I hope that the below is not too long to be useful. I just thought that I would expand upon my above comment. But, then I realized that the literal words weren't helpful without some explanation of what they themselves mean geometrically. Things spiraled out of control, and I ended up writing a novella.
Some background on étale morphisms:
A morphism of schemes (or varieties if you'd prefer) $f:X\to Y$ is étale if any of the following equivalent things are satisfied:
- $f$ is locally of finite presentation, flat, and unramified.
- $f$ is smooth of relative dimension $0$.
- $f$ is locally of finite presentation, flat, and for all geometric points $\overline{y}$ of $y$, the map $X_{\overline{y}}\to\overline{y}$ is isomorphic (as $\overline{y}$ schemes) to a finite disjoint union of $\overline{y}$.
- $f$ is locally of finite presentation and is formally étale.
Now, I think the one with the easiest geometric content to describe is 1. So, let me say some words about what each of the the three terms 'locally of finite presentation', 'flat', and 'unramified' mean.
First, locally of finite presentation means geometrically (ignoring Noetherian complications) that the fibers of $f$ (as varieties) are finite dimensional. So, you can't get something like $\mathrm{Spec}(k[x_1,x_2,\ldots])=\mathbb{A}^\infty_k$ as a fiber of $f$.
Flatness is an algebraic condition, but can be largely intuited as the statement that 'the fibers of $f$ vary continuously'. As a good example of this, one might imagine that if $f:X\to Y$ has continuously varying fibers then this should mean, in particular, that the fibers $f^{-1}(y)=:X_y$ have constant dimensions (fibers shouldn't vary continuously if they jump dimensions). This is roughly true, and under some mild conditions on $X$ and $Y$, the converse is also true.
Finally, we need to talk about the geometric content of unramifiedness. But, before I say something about that, let me cut to the punchline (since this will help us intuit unramifiednes). What are étale morphisms supposed to do? Namely, what is their intuitive geometric content. Well, the rough slogan is
étale morphisms are like local isomorphism or, said differently, covering maps.
The usefulness of having such a notion is clear. Not only do we often times want to talk about covering maps in any geometric setting, but if we could have a reasonable theory of covering maps for a scheme $X$ we might be able to define a notion of, say, the étale fundamental group of $X$. Indeed, in topology $\pi_1$ is the group which controls covers. So, in algebraic geometry we might hope that '$\pi_1$' (in étale terms) is the group which controls 'covers' (i.e. étale maps).
So, back to the geometric content of unramifiedness. Now, as of now, neither of the conditions that we have imposed (locally of finite presentation and flat) discount the algebro-geometric analogue of a ramified covering of curves $C\to\mathbb{P}^1$ (in complex geometry). Indeed, every non-constant map of (smooth projective) curves is automatically of finite presentation and flat!
And, recall that a ramified covering is a covering map outside of the finitely many ramified points. So, our goal now should be to think of an algebro-geometric notion which discounts these 'ramification points'.
So, let's imagine what a ramified covering is to look like. It should like like a couple of sheets (what should be the sheets of the covering map), but that they come together at these ramification points (stopping the map from being an actual covering map). The following image borrowed from Wikipedia nicely illustrates the concept:
(I can't figure out how to center it!)
So, what is a geometric condition which we can impose on $f$ which will discount these ramification points, these points where the sheets collide. Well, there are several equivalent possibilities:
- $f$ is unramified if $\Omega^1_{X/Y}=0$.
- $f$ is unramified if for all $x\in X$ and $y=f(x)$ we have that $\mathcal{m}_y\mathcal{O}_{X,x}=\mathcal{m}_x$, and $k(x)/k(y)$ is finite separable.
- $f$ is unramified if the diagonal map $\Delta$ is an open embedding.
If you are coming from complex curve theory, then it is 2. that you are most likely familiar with. Namely, in complex curve theory we don't have any extension of residue fields (they're both $\mathbb{C}$ for closed points!) but, in general, $\mathcal{m}_y\mathcal{O}_{X,x}$ will be a power of $\mathcal{m}_x$. This power is the integer $e_x$ (this is the same $e_x$ as in Dr. Lubin's post). Thus, $f$ should be unramified if and only if $e_x=1$ for all $x$.
