Proof of the Galois correspondence being a bijection.

Question

One claim of the Galois theorem I'm working with is the following:

For $T \leq U$, $U$ a Galois extension, there exists an anti-isomorphism between the lattice of intermediate fields $V$, ($V$ such that $T \leq V \leq U$), and the lattice of subgroups of $Gal(U,T)$.

As the maps, we are using $Gal(U,V)$ - all automorphisms of $U$ that fix $V$, and $Fix(U,G)$ - the field of all elements that every $g$ fixes.

We can use the concept of the Galois correspondence etc. to show that these maps give us our wanted anti-isomorphism. I understand all the steps, besides this - we need to show that $Gal(U,\cdot)$ and $Fix(U, \cdot)$ are onto.

Because for finite $G$ it's true that $Gal(U,Fix(U,G))=G$, we see $Gal(U,\cdot)$ is onto. We also know that $Fix(U,Gal(U,V))=V$ for $V$ a Galois extension of $T$ - but not all the fields in question have to be normal (Galois theorem makes an additional claim that the normal fields correspond to normal subgroups) - therefore I'm not sure how to show that $Fix(U,\cdot)$ is onto.

Why is $Fix(U,\cdot)$ onto?

Answer 1

The adjective "normal" is only ever used to a field extension, i.e. it is a property of a pair of fields. Therefore, it is acceptable to say something like "$T \subseteq U$ is normal / not normal", but it does not make sense to say something like "$U$ is normal."

Since we know that $T \subseteq U$ is a normal extension, it must be the case that $V \subseteq U$ is a normal extension. The reason for this is that if $T \subseteq U$ is a normal extension, then $U$ is by definition the splitting field over $T$ for some polynomial $p(X) \in T[X]$ - but then, $U$ is also the splitting field for $p(X)$ over $V$. So we can validly conclude that $Fix(U, Gal(U,V)) = V$.

What seems to be causing confusion is that, while $V \subseteq U$ is always normal, it is not necessarily true that $T \subseteq V$ is normal. $T \subseteq V$ is normal if and only if $Gal(U,V)$ is a normal subgroup of $Gal(U,T)$.

Take this example: $$ T = \mathbb Q, \ \ \ \ V = \mathbb Q(\sqrt[3]{2}), \ \ \ \ U = \mathbb Q(\sqrt[3]{2}, \exp(\tfrac{2\pi i }{3})).$$ Here, $T \subseteq U$ and $V \subseteq U$ are both normal extensions, since $U$ is the splitting field of $X^3 - 2$ over both $T$ and $U$. But $T \subseteq V$ is not a normal extension, since $X^3 - 2$ is an example of an irreducible polynomial in $T[X]$ with one root in $V$ that doesn't split completely in $V$. $Gal(U,T)$ is the dihedral group of order six, and $Gal(U,V)$ is a subgroup of order two (consisting only of the identity automorphism and complex conjugation), and as we expect, this subgroup of order two is not a normal subgroup.