# Set of elements in a Galois group after restriction

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!

• Are you assuming $M/K$ Galois? Oct 27, 2020 at 17:37

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 $$\{\sigma_{\mid_M}\mid \sigma\in\operatorname{Gal}(\bar{K}/L)\}\overset{1}{=}\operatorname{Gal}(\bar{K}/L)_{\mid_M}\overset{2}{=}(\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}})_{\mid_{M}}\overset{3}{=}(\operatorname{Gal}(LM/L))_{\mid_M}\overset{4}{=}\operatorname{Gal}(M/M\cap L).$$
• 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? Oct 28, 2020 at 9:54
• 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). Oct 28, 2020 at 11:50
• 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? Oct 28, 2020 at 11:53
• 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. Oct 28, 2020 at 11:56