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).

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