# Matlis' dual of injective modules

Let $(R, \mathfrak{m})$ be a commutative Noetherian complete local rings ($R$ can be regular, if you need). Let $E(R/\mathfrak{p})$ be injective hull of $R/\mathfrak{p}$, if $\mathfrak{p}= \mathfrak{m}$ we simply write by $E$.

Question 1: What is $Hom (E(R/\mathfrak{p}), E)$?

By the isomorphism $Hom (Hom (M,N),E) \cong M \otimes Hom (N, E)$ provided $M$ is finitely generated we can see there are duality of injective modules and flat modules

Question 2: The Matlis' duality of injective modules and flat modules is 1-1 as Noetherian and Artinian?

Question 3: Let $x$ be an element of $R$. Find an injective module $I$ such that $Hom (I, E) \cong R_x$?

EDIT: We denote $Hom(\bullet,E)$ by $D(\bullet)$. An $R$-module $M$ is called Matlis reflexive if $M \cong D(D(M))$. Notherian and Artinian is Matlis reflexive. By (E. Enochs, Proc. AMS, 92 (1984), 179--184, Proposition 1.3) a module $M$ is Matlis reflexive iff there is a Notherian submodule L such that $M/L$ is Artinian. So the dulity of injective and flat is not as good as Notherian-Artinian.

Discussion: Assume more that $(R, \mathfrak{m})$ is a Gorenstein domain of dimension one. Then there are exactly two irreducible injective modules. Namely, $E$ and $E(R) = Q$ the field of fractions. And $0 \to R \to Q \to E \to 0$ is the minimal injective resolution of $R$. Since $E$ is Artinian we have $Q$ is Matlis reflexive.

Your first question has been answered by Schenzel in the case $\text{dim}R/\mathfrak{p} = 1$, namely $$\text{Hom}(E(R/\mathfrak{p}),E)= \widehat{R_{\frak p}},$$ if and only if $R/\mathfrak p$ is complete.