A Few Comments On Matlis Duality

For a Noetherian local ring $(R,{\mathfrak m})$ with $\mathfrak p\in \Spec(R)$ we denote $E_R(R/\mathfrak p)$ by the $R$-injective hull of $R/\mathfrak p$. We will show that it has an $\hat{R}^\mathfrak p$-module structure and there is an isomorphism $E_R(R/\mathfrak p)\cong E_{\hat{R}^\mathfrak p}(\hat{R}^\mathfrak p/\mathfrak p\hat{R}^\mathfrak p)$ where $\hat{R}^\mathfrak p$ stands for the $\mathfrak p$-adic completion of $R$. Moreover for a complete Cohen-Macaulay ring $R$ the module $D(E_R(R/\mathfrak p))$ is isomorphic to $\hat{R}_\mathfrak{p}$ provided that $\dim(R/\mathfrak p)=1$ and $D(\cdot)$ denotes the Matlis dual functor $\Hom_R(\cdot, E_R(R/\mathfrak m))$. Here $\hat{R}_\mathfrak{p}$ denotes the completion of ${R_\mathfrak p}$ with respect to the maximal ideal $\mathfrak pR_\mathfrak p$. These results extend those of Matlis (see \cite{m}) shown in the case of the maximal ideal ${\mathfrak m}$.


Introduction
Throughout this paper R is a Noetherian local ring with the maximal ideal m and the residue field k = R/m. We denote the (contravaraint) Hom-functor Hom R (·, E R (R/p)) by ∨, that is M ∨ := Hom R (M, E R (R/p)) for an R-module M and E R (R/p) is a fixed R-injective hull of R/p where p ∈ Spec(R). Moreover we denote the Matlis dual functor by D(·) := Hom R (·, E R (k)). AlsoR p (resp.R p ) stands for the completion of R p (resp. of R) with respect to the maximal ideal pR p (resp. with respect to the prime ideal p).
Our main goal is to give the structure of E R (R/p) as anR p -module. In particular we will show the following result (see Theorem 3.1): Theorem. Let (R, m) be a ring and p ∈ Spec(R). Then E R (R/p) admits the structure of anR p -module and E R (R/p) ∼ = ER p (R p /pR p ) asR p -modules.
In the case of the maximal ideal m, Matlis has shown (see [11]) that the injective hull E R (k) of the residue field k has the structure ofR-module and it is isomorphic to theR-module ER(k). Here we will extend this result to an arbitrary prime ideal p, that is E R (R/p) admits the structure of anR p -module and it is isomorphic to theR p -module ER p (R p /pR p ).
Moreover, we know that Hom R (E R (R/p), E R (R/q)) = 0 for any p, q ∈ Spec(R) such that p ∈ V (q) and p = q. It was proven by Enochs (see [5, p. 183]) that the module E R (R/q) ∨ has the following decomposition: where p ∈ Spec(R) and T p denotes the completion of a free R p -module with respect to p R p -adic completion. Here we prove the following result (see Theorem

3.3):
Theorem. Let (R, m) be a complete Cohen-Macaulay ring of dimension n. Suppose that p ∈ Spec(R) with dim(R/p) = 1. Then there is an isomorphism Recently Schenzel has shown (see [17,Theorem 1.1]) that if p is a one dimensional prime ideal, then D(E R (R/p)) ∼ =Rp if and only if R/p is complete.
Furthermore Enochs (see [5, p. 183]) has shown that the module E R (R/q) ∨ is a flat cover of some cotorsion module. In the end of the Section 3 we will show the following result (see Theorem 3.7): Theorem. Let (R, m) be a ring and p, q ∈ Spec(R) such that p ∈ V (q). Then E R (R/q) ∨ is a flat precover of (R/q) ∨ .

