Infinite direct product of semisimple rings is semisimple?

Question

It's clear that the finite direct product of semisimple rings in again semisimple because in the finite case the direct producto looks isomorphic to the direct sum, but I want to know if there is a problem considering an infinite direct product.

Answer 1

There are tons of ways to see this directly.

Given $R=\prod_{i\in \kappa} S_i$ where each $S_i$ is semisimple ($\kappa$ is an infinite well-ordered index set), then $I_i=\oplus \{S_j\mid j \leq i\}$ is obviously an infinite, strictly increasing chain of right ideals of $R$. So the ring is not right Noetherian.

You can also build a strictly decreasing chain of right ideals in the a similar way, showing it is not right Artinian.

Finally, you can show that $\bigoplus_{i\in \kappa}S_i$ is a proper essential right ideal of $R$. (In semisimple rings, the only essential right ideal is the trivial ideal consisting of the entire ring. Essential right ideals cannot split out of the ring.)

but I still don't see why a semisimple ring will be noetherian.

If $R$ is a semisimple ring, then every right ideal splits out of $R$, and is equivalent to saying that it is generated by a single idempotent. Thus every right ideal is finitely generated (in fact, every right ideal is cyclic.)

Alternatively, you could cite the Hopkins-Levitzki theorem that right Artinian rings are right Noetherian.

Alternatively, you could note that after you have written $R$ as (necessarily) finite direct sum of simple right ideals, this makes a composition series for the ring as a module over itself, and we all know that modules with composition series are Artinian and Noetherian.

Answer 2

Semisimple rings are Noetherian. A field is trivially semisimple. An infinite direct product of fields is not Noetherian.

Answer 3

An infinite direct product of nonzero rings is never semisimple. Indeed, if $R$ is semisimple, every left ideal in $R$ is principal, since it is a direct summand of $R$. But if $R=\prod R_i$ is an infinite product of nonzero rings, the ideal of elements of $R$ which are zero on all but finitely many coordinates is not finitely generated (since if you had finitely many generators, every element of the ideal they generate would be zero on all but a fixed finite set of generators).

Answer 4

It's interesting to see why an infinite direct product of simple modules does not build a semisimple ring. Let $R$ be the direct product of infinitely many copies of $Z/3$, or any other field for that matter. Verify that $R$ is a ring with $1$, and the simple $R$ modules are the components, the various copies of $Z/3$. At first it seems like $R$ is semisimple. The submodules that come to mind are the direct products of some, but not all, of the components. If $V$ is the direct product of the odd numbered components, then it has a summand, $W$, which is the product of the even numbered components. However, there is a submodule that you might not think of right away. Let $V$ be the direct sum of the component rings. Let $W$ be the disjoint summand, so that $V*W = R$. Let $x$ be a nonzero member of $W$, with a nonzero value in the $j$th component. Multiply by $1_j$, to show that the $j$th copy of $Z/3$ belongs to $W$. This simple module also belongs to $V$, hence they are not disjoint after all. There is no summand, and $R$ is not a semisimple module; not a semisimple ring.