# Finite Extension has Finitely Many Automorphisms?

Let $$\mathbb{E}\supseteq\mathbb{F}$$ be a field extension with $$\dim_\mathbb{F}(\mathbb{E})<\infty$$ and let $$G\subseteq\mathrm{Aut}(\mathbb{E})$$ be the subgroup of field automorphisms $$\sigma:\mathbb{E}\to\mathbb{E}$$ that fix $$\mathbb{F}$$. What is the easiest way to prove that $$G$$ is finite?

Edit: How about this? Every element of $$\mathbb{E}$$ is algebraic over $$\mathbb{F}$$. By induction we can write $$\mathbb{E}=\mathbb{F}[\alpha_1,\ldots,\alpha_n].$$ Let $$\Omega_i$$ be the (finite) set of roots of the minimal polynomial of $$\alpha_i$$ over $$\mathbb{F}$$. Then $$G$$ acts on the Cartesian product $$\Omega_1\times \cdots \times\Omega_n$$. Let $$\mathcal{O}$$ be the (finite) $$G$$-orbit of the element $$(\alpha_1,\ldots,\alpha_n)\in\Omega_1\times\cdots\times\Omega_n.$$ Note that the stabilizer is trivial since if $$\sigma\in G$$ fixes each $$\alpha_i$$ then it fixes any polynomial expression $$f(\alpha_1,\ldots,\alpha_n)$$, hence it fixes every element of $$\mathbb{E}$$. It follows from the orbit-stabilizer theorem that $$\#G=\#\mathcal{O}$$ is finite.

• $E = F(a_1,\ldots,a_m)$ and $\sigma$ sends each $a_j$ to one of the finitely many other roots of its $F$-minimal polynomial. – reuns Apr 5 at 17:32
• I believe one must still show that a group element is determined by its action on the generators $a_j$. – Drew Armstrong Apr 5 at 20:37
• That's the definition of $F$-algebra isomorphism – reuns Apr 5 at 20:47