If $R$ is a commutative simple ring with identity , then is any matrix ring $M_n(R)$ over $R$ of matrices of size $n$ also simple ?
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Ye. There's a bijection between their double sided ideals. – Pedro Jan 30 '15 at 15:49
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2Commutativity of $R$ is not required. By the way, a simple commutative ring (with unity) is a field. – egreg Jan 30 '15 at 16:15
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The two-sided ideals of a matrix ring $M_n(R)$, regardless of what kind of ring $R$ is, are all of the form $$M_n(\mathfrak A)$$ for some two-sided ideal $\mathfrak A\subset R$. Hence, $M_n(R)$ is simple iff $R$ is.
Olivier Bégassat
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