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Let $M$ be a finitely generated module over a Noetherian ring $R$ such that $M$ is isomorphic with its double dual $M^{**}=Hom(Hom(M,R),R)$.

Then is the natural map $j:M \to M^{**}$ defined as $j(m)(f)=f(m),\forall m\in M, \forall f\in M^*$, an isomorphism ?

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    $\begingroup$ mathoverflow.net/questions/76000/… $\endgroup$
    – Qiaochu Yuan
    Oct 1, 2018 at 5:21
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    $\begingroup$ This is easy when $M^{**}$ is reflexive (e.g. if $R$ is a product of domains [Tag 0AV3]). Indeed, if $\phi \colon M \stackrel\sim\to M^{**}$ is any isomorphism, we get a commutative diagram $$\begin{array}{ccc}M & \stackrel{\operatorname{ev}}\to & M^{**}\\ \downarrow & & \downarrow \\ M^{**} & \stackrel{\operatorname{ev}}\to & M^{****},\!\end{array}$$ where the vertical arrows $\phi$ and $\phi^{**}$ are isomorphisms. Since the bottom arrow is also an isomorphism, we conclude that the top is as well. $\endgroup$
    – R. van Dobben de Bruyn
    Oct 1, 2018 at 5:42
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    $\begingroup$ Let me copy the relevant comment to Qiaochu's linked post: "For finitely generated modules over a Noetherian ring, no such examples exist. A student of Huneke proved this around 2004, but I don't think he ever published it. (It's possible it was already known at that time, but I never found a reference.) – Graham Leuschke Sep 21 '11 at 0:23" $\endgroup$
    – YCor
    Oct 1, 2018 at 6:16

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I asked this question (or at least a similar one) here: Characterisation of reflexive modules By an answer of Jeremy Rickard it is even true for noetherian rings that are not necessarily commutative. It seems to be an open question for finitely generated modules over general (not necessarily noetherian) rings.

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The answer is "yes" as pointed out by YCor and Graham Leuschke, and there is even a published proof prior to 2004: L.W. Christensen, Gorenstein Dimensions, Springer LNM, 2000. (1.1.9.b)

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