Looking for at least one surjective ring homomorphism from $M_n(R)$ to $R$ Let $R$ be a ring , I am looking for a surjective ring homomporphism from $M_n(R)$ to $R$ . Please help . Thanks in advance . 
 A: If $R$ is a commutative ring (with unity), there is no such ring homomorphism (surjective or not), assuming it preserves unity.
Suppose that $\varphi\colon M_n(R)\to R$ is a ring homomorphism (with $n>1$). Then the kernel of $\varphi$ is of the form $M_n(I)$ where $I$ is a proper ideal of $R$. So we have the induced injective homomorphism
$$
\tilde\varphi\colon M_n(R)/M_n(I)\to R
$$
but $M_n(R)/M_n(I)$ is easily seen to be isomorphic to $M_n(R/I)$ which is non commutative when $n>1$. 
In order to find an example one must find a non commutative ring $R$ having an ideal $I$ such that $M_n(R/I)$ is isomorphic to $R$, because the above argument still works the same.
If you allow rings without unity, then it is very simple: consider a free abelian group $R$ of infinite rank with zero multiplication. Then also $M_n(R)$ is a free abelian group of the same infinite rank and it is a zero ring as well. So $R$ is isomorphic to $M_n(R)$ for all $n$.
A: When $R$ is a domain and $n\geq 2$. Let's take $\phi: M_n(R)\rightarrow R$ a ring morphism. One easily see that nilpotent elements of $M_n(R)$ should be sent to $0\in R$ (because $R$ is a domain).
Let us define $E_{i,j}$ the matrix with just a $1$ at $(i,j)$ and $0$'s elsewhere.Then if $i\neq j$ we have for all $r\in R$ :
$$rE_{i,j}\text{ nilpotent so } \phi(rE_{i,j})=0 $$
$$rE_{i,i}=rE_{i,j}E_{j,i}\text{ so } \phi(rE_{i,i})=\phi(rE_{i,j})\phi(E_{j,i})=0\text{ note that you cannot do this when }n=1$$
Now for all $M\in M_n(R)$ we have :
$$\phi(M)=\phi(\sum_{i=1}^n\sum_{j=1}^nM_{i,j}E_{i,j})=\sum_{i=1}^n\sum_{j=1}^n\phi(M_{i,j}E_{i,j})=0 $$
When $R$ is a domain and $n\geq 2$ there is no such morphism.
When $R$ is a commutative ring with a non-nilpotent element and $n\geq 2$. From the same argument, when $i\neq j$ and $\phi(rE_{i,j})$ is nilpotent. Furthermore $\phi(rE_{i,i})$ is thus a product of two nilpotent elements which commute because $R$ is commutative so it is nilpotent.
Finally $Im(\phi)$ is in the ring of nilpotent elements (it is a ring, again because $R$ is a commutative ring). So if $R$ happens to have a unit element then $1_R\notin Im(\phi)$. 
This proves that in this larger case, one cannot find such a surjective morphism.  
