If $E/F$ is a field extension, and $H$ is a subgroup of $\mathrm{Aut}(E/F)$, it is quite trivial to see that $H\subset \mathrm{Aut}(E/E^H)$.

Since the theorem only shows the inclusion relationship, I think there must be a lot of examples where $\subset$ is actually $\subsetneq$. But due to lack of knowledge I can't come up with one.

Could you help me with this? Best regards.

  • $\begingroup$ In this instance $E^H$ is the invariants of $E$ under $H$ correct? $\endgroup$ Dec 28, 2015 at 5:24
  • $\begingroup$ @SirJective Exactly $\endgroup$
    – Vim
    Dec 28, 2015 at 6:31
  • $\begingroup$ There is equality here whenever $H$ is a finite group of automorphisms. For infinite groups there are counterexamples. This follow from the linear independence of characters (due to Artin IIRC). $\endgroup$ Dec 28, 2015 at 9:10
  • $\begingroup$ @JyrkiLahtonen Could you elaborate? Why is there always equality in the case of finite $H$? And for infinite ones is there any counter-example that is not so terribly wild (like $\mathrm{Aut}(\mathbb C)$ as I heard)? $\endgroup$
    – Vim
    Dec 28, 2015 at 9:16
  • $\begingroup$ See here $\endgroup$ Dec 28, 2015 at 9:16

1 Answer 1


If $H$ is finite, we always have equality. See these notes by Noam Elkies.

There are counterexamples, when $H$ is infinite.

If we allow a transcendental extension, then the following example is easy to grasp. Let $E=\Bbb{Q}(x)$ be the field of fractions of the polynomial ring $\Bbb{Q}[x]$. Let $\sigma$ be the automorphism gotten by extending $x\mapsto x+1$ in the obvious way, and let $H=\langle\sigma\rangle$ be the infinite cyclic group. Then it is not difficult to show (for an argument see an earlier answer of mine) that $E^H=\Bbb{Q}$. But $E$ has many other $\Bbb{Q}$-automorphisms. They are all of the form $$ x\mapsto \frac{ax+b}{cx+d} $$ with $ad-bc\neq0$ (= a quotient group of invertible 2x2 matrices by scalar matrices).

If you want an example of an algebraic extension, then it is a bit more complicated. Let $E$ be the algebraic closure of $\Bbb{F}_p$, and let $H$ be the group of automorphisms generated by the Frobenius mapping $F:z\mapsto z^p$. Because a degree $p$ polynomial can have at most $p$ fixed points we see that $E^H=\Bbb{F}_p$. But $H$ is not all of $Aut(E/\Bbb{F}_p)$. The recipe for all $\Bbb{F}_p$-automorphisms of $E$ is described in this answer by Ted. Here $H$ is only a dense subgroup of $Aut(E/\Bbb{F}_p)$ (w.r.t. the Krull topology). You need to read a bit about Galois theory of infinite algebraic extensions for the use of topological concepts to make sense here.

  • $\begingroup$ Thanks again. This detailed answer is certainly going to take me some time to digest. $\endgroup$
    – Vim
    Dec 28, 2015 at 10:01

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