Is $\mathrm{Hom}_S(M,E) \otimes_R X \cong \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)$ with $R$ a Noetherian ring? Given a left Noetherian ring $R$, a ring $S$, a $R$-$S$-bimodule $M$, an injective cogenerator $E$ of right $S$-module, is there an natural isomorphism $$
\mathrm{Hom}_S(M,E) \otimes_R X \cong \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)
$$
for an arbitrary finitely generated $R$-module $X$?
Since $X$ is finitely generated, we can say $X=<e_1,\cdots,e_n>$. Define a map
$$
\sigma: \mathrm{Hom}_S(M,E) \otimes_R X \to \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)
$$
by
$$
\sigma(\sum f_i \otimes x_i)(g)= \sum f_i(g(x_i)), \text{for } g \in \mathrm{Hom}_R(X,M).
$$
However, I am stuck at proving '$\sigma$ is injective and surjective'.
I think the key is the condition 'R is Noetherian' and '$E$ is an injective cogenerator', but how to use it? Is there an easy way to prove? Thanks in advance.
 A: This is true under weaker assumptions: we only need that $X$ is finitely presented (then we can omit the Noetherian condition) and that $E$ is injective.
Let's fix $M$ and $E$. We can define a map $\sigma_X$ (depending on $X$) as you did: \begin{align*}
\sigma_X: \mathrm{Hom}_S(M,E) \otimes_R X & \longrightarrow \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E) \\
f \otimes x & \longmapsto \big( g \mapsto f(g(x)) \big).
\end{align*}
A key observation is that this map is natural in $X$, so if we have a $R$-linear map $f:X \to X'$, we get a commutative diagram
$$\require{AMScd} \begin{CD} \mathrm{Hom}_S(M,E) \otimes_R X @>>> \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)\\ @VVV @VVV\\ \mathrm{Hom}_S(M,E) \otimes_R X' @>>> \mathrm{Hom}_S(\mathrm{Hom}_R(X',M),E) \end{CD}$$
I'll leave the proof as an exercise.
Now we claim that $\sigma_X$ is an isomorphism when $R$ is free of finite rank, say $X\cong R^n$. In this case, we have an isomorphism $$\mathrm{Hom}_S(M,E) \otimes_R X \cong \mathrm{Hom}_S(M,E)^n$$ sending $f \otimes (x_1, \dots, x_n)$ to $(fx_1, \dots, fx_n)$.
Similarly, we have an isomorphism $$\mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)\cong\mathrm{Hom}_S(M,E)^n$$ sending $\varphi$ to $(\varphi \circ f_1, \dots, \varphi \circ f_n)$, where $f_i:M \to \mathrm{Hom}(X,M)$ sends $m$ to the map that sends the $i$-th basis vector to $m$ and all others to zero. Now one can check that the triangle involving these two arrows and $\sigma_X$ commutes, so $\sigma_X$ is an isomorphism if $X$ is free of finite rank.
Now assume that $X$ is finitely presented. Choose a finite presentation $R^m\to R^n \to X \to 0$. From the naturality of $\sigma$, we obtain a commutative diagram
$$\require{AMScd} \begin{CD} \mathrm{Hom}_S(M,E) \otimes_R R^m @>>> \mathrm{Hom}_S(M,E)\otimes_R R^n @>>> \mathrm{Hom}_S(M,E) \otimes_R X @>>> 0 @>>> 0\\ @V\sigma_{R^m}VV  @V\sigma_{R^n}VV @V\sigma_XVV\\ \mathrm{Hom}_S(\mathrm{Hom}_R(R^m,M),E) @>>> \mathrm{Hom}_S(\mathrm{Hom}_R(R^n,M),E) @>>> \mathrm{Hom}_S(\mathrm{Hom}_R(X,M),E)@>>> 0 @>>> 0\end{CD}$$
Note that the rows are exact using standard properties of $\mathrm{Hom}$, $\otimes_R$ and the injectivity of $E$. Because $\sigma_{R^m}$ and $\sigma_{R^n}$ are isomorphisms from what we have done before, we conclude that $\sigma_X$ is an isomorphism too (by the five lemma).
Note that this is a general strategy for proving statements about finitely presented modules: first prove it for finite free modules, then reduce to that case.
