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Let $R$ be a ring and $n$ a positive integer. How do I prove that if $I$ is an ideal of the ring $M_n(R)$, then $I= M_n(J)$ for some ideal $J$ of $R$?

By the way, $M_n(R)$ is the set of all $n\times n$ matrices with entries in $R$.

Phalesh
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    Look here: http://math.stackexchange.com/questions/22629/why-is-the-ring-of-matrices-over-a-field-simple – Alex Youcis Sep 26 '13 at 02:47
  • Thank you. I looked at the proof from Grillet's $\emph{Abstract Algebra}$, but I don't know why the set of all $(1,1)$ entries of the matrix $J$, an ideal of $M_n(R)$, is an ideal. – Phalesh Sep 26 '13 at 03:08
  • It's obviously a subgroup, and if you want to see why it is stable, consider multiplying it on the right by a matrix whose only non-zero entry is the top-left most. – Alex Youcis Sep 26 '13 at 03:14

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