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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.

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

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