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$\DeclareMathOperator\Hom{Hom}$Let $M$ be a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$. Denote $(\_)^*:=\Hom_R(\_,R)$. There is a natural map \begin{align} \Phi_M: M \otimes_R M^* & \longrightarrow \Hom_R(M^*,M^*) \\ x \otimes y & \longmapsto (f\mapsto f(x)y). \end{align}

My question is:

If there exists an isomorphism $M\otimes_RM^*\cong \Hom_R(M^*,M^*)$, then is it true that $\Phi_M$ is an isomorphism?

My reason for asking this question:

It is well known that if $M$ is a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$ such that $M\cong M^{**}$, then the natural evaluation map \begin{align} M & \longrightarrow M^{**} \\ x & \longmapsto (f \mapsto f(x)) \end{align} is an isomorphism (see the book Gorenstein dimensions, Proposition (1.1.9)). The map defined in my question is kind of similar, so I wonder if that is true as well.

My thoughts:

If we have isomorphic finitely generated modules $X \cong Y$ and a linear surjection $f:X\to Y$, then $f$ is an isomorphism. Indeed, let $g:Y\to X$ be an isomorphism. Then, $g\circ f: X\to X$ is a surjection, hence it is an isomorphism (by Nakayama), so $f$ is injective, and we are done. Hence, under the hypothesis of my question, it would be enough to show $\Phi_M$ is surjective….

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3 Answers 3

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Here is a counterexample to the question, with steps left to the reader to fill in. Fixing a prime $p$, let $R = \mathbf Z/p^2\mathbf Z$ and $M = R/pR$ viewed as an $R$-module in the natural way. Note $M \cong \mathbf Z/p\mathbf Z$ as an $R$-module, where $R$ acts by multiplication on $\mathbf Z/p\mathbf Z$ in the obvious way. This fits the OP's conditions: $R$ is a Noetherian local ring and $M$ is a finitely generated $R$-module. (Note $R$ is not an integral domain and $M$ is not a free $R$-module.)

Step 1: Show $M^* \cong M$ as $R$-modules.

Step 2: Show $M \otimes_R M$ and ${\rm Hom}_R(M,M)$ as both isomorphic to $M$ as $R$-modules, so Step 1 tells us the domain and codomain of the $R$-linear map $\Phi_M$ are isomorphic to $M$ as $R$-modules and thus they are isomorphic to each other.

Step 3: Show $\Phi_M$ vanishes on all elementary tensors, so $\Phi_M$ is identically zero. In particular, $\Phi_M$ is not surjective, which was identified by the OP as a possible stumbling block. I don't think a counterexample could be much simpler than this one, since all $R$-modules here have prime order.

By the way, you can check the natural mapping $M \to M^{**}$ is an $R$-module isomorphism.

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This is not a complete answer, but it's too big for a comment. Surjectivity of $\Phi_M$ in general might be too much to hope for here, but did you know the following result?

Theorem: Suppose that $(R,\mathfrak{m})$ is a local Noetherian integral domain, and let $M$ be a finitely generated $R$-module. Then the map $$\Psi_M\colon M\otimes_RM^*\rightarrow \operatorname{Hom}_R(M,M)$$ defined by $\Psi(m\otimes f)(m')=m\cdot f(m')$ is an isomorphism if and only if $M$ is free.

There is an old paper of Auslander where he discusses this map (in particular Proposition 3.3), that you might find interesting: Auslander, "Modules Over Unramified Regular Local Rings" Illinois Journal of Mathematics (1961).

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  • $\begingroup$ The theorem you mention is interesting ... I will think about it to see if I can come up with a proof without seeing any reference ... $\endgroup$
    – strat
    Oct 2, 2021 at 8:54
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It should be noted that the answer is yes if $R$ is normal and $M$ is torsion-free. That is because of the:

Fact: if a map $f:A \to B$ of reflexive modules is locally an isomorphism in codimension one, then it is an isomorphism.

Since $M$ is torsion-free and $R$ is normal, locally at a codimension one prime $P$ the map $\Phi_P$ is an isomorphism because $M_P$ is free. The RHS is reflexive, so if the LHS is isomorphic to RHS, then both are reflexive, and one can use the Fact above to conclude.

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  • $\begingroup$ Yes ... if two modules satisfy $(S_2)$, then given any morphism between them, it is an Isomorphism if it is locally Isomorphism in co-dimension $1$ ... but that sort of takes the fun away if one considers one-dimensional rings ... $\endgroup$
    – strat
    Oct 2, 2021 at 8:46

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