Let $E/K$ and $L/K$ be finite extension. E/K is galois. prove that $$\sigma\in Gal(EL/L) \rightarrow \sigma|_{E}\in Gal(E/E\cap L)$$ is a group isomorphism.

First we have show this is well-defined, for $\sigma\in Gal(EL/L)$, we must prove that $\sigma(x)\in E$ $,\forall x\in E$. This follows from that $E/K$ is normal.

For injectivity, if $\sigma|_{E}$ fixes $E$, then $\sigma$ fixes $EL$

For surjectivity, we have to show that any automorphism in $Gal(E/E\cap L)$ can be extended to an automorphism in $Gal(EL/L)$. I'm stuck in this part.


For surjectivity, I would be inclined to use the Fundamental Theorem of Galois Theory.

The main idea is as follows. We have a group homomorphism $$ \rho : Gal(EL/L) \to Gal(E/K), \ \ \ \ \ \sigma \mapsto \sigma |_E.$$ The image $G = \rho(Gal(EL/L))$ is a subgroup of $Gal(E/K)$. Our task is to show that $G = Gal(E/E \cap L)$.

Let $F$ be the subfield of $E$ containing all elements of $E$ that are fixed by $G$.

Since $E/K$ is a Galois extension, the Fundamental Theorem of Galois Theory tells us that $G = Gal(E/F)$.

So it remains to show that $F = E \cap L$. I'll leave that to you...

[Don't forget that $EL/L$ is also a Galois extension - we showed this in a previous question on math.SE. You may wish to use the Fundamental Theorem a second time.]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.