The exterior algebra $\Lambda^*_kM$ can be definedcan 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:=\text{Hom}_k(M,k)$$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 $V$ over$k$ a field $\mathbb{k}$, 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 algebrasDifferential forms on regular affine algebras, Hochschild-Konstant-RosenbergHochschild–Konstant–Rosenberg state that "dual of exterior algebra $\simeq$ exterior algebra over dual". Similarly stated in Fulton-Harris'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 11, 22, 33, this Math Overflow questionMath Overflow question, Konrad's reviewConrad's review.
