Math 615 (Winter 2005)  Homework Set #3  Solutions

Problem 1. Let R be a Noetherian ring of prime characteristic p. Show that R has a weak test element if and only if R/\sqrt{0} has a weak test element. (Here a weak test element is, by definition, a q-weak test element for some q.)

Proof. If c ∈ R^{∘} is a q_{1}-weak test element for R, then it is straightforward to verify that c + \sqrt{0} is a q_{1}-weak test element for R/\sqrt{0}. Conversely, suppose R/\sqrt{0} has a q_{2}-weak test element, say d + \sqrt{0} ∈ (R/\sqrt{0})^{∘} so that d ∈ R^{∘}. Say \sqrt{0}^{[q_{3}]} = 0. Then direct checking shows that d^{q_{3}}, which is in ∈ R^{∘}, is a (q_{2}q_{3})-weak test element for R. □

Problem 2. Let R be a Noetherian ring of prime characteristic p and M an R-module. Recall that eM is the derived R-module structure on M via the Frobenius homomorphism F^{e} : R → R.

(1) If ^{e_{0}}M is a faithful R-module for some e_{0} > 0, then R is reduced and ^{e}M (including M = ^{0}M) are faithful for all e ∈ ℕ.

(2) Show that Ass_{R}(M) = Ass_{R}(^{e}M) for every e ∈ ℕ.

Proof. (1). Denote q_{0} = p^{e_{0}}, which is ≥ p. Suppose R is not reduced. Then there exists 0 ≠ x ∈ \sqrt{0} such that x^{q_{0}} = 0. Then we see that x ∈ Ann_{R}(^{e_{0}}M), a contradiction. Now that R is reduced, the claim that ^{e}M is faithful for all e follows immediately from the easy assertion that, quite generally, Ann_{R}(^{e_{1}}M) ⊆ Ann_{R}(^{e_{2}}M) ⊆ \sqrt{Ann_{R}(^{e_{1}}M)} for every e_{1} ≤ e_{2} and any R-module M.

(2). Firstly, we observe an easy claim that Ann_{R}(x ∈ M) ⊆ Ann_{R}(x ∈ ^{e}M) ⊆ \sqrt{Ann_{R}(x ∈ M)} for any x ∈ M and any e ∈ ℕ. Then for any P ∈ Ass_{R}(M), there is y ∈ M such that Ann_{R}(y ∈ M) = P. Hence Ann_{R}(y ∈ ^{e}M) = P and therefore P ∈ Ass_{R}(^{e}M). Conversely, suppose P ∈ Ass_{R}(^{e}M), i.e. there is z such that Ann_{R}(z ∈ ^{e}M) = P. Let Rz be the R-submodule of M generated by z. Then P ∈ min(R/ Ann_{R}(z ∈ M)) = min(Rz) ⊆ Ass_{R}(Rz) ⊆ Ass_{R}(M). □

Problem 3. Let (R, 𝔪, k) be a Noetherian equidimensional catenary local ring of prime characteristic p with dim(R) = d. Suppose q^{d} < ℓ_{R}(R/𝔪^{[q]}) < q^{d} + q for some q = p^{e} ≥ p. Prove Sing(R) = {𝔪}, where Sing(R) = {P ∈ Spec(R) | R_{P} is not regular} is the singular locus of R.

Proof. The assumption of (R, 𝔪, k) being equidimensional catenary guarantees that dim(R/P) + dim(R_{P}) = dim(R) for every P ∈ Spec(R). And R is not regular as q^{d} < ℓ_{R}(R/𝔪^{[q]}).

If dim(R) = 0, then there is nothing to prove. So we assume dim(R) ≥ 1 and it suffices to show R_{P} is regular for any prime ideal P such that dim(R/P) = 1 (and hence dim(R_{P}) = d−1). For any such P, ℓ_{R_{P}} (R_{P} /P_{P}^{[q]}) ≤ \frac{1}{q}ℓ_{R}(R/𝔪^{[q]}) < q^{d−1} + 1 = q^{dim(R_{P})} + 1, which implies R_{P} is regular. □

Problem 4. Let R be a ring (not necessarily of characteristic p). Given R-modules M, N and f ∈ Hom_{R}(M, N), we say f is pure if the induced map f ⊗_{R} 1_{L} : M ⊗_{R} L → N ⊗_{R} L is injective for every R-module L. (Denote by m-Spec(R) the set consisting of all maximal ideals of R.)

