# To intermediate field is Galois$\iff$ Galois group over intermediate field is a normal subgroup

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.

• 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 $\text{Gal}(F_2/F_0)$? – Lubin Jan 19 '15 at 23:54
• 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. – Arch Jan 20 '15 at 6:01
• Not likely that a text by Artin has anything wrong with it (he says, worshipfully). – Lubin Jan 20 '15 at 23:17

## 1 Answer

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.

• Thank you so much! I will write the proof my own words. – Arch Jan 21 '15 at 6:00