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.


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.

  • 1
    $\begingroup$ Why is $E(V)$ local? $\endgroup$ – Martin Brandenburg Apr 22 '12 at 8:14
  • 1
    $\begingroup$ Thanks . So Kaplansky's Theorem is also valid for noncommutative rings? $\endgroup$ – Martin Brandenburg Apr 22 '12 at 8:40
  • 1
    $\begingroup$ 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...) $\endgroup$ – Georges Elencwajg Apr 22 '12 at 8:42
  • $\begingroup$ @Martin: yes, Kaplanski's theorem is valid for non commutative rings. $\endgroup$ – Georges Elencwajg Apr 22 '12 at 9:18
  • $\begingroup$ 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. $\endgroup$ – Adelaide Dokras Apr 25 '12 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.