I am considering $x^2-x-1$ over $F=\mathbb{Z}/2=\{0,1\}$ as the minimum polynomial of a $3\times 3$ matrix $A$ with entries from $F$.
I know that the characteristic polynomial (in this case a degree 3 polynomial) is supposed to be able to divide some power of the minimum polynomial, but I have come across the statement that no power of $x^2-x-1$ is divisible by any polynomial of degree 3 over $F$ with no proof or hint.
Is this an obvious result and I am just being dense? How would one go about proving a statement like this?
Since $F$ is a field, $F[x]$ is a UFD. Suppose a polynomial of degree 3, $p(x)$ divides $(x^2-x-1)^n$ for some $n\geq 1$. Then $$p(x)k(x) = (x^2-x-1)^n$$ for some $k(x)$. However, the only factor of the right-hand side is $x^2-x-1$, so this can be the only factor of the left-hand side. In particular, if x^2-x-1 divides $p(x)$, then the remainder would be a linear polynomial, which gives a contradiction.
Note: The minimal polynomial divides the characteristic polynomial, not the other way around.