Difference between revisions of "Minimal polynomial"

Latest revision as of 22:02, 1 March 2018

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 $F$ .

@@ Line 2: / Line 2: @@
 == Proof of existence/uniqueness ==
-First note that as <math>\alpha</math> is algebraic over <math>F</math>, there do exist polynomials <math>f(x)\in F[x}</math> with <math>f(\alpha)=0</math>, and hence there must exist at least one such polynomial, say <math>g(x)</math>, of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero <math>a\in F</math> such that <math>m(x) = ag(x)</math> is monic. Now by definition it follows that <math>m(x)</math> is ''a'' minimal polynomial for <math>\alpha</math> over <math>F</math>. We now show that is is the only one.
+First note that as <math>\alpha</math> is algebraic over <math>F</math>, there do exist polynomials <math>f(x)\in F[x]</math> with <math>f(\alpha)=0</math>, and hence there must exist at least one such polynomial, say <math>g(x)</math>, of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero <math>a\in F</math> such that <math>m(x) = ag(x)</math> is monic. Now by definition it follows that <math>m(x)</math> is ''a'' minimal polynomial for <math>\alpha</math> over <math>F</math>. We now show that is is the only one.
 Assume that there is some other monic polynomial <math>m'(x)\in F[x]</math> such that <math>m'(\alpha)=0</math> and <math>\deg m' = \deg m</math>. By the division algorithm there must exist polynomials <math>q(x),r(x)\in F[x]</math> with <math>\deg r<\deg m</math> such that <math>m(x) = m'(x)q(x)+r(x)</math>. But now we have <math>r(\alpha) = m(\alpha)-m'(\alpha)q(\alpha) = 0</math>, which contradicts the minimality of <math>m(x)</math> unless <math>r(x) = 0</math>. It now follows that <math>m(x) = q(x)m'(x)</math>. And now, as <math>m(x)</math> and <math>m'(x)</math> are both monic polynomials of the same degree, it is easy to verify that <math>q(x)=1</math>, and hence <math>m(x) = m'(x)</math>. So indeed, <math>m(x)</math> is the ''only'' minimal polynomial for <math>\alpha</math> over <math>F</math>.
 [[Category:Field theory]]

Page

Toolbox

Search

Difference between revisions of "Minimal polynomial"

Latest revision as of 22:02, 1 March 2018

Proof of existence/uniqueness