just looking for feedback and/or hints about this proof I've been working on. No answers please, but I'd like to know if I'm on the right track here.
So I'm given a field $F$, and a non-zero $n \times n$ matrix $M$ with entries in $F$, and that $ev_M:F[x] \rightarrow M_{n\times n}(F)$ is a ring homomorphism where $ev_M(f(x)) = f(M) = ev_M(a_0+a_1x+a_2x^2+...+a_nx^n)$ is defined to be $a_0+a_1M+a_2M^2+...a_nM^n$, and I am asked to show that there is a unique polynomial $h(x) \in F[x]$ such that $h(M) = 0_M$ and $deg$ $h(x) \leq deg$ $f(x)$ whenever $f(M) = 0_M$.
Edited solution: Note that the kernel of $ev_M$ is non-trivial, because $ev_M$ is not injective. Since the kernel is a subset of $F[x]$, and every ideal of $F[x]$ is principal, $ker$ $ev_M =$ $<h(x)>$, for some unique irreducible polynomial of lowest degree.
If there is a polynomial $f(x) \in F[x]$ such that $f(M)=0$, then $f(x)\in ker$ $ev_M$. Since $h(x)$ generates the kernel, either $f(x) = h(x)$ or $f(x) = h(x)g(x)$, for some $g(x)\in ker$ $ev_M$.
Finally, we conclude by noting that $h(M)=0$, and $deg$ $h(x)$ is always less than or equal to $deg$ $f(x)$.