(1) If f ∈ Hom_{R}(M, N) is pure, then f is injective. (Therefore, f ∈ Hom_{R}(M, N) is pure if and only if f is injective and the inclusion map f(M) ⊆ N is pure.)

(2) f ∈ Hom_{R}(M, N) is pure if and only if f_{P} : M_{P} → N_{P} is pure for every P ∈ Spec(R) if and only if f_{𝔪} : M_{𝔪} → N_{𝔪} is pure for every 𝔪 ∈ m-Spec(R).

(3) Show (A) f ∈ Hom_{R}(M, N) is pure if and only if f ⊗_{R} 1_{L} : M ⊗_{R} L → N ⊗_{R} L is injective for every finitely generated R-module L; and (B) If R is Noetherian and M is finitely generated, then f ∈ Hom_{R}(M, N) is pure if and only if f ⊗_{R} 1_{L} : M ⊗_{R} L → N ⊗_{R} L is injective for every finitely generated R-module L such that Ass_{R}(L) = {𝔪} for some 𝔪 ∈ m-Spec(R).

(4) Suppose R is Noetherian and M, N are finitely generated R-modules. Then f ∈ Hom_{R}(M, N) is pure if and only if f̂_{𝔪} : M̂_{𝔪} → N̂_{𝔪} is pure for every 𝔪 ∈ m-Spec(R) if and only if f : M → N splits (meaning there exists g ∈ Hom_{R}(N, M) such that g ∘ f = 1_{M}.)

(5) Suppose R is Noetherian and F is a free R-module. Then f ∈ Hom_{R}(F, N) is pure if and only if the induced map f⊗_{R}1_{E} : F ⊗_{R}E → N⊗_{R}E is injective, where E = ⊕_{𝔪∈m-Spec(R)}E_{R}(R/𝔪).

Proof. (1). This follows from the injectivity of f ⊗_{R} 1_{R} : M ⊗_{R} R → N ⊗_{R} R.

(2). This is standard.

(3). In both (A) and (B), we only need to show ‘if’. Suppose f : M → N is not pure. Then there exists an R-module L such that f ⊗_{R} 1_{L} has a non-zero kernel. Then, by property of tensor product, there exists a (sufficiently large) finitely generated R-submodule L′ ⊆ L such that f ⊗_{R} 1_{L′} is not injective, which proves (A). If, moreover, R is Noetherian and M is finitely generated, then 0 ≠ ker(f ⊗_{R} 1_{L′}) ⊆ M ⊗_{R} L′ are all finitely generated R-modules. Choose 𝔪 ∈ m-Spec(R) such that 0 ≠ ( ker(f ⊗_{R} 1_{L′}))_{𝔪}. Then, by Krull intersection theorem, Artin-Rees Lemma etc., ker(f ⊗_{R} 1_{L′/𝔪^{n}L′}) ≠ 0 for some integer n ≫ 0, which proves case (B) as Ass_{R}(L′/𝔪^{n}L′) = {𝔪}.

(4). Without loss of generality, we assume (R, 𝔪, k) is local. Denote −^{∨} = HomR̂(−, ER̂(k)). Then f is pure ⟹ f ⊗_{R}L is injective for all R-module L such that ℓ_{R}(L) < ∞ ⟹ f̂⊗R̂L is injective for all R̂-module L such that ℓR̂(L) < ∞ ⟹ f̂ is pure ⟹ f̂⊗R̂ M̂^{∨} : M̂ ⊗R̂ M̂^{∨} → N̂ ⊗R̂ M̂^{∨} is injective ⟹ (f̂ ⊗R̂ M̂^{∨})^{∨} : (N̂ ⊗R̂ M̂^{∨})^{∨} → (M̂ ⊗R̂ M̂^{∨})^{∨} is surjective ⟺ HomR̂(f̂, M̂) : HomR̂(N̂, M̂) → HomR̂(M̂, M̂) is surjective ⟹ f̂ splits ⟹ f splits ⟹ f is pure.

