Math 615 (Winter 2005)  Homework Set #4  Solutions

Problem 1. Let R be a Noetherian ring of characteristic p, I an ideal, and M an R-module. (1) For any given e ∈ ℕ, show that H_{I}^{i}(^{e}M) ≅ ^{e}(H_{I}^{i}(M)) for every i.

(2) Suppose that ^{e_{0}}M is finitely generated over R for some e_{0} ≥ 1. Show that R/ Ann_{R}(M) is an F -finite ring. Consequently, ^{e}M is finitely generated over R for every e.

Proof. (1). Given any ring homomorphism ϕ : S → T of Noetherian rings, any T -module M (which is naturally an S-module), and any ideal I of S, we always have H_{I}^{i}(M) ≅ H_{ϕ(I)T}^{i} (M) as S-modules (and as T -modules as well) for all i. To prove (1), apply the above with S = R, T = R, and ϕ being the Frobenius ring homomorphism F^{e} : S → T, which gives H_{I}^{i}(^{e}M) ≅ ^{e}(H_{I^{[q]}}^{i}(M)) for every i. Now notice that H_{I^{[q]}}^{i}(M)) ≅ H_{I}^{i}(M)) for every i.

(2). Since e_{0} ≥ 1, we have that both M = ^{0}M and ^{1}M are finitely generated over R. Say M is generated by x_{1}, …, x_{n}. Then there is an injective R-linear map R/ Ann_{R}(M) → M^{n} sending the class of 1 to (x_{1}, …, x_{n}) ∈ M^{n}. This induces an injective R-linear map ^{1}(R/ Ann_{R}(M)) → 1(M^{n}) ≅ (^{1}M)^{n}, which forces ^{1}(R/ Ann_{R}(M)) to be finitely generated over R, i.e. R/ Ann_{R}(M) is an F -finite ring. Consequently, ^{e}(R/ Ann_{R}(M)) is finitely generated over R for every e. Also notice that there is a surjective R-linear map (R/ Ann_{R}(M))^{n} → M, which induces a surjective R-linear map (^{e}(R/ Ann_{R}(M)))^{n} → ^{e}M for every e. Thus ^{e}M is finitely generated over R for every e. □

Problem 2. Let (R, 𝔪, k) be a Noetherian local ring of prime characteristic p and P ∈ Spec(R) be any prime ideal of R. Suppose R is F -finite and say [k : k^{p}] = p^{a}.

(1) Prove that dim(R̂/Q) = dim(R/P) for every Q ∈ AssR̂(R̂/P R̂) = minR̂(R̂/P R̂).

(2) Show that [(R/P)_{P} : ((R/P)_{P})^{p}] = p^{a+dim(R/P)}. (We have proved this when R is complete.)

Proof. Without loss of generality, we assume R is a domain and P = 0. Then, as proved in class, R̂ is reduced so that AssR̂(R̂) = minR̂(R̂) and R̂_{Q} is the fraction field of R̂/Q for every Q ∈ minR̂(R̂). By going-down, Q ∩ R = P = 0 for every Q ∈ minR̂(R̂). Thus there is a natural isomorphism R̂_{Q} ≅ R_{P} ⊗_{R_{P}} R̂_{Q} for every Q ∈ minR̂(R̂). We have shown in class that R̂ ⊗_{R} ^{1}R ≅ ^{1}(R̂), which is the same as that R^{1/p} ⊗_{R} R̂ ≅ (R̂)^{1/p}. Say [R_{P} : (R_{P})^{p}] = n, i.e. (R_{P})^{1/p} = (R^{1/p})_{P} ≅ (R_{P})^{n}. Then (R̂_{Q})^{1/p} ≅ ((R̂)^{1/p})_{Q} ≅ (R^{1/p} ⊗_{R} R̂)_{Q} ≅ R^{1/p} ⊗_{R} R̂_{Q} ≅ (R^{1/p})_{P} ⊗_{R_{P}} R̂_{Q} ≅ (R_{P})^{n} ⊗_{R_{P}} R̂_{Q} ≅ (R_{P} ⊗_{R_{P}} R̂_{Q})^{n} ≅ (R̂_{Q})^{n} for any Q ∈ minR̂(R̂). In other words, [(R̂_{Q})^{1/p} : R̂_{Q}] = n = [R_{P} : (R_{P})^{p}] for all Q ∈ minR̂(R̂).

