Exterior power "commutes" with direct sum

Question

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ which holds in a canonical way : for $0 \le k \le n$, the maps $$ \Lambda^k(V) \times \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W), \\ (v_1 \wedge\cdots \wedge v_k,w_{k+1} \wedge \cdots \wedge w_n) \mapsto (v_1 \wedge \cdots \wedge v_k) \wedge (w_{k+1} \wedge \cdots \wedge w_n) $$ factors through the tensor product's universal properties, which gives us a map $\varphi_k : \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W)$, and since the category of $K$-modules has biproducts, we get a canonical map $$ \varphi = \bigoplus_{k=0}^n \varphi_k : \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \longrightarrow \Lambda^n(V \oplus W) $$

What if I replace $K$ by a (commutative ring with $1$) ring $A$? Does this map have a kernel? I know it is surjective (it suffices to "look at it", i.e generators of the codomain are obviously all hit by $\varphi$). I don't know if it helps, but I am working in integral domains. In the field case, this is obvious if $V$ and $W$ are finitely generated, since one can use dimension arguments. I wouldn't mind if the above identity held only in the finitely generated case either. If it worked in general it would be cool though.

Note that when trying to compute the inverse map with the universal property, there seems to be a problem... my guess would have been $$ (v_1 + w_1,\cdots,v_n + w_n) \mapsto \bigoplus_{k=0}^n (v_1 \wedge \cdots \wedge v_k) \otimes (w_{k+1} \wedge \cdots \wedge w_n) $$ and this map is obviously $A$-multilinear, but there is a problem with the alternating property ; I get some terms left. I'm guessing this is the wrong map... if there is one.

Answer 1

Let $A$ be a commutative ring. Then $\Lambda(-)$ is a functor from $A$-modules to graded-commutative $A$-algebras which is left adjoint to the functor which takes the degree $1$ part. Because it is left adjoint, it preserves colimits, in particular coproducts. It follows $\Lambda(M \oplus N) \cong \Lambda(M) \otimes \Lambda(N)$. Looking at the $n$th degree part, we obtain $\Lambda^n(M \oplus N) \cong \bigoplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$.

For a more direct proof, consider the left hand side as a quotient of $(M \oplus N)^{\otimes n}$ and the right hand side as a quotient of $\bigoplus_{p+q=n} M^{\otimes p} \otimes N^{\otimes q}$. We have the "binomial theorem" $(M \oplus N)^{\otimes n} \cong \bigoplus_{p+q=n} \binom{n}{p} \cdot M^{\otimes p} \otimes N^{\otimes q}$. One easily checks that the quotients agree.

Both arguments work in great generality, $A$ can be any commutative monoid object in a cocomplete linear symmetric monoidal category (this is explained, for example, in my thesis). As usual, there is no need to fiddle around with elements.

Answer 2

It is an isomorphism, where $A$ is a commutative ring, and $V$ and $W$ are $A$-modules. I think we need $V$ and $W$ to be finitely generated and projective (but I'm not sure about this; perhaps someone can opine conerning this).

See Theorem 7 here in Bergman's notes for more details. His proof includes a description of the inverse map. (In his notes, $k$ denotes a commutative ring, not necessarily a field.)

Briefly, consider an element in $\Lambda^n(V \oplus W)$ consisting of $k$ elements from $V$ and $n-k$ elements from $W$, in some order, e.g. $(v_1, w_1, w_2, v_2, w_3, w_4, \dots)$. Let $\sigma$ be the permutation that rearranges that tuple to $(v_1, \dots, v_k, w_1, \dots, w_{n-k})$. Then we map $$\Lambda^n(V \oplus W) \longrightarrow \Lambda^k V \otimes \Lambda^{n-k} W$$ $$(v_1, w_1, w_2, v_2, w_3, w_4, \dots) \longmapsto \operatorname{sgn}(\sigma)(v_1, \dots, v_k)\otimes(w_1, \dots, w_{n-k}).$$

We extend this multilinearly to get an alternating map $$\Lambda^n(V \oplus W) \longrightarrow \bigoplus_{k=0}^n (\Lambda^k V \otimes \Lambda^{n-k} W).$$