That said, I would like to quickly say why 3. (since it's the simplest) makes geometric sense. Look at the picture above of a ramified map. I want to explain why this map does not have an open diagonal map. Namely, let's look, in $X$, at the first (from the left) ramification point $x_0$. And, let's choose a sequence of points $(x_n,y_n)$ (this is all intuition) on the top two sheets coming together at $x_0$, subject to the condition that $f(x_n)=f(y_n)$ (i.e. that they are in the same vertical position relative to $Y$). Note then that we have chosen a sequence of points $(x_n,y_n)\in X\times_Y X$ which are NOT in the diagonal $\Delta$. That said, $(x_n,y_n)$ converges to $(x_0,x_0)$, a point IN the diagonal. Thus, $X\times_Y X-\Delta$ is NOT closed and so $\Delta$ is NOT open. This is the geometric intuition for $3$.
Let me end this section with two examples:
- Let me first state rigorously this complex geometric analogue I've been mentioning. Namely, let's assume that $X$ and $Y$ are varieties over $\mathbb{C}$. Then, $f:X\to Y$ is étale if and only if $f^\mathrm{an}:X^\mathrm{an}\to Y^\mathrm{an}$ (here I am thinking of the analytifications) is a local biholomorphism. A simpler case shows that if $X$ and $Y$ are (smooth projective) curves over $\mathbb{C}$, then $f:X\to Y$ (non-constant) is étale if and only if the map $f^\mathrm{an}:X^\mathrm{an}\to Y^\mathrm{an}$ is a covering map.
- Suppose now that $X$ And $Y$ are just spectra of fields, and so we have a map $f:\mathrm{Spec}(L)\to\mathrm{Spec}(K)$ correspond to an extension of fields $L/K$. I claim that $f$ is unramified if and only if $L/K$ is finite separable. Indeed, it's clear that if $f$ is étale then $L/K$ is finite separable (look at condition 2. in the unramified condition). Conversely, if $L/K$ is finite separable, then 2. it is clearly flat, of finite presentation, and satisfies 2. of the unramifiedness condition.
But, let me give some intuition about why this second bullet should should make sense in more simple terms. If $\mathrm{Spec}(L)\to\mathrm{Spec}(K)$ is going to be a 'covering map' then it should just be a finite disjoint union of points—it's a covering map of a point! But, as all things in algebraic geometry, the 'obvious' geometry only happens over $\overline{K}$. So, really what we should say is that if $L/K$ corresponds to a covering map, then we'd expect $\mathrm{Spec}(L\otimes_{K}\overline{K})\to\mathrm{Spec}(\overline{K})$ to be just a finite trivial covering of points. And, indeed, for an extension $L/K$ of fields, it's étale if and only if $L\otimes_K \overline{K}$ is isomorphic to $\overline{K}^{[L:K]}$ (which, geometrically, is a disjoint union of $[L:K]$ points).
If you want to learn more about flat or unramified morphisms, I'll give a shameless plug for my blog. This post and this post in particular are relevant to the above.
Finite Galois covers
So, now we have given a geometric face to a finite separable extension of fields, but you actually asked about Galois, not just separable. So, let's now discuss the geometric content of a Galois cover.
Let's first mention that we only ever talk about finite étale covers when we're trying to think about covering spaces. Why? Well, in simple terms, it's because they're the algebraic ones. Namely, consider the complex manifold $\mathbb{C}^\times$. This has a geometric analogue, namely $\mathbf{G}_m=\mathbb{A}^1_\mathbb{C}-\{0\}$. So, we might hope to glean some of the 'coverings' of $\mathbf{G}_m$ from $\mathbb{C}^\times$. This is totally fine for all finite coverings of $\mathbb{C}^\times$ which, essentially, are just the maps $\mathbb{C}^\times\to\mathbb{C}^\times$ given by $z\mapsto z^n$. These are algebraic, and so we can translate them over to the algebraic world of $\mathbf{G}_m$. But, the infinite covers, like the universal cover $\exp:\mathbb{C}\to\mathbb{C}^\times$ are now no longer algebraic, and so can't be described in the world of varieties/schemes. So, this is why we only discuss the finite covers.
So, what is our goal? We want to define what it means for a morphism $f:X\to Y$ (where, for comfort, we assume that $X$ and $Y$ are both connected) to be a finite étale Galois cover. What should the topological analogue of this be? Well, what we're after is trying to get the right analogy of a (finite) Galois cover of topological spaces.
Recall that a covering space $f:X\to Y$ is called Galois if the group of deck transformations $\mathrm{Aut}(X/Y)$ acts (simply) transitively on the fibers $f^{-1}(y)$ for all $y$. These are the 'homogenous covers' where all the points are permuted by the action of $\mathrm{Aut}(X/Y)$. They are also the covers which correspond to normal subgroups of $\pi_1(X)=\mathrm{Aut}(\widetilde{X}/X)$.
