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Let $V$ be a vector space of countably infinite dimension over a field $k$ and put $R = \text{End}_k(V)$. Then it is not hard to see that $R$ has a unique proper nonzero two-sided ideal $I$, which consists of those operators with finite rank. So in particular $R/I$ is a simple ring, but I'm pretty sure it's not semisimple. How can I show this? It would be enough to prove that $R/I$ is not artinian.

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2 Answers 2

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There are a lot of ways to see this. In $\S 1.9.2$ of my noncommutative algebra notes, I discuss the Invariant Basis Number property of a ring $R$, namely that the rank of a finitely generated free (say left) $R$-module is a well-defined invariant. I show the following:

1) The ring $R = \operatorname{End}_k(V)$ does not satisfy IBN. (This is probably everyone's favorite example of a ring not satisfying IBN.)

2) If $R$ is a ring not satisfying IBN and $f: R \rightarrow S$ is any (unital!) ring homomorphism, then $S$ also does not satisfy IBN. [I found this very counterintuitive and had to read it several times to believe it was stated correctly.] Therefore your simple quotient ring $R/I$ does not satisfy IBN.

3) A left Noetherian ring satisfies IBN (and in fact something much stronger: the "strong rank condition".)

4) (Akizuki-Hopkins) A left Artinian ring is left Noetherian.

Putting these results together gives you what you want. There is some fairly substantial overkill here, but perhaps it is interesting overkill. For instance, you can avoid Step 4: it is just as easy to see that a semisimple ring must be Noetherian as it is to see that it must be Artinian (Corollary 32 of my notes), so Akizuki-Hopkins need not be invoked.

I hasten to add though that I got all of this material from the early pages of T.Y. Lam's A First Course in Noncommutative Rings. He proves the result you want several times over, and some of the ways are more direct. As with every book written by T.Y. Lam that I have read, it comes with my highest recommendation.

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I just found a copy of Lam's book, and it is excellent. I had been using Dennis and Farb's Noncommutative Algebra, which is also good, but Lam covers the basic material in greater detail. In particular, he gives a direct and elementary proof of the result I asked about. Thanks for the recommendation! –  Justin Campbell Jul 22 '11 at 4:24

Suppose $\{e_n:n\geq1\}$ is a basis of $V$. Let $p_1$, $p_2$, $\dots$ be an enumeration of the prime numbers, and for each $m\geq1$ let $S_m$ be the subset of $\mathbb N$ of those natural numbers whose prime factors are among $p_1$, $\dots$, $p_m$. Then $S_1\subsetneq S_2\subsetneq S_3\subsetneq\cdots$ is a increasing sequence of sets each of which has infinitely many more elements than the one preceding it. For each $m\geq1$ let $W_m$ be the span in $V$ of $\{e_i:i\in S_m\}$, and let $J_m\subseteq R/I$ be the image in $R/I$ of the elements of $R$ whose image is contained in $W_m$. We have an increasing sequence of right ideals $J_1\subseteq J_2\subseteq J_3\subseteq\cdots$ Can you go on?

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