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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.

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See Thm 7 here. Proof includes a description of the inverse map. –  John M Jun 6 at 4:10
    
@JohnM : I see. So what I did wrong is I didn't care about the fact that $(w,v)$ should map to $- v \otimes w$ instead of $0$. –  Patrick Da Silva Jun 6 at 4:14
    
Yes that pretty much sums it up. –  John M Jun 6 at 4:15
    
@JohnM : Feel free to put your comment as an answer! –  Patrick Da Silva Jun 6 at 4:15

2 Answers 2

up vote 3 down vote accepted

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 $$\Lambda^n(V \oplus W) \longrightarrow \bigoplus_{k=0}^n (\Lambda^k V \otimes \Lambda^{n-k} W).$$

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I worked out the inverse map, we do not use finitely-generated nor projective one second. The inverse map just needs to be the right one, everything follows. –  Patrick Da Silva Jun 6 at 5:10
    
I think I was essentially scared of finding the right map in the general case because everything I looked at concerned vector spaces, and I feared something would go wrong while "trying to throw an exterior power in a tensor product" because of this $1/k!$ which appears when you embed the exterior power in the tensorial power in characteristic zero for fields. If I wouldn't have been scared this was probably a very natural statement. –  Patrick Da Silva Jun 6 at 5:13

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.

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Thanks for that answer! I knew that if I simply used the universal properties to "fiddle with elements" it would probably work in such generality, but it's nice to see the elements going around at least once in your life. Nonetheless, I am still not ready to throw myself into very categorical arguments, but it's always nice to hear about them. (I did understand your functor/left-adjoint/colimits thing though after picking up my category theory textbook :D )+1! –  Patrick Da Silva Jun 8 at 20:34
    
@Martin, the binomial is missing from the binomial theorem –  krey Oct 24 at 5:28
    
@krey: Thank you; it's corrected. –  Martin Brandenburg Oct 24 at 6:19

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