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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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    Thank you for your vigilance, @navigator23, you are absolutely right. Corrected.2012-07-07