If I is an ideal of $R$ ($R$ a ring) prove that $M_{n}(R/I)$ is isomorphic to $M_{n}(R)/M_{n}(I)$

Question

If $I$ is an ideal of $R$ ($R$ a ring) prove that $M_{n}(R/I)$ is isomorphic to $M_{n}(R)/M_{n}(I)$

I proved that $M_{n}(I)$ is an ideal of $M_{n}(R)$ but I don't know how to prove this. Thanks for your hints.

Answer 1

Consider the map $\pi \colon M_n(R) \to M_n(R/I)$ defined by $$ (a_{ij}) \mapsto (a_{ij} \mod{I}) $$ which is a ring morphism because reduction modulo $I$ is a ring morphism. Then clearly $M_n(I) \subseteq \ker(\pi)$. On the other hand, if $(a_{ij}) \in \ker(\pi)$ then $a_{ij} \equiv 0 \pmod{I}$ for every $1 \leq i,j \leq n$, i.e. $a_{ij} \in I$. Hence $\ker(\pi) \subseteq M_n(I)$, too.

Since $\pi$ is clearly surjective, by the first isomorphism theorem it induces an isomorphism between $M_n(R/I)$ and $M_n(R)/\ker(\pi) = M_n(R)/M_n(I)$.