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For a vector space $V$, is the grassman algebra $\bigwedge(V)$ always an injective module over itself? Is there a proof, or even just a brief explanation?

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up vote 2 down vote accepted

If $V$ is finite dimensional, then yes, owing to the existence of a special bilinear form from the algebra into the basefield. If $V$ is $n$ dimensional, the form is given by f(a,b)=coefficient of the grade n part of a*b. You can find this in a paper entitled Annihilators of Principal Ideals in the Exterior Algebra by Koc and Esin.

Algebras with such a functional are Frobenius algebras, which are self-injective on both sides.

I don't immediately know the answer if $V$ is infinite dimensional, but I'll try to think.

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Thanks rschwieb, that was helpful. –  Cenk Apr 26 '12 at 1:46
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