Finite Galois group of Galois extension implies that the extension is finite?

Question

Assume that the field extension $K \subset L$ is a Galois (in other words: normal and separable, possibly infinite) extension with finite Galois group. If one starts with a finite Galois extension, then one can show $[L:K]=|\mathrm{Gal}(L/K)|$, of course. I'm interested in the other way round: If we first consider the Galois group finite, how can we show (preferably without the use of Zorn's lemma) that the Galois extension must be finite (and thus of course have this equality), too? In other words: I want to show that infinite Galois extensions have always infinite Galois groups.

My idea so far is to show that for a finite Galois group $\mathrm{Gal}(L/K)$ the degree of every irreducible polynomial over $K$ is bounded by the order of $\mathrm{Gal}(L/K)$, let's say $n$, and hence the extension must be finite. The latter part is pretty easy to prove (under the condition of the former part), but how can the former part be proved? If I assume that an irreducible polynomial over $K$ has at least degree $n+1$, then it has at least $n+1$ distinct roots in $L$ (normality and separability). This means that there are at least $n+1$ possibilities to send such a root, let's say $a$, to another one. But how can such a monomorphism $K(a) \to L$ be extended to $L \to L$, preferably without the use of Zorn's lemma, in order to get at least $n+1$ elements in $\mathrm{Gal}(L/K)$ to get a contradiction?

Answer 1

Judging by your question, this is the lemma you want to prove?

Let $G := \text{Gal}(L/K)$, and let $f(x) \in K[x]$ be an irreducible polynomial that has a root in $L$. Then, $\deg(f) \leq |G|$.

Proof: Let $\alpha \in L$ be a root of $f$, then consider the polynomial $$ g(x) = \prod_{\sigma \in G} (x-\sigma(\alpha)). $$ Note that $g$ is invariant under $G$. Hence, every coefficient of $g$ lies in the fixed field $L^G = K$ (because of your assumption that it is a normal extension - I assume you can use this fact?).

Therefore, $g\in K[x]$ is a polynomial which has $\alpha$ as a root. Therefore, $f\mid g$ in $K[x]$, so it follows that $\deg(f) \leq \deg(g) = |G|$.

Answer 2

I think it’s easier to work with intermediate fields rather than polynomials here.

Let $L/K$ be infinite.

1. Show that there is no bound on the degree $[M : K]$ of finite intermediate Galois extensions $M/K$. (You should be able to reuse your argument for polynomials here.)

2. By the fundamental theorem of Galois theory, $$\operatorname{Gal}(M/K) = \operatorname{Gal}(L/K) / \operatorname{Gal}(L/M).$$ In particular, $\left|\operatorname{Gal}(L/K)\right| \geq \left|\operatorname{Gal}(M/K)\right|$ for every intermediate Galois extension $M/K$.

Answer 3

The easiest way to prove Galois group is finite implies its Galois extension is also finite, in my opinion, is to use a bit of knowledge of topological groups which is basically a group satisfying a few more topological properties. From that, you can prove that there is a bijection between an intermediate field $K \subset E \subset L$ and a closed subgroup of $Gal(L/K)$, then your statement can be easily proved.

However, I'll give you another approach that finishes your last question without using Zorn's lemma. Concretely, I'll prove the stronger lemma as follows.

Lemma: Given field $K$ with Galois extension $L$, if $E$ is subfield of $L$ containing $K$, then any embedding of $E$ into $\bar{K}$ which fixes $L$ is of the form $\left.\phi\right|_{E}$ for some $\phi \in Gal(L/K)$.

In the case $K/L$ is finite, it's obvious. Otherwise, as far as I know, the standard way to prove is to use Zorn's lemma (which appears in most of the textbooks). But I have another solution without using Zorn's lemma (as you preferred), and that's why my solution is kinda irreverent to the main point of your question. Note that Zorn's lemma is equivalent to the axiom of choice, and for many years some mathematicians rejected the choice axiom as a result. A theorem was considered to be on somewhat shaky ground if one had to use the choice axiom in its proof. Present-day mathematicians, by and large, do not have such qualms. Roughly speaking, the lemma above is true only if we accept the axiom of choice, so when you try to find a different proof of a statement that involves Zorn's lemma in its original proof, you have to think about something related to axiom of choice, in my case is well-ordering theorem which states that every set has an order relation that makes the set itself is a well-ordering.

Main Idea of Proof: Let $\tau: E \to \bar{K}$ be an embedding fixing $K$. It's obvious that $\tau(E) \subset L$ (why?). Let $B$ be the infinite set of basic of vector space $L$ over field $K$ and $C = B \cap E$, any automorphism of $L$ is uniquely determined by the actions on the basic elements. So to extend $\tau$ to an automorphism of $L$, what we need to do is find another set $B'$ of basic elements of $L$ over $K$ such that $C \subset B'$. Then the automorphism that sends $B \to B'$ and fixes $C$ is an extension of $\tau$ that satisfies the requirement. To do this, firstly, we well-order $B-C$, once for all, in such a way that $B-C$ has a largest element. Then we construct $B'$ using transfinite induction, initially $B'=B-C$. For $b_\alpha$ is the smallest element in $B-C$ (of course, after well-ordering), it's easy to prove $\tau$ can be extended to $E(b_\alpha)/K$ by considering $E(b_\alpha) = E[x]/(p(x))$ for some $p(x)$, call the extension $\sigma_\alpha$, then add $\sigma_\alpha(b_\alpha)$ to $B'$. Supposing $B'$ has been constructed for all $b_\alpha$ with $\alpha < \beta$. We can extend $\sigma : E(B' \cup {b_\beta}) \to \bar{K}$ by the inductive hypothesis, then add $\sigma(b_\beta)$. Continuing the process until we reach the largest element in $B$, then we have set $B'$ which together with $B$ form an automorphism fixing $E$ (note that $E(B)=E(B')=L$).

In your question, $K(\alpha) \to L$ is actually an embedding, so we can apply the lemma above to extend it to $L \to L$ as desired.

Hope this helps!