So, a naive guess would be that a morphism of varieties/schemes $f:X\to Y$ should be a finite Galois cover if for all points $y\in Y$ we have that $\mathrm{Aut}(X/Y)$ acts transitively on $f^{-1}(y)$. But, this is bad for the same reason as why a separable extension $L/K$ is not a trivial cover at face value—real geometry only occurs over algebraically closed fields.
For this reason, we want to replace literal fibers with so-called geometric fibers. Namely, recall that a geometric point of $Y$ is a morphism $\overline{y}:\mathrm{Spec}(\Omega)\to Y$, where $\Omega$ is an algebraically closed field. We then define, for any geometric point $\overline{y}$ of $Y$ the geometric fiber $X_{\overline{y}}$ to be all geometric points lifting $\overline{y}$. Namely, all geometric points $\overline{x}:\mathrm{Spec}(\Omega')\to X$ on $X$ such that $f\circ\overline{x}=\overline{y}$.
We then have the following definition of a finite Galois cover $f:X\to Y$. It is a finite surjective étale morphism such that $\mathrm{Gal}(X/Y):=\mathrm{Aut}(X/Y)$ acts transitively on any (equiv. one) geometric fiber of $Y$.
Let me make two comments. First, $\mathrm{Gal}(X/Y)$ already acts faithfully on a geometric fiber by an analogue of the 'rigidity of covering maps', and so we can replace 'transitively' by 'simply transitively'. Second, note that we require finite in the second of scheme theory here. Namely, we require that $f$ is an affine map. Why should this make sense? Well, if we're trying to capture the geometry of a finite covering map, then we'd hope (since this is true for finite covering maps) that $f$ is proper and quasi-finite (finite fibers), but these two things (by Zariski's main theorem) actually imply (scheme-theoretic) finiteness.
There is another useful characterization of when a finite étale map $f:X\to Y$ is Galois. Namely, one can show that $G:=\mathrm{Aut}(X/Y)$ acts so-called 'admissibly' (under very mild assumptions) on $X$, and thus it make sense to define the variety/scheme $X/G$. One can then show that $X\to Y$ is a finite Galois cover if and only if the induced map $X/G\to Y$ is an isomorphism. In fact, one then has a 'Galois correspondence' between the intermediary étale maps $X\to Z\to Y$ and the subgroups of $G$, the Galois ones corresponding to normal subgroups.
So, let us now give the two examples which, at this point, are probably pretty obvious.
- Let $X$ and $Y$ be (smooth projective) curves over $\mathbb{C}$. Then, a map $f:X\to Y$ is a finite étale Galois cover if $f^\mathrm{an}:X^\mathrm{an}\to Y^\mathrm{an}$ is a covering space. And, $f$ is Galois if and only if the covering space $X^\mathrm{an}\to Y^\mathrm{an}$ is a Galois covering space.
- Let $\mathrm{Spec}(L)\to\mathrm{Spec}(K)$ be an extension of fields. Then, it is a finite étale cover if and only if $L/K$ is finite separable, and it is a Galois cover if and only if $L/K$ is finite Galois.
Since it was the main question of your post, let me explain in more detail why the second bullet is true.
First, let us figure out what this geometric fiber means. Namely, we said we can check for any geometric fiber on $\mathrm{Spec}(K)$. But, there is an obvious one. Namely, choose an embedding $K\hookrightarrow \overline{K}$, which gives a geometric point $\overline{y}\mathrm{Spec}(\overline{K})\to\mathrm{Spec}(K)$. What then is the geometric fiber of $f:\mathrm{Spec}(L)\to\mathrm{Spec}(K)$ over this geometric point? Well, it is:
$$\begin{aligned}\left\{\overline{x}:\mathrm{Spec}(\overline{K})\to\mathrm{Spec}(L):f\circ\overline{x}=\overline{y}\right\} &= \left\{L\to\overline{K}:K\to L\to\overline{X}=\overline{y}\right\}\\ &= \mathrm{Hom}_K(L,\overline{K})\end{aligned}$$
The action of $\mathrm{Aut}(\mathrm{Spec}(L)/\mathrm{Spec}(K))=\mathrm{Aut}(L/K)$ on $\mathrm{Hom}_K(L,\overline{K})$ is the obvious one. And, as is classical, $L/K$ should Galois if and only if this action is transitive.
