Is that true that any homomorphic image of $M_n(R)$ ($R$ a ring with 1) is of the form $M_n(\overline{R})$, where $\overline{R}$ is an homomorphic image of $R$?
It's clear for me that if $\overline{R}$ is an homomorphic image of $R$, then $M_n(\overline{R})$ is an homomorphic image of $M_n(R)$.
Thanks in advance.
The ideals of $M_n(R)$ are all of the form $M_n(I)$, where $I$ is an ideal of $R$. Now the obvious map $$ M_n(R)\to M_n(R/I) $$ is a surjective ring homomorphism. What's its kernel?
The kernel is obviously $M_n(I)$. In particular, every ring of the form $M_n(R/I)$ is a homomorphic image of $M_n(R)$.
Conversely, if $\alpha\colon M_n(R)\to S$ is a surjective ring homomorphism, then $S\cong M_n(R)/\ker\alpha$. On the other hand, $\ker\alpha=M_n(I)$ for a unique ideal $I$ of $R$. Therefore $S\cong M_n(R)/M_n(I)\cong M_n(R/I)$.