I completely edited my question and with the help of Lubin, I added my proof. Any comments are welcome.

My question was:

Let $F_0 \subset F_1 \subset F_2$ be fields, and suppose $F_2/F_0$ is a Galois group.

Then, Galois group Gal($F_2/F_1$) is a normal subgroup $\iff$ $F_1 /F_0$ is a Galois extension.

How to prove normal $\Rightarrow$ Galois?

This is my revised proof:

First, $F_1=$ $ F_2^{\text{Gal}(F_2/F_1)}$, where $ F_2^{\text{Gal}(F_2/F_1)}$ is ($F_2$ fixed by Gal$(F_2/F_1)$) ,

Since $F_1 \subset F_2^{\text{Gal}(F_2/F_1)}$, and $[F_2 : F_1] = |\text{Gal}(F_2/F_1)| = |F_2 : F_2^{\text{Gal}(F_2/F_1)}|$, by using hypothesis and the fixed field theorem.

Now, suppose Gal$(F_2/F_1)$ is a normal subgroup of Gal$(F_2/F_0)$,

and take any $t \in F_2^{\text{Gal}(F_2/F_1)}$.

All conjugations of $t$ can be written as $\sigma (t)$, where $\sigma \in $ Gal$(F_2/F_0)$, since $F_2/F_0$ is a Galois extension.

I want to say $\sigma (t) \in F_2^{\text{Gal}(F_2/F_1)}$(unchanged by Gal($F_2/F_1$)).

Since Gal$(F_2/F_1)$ is a normal subgroup,

$\forall \tau \in \text{Gal}(F_2/F_1), \sigma^{-1} \tau \sigma \in \text{Gal}(F_2/F_1)$, so that

$\exists \tau' \in \text{Gal}(F_2/F_1), $ s.t. $\sigma^{-1} \tau \sigma = \tau'$.

Therefore $\sigma^{-1} \tau \sigma (t)= \tau' (t) = t$, so that

$ \tau \sigma (t)= \sigma (t)$.

Hence, $\sigma (t) \in F_2^{\text{Gal}(F_2/F_1)}$, so that $F_2^{\text{Gal}(F_2/F_1)}/F_0$ is a Galois extension.

Then, noting that $F_2^{\text{Gal}(F_2/F_1)}=F_1$, $F_1/F_0$ is a Galois extension. $\square$

thank you for Lubin.

  • $\begingroup$ Is this an exercise from the text? If $F_2\supset F_0$ wasn’t Galois to start with, do you have a definition of $\text{Gal}(F_2/F_0)$? $\endgroup$
    – Lubin
    Jan 19, 2015 at 23:54
  • $\begingroup$ Thank you. $F_2 \supset F_0$ was Galois. I will edit the post. This is a part of theorem in the book, which I think has gaps. $\endgroup$
    – Arch
    Jan 20, 2015 at 6:01
  • 1
    $\begingroup$ Not likely that a text by Artin has anything wrong with it (he says, worshipfully). $\endgroup$
    – Lubin
    Jan 20, 2015 at 23:17

1 Answer 1


This is a standard fact, part of the Fundamental Theorem of Galois Theory.

Assuming separability throughout, let’s take this as a definition of $E$ being Galois over a field $k$: that every $k$-morphism $\varphi$ of $E$ into a separably closed extension $\Omega$ of $E$ actually sends $E$ into $E$.

So let’s suppose that $F_2$ is Galois over $F_0$ and that $\text{Gal}(F_2/F_1) $ is a normal subgroup of $\text{Gal}(F_2/F_0)$. Now let $\varphi$ be an $F_0$-morphism of $F_1$ into $\Omega$. It can be extended to $\bar\varphi\colon F_2\to\Omega$, by a standard theorem, and by the hypothesis that $F_2$ is Galois over $F_0$, it sends $F_2$ to $F_2$, and thus is an element of $\text{Gal}(F_2/F_0)$. I want to show that $\bar\varphi(F_1)\subset F_1$, by showing that any element $\bar\varphi(x))$ of this field is fixed under $\text{Gal}(F_2/F_1)$. Now let $g$ be any element of this group. I want $g(\bar\varphi(x))=\bar\varphi(x)$. But by normality of the subgroup, $g\circ\bar\varphi=\bar\varphi\circ g'$, for an (perhaps other) element $g'$ of the subgroup. But since $g'$ leaves $x$ fixed, we have $g(\bar\varphi(x))=\bar\varphi(x))$, as desired.

This argument is probably unnecessarily wordy; I’m sure you can improve it.

  • $\begingroup$ Thank you so much! I will write the proof my own words. $\endgroup$
    – Arch
    Jan 21, 2015 at 6:00

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.