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. 
 A: 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.
A: 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!
A: 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.  