We have shown in class that [(R̂_{Q})^{1/p} : R̂_{Q}] = p^{a+dim(R̂/Q)} for any Q ∈ Spec(R̂). Thus dim(R̂/Q) is constant for all Q ∈ minR̂(R̂). Since dim(R̂/Q) = dim(R̂) = dim(R/P) for some Q ∈ minR̂(R̂), we conclude that (1) dim(R̂/Q) = dim(R/P) for all Q ∈ minR̂(R̂) and (2) [R_{P} : (R_{P})^{p}] = p^{a+dim(R)}. □

Problem 3. Let R be a Noetherian ring of prime characteristic p, M a finitely generated R-module such that Ann_{R}(M) ⊆ \sqrt{0}. For any R-modules N ⊆ L and x ∈ L, prove x ∈ N_{L}^{∗} ⟺ there exists c ∈ R^{∘} such that Image(x ⊗_{R} ^{e}(cM) → L ⊗_{R} ^{e}M) ⊆ Image(N ⊗_{R} ^{e}M → L ⊗_{R} ^{e}M) for all e ≫ 0.

Proof. The direction ‘⇒’ is straightforward. To show the implication ‘⇐’, it suffices to prove it for (N + \sqrt{0}M)/\sqrt{0}M ⊆ M/\sqrt{0}M ∋ x + \sqrt{0}M over the reduced ring R/\sqrt{0}M. In other words, we may assume R is reduced (hence Ann_{R}(M) = 0) without loss of generality. Let W = R^{∘}. Then there exists a surjective (W^{−1}R)-map W^{−1}M → W^{−1}R, which implies the existence of an R-linear map ϕ : M → R such that ϕ(M) ∩ R^{∘} ≠ ∅. Say c′ ∈ ϕ(M) ∩ R^{∘}. Observe that ϕ ∈ Hom_{R}(M, R) implies that ϕ ∈ Hom_{R}(^{e}M, ^{e}R) for every e. Then the assumption that Image(x ⊗_{R} ^{e}(cM) → L ⊗_{R} ^{e}M) ⊆ Image(N ⊗_{R} ^{e}M → L ⊗_{R} ^{e}M) for all e ≫ 0 would imply that x ⊗_{R} (cc′) ∈ Image(N ⊗_{R} ^{e}R → L ⊗_{R} ^{e}R) = N_{L}^{[q]} for all e ≫ 0, which shows that x ∈ N_{L}^{∗} as cc′ ∈ R^{∘}. □

Problem 4. Let R be a Noetherian F -finite ring of prime characteristic p, M a finitely generated R-module with FFRT by finitely generated R-modules M_{1}, M_{2}, …, M_{r}, and L a finitely generated R-module. Show that ∪_{e∈ℕ} Ass(L⊗_{R} ^{e}M) is a finite set and, moreover, there exists an integer k ∈ ℕ such that the following are satisfied.

(1) For every e ∈ ℕ, there exists a primary decomposition

0 = Q_{e1} ∩ Q_{e2} ∩ ⋯ ∩ Q_{es_{e}} of 0 in L ⊗_{R} ^{e}M,

where Ass(L⊗^{e}M) = {P_{ej} | 1 ≤ j ≤ s_{e}} and Q_{ej} are P_{ej}-primary components of 0 ⊆ L⊗_{R} ^{e}M satisfying P_{ej}^{k}(L ⊗_{R} ^{e}M) ⊆ Q_{ej} for all 1 ≤ j ≤ s_{e}.

(2) We have J^{k}(0 :_{L⊗_{R} ^{e}M} J^{∞}) = 0, i.e., J^{k} H_{J}^{0}(L ⊗_{R} ^{e}M) = 0 for all J ⊆ R and for all e ∈ ℕ. (In case L = R/I, the above may be stated in terms of ∪_{e∈ℕ} Ass(M/I^{[q]}M) and H_{J}^{0}(M/I^{[q]}M).)

Proof. For each i = 1, 2, …, r, choose a primary decomposition of 0 in L ⊗_{R} M_{i} (ignore the possible cases of i where L ⊗_{R} M_{i} = 0) as follows

0 = Q′_{i1} ∩ Q′_{i2} ∩ ⋯ ∩ Q′_{it_{i}},

where Q′_{ij} are P′_{ij}-primary components of 0 ⊂ L ⊗_{R} M_{i} for 1 ≤ i ≤ r, 1 ≤ j ≤ t_{i}. Since ^{e}M is a direct sum of the M_{i} (implying that L ⊗_{R} ^{e}M is a sum of the L ⊗_{R} M_{i}), we naturally get an induced primary decomposition 0 = Q_{e1} ∩ Q_{e2} ∩ ⋯ ∩ Q_{es_{e}} of 0 in L ⊗_{R} ^{e}M for every e. Therefore ∪_{e∈ℕ} Ass_{R}(L ⊗_{R} ^{e}M) is finite as it is contained in ∪_{i=1}^{r} Ass_{R}(L ⊗_{R} M_{i}). Choose k ∈ ℕ such that P′_{ij}^{k}(L ⊗_{R} M_{i}) ⊆ Q′_{ij} for all i = 1, 2, …, r and all j = 1, 2, …, t_{i}. Then (1) is evidently true.