For example, $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is finite separable, and so $\mathrm{Spec}(\mathbb{Q}(\sqrt[3]{2}))\to\mathrm{Spec}(\mathbb{Q})$ is a finite étale cover. But, it's NOT Galois since its automorphism group (the trivial group) does not act transitively on the geometric fiber
$$\mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[3]{2}),\overline{\mathbb{Q}})=\{\sqrt[3]{2}\mapsto \sqrt[3]{2},\sqrt[3]{2}\mapsto \zeta_3\sqrt[3]{2},\sqrt[3]{2}\mapsto \zeta_3^2\sqrt[3]{2}\}$$
In fact, the example shows you that if $\mathrm{Spec}(L)\to\mathrm{Spec}(K)$ is a finite étale cover, so $L=K(\alpha)$ (by the primitive element theorem), that the geometric fiber corresponds to the roots of the minimal polynomial of $\alpha$, and transitivity of the action corresponds precisely to all of these roots being in $L$—in other words, $L/K$ being normal.
Also, one can interpret this in terms of the $X/G=Y$ construction. Namely, if $G$ is a finite group acting on an affine scheme $\mathrm{Spec}(A)$ ,then $\mathrm{Spec}(A)/G$ is just $\mathrm{Spec}(A^G)$. Thus, for $\mathrm{Spec}(L)/G$ to be equal to $\mathrm{Spec}(K)$ is just the statement that $L^G=K$, as per usual. Also, the Galois correspondence mentioned above is the usual Galois correspondence.
Bonus: étale fundamental group
Having written all of the above, I would feel remiss to not give the 'punchline' of the above.
Namely, once one has gone through all the effort of defining the notion of a 'connected Galois cover' $X\to Y$, which is supposed to be like finite Galois covering spaces, one better use it to define a notion of the fundamental group. Namely, one can define the étale fundamental group of $(X,\overline{x})$ (i.e. we need to choose a geometric 'base point') to be
$$\pi_1^\mathrm{\acute{e}t}(X,\overline{x})=\varprojlim \mathrm{Gal}(X'/X)$$
where $(X',\overline{x'})$ runs over the 'pointed' connected finite Galois covers of $X$. You can largely ignore the pointing.
This is analogous to the usual fundamental group in two ways. Namely, one can define $\pi_1(X,x)$ to be the projective limit over $\mathrm{Aut}(X'/X)$, connected Galois covering spaces. But, in the topological world, this projective system has a final object, namely, the universal cover $\overline{X}\to X$ and so this inverse limit is just $\mathrm{Aut}(\overline{X}/X)$ which is classically known to be $\pi_1(X,x)$. For reasons mentioned above, in the algebraic world we don't have access to a 'universal cover' (it won't be given by an algebraic map) and so we can't simplify this inverse limit in the same way.
Also, one can define a fancier notion of a 'fiber functor' which associates to any finite connected covering spaces $f:X'\to X$ the fiber $f^{-1}(x)$ (note the base point has a role here). This then defines a functor $\mathrm{Cov}(X)\to\mathsf{Set}$ ($\mathrm{Cov}(X)$ is the category of finite covering spaces of $X$). One can then prove that $\pi_1(X,x)$ is the automorphism group of this functor. If one makes the exact analogous definitions in the étale world, one recovers the étale fundamental group as the automorphisms of the geometric fiber functor.
Let me end by giving the same two examples:
- If $X/\mathbb{C}$ is a variety, then the Riemann Existence Theorem implies that $\pi_1^\mathrm{\acute{e}t}(X,\overline{x})$ is $\widehat{\pi_1(X^\mathrm{an})}$—the profinite completion of the topological fundamental group of the analytification. This makes sense since the étale fundamental group is the inverse limit over $\mathrm{Aut}$'s of finite covers, which corresponds to taking the direct limit (on the topological side) of the Aut's of finite covers which are the finite quotients of the topological fundamental group—thus the profinite completion.
- For a field $X=\mathrm{Spec}(K)$, choosing a geometric base point $\overline{x}$ is the same thing as choosing an algebraic closure $\overline{K}$. Then, one can check (it's a nice exercise in the definitions) that $\pi_1^{\mathrm{\acute{e}t}}(X,\overline{x})=\mathrm{Gal}(K^\mathrm{sep}/K)$. Thus, in some sense, we can think of the separable closure $K^\mathrm{sep}$ of $K$ as being the 'universal cover' of $K$.
EDIT: This is a cursory answer to Dorebell's question in the comments. I'll come back later to answer more in-depth.
Namely, when is a map of, say, compact Riemann surfaces $X\to Y$ correspond to a Galois cover $K(X)/K(Y)$. For me, a Riemann surface is automatically connected.
