# On the map $\Phi_M: M\otimes_RM^*\xrightarrow{x\otimes y\mapsto \left\{f\mapsto f(x)y\right\}}\text{Hom}_R(M^*,M^*)$

$$\DeclareMathOperator\Hom{Hom}$$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….

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.

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).

• 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 ... Oct 2, 2021 at 8:54

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.

• 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 ... Oct 2, 2021 at 8:46