Noetherian modules and Noetherian rings

Question

I want to show that if $R$ is a Noetherian ring then $Mat_n(R)$ is also a Noetherian ring. It is obvious that $Mat_n(R)$ is a finitely generated $R$-module. So $Mat_n(R)$ is a Noetherian R-module.

How can I conclude that $Mat_n(R)$ is a Noetherian ring?

Answer 1

I'll assume $R$ is commutative.

Every left (or right) ideal of $\textrm{Mat}_n(R)$ is an $R$-submodule of $\textrm{Mat}_n(R)$. Therefore any ascending chain of left (or right) ideals of $\textrm{Mat}_n(R)$ stabilises, since $\textrm{Mat}_n(R)$ is a Noetherian $R$-module. Therefore $\textrm{Mat}_n(R)$ is left (and right) Noetherian.