Proof that the minimal polynomial divides every polynomial vanishing at the element

Proof

1. Let $L/K$ be a field extension, let $a \in L$ be algebraic over $K$, and let $m_a$ be its minimal polynomial. Let $f \in K[x]$ with $f(a) = 0$.
2. The polynomial $m_a$ is monic. By polynomial division, there are $q, r \in K[x]$ with $f = q m_a + r$ and $\deg r < \deg m_a$.
3. Evaluation at $a$ is a ring homomorphism. So $f(a) = q(a) m_a(a) + r(a)$, and $0 = q(a) \cdot 0 + r(a)$. Hence $r(a) = 0$.
4. Suppose that $r \ne 0$, and let $c \ne 0$ be its leading coefficient. Then $c^{-1} r$ is monic, vanishes at $a$, and has degree less than $\deg m_a$.
5. This contradicts the minimality of $\deg m_a$. So $r = 0$.
6. Then $f = q m_a$, so $m_a$ divides $f$ in $K[x]$.