$E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$. Prove $\sigma = 1_E$

Question

Let $E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$, then $\sigma = 1_E$.

I think that the way is to see that $K$ and the elements $a_i$ are in the field fixed by $\sigma$, $C_E(\sigma)=\left\lbrace e\in E \vert \sigma (e)=e \right\rbrace$.

By definition of Galois Group, since $\sigma \in Gal (E/K)$, then $\sigma(k)=k \,\, \forall k\in K$ hence $K\subset C_E(\sigma)$ and since $\sigma (a_i)=a_i$, all the elements $a_i$ are in $C_E(\sigma)$.

How can I use that for prove that $\sigma$ is the identity?

Answer 1

Let $H$ be a cyclic subgroup of $\operatorname{Gal}(E/K)$ generated by $\sigma$, i.e. $H = \langle \sigma \rangle$. Let us consider the fixed field by $H$, i.e. $E^H$. As you have mentioned, every element of $K$ as well as all $a_i$ are fixed by $\sigma$, thus $E \subset E^H$.

On the other hand, clearly $E^H \subset E$. So we have equality $E^H = E$. Now apply the Galois correspondence, which says that there is a bijection between subgroups $H$ of $\operatorname{Gal}(E/F)$ to intermediate subfields of $E/F$ by considering the operatorname $H \mapsto E^H$.

Clearly the trivial subgroup $\{e\}$ satisfies $E^{\{e\}} = E$. So by the correspondence, $H = \{e\}$ is trivial, i.e. $\sigma = e$.

Answer 2

Every element in E may be written as $c_0+\sum_{i=1}^n c_ia_i$ with $c_i\in K$, so $\sigma(c_0+\sum_{i=1}^n c_ia_i)=\sigma(c_0)+\sum_{i=1}^n\sigma(c_i)\sigma(a_i)=c_0+\sum_{i=1}^n c_ia_i$.