Show set of matrices forms a finite field of order 27 Given the matrix
$$
M=\begin{pmatrix}0 & 0 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}\in\mathbb{F}_3^{3\times3},
$$
show that
$$
\mathbb{F}_3[M]:=\big\{\sum_{i=0}^na_iM^i\;\big|\;n\in\mathbb{N}_0,a_i\in\mathbb{F}_3\big\}
$$
is a field of order 27. I have seen similar versions of this question on here that use the fact that the multiplicative group of a finite field is cyclic, but they all assume we are starting with a field in the first place. How can I prove that $\mathbb{F}_3[M]$ actually is a field at all? Maybe convert the problem into something involving polynomials?
 A: $M$ has characteristic polynomial $t^3 - t - 2,$ and this is irreducible in the field with 3 elements. Indeed, I suggest you carefully confirm that $M^3$
This is easier than appears. As you were told, the elements of the new ring are 
$$ xI + y M + z M^2  $$
Addition needs nothing new ( except using mod 3), but multiplication needs 
$$ M^3 = M + 2 I \equiv M-I  $$
$$  M^4 = M^2 + 2 M \equiv M^2 - M$$
MORE to come
We just need a multiplicative inverse for all 26 nonzero matrices. For example, we can dump the distracting 2 by changing it to $-1.$ First
$$  M^3-M + I \equiv 0  $$ 
If you check this with ordinary integers you just get $3I$
So
$$ I \equiv M - M^3 \equiv M (I-M^2), $$
so that the inverse of $M$ is $I-M^2$ and vice versa.
EXERCISE: LET
$$ (xI + yM + z M^2)(pI + q M + rM^2) = (uI + vM + wM^2)   $$
Calculate $u,v,w$ 
A: First understand the isomorphism $\mathbb F_3[M] \cong \mathbb F_3[X]/(f)$ where $f$ is the minimal polynomial of $M$. Then so long as $f$ is irreducible, $(f)$ is a prime ideal and $\mathbb F_3[X]/(f)$ must be a field (as all finite integral domains are fields). Then show that $1, X, X^2$ is a basis for $\mathbb F_3[X]/(f)$ over $\mathbb F_3$. Or equivalently that $1, M, M^2$ is a basis for $\mathbb F_3[M]$.
