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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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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.

Edit
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?2012-04-22
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    Thanks . So Kaplansky's Theorem is also valid for noncommutative rings?2012-04-22
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    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...)2012-04-22
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    @Martin: yes, Kaplanski's theorem is valid for non commutative rings.2012-04-22
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    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.2012-04-25
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    Dear @Adeal, no, projective modules over $E(V)$ needn't be finitely generated even if $V$ is finite dimensional: just consider infinite dimensional free modules $E(V)^{(B)}$ (for some infinite set $B$) to get counterexamples.2012-04-25
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    Oops, of course. Thank you Georges for this nice answer, and introducing me to another useful result.2012-04-29
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    Thank you for your vigilance, @navigator23, you are absolutely right. Corrected.2012-07-07