Understanding the Galois Extension

Question

I'm currently learning a little bit of Galois Theory at the end of my Abstract Algebra course and am having a little trouble understanding the Galois Extension. I will provide the definition given to me in the book below and then my questions/concerns below.

---

Definition

Suppose that $\mathbb{F} \subset \mathbb{K}$ is a finite extenstion field and let $G = Aut(\mathbb{K}, \mathbb{F})$. For any subgroup $H < G$ the fixed field of $H$ is defined to be: $$H' = \text{Fix}_{\mathbb{K}}(H) = \{k \in \mathbb{K} \mid h(k) = k,\ \forall h \in H \}$$ For any subfield $\mathbb{L} \subset \mathbb{K}$, set $$\mathbb{L}' = \bigcap_{k \in \mathbb{L}} \text{stab}_G(k) = \{g \in G \mid g(k) = k,\ \forall k \in \mathbb{L} \}$$ Finally, we say that $\mathbb{K}$ is a Galois Extension of $\mathbb{F}$ if $$G' = \mathbb{F}$$ That is, if $\mathbb{F}$ is precisely the set of elements fixed by $G$.

---

Questions

Okay, so my main question here is regarding the final statement in the definition "$\mathbb{F}$ is precisely the set of elements fixed by $G$". Isn't this always the case if $G$ exists? By definition of the Galois group $G = \text{Aut}(\mathbb{K},\mathbb{F})$, $G$ is the group of automorphisms from $\mathbb{K}$ to $\mathbb{K}$ that fix every element of $\mathbb{F}$? So, in that case, wouldn't any extension for which you could construct a Galois Group then be a Galois extension? It just seems like, from my naive perspective, that almost any extension would then be a Galois extension, but perhaps it's not so easy to construct a Galois Group for an arbitrary field and extension, and so the distinction of being a Galois Extension would be more meaningful than I am imagining it to be right now.

Answer 1

You are correct that every element of $\mathbb{F}$ is fixed by $G$, but in principle there could be other fixed points. That is, the definitions immediately imply $G' \supseteq \mathbb{F}$, but not necessarily $G' = \mathbb{F}$.

For example, consider $\mathbb{F} = \mathbb{Q}$ and $\mathbb{K} = \mathbb{Q}(\sqrt[3]{2})$. Suppose that $g \in Aut(\mathbb{K}, \mathbb{F})$. Then $g(\sqrt[3]{2})$ must be a cube root of $2$ in $\mathbb{K}$, but $\mathbb{K}$ is a subfield of $\mathbb{R}$, so there is only one cube root of $2$ in $\mathbb{K}$. Thus $g(\sqrt[3]{2}) = \sqrt[3]{2}$, and so (since $g$ also fixes $\mathbb{F}$) $g$ is the identity. Thus in this case $Aut(\mathbb{K}, \mathbb{F})' = \mathbb{K}$, not $\mathbb{F}$. You won't always have $Aut(\mathbb{K}, \mathbb{F})' = \mathbb{K}$ in non-Galois extensions, that's just an artifact of this particular example.

Answer 2

It is certainly true that $\mathbb F\subseteq G'$. But there are field extensions where $\mathbb F\subsetneq G'$. For instance, take $\mathbb Q$ and the extension field $\mathbb Q[\sqrt[3]2]$. The element $\sqrt[3]2$ is in $G'$ for the following reason: Any homomorphism $\sigma:\mathbb Q[\sqrt[3]2]\to\mathbb Q[\sqrt[3]2]$ that fixes every element of $\mathbb Q$ must send every root $a$ of the polynomial $X^3-2$ to another root of the same polynomial. This is because $$0=\sigma(0)=\sigma(a^3-2)=\sigma(a^3)-\sigma(2)=\sigma(a)^3-2,$$ meaning that $\sigma(a)$ is also a root of the polynomial. However, $\mathbb Q[\sqrt[3]2]$ only contains one root of that polynomial. The other two roots are complex, so not elements of $\mathbb Q[\sqrt[3]2]$. This means that any such homomorphism must send $\sqrt[3]2$ to itself, since there is no other root to send it to. So the fixed field of $G$ in this case is larger than the base field, not equal to it.