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The exterior algebra $\Lambda^*_kM$ can be defined for a $k$-module $M$, where $k$ is a commutative ring. A number of sources mention, without condition or proof, a (canonical) isomorphism $$(\Lambda^*_kM)^\vee\cong\Lambda^*_k(M^\vee),$$ where $M^\vee:=\operatorname{Hom}_k(M,k)$ is the dual of $M$. Any proofs I can find, however, are for $M$ a finitely-generated vector space and $k$ a field, with no discussion of other cases. Which, if any, of these conditions are needed for the isomorphism? Several proofs construct a homomorphism without these conditions, but argue in terms of a finite basis to show isomorphy.

Examples of the result stated without conditions or proof: in their 1961 paper Differential forms on regular affine algebras, Hochschild–Konstant–Rosenberg state that "dual of exterior algebra $\simeq$ exterior algebra over dual". Similarly stated in Fulton–Harris' Representation Theory. They work over a field, but make no mention of finite-generation.

Examples of proofs/proof sketches over a finitely-generated vector space over a a field: Stack Exchange questions 1, 2, 3, this Math Overflow question, Conrad's review.

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    $\begingroup$ I think that the linked MO question says that this also holds for finite projective $k$-modules for any commutative ring $k$. $\endgroup$
    – Z. M
    Commented Sep 24 at 14:27
  • $\begingroup$ I think the field case is ok, you can work with a Hamel basis. $\endgroup$
    – Antonius
    Commented Sep 24 at 14:29
  • $\begingroup$ When you refer to several proofs constructing a homomorphism, do they all go the same way? Which way? (I assume right to left, i.e., $\Lambda_k^*(M^\vee) \to (\Lambda_k^*(M))^\vee$.) $\endgroup$
    – LSpice
    Commented Sep 24 at 14:56
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    $\begingroup$ You certainly need finite generation, even in the field case. Or at least, if $M$ is graded, finite generation in each degree. The canonical map $M^\vee\otimes M^\vee\to (M\otimes M)^\vee$ will not be surjective otherwise. $\endgroup$
    – Tom Goodwillie
    Commented Sep 24 at 16:15
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    $\begingroup$ @LSpice the map is usually constructed as a perfect pairing $(\Lambda^*_kM)\times(\Lambda^*_kM^\vee)$. See Theorem 3.1 in Conrad's virtualmath1.stanford.edu/~conrad/diffgeomPage/handouts/… $\endgroup$
    – Cameron
    Commented Sep 25 at 8:24

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