Minimal polynomial

Given a field extension $F\subseteq K$ , if $\alpha\in K$ is algebraic over $F$ then the minimal polynomial of $\alpha$ over $F$ is defined the monic polynomial $f(x)\in F[x]$ of smallest degree such that $f(\alpha)=0$ . This polynomial is often denoted by $m_{\alpha,F}(x)$ , or simply by $m_\alpha(x)$ if $F$ is clear from context.

Proof of existence/uniqueness

First note that as $\alpha$ is algebraic over $F$ , there do exist polynomials $f(x)\in F[x]$ with $f(\alpha)=0$ , and hence there must exist at least one such polynomial, say $g(x)$ , of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero $a\in F$ such that $m(x) = ag(x)$ is monic. Now by definition it follows that $m(x)$ is a minimal polynomial for $\alpha$ over $F$ . We now show that is is the only one.

Assume that there is some other monic polynomial $m'(x)\in F[x]$ such that $m'(\alpha)=0$ and $\deg m' = \deg m$ . By the division algorithm there must exist polynomials $q(x),r(x)\in F[x]$ with $\deg r<\deg m$ such that $m(x) = m'(x)q(x)+r(x)$ . But now we have $r(\alpha) = m(\alpha)-m'(\alpha)q(\alpha) = 0$ , which contradicts the minimality of $m(x)$ unless $r(x) = 0$ . It now follows that $m(x) = q(x)m'(x)$ . And now, as $m(x)$ and $m'(x)$ are both monic polynomials of the same degree, it is easy to verify that $q(x)=1$ , and hence $m(x) = m'(x)$ . So indeed, $m(x)$ is the only minimal polynomial for $\alpha$ over $F$ .

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Minimal polynomial

Proof of existence/uniqueness