Is $Hom_R(S_X^{-1}R, E)$ the minimal injective cogenerator of $S_X^{-1}R$?

Question

Assume that $R$ is a commutative Noetherian ring with minimal injective cogenerator $E$. For a finite set of maximal ideals $X$ of $R$, define the multiplicative set $$S_X=R-\bigcup_{\mathfrak{m}\in X}\mathfrak{m}.$$

The question is: $$Hom_R(S_X^{-1}R, E) =~\!\! ?$$

I know that this module is an injective cogenerator of $S_X^{-1}R$. I want to know if it is its minimal one.

EDIT: For any commutative ring with identity $R$, the $R$-module $E$ is said to be an injective cogenerator if the functor $Hom_{R}(-,E)$ is exact and faithful; that is is, $E$ is an injective $R$-module such that, given any non-zero element $x$ in a module $M$, there exists a homomorphism of $R$-modules $\varphi:M\rightarrow E$ such that $\varphi(x)\neq0$. A minimal injective cogenerator of $R$ is an injective cogenerator which is contained in any injective cogenerator. If $R$ is Noetherian, the minimal injective cogenerator of $R$ can be expressed as the module $\bigoplus E_{R}(R/\mathfrak{m})$ where $\mathfrak{m}$ runs over all the maximal ideals of $R$.

Answer 1

I give a counter-example. Let $R$ be a semi-local domain with two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$. So the minimal injective generator module is $E = E(R/\mathfrak{m}) \oplus E(R/\mathfrak{n})$. Choose $X = \{\mathfrak{m}\}$. If your question holds true then $$\mathrm{Hom}_R(R_{\mathfrak{m}}, E) \cong \mathrm{Hom}_R(R_{\mathfrak{m}}, E(R/\mathfrak{m})) \oplus \mathrm{Hom}_R(R_{\mathfrak{m}}, E(R/\mathfrak{n}))$$ is the minimal injective generator of $R_{\mathfrak{m}}$. However the minimal injective generator of $R_{\mathfrak{m}}$ is just $E(R/\mathfrak{m})$, an $\mathfrak{m}$-torsion module (see Matsumua, commutative ring theory). Since $E(R/\mathfrak{n})$ is an $R_{\mathfrak{n}}$-module, $\mathrm{Hom}_R(R_{\mathfrak{m}}, E(R/\mathfrak{n}))$ is not $\mathfrak{m}$-torsion provided it is non-zero. The later easily follows from the short exact sequence $$0 \to R \to R_{\mathfrak{m}} \to R_{\mathfrak{m}}/R \to 0$$ and the fact $\mathrm{Hom}_R(R, E(R/\mathfrak{n})) = E(R/\mathfrak{n})$.