Math 615 (Winter 2005)  Homework Set #2  Solutions

Problem 1. Let (R, 𝔪) be a homomorphic image of a Noetherian Cohen-Macaulay local ring (S, 𝔫). Assume that R is equidimensional with dim(R) = d. For any system of parameters x_{1}, …, x_{d} of R, show that ((x_{1}, …, x_{i})_{R}^{∗} :_{R} x_{i+1}) = (x_{1}, …, x_{i})_{R}^{∗} for every i < d. (The case i = 0 may be interpreted as (0^{∗} :_{R} x_{1}) = 0^{∗}, the proof of which should be well incorporated into the general case.)

Proof. Say R = S/Q with height(Q) = h. As shown in class, there exist \underline{y} = y_{1}, …, y_{d} ∈ S, \underline{z} = z_{1}, …, z_{h} ∈ Q, and c ∈ S such that (a) the images of y_{i} are x_{i} for i = 1, …, d; (b) y_{1}, …, y_{d}, z_{1}, …, z_{h} form part of a system of parameters of S; (c) the image of c is c̄ ∈ R^{∘}; and (d) there exists q_{0} ∈ ℕ such that cQ^{[q_{0}]} ⊆ (\underline{z})S.

For any i < d and any x ∈ ((x_{1}, …, x_{i})_{R}^{∗} :_{R} x_{i+1}), we have xx_{i+1} ∈ (x_{1}, …, x_{i})_{R}^{∗}, meaning that there exists b̄ ∈ R^{∘} and q_{1} such that b̄(xx_{i+1})^{q} ∈ (x_{1}, …, x_{i})^{[q]} for all q ≥ q_{1}. Say y, b ∈ S maps to x, b̄ ∈ R respectively. Then b(yy_{i+1})^{q} ∈ (y_{1}, …, y_{i})^{[q]} + Q for all q ≥ q_{1}. Thus b^{q_{0}}(yy_{i+1})^{q_{0}q} ∈ (y_{1}, …, y_{i})^{[q_{0}q]} + Q^{[q_{0}]} for all q ≥ q_{1}. Hence cb^{q_{0}}y^{q_{0}q}y_{i+1}^{q_{0}q} ∈ (y_{1}, …, y_{i})^{[q_{0}q]} + (\underline{z}), which implies cb^{q_{0}}y^{q_{0}q} ∈ (y_{1}, …, y_{i})^{[q_{0}q]} + (\underline{z}) for all q ≥ q_{1} (as S is Cohen-Macaulay). Applying the homomorphism S → R, we get c̄b̄^{q_{0}}x^{q_{0}q} ∈ (x_{1}, …, x_{i})^{[q_{0}q]} for all q ≥ q_{1}. In other words, we have c̄b̄^{q_{0}}x^{q} ∈ (x_{1}, …, x_{i})^{[q]} for all q ≥ q_{0}q_{1}. As c̄b̄^{q_{0}} ∈ R^{∘}, we conclude that x ∈ (x_{1}, …, x_{i})_{R}^{∗}. Hence ((x_{1}, …, x_{i})_{R}^{∗} :_{R} x_{i+1}) ⊆ (x_{1}, …, x_{i})_{R}^{∗}. Consequently, ((x_{1}, …, x_{i})_{R}^{∗} :_{R} x_{i+1}) = (x_{1}, …, x_{i})_{R}^{∗} as the other containment, ((x_{1}, …, x_{i})_{R}^{∗} :_{R} x_{i+1}) ⊇ (x_{1}, …, x_{i})_{R}^{∗}, is obvious. □

Problem 2. Let R ⊆ S be an extension of Noetherian domains of prime characteristic p such that R is complete local and S is weakly F -regular. Moreover, assume that (IS) ∩ R = I for all ideals I of R (e.g. R is a direct summand of S as R-modules). Show R is Cohen-Macaulay.

Proof. Choose a system of parameters x_{1}, …, x_{d} of R (with d = dim(R)). Observe that R has ‘colon-capture’ property: ((x_{1}, …, x_{i}) :_{R} x_{i+1}) ⊆ (x_{1}, …, x_{i})_{R}^{∗} for every i < d. Hence it suffices to show that (x_{1}, …, x_{i})_{R}^{∗} = (x_{1}, …, x_{i})R for every i < d. We are actually to prove I^{∗} = I for every ideal I of R. Indeed, since R^{∘} ⊆ S^{∘} and S is weakly F -regular, we have I_{R}^{∗}S ⊆ (IS)_{S}^{∗} = IS and therefore I_{R}^{∗} = (I_{R}^{∗}S) ∩ R = (IS) ∩ R = I for every ideal I of R. □

Problem 3. Let R ⊆ S be any integral extension of commutative rings of prime characteristic p in which R is Noetherian.