To see (2), recall that, for any ideal J, (0 :_{L⊗_{R} ^{e}M} J^{∞}) = ∩_{J⊈P_{ej}}Q_{ej}. Thus, for any ideal J and any e ∈ ℕ, we have J^{k}(0 :_{L⊗_{R} ^{e}M} J^{∞}) ⊆ ( ∩_{J⊈P_{ej}} Q_{ej}) ∩ ( ∩_{J⊆P_{ej}} Q_{ej}) = 0.

(In case L = R/I, the result implies that ∪_{e∈ℕ} Ass(M/I^{[q]}M) is finite (cf. Homework Set #3 Problem 2(2)) and, for any ideal J, J^{(k+μ(J))q} H_{J}^{0}(\frac{M}{I^{[q]}M}) ⊆ (J^{k})^{[q]} H_{J}^{0}(M/I^{[q]}M) = 0, where μ(J) is the least number of generators of the ideal J.) □

Problem 5. Let R → S be a homomorphism of Noetherian rings of prime characteristic p and M a finitely generated R-module. (In this problem, we treat ^{e}M as an R-R-bimodule where r_{1} ·x·r_{2} =

r_{1}^{q}r_{2}x for any r_{1}, r_{2} ∈ R, x ∈ M. Also recall that #_{R}(^{e}M) = ℓ_{R}^{r}(Image(k ⊗_{R} ^{e}M ⟶^{ψ⊗1} E_{R}(k) ⊗_{R} ^{e}M))

where ψ : k → E_{R}(k) is any injective R-map and ℓ_{R}^{r}(−) denotes length as a right R-module.)

(1) For any R-module E, there is an isomorphism (E ⊗ ^{e}M) ⊗_{R} S ≅ (E ⊗_{R} S) ⊗_{S} ^{e}(M ⊗_{R} S). Moreover, the isomorphism is natural in the sense that for any R-linear map E_{1} → E_{2}, the following diagram commutes:

\begin{aligned}(E_{1} ⊗ ^{e}M) ⊗_{R} S & ⟶^{≅} & (E_{1} ⊗_{R} S) ⊗_{S} ^{e}(M ⊗_{R} S) \\ ↓ & & ↓ \\ (E_{2} ⊗ ^{e}M) ⊗_{R} S & ⟶^{≅} & (E_{2} ⊗_{R} S) ⊗_{S} ^{e}(M ⊗_{R} S) \end{aligned}

(2) Assume, furthermore, that (R, 𝔪, k) → (S, 𝔫, l) is a flat homomorphism of local rings such that 𝔪S = 𝔫. Show that (a) E_{R}(k) ⊗ S ≅ E_{S}(l) and (b) #_{R}(^{e}M) = #_{S}(^{e}(M ⊗_{R} S) for all e.

Proof. (1). For every e ∈ ℕ, we have a series of natural isomorphisms

(E ⊗_{R} ^{e}M) ⊗_{R} S ≅ E ⊗_{R} ^{e}(M ⊗_{R} S) ≅ E ⊗_{R} (S ⊗_{S} ^{e}(M ⊗_{R} S)) ≅ (E ⊗_{R} S) ⊗_{S} ^{e}(M ⊗_{R} S).

And it is routine to verify that the diagram commutes.

(2). Since every element of E_{R}(k) is killed by a power of 𝔪, every element of E_{R}(k) ⊗ S is killed by a power of 𝔪S = 𝔫. Thus E_{R}(k) ⊗ S is an essential extension of its socle (0 :_{E_{R}(k)⊗S} 𝔫) while

(0 :_{E_{R}(k)⊗S} 𝔫) = (0 :_{E_{R}(k)⊗S} 𝔪S) ≅ (0 :_{E_{R}(k)} 𝔪) ⊗_{S} S ≅ (R/𝔪) ⊗_{S} S ≅ S/𝔪S = S/𝔫 = l.

This implies that E_{R}(k) ⊗ S is isomorphic to an S-submodule of E_{S}(l). To prove (a), it will then suffice to show Ann_{S}(E_{R}(k) ⊗_{R} S) = 0. For any i ∈ ℕ, let E_{i} = (0 :_{E_{R}(k)} 𝔪^{i}), which is a finitely generated R-module with Ann_{R}(E_{i}) = 𝔪^{i}. Then Ann_{S}(E_{R}(k) ⊗_{R} S) ⊆ Ann_{S}(E_{i} ⊗_{R} S) = Ann_{R}(E_{i})S = 𝔪^{i}S = 𝔫^{i} for all i ∈ ℕ, which forces Ann_{S}(E_{R}(k) ⊗_{R} S) = 0. Finally, (b) follows from (a) and (1) immediately: Simply apply the commutative diagram in (1) with E_{1} = k, E_{2} = E_{R}(k), and ψ : E_{1} = k → E_{R}(k) = E_{2} being the injective R-linear map as in the definition of #(^{e}M). □