Isn't the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}(\sqrt3))$ just the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2))$? Isn't the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}(\sqrt3))$ just the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2))$?
    I'm getting confused about common algebra notation. What does it mean when a field extension is "divided" by another, smaller field extension (see title)? I know divided isn't the right word for what is really happening here, but I'm not sure of the correct way to phrase it.
I know what I am supposed to do when finding the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3))$, but I am confused about notation. It's a small thing I should already know, but now that I don't it is causing me problems.
 A: When $K\subset L$ is a subfield (equivalently, $L/K$ is a field extension), then the group
$$\mathrm{Gal}(L/K)=\{\phi\in\mathrm{Aut}(L)\mid \phi|_{K}=1_K\}.$$
In words, these are the automorphisms of $L$ that fix $K$ pointwise.
In the case $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $K=\mathbb{Q}(\sqrt{3})$, you are correct that $\mathrm{Gal}(L/K)\cong\mathrm{Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})\cong\mathbb{Z}_2$.
Note that in this context, $\mathrm{Gal}(L/K)$ is a subgroup of $\mathrm{Gal}(L/\mathbb{Q})\cong\mathbb{Z}_2\times\mathbb{Z}_2$ (the automorphisms of $L$ that fix $K$ pointwise forms a subset of the set of all automorphisms of $L$), whereas $\mathrm{Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})$ is not a subgroup (the domains of the relevant automorphisms are different).
A: There’s a theorem sometimes called the Theorem on Natural Irrationalities—it says that if $K$ and $L$ are extensions of a field $k$, and if $L$ is Galois over $k$, then the Galois group of $KL$ over $K$ is isomorphic to the group of $L$ over $K\cap L$, and in a very natural way. Looks like a tall order to satisfy, but the case $k=\Bbb Q$, $K=\Bbb Q(\sqrt3)$, $L=\Bbb Q(\sqrt2)$ fits, since the intersection is $\Bbb Q$. So in that sense, the extension $\Bbb Q(\sqrt2,\sqrt3)\supset\Bbb Q(\sqrt3)$ is, in that sense “just like” the extension $\Bbb Q(\sqrt2)\supset\Bbb Q$.