(1) Assume that S is module-finite over R. Then (IS)_{S}^{∗} ∩ R ⊆ I_{R}^{∗} for any ideal I of R.

(2) Show that (IS) ∩ R ⊆ I_{R}^{∗} for any ideal I of R.

Proof. Say min(R) = {P_{1}, …, P_{r}}. By lying over, there exists Q_{i} ∈ Spec(S) such that Q_{i} ∩ R = P_{i} for every 1 ≤ i ≤ r. By incomparability, each Q_{i} is a minimal prime of S. For every 1 ≤ i ≤ r, there is an induced embedding R/P_{i} → S/Q_{i}. For this reason, we simply treat R_{i} := R/P_{i} as a subring of S_{i} := S/Q_{i} for every 1 ≤ i ≤ r. Remember that we have proved the claims of this problem in case of extensions of domains.

(1). For any x ∈ (IS)^{∗} ∩R, we have x+Q_{i} ∈ (IS_{i})_{S_{i}}^{∗} over S_{i}, which implies x+P_{i} ∈ (IS_{i})_{S_{i}}^{∗} ∩R_{i} = (IR_{i})_{R_{i}}^{∗} over R_{i} for every 1 ≤ i ≤ r. This shall force x ∈ I_{R}^{∗} over R.

(2). For any x ∈ (IS) ∩ R, we can write x = x_{1}s_{1} + ⋯ + x_{n}s_{n} for some x_{i} ∈ I, s_{i} ∈ S. Then we may apply part (1) to R ⊆ R[s_{1}, …, s_{n}] and conclude x ∈ I_{R}^{∗}. Alternatively, we observe that x ∈ (IS) ∩ R implies that x + Q_{i} ∈ IS_{i}, which implies x + P_{i} ∈ (IS_{i}) ∩ R_{i} ⊆ (IR_{i})_{R_{i}}^{∗} over R_{i} for every 1 ≤ i ≤ r. This shall force x ∈ I_{R}^{∗} over R. □

Problem 4. Let R be a Noetherian ring of prime characteristic p and N ⊊ M be finitely generated R-modules. Let Λ = \{L | N ⊆ L ⊆ M and \sqrt{Ann_{R}(M/L)} is a maximal ideal\}.

(1) Show that N = ∩_{L∈Λ}L. (This claim does not depend on characteristic p.)

(2) Suppose that J^{∗} = J for all ideals J of R such that \sqrt{J} are maximal ideals. Show that R is weakly F -regular, i.e. every ideal is tightly closed.

(3) Show that R is weakly F -regular if and only if R_{𝔪} is weakly F -regular for all maximal ideals 𝔪 of R.

Proof. (1). We assume N = 0 without loss of generality. For any 0 ≠ x ∈ M, choose a maximal ideal 𝔪 of R such that (0 :_{R} x) ⊆ 𝔪. Then x/1 is nonzero in M_{𝔪}. But we have ∩_{n>0}𝔪^{n}M_{𝔪} = 0, which implies that x/1 ∉ 𝔪^{n}M_{𝔪} and, hence, x ∉ 𝔪^{n}M for some 0 < n ∈ ℕ. Let L = 𝔪^{n}M and, clearly, \sqrt{Ann_{R}(M/L)} = 𝔪. (The argument actually implies that N = ∩_{𝔪} ∩_{n∈ℕ} (N + 𝔪^{n}M), in which 𝔪 runs through all maximal ideals of R.)

(2). For any ideal I of R, we have I = ∩_{J∈Λ}J with Λ = \{J | I ⊆ J and \sqrt{J} is maximal\} by part (1). Therefore I^{∗} ⊆ ∩_{J∈Λ}J^{∗} = ∩_{J∈Λ}J = I. So I^{∗} = I for every ideal I of R.

