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Let $E(V)$ be the exterior algebra of a vector space $V$ (I've also seen this denoted $\Lambda(V)$).Is it true that any projective $E(V)$-module is necessarily free? If it's any easier, is it at least true if we assume $V$ has finite dimension?

This popped into my head for some reason while I was experimenting with projective complexes. I couldn't tell either way, but I hope it's not embarrassingly simple. Thanks.

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up vote 8 down vote accepted

Yes, if $V$ is a vector space, every projective $E(V)$-(right) module is free, because $E(V)$ is a local ring and (right) projective modules over a local ring are free according to a theorem of Kaplansky.

Since Martin asks, here is the reason why $E(V)$ is local.

Consider the vector subspace $\mathfrak m=\wedge ^1V\oplus \wedge ^2V\oplus...\subset E(V)$
It is a two-sided ideal but also the unique maximal right ideal of $E(V)$.
Indeed every $x\in E(V)\setminus \mathfrak m$ is invertible since it can be written as $x=q+m$ with $q\in k^*$ and $m\in \mathfrak m $, and every element of $\mathfrak m$ is nilpotent .
Since $E(V)$ has a unique maximal right ideal, it is local by definition.

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Why is $E(V)$ local? –  Martin Brandenburg Apr 22 '12 at 8:14
Thanks . So Kaplansky's Theorem is also valid for noncommutative rings? –  Martin Brandenburg Apr 22 '12 at 8:40
I have added an edit to explain in detail why $E(V)$ is local (I sympathize with Martin's question since I'm not familiar at all with non commutative rings and I prefer to spell things out when venturing in that mysterious territory...) –  Georges Elencwajg Apr 22 '12 at 8:42
@Martin: yes, Kaplanski's theorem is valid for non commutative rings. –  Georges Elencwajg Apr 22 '12 at 9:18
Thank you Georges, I appreciate it! I know Kaplanski's theorem is actually considered quite difficult. However, it's not hard to show that finitely generated projective modules over local rings are free. If I assume $V$ is finite dimensional, so that $E(V)$ is finite dimensional, am I being naive, or would it follow immediately that every projective $E(V)$ module is finitely generated as well? I'm not opposed to accepting Kaplanski's theorem right now without understanding it, I'm just curious if there's a less advanced argument. –  Adelaide Dokras Apr 25 '12 at 3:28

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