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I know the result is true, however, if we are to consider $E_{ij} = \left\{\begin{pmatrix} 0 & 0 & ... & 0\\ . & . & ...& . \\ 0 & 0 & r_{ij} & 0 \\ . & . & ...& . \\ \end{pmatrix} : r_{ij}\in R \right\}$.
$E_{ij}$ is an ideal of $M_n(R)$ but doesn't appear to be of the type $M_n(I)$.

Can someone please explain what I am doing wrong here?

Ritwik Gupta
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    What makes you think this is an ideal ? – Maxime Ramzi Mar 12 '21 at 12:54
  • Does this help: https://math.stackexchange.com/questions/22629/why-is-the-ring-of-matrices-over-a-field-simple? – Rob Arthan Mar 12 '21 at 12:54
  • Well, if $P$ is a permutation matrix, then $P(r E_{ij})P^{-1}$ should lie on the "ideal". But for most $P$, this shuffles the columns so that the non-zero entry is no longer in the $(i,j)$ coordinate. – qualcuno Mar 12 '21 at 13:12

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Your set is not closed by multiplication on the right.

You can multiply a nonzero element like that by a matrix with a 1 in the $j,1$ position and zeros elsewhere, and that is not in your set.

rschwieb
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