# Prove that $\sigma(x)$ and $x$ are conjugate over $F$

Let $E$ be an algebraic extension $F$ and $x \in E$ and $\sigma: E \to E$ be an automorphism of $E$ fixing $F.$ Prove that $\sigma(x)$ and $x$ are conjugate over $F.$

I am starting to learn about extension field, automorphism and Galois theory now and there's a lot of stuffs that confuse me so any help for this question is really appreciated.

In order to prove that $\sigma(x)$ and $x$ are conjugate over $F,$ I need to find an irreducible polynomial $p(x) \in F[x]$ such that $p(x) = p(\sigma(x)) = 0.$ If $x \in F,$ then the proof completes, but I still stuck on the case when $x\notin F.$

There is a theorem in my book which says that if $F$ is a field and $\alpha$ and $\beta$ are algebraic over $F$ with $deg(\alpha, F) = n.$ The the map $$\psi_{\alpha, \; \beta}:(c_{0} + c_{1}\alpha + \dots + c_{n - 1}\alpha^{n - 1}) = c_{0} + c_{1}\beta + \dots + c_{n - 1}\beta^{n - 1}$$ is an isomorphism of $F(\alpha)$ onto $F(\beta)$ if and only if $\alpha$ and $\beta$ are conjugate over $F.$ I attempt this approach but fail to prove that $\psi_{\sigma(x), \; x}$ is an isomorphism.

In fact any polynomial $p \in F[X]$ with $p(x)=0$ satisfies $p(\sigma (x))=\sigma(p(x))=\sigma(0)=0$, in particular the minimal polynomial of $x$ is the minimal polynomial of $\sigma(x)$.
• How do you get $p(\sigma(x)) = \sigma(p(x))?$ – user298251 May 11 '16 at 7:21
• Just write down the polynomial expressions and use that $\sigma$ is a ring homomorphism, which fixes the coefficients of the polynomial. – MooS May 11 '16 at 7:23
Let $p(X)$ be an irreducible polynomial for $\alpha$ over $F$ (so that $p(X) \in F[X]$ and $p(\alpha)=0$). Suppose that $f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0$, then $a_i \in F$ and is fixed by $\sigma$, hence $$\sigma(p(x))=\sigma\left(\sum_{i=0}^n a_ix^i\right) = \sum_{i=0}^n \sigma(a_ix^i)=\sum_{i=0}^n a_i\sigma(x)^i = p(\sigma(x)).$$
Then $p(\sigma(x))=\sigma(p(x))= \sigma(0)=0$ and it follows that the degree of $\sigma(x)$ is smaller or equal to the degree of $x$. By a symmetric argument using the inverse of $\sigma$, we obtain that the degree of $\sigma(x)$ is smaller or equal to the degree of $x$ over $F$. Since $p(\sigma(x))=0$ and $p(X)$ is irreducible, it follows that $irr(\sigma(x),F)=p(X)$.