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.
[Math] simple ring which is not semisimple
linear algebranoncommutative-algebra
Related Solutions
The Weyl algebra $A_1 := k\{x, y: xy - yx = 1\}$ is simple when char $k = 0$ but is not Artinian: the left ideals $A_1x^i$ form an infinite decreasing sequence of left ideals which is never constant.
The Artin-Wedderburn theorem tells us that the maximal semisimple quotient is a product of matrix rings over finite division rings, one for each irreducible representation. Furthermore, every finite division ring is a field, and the unit group of any finite field is cyclic. The only nontrivial homomorphism from $S_3$ to a cyclic group is the sign homomorphism $S_3\to\mathbb{Z}/2$. It follows that any homomorphism from $\mathbb{Z}S_3$ to a finite field lands in the prime subfield (since elements of $S_3$ can only map to $\pm 1$).
So, writing $\mathbb{F}$ for either $\mathbb{F}_2$ or $\mathbb{F}_3$, the maximal semisimple quotient of $\mathbb{F}S_3$ is a product of matrix rings $M_n(K)$ for finite extensions $K$ of $\mathbb{F}$, one for each irreducible representation, and in all the cases where $n=1$ the $K$ is just $\mathbb{F}$. The only $1$-dimensional representations are the trivial representation and the sign representation, and the sign representation is the same as the trivial representation in the case $\mathbb{F}=\mathbb{F}_2$.
For $\mathbb{F}=\mathbb{F}_3$, dimension-counting now tells us there can be no more irreducible representations: the two $1$-dimensional representations take up $2$ dimensions of the semisimple quotient, and the Jacobson radical is nontrivial since it contains $\sum_{g\in S_3} g$, so there are at most $3$ dimensions left. Another irreducible representation would give a copy of $M_n(\mathbb{F}_{3^d})$ in the semisimple quotient for some $d$ and some $n>1$, which is impossible since there aren't enough dimensions left. We conclude that the two $1$-dimensional representations are the only irreducible representations for $\mathbb{F}=\mathbb{F}_3$, and so the maximal semisimple quotient is $\mathbb{F}_3\times\mathbb{F}_3$. The Jacobson radical is then the kernel of the map $\mathbb{F}_3S_3\to\mathbb{F}_3\times\mathbb{F}_3$; explicitly, it is the set of elements $\sum_{g\in S_3} a_g g$ such that $\sum a_g=0$ and $\sum a_g \sigma(g)=0$, where $\sigma(g)$ is the sign of $g$.
Over $\mathbb{F}_2$, on the other hand, there are up to $4$ dimensions left after accounting for the single $1$-dimensional representation and the fact that the Jacobson radical is nontrivial, so there might be a $2$-dimensional irreducible representation. To find one, note that there is a permutation representation of $S_3$ on $\mathbb{F}_2^3$, and this splits as a direct sum of a trivial subrepresentation (generated by $(1,1,1)$) and a $2$-dimensional subrepresentation (consisting of $(a,b,c)$ such that $a+b+c=0$). (Note that this splitting of the permutation representation doesn't happen over $\mathbb{F}_3$, since $(1,1,1)$ is contained in the latter $2$-dimensional subrepresentation.) This $2$-dimensional representation can easily be verified to be irreducible (for another way of seeing it, note that $\mathbb{F}_2^2\setminus\{0\}$ has three elements, and every permutation of them gives a linear map, so in fact $GL_2(\mathbb{F}_2)\cong S_3$).
So over $\mathbb{F}_2$, we conclude that there is the trivial representation and also this $2$-dimensional irreducible representation; counting dimensions, we now see that we have accounted for all $6$ dimensions of $\mathbb{F}_2S_3$. We conclude that the Jacobson radical is only $1$-dimensional (generated by $\sum_{g\in S_3} g$), and the quotient is $\mathbb{F}_2\times M_2(\mathbb{F}_2)$.
Best Answer
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:
The ring $R = \operatorname{End}_k(V)$ does not satisfy IBN. (This is probably everyone's favorite example of a ring not satisfying IBN.)
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.
A left Noetherian ring satisfies IBN (and in fact something much stronger: the "strong rank condition".)
(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.