# nonsemisimple $k$-algebra

Say $k$ is a field and is the $k$-algebra $A:=\prod_{i\in \mathbb N} k$ (multiplication is defined componentwise) semisimple? If not, what would be a submodule of the regular representation , that is not a direct summand?

As an application I wonder whether the semisimple quotient of the group algebra of the general linear group that only has the rational representations is just the product of the corresponding matrix algebras.

Edit note: after noticing the problems with my question through the comments, I revised it. Sorry for the late reply.

• Your algebra $A$ is not unital, so Artin-Wedderburn can't be applied directly to it. You need to adjoin a unit at least. – Qiaochu Yuan Apr 30 '14 at 14:30
• Usually Artin-Wedderburn is stated as a theorem about artinian rings, and this ring isn't artinian. Maybe it helps if you state exactly the version of A-W you are thinking of. – Matthew Towers Apr 30 '14 at 14:31
• @mt_: Artin-Wedderburn is a classification theorem for semisimple algebras, and this algebra (after adjoining a unit) doesn't fall in that classification, hence it's not semisimple. What is the issue? (Are you using semisimple to mean semiprimitive?) – Qiaochu Yuan Apr 30 '14 at 14:39
• @Peter: please do not edit questions in such a way that existing answers become irrelevant. Also, arbitrary rings don't have semisimple quotients (e.g. your algebra $A$), only Artinian rings. Arbitrary rings have semiprimitive quotients (and $A$ is already semiprimitive). – Qiaochu Yuan Apr 30 '14 at 17:53

## 3 Answers

Unfortunately, you have changed the question to an extent that makes the existing answers irrelevant (a new question would have been better). I am answering the question whether the algebra $A=\prod_{\mathbb{N}}k$ is semi-simple.

The answer is no: take the submodule $U=\bigoplus_{\mathbb{N}}k$ of the regular module. I claim that any non-zero submodule of the regular module intersects $U$ non-trivially, which will imply that $U$ is not a direct summand. Indeed, if $V\neq 0$ is any submodule, let $\underline{v}\in V$ have non-zero entry in $i$-th position. Then $a_i\cdot \underline v\in U\cap V$, where $a_i$ has a $1$ in the $i$-th entry, and $0$s elsewhere.

• Nice answer to the new question. And I agree that Peter Patzt should have written a new question. I have to edit my answer now just to avoid it looking wrong. – Seth Apr 30 '14 at 17:09
• One can also proceed here by classifying all of the idempotents in $A$; these are precisely the sequences of $0$s and $1$s, and hence all direct summands have the form "sequences whose support lies in some fixed set $S$." The submodule $U$ does not have this form (although it is an increasing union of submodules with this form). – Qiaochu Yuan Apr 30 '14 at 17:54
• @QiaochuYuan: Nice! – Alex B. Apr 30 '14 at 20:34

A bit tangential, but there's a really cool old theorem of Barbara Osofsky about the global dimension of $\prod_{\mathbb{N}}k$ (somewhat relevant to this question because global dimension zero is equivalent to semisimple).

She proved that the global dimension is at least two, with equality if and only if the Continuum Hypothesis is true!

EDIT: This was an answer to the original question posed by Peter Pantzt, which has since been changed. The question was: "Let $A=\bigoplus_{i\in\mathbb{N}}k$. By Wedderburn $A$ is not semisimple. Give an example of a submodule which is not a direct summand." (note that the question was "incorrect" as I explain below).

ORIGINAL ANSWER:

Note that while $A$ does not have a unit, $A$ viewed a regular $A$-module is a direct sum of irreducible submodules. Each copy of $k$ in the sum is an irreducible submodule. This implies every submodule is a direct summand.

However, the standard definition of semi simple requires that the algebra have a unit, which does not hold in this case so Wedderburn does not apply.

Note also that the unit is the reason that a semisimple algebra viewed with the regular module structure has to be a finite direct sum of it's irreducible submodules: the unit is a finite sum so the finitely many summands containing the elements in the sum must generate the whole algebra.