Semisimplicity of a polynomial ring Given a ring $R$ (with 1 and not necessarily commutative) when is the polynomial ring $R[x]$ semisimple?
For example if R is a Noetherian integral domain then R[x] is not semisimple. Indeed, $R[x]$ is Noetherian but then if we assume that $R[x]$ is semisimple this would imply that $R[x]$ is Artinian which is not.
Question: is there a ring $R$ (with $1$ and not necessarily commmutative) such that $R[x]$ is semisimple? or is $R[x]$ never semisimple for any ring $R$?
 A: Your post already contains the answer. Polynomial rings are not Artinian, since $$(x)\supset(x^2)\supset(x^3)\supset\ldots$$
But semisimple rings are Artinian.
In fact, the Wedderburn-Artin Theorem states that all semisimple rings are direct sums of matrix rings over division algebras. Polynomial rings are certainly not of this form.
A: If we understand semisimple to mean all modules are projective then $R[X]$ is never semisimple. 
Indeed, if we let $S=R[X]/(X)$, the canonical $R[X]$-linear projection map $f:R[X]\to S$ is not split: there is no $R[X]$-linear map $s:S\to R[X]$ such that $f\circ s$ is the identity of $S$. To see this notice that the element $u=f(1_R)\in S$ is an element such that $X\cdot u=0$ and that $s$ would necessarily be injective, so that we would have an element $v=s(u)\in R[X]$ such that $X\cdot v=0$. 
A: Yay! Let's do all the characterizations. All semisimple rings are Von Neumann regular. But, if you choose $x\in R[x]$ then you need to be able to find a $p(x)\in R[x]$ such that $x=x^2p(x)$ but this is clearly impossible. 
