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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 $R$?

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$\mathbb Z[X]$ is not a PID (for example, the ideal $(2,X)$ is not principal) so there is something wrong with your second paragraph! –  Mariano Suárez-Alvarez Apr 19 '12 at 3:42
    
It would be useful if you make explicit what definition of semisimplicity you have in mind... –  Mariano Suárez-Alvarez Apr 19 '12 at 3:42
    
@Mariano Suárez-Alvarez: sorry, just edited. Well I've read several equivalences: i.e every submodule is a direct summand, isomorphic to a direct sum of simple modules,etc. –  user6495 Apr 19 '12 at 3:44
    
Please add that information to the question itself. –  Mariano Suárez-Alvarez Apr 19 '12 at 3:47

3 Answers 3

up vote 2 down vote accepted

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 $X\cdot v=0$.

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If you don't assume $R$ has a $1$, this argument still works taking $u=f(w)$ for any non-zero $w\in R$. Artinianness follows, in the usual situation, by the existence of $1$, so this is indeed a separate obstacle to semisimplicity. –  Mariano Suárez-Alvarez Apr 19 '12 at 3:51
    
cheers, very nice! –  user6495 Apr 19 '12 at 3:52

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.

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thanks, can't believe I forgot that fact! –  user6495 Apr 19 '12 at 3:51

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.

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