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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

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(I will be assuming $k$ is a field; if it is not, then you will need some hypothesis on it for your statement to be true)

Suppose $R$ is a ring and that $S$ is a non-zero simple left $R$-module. Pick a non-zero element $s_0\in S;$ then the map $\phi:r\in R\mapsto sr_0\in S$ is a surjective map of left $R$-modules. Let $I\subseteq R$ be its kernel, a left ideal.

Now suppose $t\in R$ is an homogeneous element of positive degree. It is easy to see that the subset $tS=\{ts:s\in S\}$ is a submodule of $S$. Since $S$ is simple, then either $tS=0$, and in that case $t\in T$, or $tS=S$.

I want to check that the second cannot occur. Suppose otherwise: then there exists an $s\in S$ such that $ts=s_0$ and, since $s_0$ generates $S$, there also exists an $r\in R$ such that $s=rs_0$; it follows that $s_0=rts_0$ or, in other words, that $(1-rt)s_0=0$. Now it turns out that the element $1-rt$ is invertible in $R$ (since $t$ has positive degree, $rt$ is a sum of homogeneous elements of positive degree, and then $(rt)^d=0$: it follows that $\sum_{n\geq0}(rt)^n$ is a finite sum and the usual argument shows that it is the inverse of $1-rt$), so we see that $s_0=0$, which is absurd.

We thus conclude that every homogeneous element of positive degree in $R$ is in fact contained in $I$. Now we observe that there is exactly one left ideal in $R$ which contains all the homogeneous elements of positive degree, so there is in fact, as we wanted, exactly one isomorphism class of simple modules.

NB: if you know a little more about ring theory---specifically, of Artinian algebras---we can recast this argument as follows: let $R$ be your exterior algebra on $d$ generators, and let $I$ be the ideal generated by all homogeneous elements of positive degree. It is pretty obvious that $I^d=0$, so that $I$ is a nilpotent ideal; on the other hand, it is also clear that $R/I$ is isomorphic to $k$ as a ring, so that it is in particular semisimple. It follows from a well-known characterization of the Jacobson radical that $I$ is the Jacobson radical of $R$. As a consequence, there is a bijection between the isoclasses of simple $R$-modules and the isoclasses of simple $R/I$-modules, and there is only one of the latter.

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    This time, I will forgo your suggestion: good typography is something I enjoy very much and, in my experience, it is quite not at contraposition with math. I have absolutely no problem with my posts being edited —typing on my phone leads systematically to innumerable typos, sadly, and just as I am particular about the font used for punctuation, I appreciate the love of detail in others which results in them sometimes being helpfully corrected. By the way, I do not understand what is absurd about this conversation, probably because I have not spent on it more that a couple of minutes total!2012-06-28
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Assuming that $k$ is a field for the same reasons Mariano gave, the short answer is:

The exterior algebra is local, and so any simple module over it looks like $R/M$ where $M$ is the unique maximal right ideal.

This is detailed in a similar question here.

This isoclass of module is clearly not faithful since $M$ annihilates it.