Proof that the minimal polynomial divides every annihilating polynomial

Proof

We divide with remainder and use the minimality of the degree.

1. Let $m_T$ be the minimal polynomial of $T$. It is monic, and it satisfies $m_T(T) = 0$. No nonzero polynomial of smaller degree annihilates $T$, since such a polynomial could be scaled to a monic one.
2. Let $p \in F[x]$ with $p(T) = 0$, so $p$ is an annihilating polynomial of $T$.
3. By the division algorithm for polynomials, applied with the monic divisor $m_T$, there are $q, r \in F[x]$ with $p = q m_T + r$ and $\deg r < \deg m_T$.
4. The map $f \mapsto f(T)$ of evaluation at $T$ respects sums and products, because powers of $T$ commute with each other. So $p(T) = q(T) m_T(T) + r(T)$.
5. Since $p(T) = 0$ and $m_T(T) = 0$, step 4 gives $r(T) = 0$.
6. If $r \ne 0$, then $r$ would be a nonzero annihilating polynomial with $\deg r < \deg m_T$. This contradicts step 1.
7. Hence $r = 0$ and $p = q m_T$. So $m_T$ divides $p$.