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I'm stuck with this algebra question.

I try to prove that the exterior algebra $R$ over $k^d$, that is, the $k$-algebra that is generated by $x_1,\ldots,x_d$ and $x_ix_j=- x_jx_i$ for each $i,j$, has just one simple module which is not faithful.

I think the only simple module is $k$, but I am not really sure if my idea does work or not.

Can I use the fact $(x_i)^2=0$ for all $i$, then $k$ has cyclic subrings? If yes, then HOW?

Also, one more question: if $k$ is finitely generated, is that enough to say that $R$ is Artinian? Thank you

2 Answers 2