0
$\begingroup$

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

1

(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.

  • 0
    @navigetor23: you «corrected» a typo from `$s_0\in S$;` to `$s_0\in S;$` which is wrong :)2012-06-27
  • 0
    My point is that those aesthethic reasons are wrong. The font to be used for the punctuation belonging to the sentence surrounding a formula has to be in the font used for the text. This matters when the fonts used for math and for text are different. Moreover, the spacing in the two variants is quite different. This is discussed in many places. Of course, here it is pretty much irrelevant.2012-06-28
  • 0
    Well, I am working under the hope that we will not have this conversation ever again, because you will not make the same mistake :) So think of this as an investment!2012-06-28
  • 0
    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
0

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.