_{2} … 𝔪_{r}. A maximal ideal is prime, so 𝔪 ⊃ 𝔪_{i} for some i, but since 𝔪_{i} is maximal so 𝔪 = 𝔪_{i}.

Date: April 1, 2003.

XVII: 12 (first sentence); X: 9, 10, 11.

Problem X.9(c). The ideal N of nilpotent elements in A is nilpotent, that is there exists a positive integer k such that N^{k} = (0). [Hint: Let k be such that N^{k} = N^{k+1}. Let 𝔞 = N^{k}. Let 𝔟 be a minimal ideal such that 𝔟𝔞 ≠ 0. Then 𝔟 is principal and 𝔟𝔞 = 𝔟.]

Solution. Let k be such that N^{k} = N^{k+1}. Suppose that N^{k} ≠ 0; let S be the set of ideals 𝔟 of A such that 𝔟N^{k} ≠ 0. The set S is nonempty because N^{k} ∈ S, as N^{k}N^{k} = N^{2k} = N^{k} ≠ 0. Since A is artinian, S has a minimal element 𝔟. There is an element b ∈ 𝔟 such that bN^{k} ≠ 0; therefore (b) ∈ S and (b) ⊂ 𝔟 so by minimality (b) = 𝔟, in particular, 𝔟 is finitely generated. But 𝔟N^{k} ⊂ 𝔟 and (𝔟N^{k})N^{k} = 𝔟N^{2k} = 𝔟N^{k}, so again by minimality, 𝔟N^{k} = 𝔟. But every element of N^{k} is nilpotent hence contained in every maximal ideal, so by Nakayama’s lemma, 𝔟 = 0, a contradiction.

Problem X.9(d). A is noetherian.

Solution. Let k be an integer such that N^{k} = 0 as in part (c). Then ⋂_{𝔭} 𝔭 = \sqrt{0} = N, but by part (a) this implies N = ⋂_{i} 𝔪_{i}. Let k be an integer such that N^{k} = 0. Then

N^{k} = 0 = (⋂_{i}𝔪_{i})^{k} ⊃ (𝔪_{1} … 𝔪_{r})^{k} = 𝔪_{1}^{k} … 𝔪_{r}^{k}.

Consider A as a module over itself; A is noetherian as an A-module if and only if A is noetherian as a ring. We have a filtration

A ⊃ 𝔪_{1} ⊃ 𝔪_{1}^{2} ⊃ ⋯ ⊃ 𝔪_{1}^{k} … 𝔪_{r}^{k} = 0

of A. At the step E ⊃ E𝔪_{i} in the filtration, E/E𝔪_{i} is a vector space over the field A/𝔪_{i} which is finite-dimensional as A is artinian (as in Exercise XVII.2(a)). Therefore A has a finite simple filtration, so by Proposition 7.2, A is noetherian as well as artinian.

Problem X.9(e). There exists an integer r such that

A ≅ ∏_{𝔪} A/𝔪^{r}

where the product is taken over all maximal ideals.

Solution. Let r be such that N^{r} = 0, and let 𝔪_{i} be the maximal ideals of A. Since 𝔪_{i} + 𝔪_{j} = A for i ≠ j, we also have 𝔪_{i}^{r} + 𝔪_{j}^{r} = A for i ≠ j (otherwise, 𝔪_{i}^{r} + 𝔪_{j}^{r} ⊂ 𝔪 for some maximal ideal 𝔪; then 𝔪_{i}^{r} ⊂ 𝔪 so 𝔪_{i} ⊂ 𝔪, and similarly 𝔪_{j} ⊂ 𝔪, a contradiction). By the Chinese remainder theorem, then,

A → ∏_{𝔪} A/𝔪_{i}^{r}

is surjective. It is also injective, since ⋂_{i} 𝔪_{i}^{r} = N^{r} = 0 (as in part (d)), therefore it is an isomorphism.

Problem X.9(f). We have

A ≅ ∏_{𝔭} A_{𝔭}

where again the product is taken over all prime ideals 𝔭.

Solution. It is enough to show that this map is an isomorphism considered as a map of A-modules. Let 𝔭_{i} be the primes (maximal ideals) of A. Since localization preserves exact sequences (it is flat), it is enough to show that the map A → ∏_{i} A_{𝔭_{i}} is an isomorphism after localization at every prime ideal 𝔭 of A. But in this circumstance we have the map

A_{𝔭} → ∏_{i}(A_{𝔭_{i}})_{𝔭} = ∏_{i}(A_{𝔭})_{𝔭_{i}}.

Now A_{𝔭} is artinian (descending chains of ideals of A_{𝔭} are descending chains of ideals of A contained in 𝔭) and a local ring with maximal ideal 𝔭A_{𝔭}, so N = 𝔭A_{𝔭} and (𝔭A_{𝔭})^{r} = 0. Then for 𝔭 ≠ 𝔭_{i}, if x_{i} ∈ 𝔭 \ 𝔭_{i}, x_{i}^{r} = 0, so (A_{𝔭})_{𝔭_{i}} = 0 and the map is an isomorphism.

Problem X.10. Let A, B be local rings with maximal ideals 𝔪_{A}, 𝔪_{B}, respectively. Let f : A → B be a homomorphism. Suppose that f is local, i.e. f^{−1}(𝔪_{B}) = 𝔪_{A}. Assume that A, B are noetherian, and assume that:

(1) A/𝔪_{A} → B/𝔪_{B} is an isomorphism;

(2) 𝔪_{A} → 𝔪_{B}/𝔪_{B}^{2} is surjective;

(3) B is a finite A-module, via f. Prove that f is surjective.

Solution. First, 𝔪_{B} is a finitely generated B-module (since B is noetherian) and f(𝔪_{A}) is a finitely generated B-submodule of 𝔪_{B} with 𝔪_{B} = f(𝔪_{A}) + 𝔪_{B}^{2} by (2). By Nakayama’s lemma (X.4.2), f(𝔪_{A}) = 𝔪_{B}.

Second, since B is finite over A and f(A) is a B-submodule with B = f(A)+𝔪_{B}B by (1), since A → B/𝔪_{B} is surjective. But 𝔪_{B}B = 𝔪_{A}B treating B as an A-module by f, so by Nakayama’s lemma, f(A) = B, so f is surjective.

Problem X.11. Let A be a commutative ring and M an A-module. Define the support of M by

Supp(M) = {𝔭 ∈ Spec A : M_{𝔭} ≠ 0}.

If M is finite over A, show that Supp M = V (Ann(M)), where V (𝔞) = {𝔭 ∈ Spec A : 𝔭 ⊃ 𝔞} and Ann(M) = {a ∈ A : aM = 0}.

Solution. Let 𝔭 ∈ Spec A be so that M_{𝔭} ≠ 0. If aM = 0 then aM_{𝔭} = 0 so if a ∉ 𝔭 then M_{𝔭} = 0; hence Ann(M) ⊂ 𝔭.

Conversely, suppose Ann(M) ⊂ 𝔭. Let m_{1}, …, m_{r} generate M over A. Suppose that M_{𝔭} = 0; then for all i there exists an a_{i} ∉ 𝔭 such that a_{i}m_{i} = 0 ∈ M. Then a = ∏_{i} a_{i} has aM = 0; therefore a ∈ 𝔭, a contradiction.