A finite ring of matrices is a finite field i have a prime number $p$ and an irreducible polynomial $R(x)$ of degree $n$ , $\alpha$ a root of $R$ then it's known that the field $\mathbb{F}_{p^n}$ is isomorphic to the field $\mathbb{F}_{p}[\alpha]$.
let $M$ be the companion matrix of $R$ in particular .
i wish to proof that $\mathbb{F}_{p}[M]$ is a field isomorphic to $\mathbb{F}_{p^n}$.
i've noticed that $R(M)=O$ (Carley-Hamilton theorem) but i don't know whether or not $M$ lies in the algebraic closure of $\mathbb{F}_{p}$ if it dose that will give the result by construction of finite fields so i got stuck there...
Next i tried to give an explicit isomorphism (ring isomorphism)
\begin{array}{ccccc}
\psi : & \mathbb{F}_{p^{n}} & \rightarrow  & \mathbb{F}_{p}[M] &  \\ 
& 0 & \rightarrow  & O &  \\ 
& \alpha ^{i} & \rightarrow  & M^{i} & 0<i<q^{n}-1%
\end{array}
with the assumption that $\alpha$ is a primitive element of  $\mathbb{F}_{p^{n}}$
it's easy to poof that $\psi(a*b)=\psi(a)*\psi(b)$ but i couldent proof that $\psi(a+b)=\psi(a)+\psi(b)$
UPDATE:
can i say that $M$ is algebraic over  $\mathbb{F}_{p}$ ?
 A: Another approach is to prove that $\mathbb{F}_{p}[M]$ is a field. Since it has $p^n$ elements, it must be isomorphic to $\mathbb{F}_{p^n}$, by uniqueness of finite fields.
A typical element of $\mathbb{F}_{p}[M]$ is given by $f(M)$ with $\deg f < \deg R$. If $f(M)\ne0$, then $\gcd(f,R)=1$, because $R$ is irreducible and $\deg f < \deg R$.
Then, we can write $1=a(x)f(x)+b(x)R(x)$ in $\mathbb{F}_{p}[x]$.
Therefore, $1=a(M)f(M)$ and $f(M)$ is invertible.
A: First, you should define $\psi\left(\sum_ia_i\alpha^i\right)=\sum_ia_iM^i$ so that $\psi(a+b)=\psi(a)+\psi(b)$ holds automatically. However, you will also need to prove that $\psi$ is well defined, e.g., if $\sum_ia_i\alpha^i=0$, then you should have $\sum_ia_iM^i=0$ as well.
You can get around this issue by using the 1st isomorphism theorem. Let
$\psi:\mathbb{F}_p[x]\to\mathbb{F}_p(M)$ be the unique ring homomorphism given by $\psi(x)=M$ (i.e. $\psi\left(\sum_ia_ix^i\right)=\sum_ia_iM^i$. This map is clearly surjective. Since $R(M)=0$, we know that $\langle R(x)\rangle\subseteq\ker\psi$. But $R(x)$ is irreducible, so $\langle R(x)\rangle$ is maximal and equality holds.
Finally, by the 1st isomorphism theorem $\mathbb{F}_p/\langle R(x)\rangle\cong\mathbb{F}_p[M]$. Since $\mathbb{F}_{p^n}\cong\mathbb{F}_p/\langle R(x)\rangle$, we are done.
