$R$ is simple iff $M_n(R)$ is simple for rings without identity

Asked Dec 08 '17 at 19:37

Active Dec 08 '17 at 19:45

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Let $R$ be a ring without identity and $n$ be a positive integer. Prove that the ring $R$ is simple iff $M_n(R)$ is simple.

When $R$ is unital ring (i.e. with identity) it is easy because we know the structure of two-sided ideals in $M_n(R)$. So we can consider the case, when $R$ is non-unital.

edited Dec 08 '17 at 19:45

asked Dec 08 '17 at 19:37

Mikhail Goltvanitsa

1,394

this question differs from https://math.stackexchange.com/questions/654451/for-any-ring-a-the-matrix-ring-m-na-is-simple-if-and-only-if-a-is-simple because the last is for unital rings – Mikhail Goltvanitsa Dec 08 '17 at 19:48
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What exactly is your definition of simple rings, please? – rschwieb Dec 08 '17 at 20:13
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$ R $ is simple if $ R^2\neq 0$ and there are no non-trivial two-sided ideals in $ R $. – Mikhail Goltvanitsa Dec 09 '17 at 04:11

$R$ is simple iff $M_n(R)$ is simple for rings without identity

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