Preliminaries
In this section we will fix the notation of the paper and summarize a few preliminaries and auxiliary results. Notice that the following two propositions are known for the maximal ideal m we will extend them to any prime ideal p ∈ Spec(R).
Proposition 2.1. Let (R, m) be a ring and p ∈ Spec(R). Then we have: (b) Suppose that M is an R-module and p ∈ Supp R (M ). Then M ∨ has a structure of an R p -module.
Proof. For the proof of the statement (a) we refer to [12,Theorem 18.4]. We only prove the last claim. For this purpose note that E R (R/p) is an R p -module. Now we define ( r s · f )(m) := r s f (m) where r s ∈ R p , m ∈ M and f ∈ M ∨ . Then it defines the structure of M ∨ to be an R p -module which completes the proof. Proposition 2.2. Let (R, m), (S, n) be local rings and R → S be a surjective local homomorphism i.e. S = R/I for some ideal I ⊆ R. Suppose that M is an R-module and p ∈ Spec(R). Then the following are true: (a) Suppose that N is an S-module. Then for any p ∈ V (I) Proof. For the proof see [9, Example 3.6 and Excercise 13]. Lemma 2.3. For a local ring R let M, N be any R-modules. Then for all i ∈ Z the following hold: Proof. For the proof see [9, Example 3.6] and [18].
In order to prove the next results we need a few more preparations. Let I ⊆ R Hom R (m α , ·). Also note that the local cohomology functor with respect to I is denoted by H i I (·), i ∈ Z, see [3] for its definition and applications in local algebra.
Lemma 2.4. Let R be a local ring and I ⊆ R be an ideal. Then there is an exact sequence: whereR I denotes the I-adic completion of R.
Proof. Note that for each α ∈ N there is an exact sequence For s ∈ N apply the functor Hom R (·, R/I s ) to this sequence. Then it induces the following exact sequence Now take the direct limit of this we again get an exact sequence It induces the following two short exact sequences Note that the inverse systems at the left hand side of these short exact sequences satisfy the Mittag-Leffler condition. So if we take the inverse limit to them, then the resulting sequences will be exact also (see [1,Proposition 10.2]). If we combined these resulting sequences, we get the following exact sequence Then the exactness of the last sequence provides the required statement.
In the next context we need the definition of the canonical module. To do this we recall the Local Duality Theorem (see [8]). Let (S, n) be a local Gorenstein ring of dimension t and N be a finitely generated R-module where R = S/I for some ideal I ⊆ S. Then It was shown by Grothendieck (see [8]) that there is an isomorphism for all i ∈ N (see [8] Definition 2.5. With the notation of the above Local Duality Theorem we define as the canonical module of N . It was introduced by Schenzel (see [15]) as the generalization of the canonical module of a Cohen-Macaulay ring (see e.g. [2]).
Corollary 2.6. With the notation of Lemma 2.4, suppose that I is a one dimensional ideal. Then the following are true: (a) There is an exact sequence where p i ∈ Ass R (R/I) such that dim(R/p i ) = 1 for i = 1, . . . , s.
(b) Suppose in addition that R is complete Cohen-Macaulay. Then there is an Proof. Note that for the proof of the statement (a) it will be enough to prove the following isomorphism . Then x ∈ S and by Local Global Principal for each s ∈ N there is an isomorphism Since R S is a semi local ring, there is an isomorphism (by Chinese Remainder Theorem). Since dim(R/p i ) = 1 for i = 1, . . . , s, it follows that Rad(IR p i ) = p i R p i . By passing to the inverse limit of the last isomorphism we have which is the required isomorphism. To prove (b), note that the canonical module of K R of R exists (since R is complete Cohen-Macaulay). So by [14,Lemma 3.1] there is an isomorphism is an Artinian R-module and R is complete, by Matlis Duality, its double Matlis dual is itself. That is for each s ∈ N there is an  3)). For i = 1, by the exact sequence in (a), we can get the exact sequence of (b) (since R is complete). This finishes the proof of the Corollary.

On Matlis Duality
In this section, we will prove one of the main results. Actually this result was proved by Matlis for the maximal ideal m. Here we will extend this to any prime ideal p ∈ Spec(R).
Theorem 3.1. Let (R, m) be a ring and p ∈ Spec(R). Then E R (R/p) admits the structure of anR p -module and E R (R/p) ∼ = ER p (R p /pR p ).
Proof. If x ∈ E R (R/p), then by [11] Ass R (Rx) = {p}. It follows that some power of p annihilates x.
Now letr ∈R p . Thenr = (r 1 + p, . . . , r n + p n , . . .) such that r n+1 − r n ∈ p n where r i + p ∈ R/p i for all i ≥ 1. Since x ∈ E R (R/p), then there exists a fixed n ∈ N such that p n x = 0. Choose r ∈ R such thatr − r ∈ p n (by definition of the completion). Definerx = rx. Then it is clear that this gives a well-defined R p -module structure to E R (R/p) which agrees with its R-module structure. Since E R (R/p) is an essential extension of R/p as an R-module then it necessarily is also an essential extension ofR p /pR p as anR p -module. So E R (R/p) ⊆ ER p (R p /pR p ).
To show that E R (R/p) is an injective hull ofR p /pR p as anR p -module, it is enough to prove that E R (R/p) = ER p (R p /pR p ). To do this it suffices to see that is an essential extension of R/p as an R-module since E R (R/p) is an extension of R/p as an R-module (see [11]). Let x ∈ ER p (R p /pR p ) and choose an elementr ∈R p such that 0 =rx ∈R p /pR p .
By the argument used in the beginning of the proof applied toR p , ER p (R p /pR p ) and x there is an n ∈ N such that p n x = 0. Choose r ∈ R such thatr − r ∈ p n . Thenrx = rx ∈ R/p ∼ =R p /pR p (see [1]) and rx = 0. Hence ER p (R p /pR p ) is an essential extension of R/p as an R-module and E R (R/p) = ER p (R p /pR p ) which completes the proof.

