Minimal polynomial

Revision as of 23:02, 1 March 2018 by Makar (talk | contribs) (Proof of existence/uniqueness)
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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 $F$.