If $L$ is a Galois extension of $K$ and $M$ is a finite Galois subextension of $L \mid K$, then a standard lemma says that any automorphism of $M \mid K$ can be extended to an automorphism of $L \mid K$; in other words, the restriction homomorphism $\textrm{Gal}(L \mid K) \to \textrm{Gal}(M \mid K)$ is surjective.

Question. Is this true for infinite Galois subextensions as well? If it is, where can I find a proof? If not, what is a counterexample?


Take an algebraic closure $\overline{L}$ of $L$. Note that since $L/K$ is Galois, it is algebraic, separable and normal over $K$. Any element $\sigma \in Gal(M|K)$ gives us a homomorphism $\sigma: M \rightarrow \overline{L}$ (just compose with the inclusion $M \hookrightarrow \overline{L}$). Since $L$ is algebraic over $M$, $\sigma$ can be extended to a homomorphism $\overline{\sigma}: L \rightarrow \overline{L}$ (This is proved for example in Keith Conrad's online notes, Zorn's Lemma and some Applications, II).

Note that $\overline{\sigma}$ is a $K$-homomorphism. As $L/K$ is a normal, algebraic extension of $K$, this means that $im(\overline{\sigma}) \subset L$. But, since $L/K$ is algebraic, any $K$-endomorphism $\overline{\sigma}: L \rightarrow L$ must be an automorphism (Lang's Algebra, Chapter 5, Proposition $2.1$). Then $\overline{\sigma}$ is a desired extension.

  • $\begingroup$ Ah, so it doesn't so much follow from the finite case as it does have the same proof. Pity, I was hoping it was just abstract nonsense... $\endgroup$ – Zhen Lin Dec 16 '12 at 9:57

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