Math 615 (Winter 2005)  Homework Set #1  Solutions

Problem 1. Let (R, 𝔪) be a 0-dimensional Noetherian (i.e. Artinian) local ring of prime characteristic p and M be a finitely generated R-module. Show that there exists e_{0} ∈ ℕ such that F^{e}(M) ≅ F^{e_{0}}(M) for all e ≥ e_{0}. (Actually, this is true as long as R is 0-dimensional Noetherian, the proof of which reduces to local case.)

Proof. Write down an representation R^{m} ⟶^{(a_{ij})} R^{n} ⟶ M ⟶ 0 such that a_{ij} ∈ 𝔪. Then, for any

q = p^{e}, F^{e}(M) is represented by exact sequence R^{m} ⟶^{(a_{ij}^{q})} R^{n} ⟶ F^{e}(M) ⟶ 0. By assumption, there exists q_{0} = p^{e_{0}} such that 𝔪^{[q]} = 0 for all q ≥ q_{0}. Therefore F^{e}(M) ≅ R^{n} for all e ≥ e_{0}.

(In case R is Artinian (not necessarily local), R = ∏_{i=1}^{s} R_{i} is a direct product of Artinian local rings R_{i}. So the claim remains true although the stabled F^{e}(M) is not necessarily free over R. Instead, F^{e}(M) is isomorphic to ⊕_{i=1}^{s}R_{i}^{b_{i}} for all e ≫ 0.) □

Problem 2. Let R be a Noetherian ring of prime characteristic p. It is known that I^{∗} ⊆ \sqrt{I} for every ideal I ⊆ R (by comparing tight closure with integral closure). Here we show this from definition of tight closure.

(1) If (R, 𝔪) is local (with maximal ideal 𝔪), show 𝔪^{∗} = 𝔪.

(2) Show P^{∗} = P for any prime ideal of R, which is not assumed to be local.

(3) Show I^{∗} ⊆ \sqrt{I} for every ideal I of R.

(4) Compute 0_{R}^{∗}. Here 0 refers to the zero ideal of R.

Proof. (1). Suppose 𝔪^{∗} ⊋ 𝔪. Then 1 ∈ 𝔪^{∗}, meaning that there exists c ∈ R^{∘} such that c1^{q} ∈ 𝔪^{[q]} for all q ≫ 0. But this forces c ∈ ∩_{q≫0}𝔪^{[q]} = 0, a contradiction.

(2). By (1), the ideal P_{P} is tightly closed in R_{P} for any P ∈ Spec(R). This gives the conclusion P^{∗} = P as (P_{R}^{∗})R_{P} ⊆ (P_{P})_{R_{P}}^{∗} for any P ∈ Spec(R).

(3). For any ideal I of R, say \sqrt{I} = ∩_{i=1}^{n}P_{i} for P_{i} ∈ Spec(R). Then I^{∗} ⊆ ∩_{i=1}^{n}P_{i}^{∗} = ∩_{i=1}^{n}P_{i} = \sqrt{I}. (4). We always have \sqrt{0} ⊆ 0^{∗}. By (3), we also have 0^{∗} ⊆ \sqrt{0}. Thus 0^{∗} = \sqrt{0}. □

Problem 3. Let k be a field of characteristic 2, S = k[X, Y] be a polynomial ring over k with indeterminates X, Y, and R = S/(X^{3})S. Let I = (X^{2})R ⊂ (X)R = J be ideals of R. (Then I ⊂ J are also modules over S under the natural ring homomorphism S → R.)

(1) Let F_{R}^{e}(−) be the Frobenius functor over R. Up to isomorphism, how many distinct Rmodules are there among 0, I, J, R, F_{R}(I), F_{R}(J), I_{R}^{[2]}, J_{R}^{[2]}, I_{J}^{[2]}? (Everything is considered as an R-module, including I as in the notation F_{R}(I), for example.) Group isomorphic R-modules together.

(2) Determine I_{R}^{∗} and I_{J}^{∗} over R. (Everything is considered as an R-module.)

(3) Determine I_{R}^{∗} and I_{J}^{∗} over S. (Everything is considered as an S-module, including R as in I_{R}^{∗}.)

Proof. Denote the images of X, Y in R by x, y. Therefore x^{3} = 0 in R.

(1). As R-modules, I = xJ ≅ R/xR and J ≅ R/x^{2}R. Hence F_{R}(I) ≅ R/x^{2}R, F_{R}(J) ≅ R/x^{4}R = R and I_{J}^{[2]} = x^{2}F_{R}(J) ≅ x^{2}R = I. We also have I_{R}^{[2]} = x^{4}R = 0 and J_{R}^{[2]} = x^{2}R = I. To sum up, there are four (4) distinct R-modules up to isomorphism, namely 0 ≅ I_{R}^{[2]}, I ≅ J_{R}^{[2]} ≅ I_{J}^{[2]}, J ≅ F_{R}(I) and R ≅ F_{R}(J).

(2). First, I_{R}^{∗} = \sqrt{0} = J because I ⊂ \sqrt{0}. Then notice that J_{R}^{∗} = J as J = \sqrt{0}, which means that 0_{R/J}^{∗} = 0. But R/J ≅ J/I. So 0_{J/I}^{∗} = 0, meaning I_{J}^{∗} = I.

