# Show that every ideal of the matrix ring $M_n(R)$ is of the form $M_n(I)$ where $I$ is an ideal of $R$ [duplicate]

Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$ where $I$ is an ideal of $R$.

I have spent 30 minutes on this question and I still got nowhere. Can anyone give some hints ?

Let $J$ be an ideal of $M_n(R)$.
The standard procedure is to let $I=\{A_{11}\mid A\in J\}\subseteq R$ and then show that $I$ is an ideal of $R$. You will be able to show that $J=M_n(I)$ by using permutation matrices and sums of matrices. What I mean is, if $E_{ij}$ denotes a standard matrix unit, you can consider $A\in J$ and $E_{ij}A{E_{k,l}}\in J$, and sums of such products.
In the other direction, it's clear that if you multiply anything in $M_n(R)$ by an element of $M_n(I)$, it will land back in $M_n(I)$.
• what is a standard matrix unit? I.e., what does $E_{ij}$ look like as a matrix?
• @ALannister it's 1 in the $ij$ position and zero elsewhere. Commented Mar 17, 2017 at 21:20