The minimal polynomial of a over F is unique

Let $F$ be a field, let $E$ be an extension of $F$, and let $\alpha \in E$ be algebraic over $F$. Recall that a polynomial $p(x)$ is said to be minimal for $\alpha$ over $F$ if $p(\alpha) = 0$, $p(x)$ is monic, and $p(x)$ has minimal degree among the nonzero polynomials having $\alpha$ as a root.

Prove that minimal polynomials are unique.

Suppose $p(x)$ and $q(x)$ are minimal polynomials of $\alpha$ over $F$. By definition, $p(x)$ and $q(x)$ have the same degree and are both monic. Now $\alpha$ is a root of $(p-q)(x)$, and $p-q$ has strictly smaller degree than do $p$ and $q$. Multiplying by a unit if necessary, $p-q$ is monic. If $p-q \neq 0$, then we contradict the minimalness of $p$ (and $q$). Thus $p - q = 0$, and thus $p(x) = q(x)$.