Matlis dual of injective module over complete local ring

Question

Let $M$ be an injective module over a Noetherian complete local ring $(R,\mathfrak m,k)$. Let $E(k)$ denote the injective hull of $k$. Then, is it true that $\text{Hom}_R(M, E(k))$ is a free $R$-module ? Or at least a flat $R$-module ?

Answer 1

There are isomorphisms $\text{Ext}_{R}^{n}(R/I,M)^{\vee}\simeq \text{Tor}_{n}^{R}(R/I,M^{\vee})$ for any $R$-module $M$, hence if $M$ is injective it is clear that $M^{\vee}$ is flat.

(This doesn't really need Matlis duality, and $E(k)$ can be replaced by any injective cogenerator)

In general this will not be a free module: for any prime $\mathfrak{p}$ one $\text{Hom}(E(R/\mathfrak{p}),E(k))$ isomorphic to the completion of a free $R_{\mathfrak{p}}$-module, which is usually not free (but clearly flat) over $R$.

See §3.3 and §3.4 of Relative Homological Algebra by Enochs and Jenda for more details and proofs.