How to show the subring generated by the image of a ring homomorphism is not a field.

Question

For the matrix $$A = \begin{pmatrix} 2 & 1 & 2 \\ 0 & 1 & 2 \\ 1 & 0 & 1\end{pmatrix}$$ let the map $\phi_{A}: \mathbb{Z}_{5}[x] \rightarrow M_{3}(\mathbb{Z}_{5})$ be the one taking $p(x)\in \mathbb{Z}_{5}[x]$ to $p(A) \in M_{3}(\mathbb{Z}_{5})$. I want to show that the subring of $M_{3}(\mathbb{Z}_{5})$ generated by the image of $\phi_{A}$ is $\textit{not}$ a field.

This is my strategy: From the Fundamental Theorem of Homomorphisms, we have $${\rm Im}(\phi_{A}) \cong \mathbb{Z}_{5}[x]/{\rm Ker}(\phi_{A})$$ and also that if $I$ is an ideal in a nonzero commutative ring $R$, then $R/I$ is a field if and only if $I$ is maximal. My approach was then to hopefully show that ${\rm Ker}(\phi_{A})$ is not a maximal ideal of $\mathbb{Z}_{5}[x]$, i.e. the kernel is not generated by an irreducible polynomial in $\mathbb{Z}_{5}[x]$.

I also happen to know that the smith normal form of $A$ is $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -x^{3}+4x^{2}-3x+2\end{pmatrix}$$ and that the rational canonical form of $A$ is $$\begin{pmatrix} 0 & 0 & -2 \\ 1 & 0 & 3 \\ 0 & 1 & -4\end{pmatrix}$$ I know very little about what these two forms actually tell me, but the problem I'm working on is the last one in a larger exercise which also asked me to find these two forms. I'm thinking it might be useful, but I just don't know how to apply them.

So to summarize, I'm struggling with solving this problem with the above-mentioned strategy. I would really appreciate some small advice on how to do it, or if my strategy is a bad one, a different one would also be appreciated.

Answer 1

The characteristic polynomial $\chi(X)\in \mathbb F_5[X]$ of $A$ is $$X^3-4X^2+3X-2=(X+1)(X^2+3)$$ Since $X+1$ and $X^2+3$ are irreducible the minimal polynomial $\mu(X)\in \mathbb F_5[X]$ of $A$ is also $(X+1)(X^2+3)$, because the minimal polynomial divides the characteristic polynomial and has the same irreducible factors.
Hence your algebra $\mathbb F_5[A]$ satisfies $\mathbb F_5[A]=\frac {\mathbb F_5[X]}{\langle \mu(X)\rangle}=\frac {\mathbb F_5[X]}{\langle(X+1)(X^2+3)\rangle}$ and is thus not a field.

Remarks
1) Actually by the Chinese Remainder Theorem $$\mathbb F_5[A]=\frac {\mathbb F_5[X]}{\langle(X+1)(X^2+3)\rangle}=\frac {\mathbb F_5[X]}{\langle X+1\rangle}\times \frac {\mathbb F_5[X]}{\langle X^2+3\rangle}=\mathbb F_5\times \mathbb F_{25}$$ but it is not necessary to know this in order to solve your problem.
2) There is no need to invoke the Smith canonical form nor the rational canonical form.

Answer 2

The characteristic polynomial of $A$ is $$ X^3-4X^2+3X-2=(X+1)(X^2+3) \ \mbox{ in } \mathbb{Z}_{5}[X] $$ By the Cayley–Hamilton theorem, $$ 0 = A^3-4A^2+3A-2I=(A+I)(A^2+3I) $$ but $A+I\ne0$ and $A^2+3I\ne0$. Therefore, $\mathbb Z_5[A]$ is not a domain.