3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$\text{Mat}_n(R)/\text{Mat}_n(J)\cong\text{Mat}_n(R/J)$$

(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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.