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?
Is $\bigwedge(V)$ self-injective?
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$\begingroup$
abstract-algebra
ring-theory
modules
exterior-algebra
1 Answers
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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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0Thanks rschwieb, that was helpful. – 2012-04-26