Find the monic generator of and ideal. Let $\mathbb{F}$ be a subfield of complex numbers, and let
$$
A = \begin{bmatrix}
       1  & -2  \\[0.1em]
       0  &  3  \\[0.1em]
   \end{bmatrix}
$$
Find the monic generator of the ideal of all polynomials $f$ in $F[x]$ such that $f(A)=0.$

We can solve this in a reasonably simple manner.
Consider some $f \in F[x]$ such that $f(A) = 0$. Suppose that $deg(f) = 1$. Then, $f$ is of the form $$ f(x) = cx + d. $$ However, we need $f(A)$ to be zero and all values are in $\mathbb R^{2x2}$, so
\begin{align*}
    f(A) &= cA + d \\
    &= c\begin{bmatrix}1&-2\\0&3\end{bmatrix} + \begin{bmatrix}d&0\\0&d\end{bmatrix} \\
    &= \begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}
\end{align*}
So, unless we set $c=0$ and $d=0$, we cannot scale or add by any $c$ and $d$ such that $f(A)=0$. We can, however, try a function of degree 2, written as $$f(x)=x^2+cx+d.$$
Rewriting a bit, we see that
\begin{align*}
    0 &= f(A) \\
    &= A^2 + cA + d \\
    &= \begin{bmatrix}1&-8\\0&9\end{bmatrix}+\begin{bmatrix}c+d&-2c\\0&3c+d\end{bmatrix}\\
    &= \begin{bmatrix}1+c+d&-8-2c\\0&9+3c+d\end{bmatrix}
\end{align*}
Because each entry of $f(A)$ needs to be zero, we can find $c$ and $d$ by solving this system of linear equations:
$$
\begin{cases}
-1=c+d \\-8=2c \\-9=3c+d
\end{cases}
$$
This is solved by $c=-4$ and substituting into either equation to find that $d=3$. Thus, we have that $$ f(x)=x^2-4x+3 $$ and that $$ f(A)=A^2-4A+3 = 0, $$ as desired. Thus, $f$ is the monic polynomial that generates the ideal $fF[x]$; as such, $f$ is the gcd for all polynomials $g$ such that $g(A)=0$.
(Note that I am implicitly applying the isomorphism from the set of polynomials over $F$, denoted as $F[x]$, to the set of all polynomial functions over $F$, which I usually denote as $\mathbb P_\infty$.)
