$M_n(I_1)M_n(I_2)= M_n(I_1I_2)$ or not? Let $R$ be a ring with identity and $M_n(R)$ be a matrix ring over $R$. I know that if $J$ is an ideal of matrix ring then $J=M_n(I)$ where $I$ is an ideal of $R$. 

The question is whether $M_n(I_1)M_n(I_2)= M_n(I_1I_2)$ or not (where $I_1,I_2$ are ideals of $R$)? 

I have a problem that $M_n(R)$ is a prime ring if and only if $R$ is a prime ring,( where prime ring mean ideal $0$ be a prime ideal). So if $M_n(I_1)M_n(I_2)= M_n(I_1I_2)$ then the problem be easy to solve. 

Certainly one can check that $M_n(I_1)M_n(I_2) \subseteq M_n(I_1I_2)$. As for the other direction, let $e_{ij}$ be the standard basis of $M_n(R)$, with $e_{ij}$ the matrix that has a 1 in the $i$th row and $j$th column and every other entry 0. Since $M_n(I_1I_2) = \sum_{ij} I_1I_2e_{ij}$ as abelian groups, it suffices to show that $i_1i_2e_{ij} \in M_n(I_1)M_n(I_2)$ for $i_1\in I_1$, $i_2\in I_2$. However $(i_1 e_{ik})(i_2e_{kj})=i_1i_2e_{ij}$ for any $k$, and $i_1e_{ik}\in M_n(I_1)$ and $i_2e_{kj}\in M_n(I_2)$, so this implies $i_1i_2e_{ij}\in M_n(I_1)M_n(I_2)$. Hence $M_n(I_1I_2)\subseteq M_n(I_1)M_n(I_2)$ as well
giving the desired equality.
I have to admit I'm much less familiar with noncommutative ring theory as compared to the commutative case, so please let me know if I've overlooked anything, but I believe this should work.
