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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .