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Let $R$ be a commutative ring and let $M,N$ be free modules over $R$ and suppose we have a map $f: M \rightarrow N$ such that upon taking $\text{ Hom}(-,R)$ we get an isomorphism $f^* : N^* \rightarrow M^*.$ Must $f$ be an isomorphism as well?

This is clear if $M,N$ are free and of finite dimension, since then double dualization can be applied. But in general, what is true?

Edit. In the comments egreg has sketched an example which seems fine. This seems to involve cardinality issues, so what happens if I assume that $M$ and $N$ are both countably generated?

user26857
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Twistediso
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  • If you find two cardinal numbers $\alpha<\beta$ with $2^\alpha=2^\beta$, you have a counterexample (take suitable vector spaces over the rationals). I think this can be proved consistent with ZFC. – egreg Apr 25 '17 at 21:25
  • @egreg I believe your example. I edited the question to ask what happens when M and N have bases of the same cardinality. – Twistediso Apr 25 '17 at 21:33
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    If $M$ and $N$ have bases of the same cardinality, they are isomorphic. – egreg Apr 25 '17 at 21:40
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    @egreg: Yes, but why is $f$ an iso? – Twistediso Apr 25 '17 at 22:07
  • Oh, that's not necessarily true, I guess. – egreg Apr 25 '17 at 22:11
  • @Twistediso: I would suggest you accept Jeremy Rickard's answer instead of mine, as his is the correct answer that addresses your question in full generality. – Eric Wofsey Nov 26 '17 at 00:33

2 Answers2

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As pointed out by Jeremy Rickard, this answer is incorrect as originally stated and actually shows only that $f$ is an isomorphism iff $f^*$ is a topological isomorphism. His answer gives an example where $f^*$ is an isomorphism but not a topological isomorphism and so $f$ is not an isomorphism.

For any $M$, the dual module $M^*$ can be given the topology of pointwise convergence (that is, the product topology considering $M^*$ as a subset of $R^M$ where $R$ has the discrete topology). Note furthermore that for any homomorphism $f:M\to N$, the induced homomorphism $f^*:N^*\to M^*$ is continuous. Finally, I claim that $M$ is the continuous dual of $M^*$ with respect to its topology: that is, every continuous homomorphism $M^*\to R$ is evaluation at an element of $M$.

To prove this, suppose $f:M^*\to R$ is a continuous homomorphism. Since $f$ is continuous, $\ker(f)$ is an open neighborhood of $0$, which means that there are finitely many elements $x_1,\dots,x_n\in M$ such that for all $\alpha\in M^*$ with $\alpha(x_i)=0$ for $i=1,\dots,n$, $f(\alpha)=0$. Let $M_0\subseteq M$ be a finitely generated free direct summand of $M$ containing each $x_i$ (for instance, pick a basis for $M$ and let $M_0$ be spanned by all the basis elements which have nonzero coefficient in some $x_i$). Then $f$ vanishes on the kernel of the restriction map $M^*\to M_0^*$ and thus induces a homomorphism $g:M_0^*\to R$. Since $M_0$ is finitely generated, $g$ is given by evaluation at some element $x\in M_0$. It follows that $f$ is also given by evaluation at $x$.

It now follows easily that we can recover $f:M\to N$ from $f^*:N^*\to M^*$ by applying the continuous dual functor. Thus $f$ is an isomorphism of $R$-modules iff $f^*$ is an isomorphism of topological $R$-modules. However, beware that $f^*$ may be an isomorphism of $R$-modules without being an isomorphism of topological $R$-modules, since its inverse may not be continuous. We do get a positive answer in the special case that $R$ is finite, in which case $N^*$ and $M^*$ are compact Hausdorff and so a continuous bijection between them automatically has continuous inverse.

Eric Wofsey
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    Bravo and thanks, Eric, for answering this question which really baffled me. – Georges Elencwajg Apr 26 '17 at 06:35
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    Awesome answer indeed! It seems to me the argument also works if $R$ is not commutative. (The only change I see is that $M^$ and $N^$ become $R^{op}$-modules.) Am I right? – Pierre-Yves Gaillard Apr 26 '17 at 10:54
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    @Pierre-YvesGaillard: Yes, I believe that's correct. – Eric Wofsey Apr 26 '17 at 14:55
  • @Eric Wofsey: Thank you for your very elegant answer. I posted a follow-up here: https://math.stackexchange.com/questions/2253255/a-module-that-is-free-in-the-profinite-sense-and-double-dual Inspired by a similar question I had upon reading your answer and thinking whether it generalized to topological rings. Feel free to look at it if you're interested. – Twistediso Apr 26 '17 at 15:41
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    Why does applying the continuous dual functor to $f^$ give an isomorphism? Must $f^$ have a continuous inverse? – Jeremy Rickard Jul 02 '17 at 08:30
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Let $R=\mathbb{Z}_p$, the $p$-adic integers, and let $M=N$ be countably generated free modules. We can regard elements of $M$ as finite sequences of $p$-adic integers, and elements of $M^*$ as arbitrary sequences of $p$-adic integers.

Let $f$ be the map $$(a_0,a_1,a_2,\dots)\mapsto(a_0,-pa_0+a_1,-pa_1+a_2,\dots),$$ so that $f^*$ is the map $$(b_0,b_1,b_2,\dots)\mapsto(b_0-pb_1,b_1-pb_2,b_2-pb_3\dots).$$

Then $f$ is not invertible, since $(1,0,0\dots)$ is not in the image.

But $f^*$ is invertible, with inverse $$(c_0,c_1,c_2,\dots)\mapsto(c_0+pc_1+p^2c_2+\dots,\text{ } c_1+pc_2+p^2c_3+\dots,\text{ } c_2+pc_3+p^2c_4+\dots,\text{ }\dots).$$

Jeremy Rickard
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    By the way, $0\to M\stackrel{f}{\to}M\to\mathbb{Q}p\to0$ is a $\mathbb{Z}_p$-projective resolution of the $p$-adic rationals, and what I'm actually proving here is that $\text{Hom}{\mathbb{Z}p}(\mathbb{Q}_p,\mathbb{Z}_p)=0=\text{Ext}^1{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)$. – Jeremy Rickard Jul 02 '17 at 12:15