Let $K$ be a field and $\bar{K}$ be an algebraic closure of $K$. Let $L$ and $M$ be finite extensions of $K$.

Question: Is $\{\sigma|_M \, | \, \sigma \in \operatorname{Gal}(\bar{K}/L)\} = \operatorname{Gal}(M/M\cap L)$?

This seemed intuitive for me but I am not able to show it myself. I guess one must somehow use the Fundamental Theorem of Galois Theory (as always) but I have trouble doing that.

Any help or reference is highly appreciated!

  • $\begingroup$ Are you assuming $M/K$ Galois? $\endgroup$
    – Losten
    Oct 27, 2020 at 17:37

1 Answer 1


Assume $M/K$ is Galois. Then there is an isomorphism $$\operatorname{Gal}(LM/L)\to\operatorname{Gal}(M/M\cap L)$$ given by restricting to $M$ (this is Theorem 2.6 of https://kconrad.math.uconn.edu/blurbs/galoistheory/galoiscorrthms.pdf). Can you conclude from here?

EDIT: Now by transitivity of restriction, $$\operatorname{Gal}(\bar{K}/L)_{\mid_M}=(\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}})_{\mid_{M}},$$ and $$\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}}=\operatorname{Gal}(LM/L).$$ We conclude that $$\{\sigma_{\mid_M}\mid \sigma\in\operatorname{Gal}(\bar{K}/L)\}\overset{1}{=}\operatorname{Gal}(\bar{K}/L)_{\mid_M}\overset{2}{=}(\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}})_{\mid_{M}}\overset{3}{=}(\operatorname{Gal}(LM/L))_{\mid_M}\overset{4}{=}\operatorname{Gal}(M/M\cap L).$$

  • $\begingroup$ Do I have to apply the result twice by restricting from $\bar{K} $ to $LM$ first and then to $M$? As far I can see the result holds only for finite extensions which the algebraic closure is not. Does this still work? $\endgroup$
    – Ribbity
    Oct 28, 2020 at 9:54
  • $\begingroup$ I noted the edit. But my question still remains if restricting from $\bar{K} $ to $LM$ works. I cannot see why your suggested result from your source works here. (restricting from $LM$ to $M$ is fine though). $\endgroup$
    – Ribbity
    Oct 28, 2020 at 11:50
  • $\begingroup$ The result I cited is used only to show that $\text{Gal}(LM/L)\cong \text{Gal}(M/M\cap L)$. Then I am using the fact that taking an automorphism of $\bar{K}$ fixing $L$ and restricting it to $LM$, we get an automorphism of $LM$ fixing $L$ (that is, $\text{Gal}(LM/L)=\text{Gal}(\bar{K}/L)_{\mid_{LM}}$). Do you agree with this? $\endgroup$
    – Losten
    Oct 28, 2020 at 11:53
  • $\begingroup$ So you do not actually need to use twice the cited result, in order to deal with the infinite extension $\bar{K}/L$. It is just enough to restrict to $LM$ and then to $M$ is a second time. $\endgroup$
    – Losten
    Oct 28, 2020 at 11:56
  • $\begingroup$ 1: definition. 2: transitivity of restriction. 3: see my previous comment. 4: result in Conrad's notes. $\endgroup$
    – Losten
    Oct 28, 2020 at 12:00

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