(3). Over regular ring S, N_{M}^{∗} = N for any S-modules N ⊆ M. So I_{R}^{∗} = I and I_{J}^{∗} = I over S. □

Problem 4. Let R be a Noetherian ring of prime characteristic p. Suppose

G_{•} : 0 ⟶ G_{n} ⟶^{ϕ_{n}} G_{n−1} ⟶^{ϕ_{n−1}} ⋯ ⟶^{ϕ_{2}} G_{1} ⟶^{ϕ_{1}} G_{0}

is a complex with G_{i} free of finite rank for 0 ≤ i ≤ n. Apply Frobenius functor F^{e} to get F^{e}(G_{•}) for every e ∈ ℕ.

(1) If G_{•} is exact, show F^{e}(G_{•}) remains exact for every e ∈ ℕ.

(2) If ℓ_{R}(H_{i}(G_{•})) < ∞ for every 1 ≤ i ≤ n, show ℓ_{R}(H_{i}(F^{e}(G_{•}))) < ∞ for every 1 ≤ i ≤ n and for every e ∈ ℕ. (Here ℓ_{R}(−) represents the length of an R-module.)

Do (1) and (2) still hold if, instead, we assume G_{i} are finitely generated projective R-modules for 0 ≤ i ≤ n?

Proof. (1). Since Frobenius function F^{e} commutes with localization and a complex is exact if and only if every localization of it is exact, we may assume R is local. Then the assumption that G_{•} is exact implies that depth_{I(ϕ_{i})}(R) ≥ i and rank(G_{i}) = rank(ϕ_{i}) + rank(ϕ_{i+1}) for every 1 ≤ i ≤ n. In particular, I(ϕ_{i}) contains a non-zero-divisor for every 1 ≤ i ≤ n, which guarantees that rank(ϕ_{i}) = rank(F^{e}(ϕ_{i})) for every i and every e. This also implies that I(F^{e}(ϕ_{i})) = (I(ϕ_{i}))^{[q]} and thus depth_{I(F^{e}(ϕ_{i}))}(R) = depth_{I(ϕ_{i})}(R) for every q = p^{e}. Therefore, the complex F^{e}(G_{•}) still satisfies the ‘rank and depth’ condition. Consequently, F^{e}(G_{•}) is exact for every e ∈ ℕ. (If, for any given i, ϕ_{i} is represented by a matrix (a_{jk}), then F^{e}(ϕ_{i}) may be represented by the matrix (a_{jk}^{q}).)

(2). This follows from (1). Indeed, for any given e, ℓ_{R}(H_{i}(F^{e}(G_{•}))) < ∞ for all 1 ≤ i ≤ n if and only if F^{e}(G_{•}) ⊗_{R} R_{P} is exact for all prime ideals P that are not maximal.

Finally, (1) and (2) still hold if we assume G_{i} are finitely generated projective R-modules for 0 ≤ i ≤ n. As seen in the above proof, both (1) and (2) reduce to local case. □

Problem 5. Let R be a Noetherian ring of prime characteristic p and N ⊆ M be R-modules. Define N_{M}^{F} = {x ∈ M | x^{q} ∈ N_{M}^{[q]} ⊆ F^{e}(M) for some q = p^{e}}. (N_{M}^{F} is called the Frobenius closure of N in M.)

(1) Show that N_{M}^{F} ⊆ N_{M}^{∗}.

(2) Show that Frobenius closure commutes with localization, i.e. (W^{−1}N)_{W^{−1}M}^{F} = W^{−1}(N_{M}^{F}) for any multiplicatively closed set W ⊂ R.

Proof. (It is routine to check that N_{M}^{F} is an R-submodule of M.)

(1). For any x ∈ N_{M}^{F}, we have x^{q_{0}} ∈ N_{M}^{[q_{0}]} ⊆ F^{e_{0}}(M) for some q_{0} = p^{e_{0}}. This actually implies that x^{q} ∈ N_{M}^{[q]} ⊆ F^{e}(M) for all q ≥ q_{0}, showing x ∈ N_{M}^{∗} (with c = 1 ∈ R^{∘}).

(2). Let W ⊂ R be a multiplicatively closed subset of R. If x ∈ N_{M}^{F}, then x^{q} ∈ N_{M}^{[q]} ⊆ F^{e}(M) for some q = p^{e}. This gives (x/1)^{q} ∈ (W^{−1}N)_{W^{−1}M}^{[q]} ⊆ F^{e}(W^{−1}M). (Here we use the fact that Frobenius functor commutes with localization.) Hence x/1 ∈ (W^{−1}N)_{W^{−1}M}^{F}, showing (W^{−1}N)_{W^{−1}M}^{F} ⊇ W^{−1}(N_{M}^{F}). On the other hand, for any x/w ∈ (W^{−1}N)_{W^{−1}M}^{F} with x ∈ M and w ∈ W, there exists q = p^{e} such that (x/w)^{q} ∈ (W^{−1}N)_{W^{−1}M}^{[q]} ⊆ F^{e}(W^{−1}M). (We may simply assume w = 1 as Frobenius closure is a submodule.) Using the fact that Frobenius functor commutes (naturally) with localization, we may write (x/w)^{q} ∈ W^{−1}(N_{M}^{[q]}) ⊆ W^{−1}(F^{e}(M)) for the same q = p^{e}. Therefore there exists w′ ∈ W such that w′x^{q} ∈ N^{[q]} ⊆ F^{e}(M) and hence (w′x)^{q} ∈ N^{[q]} ⊆ F^{e}(M), which implies w′x ∈ N_{M}^{F}. This shows that (W^{−1}N)_{W^{−1}M}^{F} ⊆ W^{−1}(N_{M}^{F}). Finally, (W^{−1}N)_{W^{−1}M}^{F} = W^{−1}(N_{M}^{F}). □