I know that if $J$ is an ideal in a ring $R$ then $M_n(J)$, the set of all $n\times n$ matrices with entries in $J$, is also an ideal. How would I show that $M_n(J)$ is maximal iff $J$ is maximal? I see how to do this if $J$ is prime and I want to show $M_n(J)$ is prime, but the other condition stumps me. Thank you.


You could either prove this by proving the inclusion property, or you could think about it like this.

Step 1: Consider the reduction map $\text{Mat}_n(R)\to\text{Mat}_n(R/J)$ where $J$ is an ideal of $R$, and this map just reduces the entries modulo $J$. It's clearly a surjective ring map, and the kernel is obviously $\text{Mat}_n(J)$. So, you see that


(there are other more sophisticated ways of seeing this, but this is fine for now).

Step 2: Recall that an ideal of a ring is maximal if and only if its quotient is simple.

Step 3: Now, let $A$ be any unital commutative ring. You're last problem showed that the ideals of $\text{Mat}_n(A)$ are just those of the form $\text{Mat}_n(I)$ for $I$ an ideal of $A$. So, what is the only way that $\text{Mat}_n(A)$ could be simple?

  • $\begingroup$ Why is $Mat_n(R)/Mat_n(J) \cong Mat_n(R/J)$ based on what you said prior to that? $\endgroup$ – Hassan Sep 26 '13 at 16:30
  • $\begingroup$ @MartinaK. The first isomorphism theorem. $\endgroup$ – Alex Youcis Sep 26 '13 at 18:03

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