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Let $R$ be a ring with $1 \neq 0$, let $e$ be any nontrivial idempotent in $R$, and let $f = 1-e$. Then we can write $$R \cong \begin{pmatrix} eRe & eRf \\ fRe & fRf \end{pmatrix}.$$ If the rings $eRe$ and $fRf$ are both division rings and $eRf$ and $fRe$ are both nonzero, is the ring $R$ simple?

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1 Answer 1

Let $R$ be the path algebra of the quiver

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modulo the ideal generated by all paths of length $2$, and let $e$ be one of the two trivial paths.

This is not simple (because it has two non-isomorphic simple modules, for example), $eRe$ and $fRf$ are fields, and $eRf$ and $fRe$ are $1$-dimensional.

This algebra can be described also as that of matrices of the form $$\begin{pmatrix}a&0&0&0\\0&b&0&0\\0&c&a&0\\d&0&0&b\end{pmatrix}$$ with $a$, $b$, $c$ and $d$ in the basefield.

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I can't wait to understand quivers well enough to do stuff like that! Also, as a kind of followup, I wanted to comment on what the OP's rings look like when they are simple. Under the hypotheses, R must be semiperfect, and if it is also simple, rad(R)=0, whence R is simple, Artinian and composition length two. This means R is a 2x2 matrix ring over a division ring. (Another bonus comment is that the radical of Mariano's example is plainly the matrices with c=d=0, and so $rad(R)\neq0$, so it is not simple.) –  rschwieb Apr 24 '12 at 0:21

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