Issues understanding $|\text{Gal}(E/F)| = [E:F] \iff E/F$ is Galois.

abstract-algebragalois-extensionsgalois-theory

$\newcommand{\Gal}{\text{Gal}}$
Precursor Definitions: A finite field extension $E/F$ is called Galois if it is both normal and separable. The Galois group of $E/F$ is
$$
\Gal(E/F):= \{\sigma: E\rightarrow E \mid \sigma \text{ is an iso. and } \sigma|_F = id_F\}.
$$

I'm struggling to understand the proof of the reverse direction of the following proposition.

Proposition: Let $E/F$ be a finite field extension, then $|\Gal(E/F)| = [E:F] \iff E/F$ is Galois.

Proof: Assume that $E/F$ is Galois, hence separable and normal. Because it is separable, there exists a homomorphism $\sigma:F \rightarrow L$ which admits exactly $n=[E:F]$ many extensions $\tau_i:E \rightarrow L$. Now $\color{red}{\text{identify $E$ with a subset of $L$}}$ via (arbitrarily) $\tau_1$. Then $\tau_1$ becomes the inclusion of $E$ in $L$ and (thinking of $\sigma$ as becoming the identity) it follows that we have $\color{red}{\text{ $n$ extensions $\tau_i:E \rightarrow L$ of the identity on $F$.}}$ Because $E/F$ is normal $\tau_i(E) = E$ for all $1 \leq i \leq n$ and $|\Gal(E/F)| \geq n$. Since $|\Gal(E/F)| \leq [E:F]=n$ the claim follows.

  1. My first qualm is with the identification of $E$ with a subset of $L$. From what I understand, this has nothing to do with the normality of the extension, but just with the fact that $\tau_i$ are non-zero field homomorphisms. Because of this $\tau_1$ is injective and evidently surjects onto it's image so that $E \cong \tau_1(E) \subseteq L$ and so we identify $E$ with a subset of $L$. Is this the correct interpretation of what's occurring here? Because if it is, I don't understand how it leads to my second piece of highlighted text.
  2. How does this identification imply that the $\tau_i$ become extensions of the identity on $F$? I also don't understand why they would remark that $\sigma$ becomes the identity. I understand that for any $1 \leq i \leq n, \tau_i|_F = \sigma$ but why is $\sigma:F \rightarrow L$ now the identity on $F$? I suppose the identification $E \cong \tau_1(E)$ also means you could think of $\tau_i:\tau_1(E) \rightarrow L$, but what good is this?

Thanks in advance for the clarification.

Best Answer

I believe $L$ is an algebraic closure of $F$. Let's keep it that way. I believe for $(2)$ it is supposed to say $\tau_1$ becomes the identity on $F$, because that is your inclusion in $L$.

I will clarify what is happening. Define $X(E/F) := \text{Hom}_F(E,L)$ (field homomorphisms from $E$ to $L$ that fix $F$), then you can imbed $E$ into $L$ in exactly $[E:F]$ different ways (using that $F \subset E$ is separable). Using the normality of $F \subset E$, the image $\psi[E]= E \subset L$ for every $\psi \in X(E/F)$ is the same.

Fix one imbedding $\tau_1 : E \to L$ such that you can interpret it as an inclusion $E \subset L$. We get an identification : $X(E/F) \xrightarrow[]{\cong} \text{Aut}_F(E)$ by sending $\psi \mapsto \tau^{-1}_1\psi$ (this is well-defined), where $\tau_1$ corresponds to $\text{id}_E$.

The argument they give is that they have found $n$ automorphisms of $E$ that fix $F$ ($\tau_i$), so $\#\text{Gal}(E/F) \geq n$. But because of the inseparability you have the other inequality as well. I hope this helps.

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