Originally Posted by

**Sampras** Let $\displaystyle E $ be an algebraic extension of a field $\displaystyle F $, and let $\displaystyle \sigma $ be an automorphism of $\displaystyle E $ leaving $\displaystyle F $ fixed. Let $\displaystyle \alpha \in E $. Show that $\displaystyle \sigma $ induces a permutation of the set of zeros of $\displaystyle \text{irr}(\alpha, F) $ that are in $\displaystyle E $.

Let $\displaystyle \text{irr}(\alpha, F) = a_0+a_{1}x+ \cdots + a_{n}x^n $. Then $\displaystyle a_0+ a_{1}\alpha + \cdots + a_{n} \alpha^{n} = 0 $. So $\displaystyle a_0+ a_1 \sigma(\alpha) + \cdots + a_{n} \sigma(\alpha)^{n} = \sigma(0) = 0 $. So $\displaystyle \sigma(\alpha) $ is a zero of $\displaystyle \text{irr}(\alpha, F) $.