- For a polynomial \(f \in K[x]\), prove that if \(r \in F\) is a root of \(f\) then for any \(\sigma\in\operatorname{Aut}_K(F)\), \(\sigma(r)\) is also a root of \(f\).
Solution
1. A \(K\)-automorphism fixes every coefficient of \(f\).
Write \[ f(x)=a_0+a_1x+\cdots+a_nx^n \] with \(a_i\in K\). If \(\sigma\in\operatorname{Aut}_K(F)\), then by definition \(\sigma(a_i)=a_i\) for every \(i\).
2. If \(r\in F\) satisfies \(f(r)=0\), then \(f(\sigma(r))=0\).
Using 1 and the fact that \(\sigma\) is a field homomorphism, \[ \begin{aligned} f(\sigma(r)) &=\sum_{i=0}^n a_i\sigma(r)^i\\ &=\sum_{i=0}^n \sigma(a_i)\sigma(r^i)\\ &=\sigma\left(\sum_{i=0}^n a_ir^i\right)\\ &=\sigma(f(r))\\ &=\sigma(0)=0. \end{aligned} \] Thus \(\sigma(r)\) is again a root of \(f\).
3. Hence \(K\)-automorphisms of \(F\) permute the roots in \(F\) of every polynomial in \(K[x]\).
By 2, each automorphism sends the set of roots in \(F\) into itself. Since \(\sigma\) is bijective and \(\sigma^{-1}\) has the same property, the induced map on that root set is a permutation.