(3). If R is weakly F -regular, then I^{∗} = I for all ideals I of R. In particular, J^{∗} = J for all ideals J primary to any given maximal ideal 𝔪. For any 𝔪R_{𝔪}-primary ideal H of R_{𝔪}, H = JR_{𝔪} for some 𝔪-primary ideal J of R. Therefore H_{R_{𝔪}}^{∗} = J_{R}^{∗}R_{𝔪} = JR_{𝔪} = H, implying R_{𝔪} is weakly F -regular by part (2). Conversely, suppose R_{𝔪} is weakly F -regular for all maximal ideals 𝔪 of R. For any ideal J that is primary to a maximal ideal, say 𝔪, we have J_{R}^{∗}R_{𝔪} = (JR_{𝔪})_{R_{𝔪}}^{∗} = JR_{𝔪}. This implies that J_{R}^{∗} = J. Hence R is weakly F -regular by part (2). □

Problem 5. Let R be a commutative ring (not necessarily Noetherian or with characteristic p). Given an ideal I ⊆ R and x ∈ R, prove that x is in the integral closure of I if and only if x + P is in the integral closure of (I + P)/P ⊆ R/P for all P in min(R), the set of minimal primes of R.

Proof. Let W = {x^{n} + a_{1}x^{n−1} + ⋯ + a_{n−1}x + a_{n} | n ∈ ℤ, a_{i} ∈ I^{i}}. Then we see that W is a multiplicatively closed subset of R.

Suppose that x is not in the integral closure of I on the contrary. Then 0 ∉ W, meaning W^{−1}R is not a zero ring and hence Spec(W^{−1}R) ≠ ∅. Then, by Zorn lemma, there is a minimal prime ideal Q of W^{−1}R, in which Q has to be of the form W^{−1}P for some P ∈ min(R). Clearly P ∩W = ∅. But this contradicts to the assumption that x + P is in the integral closure of (I + P)/P ⊆ R/P. □

Problem 6. Let R be a Noetherian ring of prime characteristic p and N ⊆ M R-modules such that (N_{P})_{M_{P}}^{∗} = (N_{M}^{∗})_{P} for every P ∈ Ass(M/N_{M}^{∗}). Show that (W^{−1}N)_{W^{−1}M}^{∗} = W^{−1}(N_{M}^{∗}) for any multiplicatively closed set W ⊂ R.

Proof. Let W be any multiplicatively closed subset of R. Since W^{−1}(N_{M}^{∗}) ⊆ (W^{−1}N)_{W^{−1}M}^{∗} and, hence, (W^{−1}(N_{M}^{∗}))_{W^{−1}M}^{∗} = (W^{−1}N)_{W^{−1}M}^{∗}, we may mod out N_{M}^{∗}. In other words, we may assume N = 0 = 0_{M}^{∗} without loss of generality. Now we need to show that 0 = 0_{W^{−1}M}^{∗}.

Say Ass_{R}(M) = {P_{1}, …, P_{t}} and Ass_{W^{−1}R}(W^{−1}M) = {W^{−1}P_{1}, …, W^{−1}P_{s}} for some s ≤ t. Clearly, we have Ass_{W^{−1}R}(0_{W^{−1}M}^{∗}) ⊆ Ass_{W^{−1}R}(W^{−1}M).

Suppose, on the contrary, that 0 ⊊ 0_{W^{−1}M}^{∗}. Then Ass_{W^{−1}R}(0_{W^{−1}M}^{∗}) ≠ ∅. Say Ass_{W^{−1}R}(0_{W^{−1}M}^{∗}) = {W^{−1}P_{1}, …, W^{−1}P_{r}} for some 1 ≤ r ≤ s. Let P = P_{1}. Then M_{P} can be viewed as a further localization of W^{−1}M at W^{−1}P. Thus we have (0_{W^{−1}M}^{∗})_{W^{−1}P} ⊆ 0_{M_{P}}^{∗}.

By our choice, W^{−1}P ∈ Ass_{W^{−1}R}(0_{W^{−1}M}^{∗}), which implies that 0 ≠ (0_{W^{−1}M}^{∗})_{W^{−1}P} ⊆ 0_{M_{P}}^{∗}. On the other hand, the assumption, together with the fact that P ∈ Ass_{R}(M) (which is Ass_{R}(M/N_{M}^{∗}) since we are assuming N_{M}^{∗} = 0), forces 0_{M_{P}}^{∗} = 0. We have reached a contradiction. □