Extension of a finite field to a finite non commutative ring 
Can a finite field be extended to non-commutative finite rings so that not all elements of the field commutes with the elements of the ring? 

I have been trying this taking the examples of matrices.  
 A: Let $K(\alpha)/K$ be an algebraic extension of fields, and suppose $\alpha$ has minimal polynomial $f(x)$ over the base field $K$, with $\deg f(x)=n$ (with $n\ge2$). Let $A\in M_n(K)$ be the companion matrix associated to $f(x)\in K[x]$. Then the there is an isomorphic copy of $K(\alpha)$, call it $L$, sitting inside the matrix algebra $M_n(K)$, which the unital $K$-subalgebra generated by $A$. Then the field extension $L/K$ can be extended to $M_n(K)/K$ with $L$ not central in $M_n(K)$ (since for instance the element $A\in L$ does not commute with the diagonal matrix $\mathrm{diag}(1,0,\cdots,0)$.)
A: If you don't care about mapping $1$ to $1$, then we can embed any field $K$ into the ring $A$ of $2\times 2$ matrices over $K$, by sending $a\in K$ to $\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix}$.  This preserves addition and multiplication, and $K$ clearly this does not lie in the center of $A$.
If you care about mapping $1$ to $1$, then we have to be more careful.  If $K$ is a prime field, this is impossible.  But if we can write $K=L(\alpha)$ for some field $L$ and $\alpha$ algebraic of degree $d>1$ over $L$, then we can let $A$ be the ring of $d\times d$ matrices over $L$, with $L$ identified with multiples of the identity matrix, and map $\alpha$ to a matrix with the same minimal polynomial as $\alpha$.
In particular, a finite field of order $q=p^n$, with $n\geq 2$, has a non-central embedding into the ring of $n\times n$ matrices over $\mathbb{F}_p$.