(5). Without loss of generality, we assume (R, 𝔪, k) is local. We only need to show ‘if’. Suppose f ∈ Hom_{R}(F, N) is not pure. Then, as F is free (not necessarily of finite rank), an argument similar to the one in part (3) above shows ker(f ⊗_{R} 1_{L}) ≠ 0 for some L with ℓ(L) < ∞. Then there exists an integer n > 0 such that L is embedded into E^{n} where E = E_{R}(k). Then, as F is free, we have ker(f ⊗_{R} 1_{E^{n}}) ≠ 0 ⟹ ker(f ⊗_{R} 1_{E}) ≠ 0, a contradiction. □

Problem 5. Given a local Noetherian ring (R, 𝔪, k) of prime characteristic p (not necessarily F - finite), one could define R to be strongly F -regular if, for any c ∈ R^{∘}, there exists an integer e ≥ 1 such that the R-linear map R → ^{e}R sending 1 to c is pure. In general, one could define R is strongly F -regular if R_{𝔪} is strongly F -regular for every 𝔪 ∈ m-Spec(R). (By Problem 4, we see that the above definition agrees with the one given in class when R is F -finite.)

(1) If there exists a pure R-linear map R → ^{e}R sending 1 to c with e ≥ 1, then R is reduced and, for every e′ ≥ e, the R-linear map R → ^{e′}R sending 1 to c is pure. (Thus the above definition of strong F -regularity forces R to be reduced.)

(2) Show that (R, 𝔪, k) is strongly F -regular if and only if 0_{E_{R}(k)}^{∗} = 0.

Proof. (1). The given pure map shows ^{e}R is faithful, implying R is reduced by Problem 2(1). So we are free to identify ^{e}R with R^{1/q} as R-modules for any q = p^{e}. Thus the given pure map may

be considered as f : R → R^{1/q} sending 1 to c^{1/q}. But f = g ∘ i : R ⊆^{i} R^{1/p} →^{g} R^{1/q} in which g is the R^{1/p}-linear map sending 1 to c^{1/q}. Thus the purity of f forces the purity of inclusion map i (which is easy to check). Also the purity of f amounts to the purity of the R^{1/p}-linear map f′ : R^{1/p} → R^{1/qp} sending 1 to c^{1/qp}, which readily implies the purity of f′ as an R-linear map. Therefore the R-linear map f′ ∘ i : R → R^{1/qp} is pure (which is easy to check) and it sends 1 to c^{1/qp}. In other words, the R-linear map R → ^{e+1}R sending 1 to c is pure. This in enough to prove (1).

(2). First of all, for any c ∈ R and e ∈ ℕ, let us denote by f_{c,e} : R → ^{e}R the R-linear map sending 1 to c ∈ ^{e}R.

To show ‘only if’, suppose R is strongly F -regular. For any x ∈ 0_{E_{R}(k)}^{∗}, by the definition of tight closure, there exists c ∈ R^{∘} such that 0 = c ⊗_{R} x ∈ ^{e}R ⊗_{R} E_{R}(k) for all e ≫ 0. Thus 1 ⊗_{R} x ∈ R ⊗_{R} E_{R}(k) is in ker(f_{c,q} ⊗_{R} 1_{R_{R}(k)}) for all e ≫ 0. But, by part (1) above, we know that f_{c,q} : R → ^{e}R are pure for all e ≫ 0, which forces 0 = 1 ⊗_{R} x ∈ R ⊗_{R} E_{R}(k), implying x = 0. So 0_{E_{R}(k)}^{∗} = 0.

Finally, let us prove ‘if’. Choose 0 ≠ w ∈ (0 :_{E_{R}(k)} 𝔪) so that w generates the socle of E_{R}(k). The assumption 0_{E_{R}(k)}^{∗} = 0 implies that w ∉ 0_{E_{R}(k)}^{∗}. Thus, for any c ∈ R^{∘}, there exists an integer e ≥ 1 such that 0 ≠ c ⊗_{R} w ∈ ^{e}R ⊗_{R} E_{R}(k), which means that 1 ⊗_{R} w ∈ R ⊗_{R} E_{R}(k) is not in ker(f_{c,q} ⊗_{R} 1_{E_{R}(k)}), which implies that ker(f_{c,q} ⊗_{R} 1_{E_{R}(k)}) = 0 as every non-zero R-submodule of R ⊗_{R} E_{R}(k) contains 1 ⊗_{R} w. Thus f_{c,q} ⊗_{R} 1_{E_{R}(k)} : R ⊗_{R} 1_{E_{R}(k)} → ^{e}R ⊗_{R} 1_{E_{R}(k)} is injective and, therefore, f_{c,q} : R → ^{e}R is pure by Problem 4(5). Hence R is strongly F -regular. □