Is every field Galois over its prime subfield?
问题内容
A field extension $F/K$ is Galois if $F^{\operatorname{Aut}_K(F)}=K$ in Hungerford's Algebra, and it is Galois if it is normal and separable in Lang's Algebra. For the finite extension, these two definitions are the same. For infinite algebraic extensions or for transcendental extensions, which definition is better to pick? I have not seen this question on this website. However, I heard that we have a transcendental Galois extension. If we pick Hungerford's definition, then every field over its prime subfield is Galois, am I right? If $Q$ is the prime field in $F$ then $$F^{\operatorname{Aut}(F)}=F^{\operatorname{Aut}_Q(F)}=Q,$$ is that right? However, $F/Q$ is possible to be not normal. So, which definitions make sense in this context?
回答 (4)
For algebraic extensions, these two definitions coincide, and Galois theory works well.
For transcendental extensions, there is no sensible Galois correspondence between subgroups and intermediate fields no matter which definition you pick.
In any case, $\DeclareMathOperator\aut{Aut}F^{\aut(F)}$ is in general not the prime field $Q$ of $F$. For one thing, this fails when $F/Q$ is algebraic, but not normal, by the above. For another example, $\aut(\def\R{\mathbb R}\R)$ is trivial, hence $\R^{\aut(\R)}=\R$.
As you point out, the two definitions are equivalent for finite extensions. But finite extensions of a prime subfield are not guaranteed to be Galois; take a typical non-Galois extension like $\mathbb Q[\sqrt[3]2]/\mathbb Q$, in this case $\mathbb Q$ is the prime subfield.
For the question about which definition is better in the non-finite case:
(1) If $F/K$ is infinite algebraic, then Hungerford and Lang's definitions still agree. You can prove this from the finite case but it's a bit nontrivial, one direction is easy at least and I'll prove it below using the following two (easy to prove) facts:
$F/K$ is normal if and only if $F$ can be written as a union $F=\bigcup_i F_i$ where $F_i/K$ are finite normal extensions,
$F/K$ is separable if and only if every intermediate extension $E/K$ is separable.
Suppose $F/K$ is normal and separable; to prove $F^{\mathrm{Aut}_K(F)}=K$ suppose $\alpha\in F^{\mathrm{Aut}_K(F)}$. Invoking the first fact we write $F=\bigcup_i F_i$ as above with $F_i/K$ finite normal (and also separable by the second fact), then $\alpha\in F_i$ for some $i$. Now any $\sigma\in\mathrm{Aut}_K(F_i)$ can be extended to some $\widetilde\sigma\in\mathrm{Aut}_K(F)$ using the fact that $F/K$ is normal. But then $\alpha$ is fixed by $\widetilde\sigma$, hence by $\sigma$, and thus $\alpha\in F_i^{\mathrm{Aut}_K(F_i)}=K$ (where we have invoked the equivalence of Hungerford and Lang's definitions for $F_i/K$).
(2) There is no standard notion of a Galois extension for transcendental extensions; you will notice if you go to just about any source defining a Galois extension, they will only be talking about algebraic extensions. You can find some people attempting to make sense of the notion if you google it, but again none of this will be considered standard terminology.
An infinite extension $F/K$ is defined to be Galois if it is the union of finite Galois extensions $L/K$ (where the union is taken within an algebraic closure of $K$). Hence, by definition, the extension $F/K$ is algebraic. As a counterexample, the field $\mathbf{R}$ is not Galois over its prime subfield $\mathbf{Q}$. Neither is it normal over $\mathbf{Q}$, and moreover, it contains elements transcendental over $\mathbf{Q}$. And though the field $\mathbf{C}$ (of complex numbers) is normal and separable over $\mathbf{Q}$, it is not a Galois extension, so Lang's definition does not apply. If we consider the infinite extension $\mathbf{Q}(\pi)/\mathbf{Q}$, it is not Galois, and yet the fixed field of ${\rm Aut}_\mathbf{Q}(\mathbf{Q}(\pi))$ is equal to $\mathbf{Q}$, and so Hungerford's definition does not apply either.
Note that $$ {\rm Aut}_\mathbb{Q}(\mathbb{Q}(2^{1/3}))=\{{\rm id}\}. $$ Indeed, an automorphism $\varphi$ is completely determined by $\varphi(2^{1/3})$ and $\mathbb{Q}(2^{1/3})$ contains only one cubic root of $2$.