If $ R $ is a prime ring then $ M_{n}(R) $ is prime ring. If $ R $ is a prime ring then $ M_{n}(R) $ is prime ring.
SOLUTION
Let $ S=M_{n}(R) $ and $ aSb=0 $. For all $ r∈R $ say that $ s=re_{ij} $. So $ (are_{ij}b)_{kl}=0 $. I obtain $ (a_{k1}+a_{k2}+...+a_{kn})r(b_{1l}+b_{2l}+...+b_{nl})=0 $ and since R is prime $ (a_{k1}+a_{k2}+...+a_{kn})=0 $ or $ (b_{1l}+b_{2l}+...+b_{nl})=0 $. But, I don't know how can I say that each $ a_{ki}=0 $ whenever $ i=1,2,...,n $ 
 A: 
I mean that if for all $a,b∈R$, if $aRb=0$ then $a=0$ or $b=0$ satifies then we say that $R$ is prime ring ( I mean not as ideal ). I want to use this definition to solve problem

It's a little messy looking, but perhaps no more work than translating everything to ideals and solving it as in this solution.
Let's start with $aSb=\{0\}$ and work with the unit matrices as you were. Remember that $e_{jj}ae_{kk}$ is going to yield $a_{jk}e_{jk}$, the matrix you get when you replace everything in $a$ except the $i,j$ entry with zero, $a_{jk}\in R$.
We can say that $\{0\}=e_{ii}aSbe_{mm}\supseteq e_{ii}a(e_{jj}e_{jj}Se_{kk}e_{kk})be_{mm}$ for every $i,j,k,m$. The important stuff you want from the last expression is rewritten here:
$$
\{0\}=(e_{ii}ae_{jj})(e_{jj}Se_{kk})(e_{kk}be_{mm})=(a_{ij}e_{ij})(S_{jk}e_{jk})(b_{km}e_{km})=a_{ij}S_{jk}b_{km}e_{im}
$$
By a little abuse of notation, I'm writing $S$ both as if it were a matrix and as if it were a set of matrices. The symbol $S_{jk}$ should be accordingly read as "the set of $jk$ entries of elements in $S$". But this is just the set $\{re_{jk}\mid r\in R\}$.
Now if both $a$ and $b$ were nonzero on some element, you could pick $i,j,k,m$ such that $a_{ij}$ and $b_{km}$ are both nonzero. As discussed above, the $S_{jk}$ ranges all over $R$, so the far right of the equation is saying that $a_{ij}Rb_{km}=\{0\}$ in $R$. But this is a contradiction if $R$ is prime. 
A: A two-sided ideal of $M_n(R)$ has the form $M_n(I)$, meaning the matrices with coefficients in $I$, where $I$ is a two-sided ideal of $R$.
Prove that $M_n(I)M_n(J)=M_n(IJ)$ and you're done.
