Prove that map between modules is isomorphic.

Question

Let $R$ be a ring and $M$ and $R$-module. Let $M$ be finitely presented, i.e it is finitely generated with the surjective function $p: R^n \to M$ and $\ker(p)$ is also finitely generated.

Consider the right $R$-module $\text{Hom}_R(M,R)$ and for an $R$-module $N$ consider the map $$a:\text{Hom}_R(M,R) \otimes_R N \to \text{Hom}_R(M,N)$$ given by $f \otimes n \mapsto (m \mapsto f(m)n)$.

How can I prove that $a$ is an isomorphism if in additional to $M$ being finitely presented, I also know that it is projective?

Answer 1

Recall that a module is finitely generated projective iff it is finitely presented and flat. Thus here, projectivity of $M$ suffices.

As $M$ is projective, $R^{n}=M \oplus ker(p)$. Now the isomorphism stated above by you holds in the case where $M$ is replaced by $R^{n}$. Restrict this isomorphism to $M$.

Note - I am typing this on a phone, so could not write all the details. If you have any queries, please wait till I get home :)

Answer 2

If $M$ is finitely generated and projective, then $\operatorname{Hom}_R(M,R)$ is also projective. Indeed, if $R^n\to M$ is a finite presentation of $M$, then it splits by projectivity, so we have a direct sum decomposition $R^n\cong M\oplus N$ for some $N$. Applying the functor $\operatorname{Hom}_R(-,R)$ to this, we get a dual direct sum decomposition $R^n\cong \operatorname{Hom}_R(M,R)\oplus \operatorname{Hom}_R(N,R)$ of right $R$-modules. It follows that $\operatorname{Hom}_R(M,R)$ is projective (finite generation is needed here because if $n$ were infinite, the dual of $R^n$ would be an infinite product of copies of $R$, which need not be projective).

In particular, this means $\operatorname{Hom}_R(M,R)$ is flat, so the functor $C(N)=\operatorname{Hom}_R(M,R)\otimes_R N$ is exact. Since $M$ is projective, the functor $D(N)=\operatorname{Hom}_R(M,N)$ is also exact. You have described a natural transformation $a:C\to D$. It is straightforward to check that $a_N:C(N)\to D(N)$ is an isomorphism if $N$ is free. Now for general $N$, choose a presentation $F_1\to F_0\to N\to 0$, where $F_1$ and $F_0$ are free. Consider the diagram $$\require{AMScd} \begin{CD} C(F_1) @>>> C(F_0) @>>> C(N) @>>> 0\\ @VV{a_{F_1}}V @VV{a_{F_0}}V @VV{a_N}V \\ D(F_1) @>>> D(F_0) @>>> D(N) @>>> 0 \end{CD}$$

Since $C$ and $D$ are exact functors, the rows of this diagram are exact. Since $a_{F_1}$ and $a_{F_0}$ are isomorphisms, $a_N$ is an isomorphism by the five lemma.

More generally, this argument shows that if a natural transformation between right exact functors is an isomorphism on free modules, it is an isomorphism on all modules.