Remark 3.2.
(1) If p ∈ Spec(R), then by Theorem 3.1 for any finitely generated R-module M we have M ∨ ∼ = Hom R (M, HomR p (R p , ER p (R p /pR p ))).
By Lemma 2.3 the later module is isomorphic to HomR p (M p , ER p (R p /pR p )). Here we use thatM p ∼ = M ⊗ RR p . By Proposition 2.1 it implies that M ∨ has anR pmodule structure.
(2) Since Supp R (E R (R/p)) = V (p), where p ∈ Spec(R) (by Proposition 2.1(a)), is an isomorphism. But E R (R/p) is isomorphic to the module ER p (R p /pR p ) and it admits the structure of anR p -module (see Theorem 3.1). Moreover the above natural homomorphism is compatible with this structure. It implies that the p-adic completion of R commutes with the injective hull of R/p.
It is well known that Hom R (E R (R/p), E R (R/q)) = 0 for any p ∈ V (q) and p = q.

Moreover Enochs has proved that
where p ∈ Spec(R) and T p denotes the completion of a free R p -module with respect to p R p -adic completion. Here we will show the following theorem: Proof. Since R is complete Cohen-Macaulay, K R exists, and we have This is true, because K R is a maximal Cohen-Macaulay R-module of finite injective dimension and Supp R (K R ) = Spec(R). Let E · R (K R ) be a minimal injective resolution of K R . Then by [7, Theorem 1.1], we have Moreover, Γ q (E R (R/p)) = 0 for all p / ∈ V (q) and Γ q (E R (R/p)) = E R (R/p) for all p ∈ V (q). Then apply Γ q to E · R (K R ). It induces the following exact sequence Applying Matlis dual to this sequence yields the following exact sequence Here we use that D(E R (k)) ∼ = R (since R is complete). By the proof of Corollary 2.6 (b) and [16,Lemma 4.1], there are isomorphisms So the exact sequence (1) provides the following exact sequence Now the module D(E R (R/q)) is an R q -module, so there is a natural homomorphism R q → D(E R (R/q)). Then tensoring with R q /q s R q , this homomorphism induces the following homomorphism for each s ∈ N. Now take the inverse limit of it to get On the other side, since Supp R (E R (R/q)) = V (q), the module E R (R/q) is isomorphic to lim But again Proposition 2.3 (2) implies that the module D(Hom R (R/q s , E R (R/q))) is isomorphic to D(E R (R/q))/q s D(E R (R/q)). Therefore, there is a natural homo-morphismR q → D(E R (R/q)).
Since R is complete Cohen-Macaulay, by Corollary 2.6 (b), there is an exact se- Then this sequence together with the sequence (2) induces the following commutative diagram with exact rows Then by Five Lemma, there is an isomorphism D(E R (R/q)) ∼ =Rq which is the required isomorphism.
Corollary 3.4. Fix the notation of Theorem 3.3. Then the following hold: (a) There is an exact sequence (b) Suppose in addition that R is a domain. Then there is an isomorphism Proof. Since R is complete, we apply Corollary 2.6 (b) for I = q. Then it implies that there is an exact sequence Then the statement (a) is easily follows from Theorem 3.3. Note that the statement (b) follows from the above short exact sequence. Recall that u(q) = 0 since R is a domain.
In the next context we need the following definition of flat covers.  Enochs (see [5, p. 183]) showed that the module Hom R (E R (R/q), E R (R/p)) is a flat cover of some cotorsion module. Here we will show the following result: