Quotient Ring of $M_n(\mathbb Z)$

Question

I have an question.

How can I show that the quotient rings of $M_n(\mathbb Z)$ other than itself have finitely many elements?

If it was division ring instead of $\mathbb{Z}$, I could say something. But know I don't know how to start.

Answer 1

For $I$ a non-zero two-sided ideal of $M_n(\Bbb{Z})$ then $\Bbb{Q} I$ is a two sided ideal of $M_n(\Bbb{Q})$ thus the whole it, a $n^2$-dimensional $\Bbb{Q}$-vector space, generated by $n^2$ elements, once multiplied by the right integer they are in $I$ and they generate a finite index sub-$\Bbb{Z}$-module of $M_n(\Bbb{Z})$.