Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring $A$ (not necessarily commutative), the matrix ring $M_n(A)$ is simple if and only if $A$ is simple.

How to prove?


All ideals below are two-sided ideals. I also assume $A$ has a unit.

  • If $I$ is an ideal of $A$, then $J=M_n(I)$ is an ideal of $M_n(A)$.
    This is easy to prove from the definition of matrix addition and multiplication.

  • If $J$ is an ideal of $M_n(A)$, then $J=M_n(I)$ for some $I$ is an ideal of $A$.
    Take $I$ to be the set of all $a \in A$ such that there is $M\in J$ having $a$ in one entry. Then $I$ is an ideal of $A$ and $J=M_n(I)$ because you can pre- and post- multiply matrices in $J$ by elementary matrices $E_{ij}$ to move entries to any desired place. ($E_{ij}$ is the matrix will all zeros except for the entry $ij$ which is $1$.)

  • $\begingroup$ And what about the case, when $R$ is not unital? There are two-sided ideals not of the form $M_n(I)$. But is the following result: $R$ is simple iff $M_n(R)$ is simple valid? $\endgroup$ – Mikhail Goltvanitsa Dec 8 '17 at 20:00

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