Well, the obvious guess is the correct one, once one thinks about what $K(X)/K(Y)$ is geometrically. Namely, $K(X)$ is the stalk of $\mathcal{O}_{X,x}$ at the generic point of $X$, and similarly for $Y$. So, we'd expect that the data which is going on in $K(X)/K(Y)$ should be something like 'what happens generically'.
So, for example, we'd expect that $K(X)/K(Y)$ is separable since generically (say, on dense opens) the map $f:X\to Y$ is étale (away from the finitely many ramification points). Of course, this is a silly observation since $K(Y)$ is characteristic $0$, and so perfect. But, it's a good illustration of the concept and, in fact, can be used to prove the other direction in our thought experiment—that since $K(X)/K(Y)$ is separable, that $f:X\to Y$ should be generically étale.
So, we expect to be able to say that $K(X)/K(Y)$ is Galois if and only if $f:X\to Y$ is 'generically Galois. It's a little less clear what this means, but a little thought sorts it out. Namely, let's make our statement that $K(X)/K(Y)$ being separable means that $f:X\to Y$ is generically separable more concrete.
Namely, we can take the finite set $S_Y\subseteq Y$ of ramification values and $S_X=f^{-1}(S)\subseteq X$ and consider the holmorphic map $f':X'\to Y'$ where $X'=X-S_X$, $Y'=Y-Y_S$, and $f'=f\mid_{X'}$. Now, $f'$ is an honest to god covering map, opposed to a 'branched cover'. Moreover, $f'$ determines $f$ in the usual way—explicitly, this generically defined $f'$ still captures the generic data of $K(X)/K(Y)$ which, by the standard theory, is all one needs to capture $f:X\to Y$.
One then expects that $K(X)/K(Y)$ is Galois if and only if $f'$ is a covering map. And, indeed, this is the case as one can easily prove.
One thing to note that is that we chose the sets $S_Y$ and $S_X$'s sort of minimally. Since $K(X)/K(Y)$ is only capturing generic data, we should be able to choose any sets $T_Y\supseteq S_Y$ and $T_X=f^{-1}(T_Y)$ and have the same discussion. But, there is no difference. Namely, one can check the Galoisness of a cover at ONE point (much like one can check normality at one generator), and so the choice really doesn't matter.
For anyone who would like to see proof of the above statements about complex curves, I highly recommend Szamuely's book Galois Groups and Fundamental Groups.
Best Answer
I think this is not at all an elementary result; it really tells you something nontrivial about the absolute Galois group of $\mathbb{Q}$ (note that it's equivalent to the existence of infinitely many distinct open normal subgroups of index $d$). For example, I think it doesn't hold for $\mathbb{Q}_p$ because the absolute Galois group of $\mathbb{Q}_p$ is topologically finitely generated, which implies there are only finitely many Galois extensions with a fixed Galois group and hence finitely many Galois extensions with a fixed degree.
I think the fastest way to do this "from scratch" is to exhibit subextensions of cyclotomic extensions:
The existence of the cyclotomic extensions $\mathbb{Q}(\zeta_n)$ implies that the absolute Galois group surjects onto $\widehat{\mathbb{Z}}^{\times}$, the group of units of the profinite integers, which is an enormous group and in particular not topologically finitely generated. (When we consider subextensions of $\mathbb{Q}(\zeta_p)$ we are considering the projection of this group to $(\mathbb{Z}_p / p \mathbb{Z}_p)^{\times}$.)
We can ask for elementary methods in each small degree at least. For degree $3$ we want to exhibit infinitely many cubics with Galois group $C_3$ with nonisomorphic splitting fields. This can be done without cyclotomic extensions by proving that an irreducible cubic has Galois group $C_3$ iff its discriminant is a square and exhibiting a family of irreducible cubics with square discriminants. This is already not entirely elementary. The discriminant of $x^3 - px + q$ is $\Delta = 4p^3 - 27 q^2$ so requiring that this is a square involves solving a Diophantine equation
$$4p^3 - 27q^2 = r^2.$$
Equivalently we want to exhibit infinitely many $p$ such that $4p^3$ can be represented by the quadratic form $r^2 + 27q^2$. Recognizing this as the norm of $r + 3q \sqrt{-3} \in \mathbb{Q}(\sqrt{-3})$, we can do this by taking $p = x^2 + 3y^2 = N(x + y \sqrt{-3})$ and writing $4 = N(1 + \sqrt{-3})$, which gives
$$4p^3 = N \left( (1 + \sqrt{-3})(x + y \sqrt{-3})^3 \right) = (x^3 - 9x^2 y - 9 xy^2 + 9y^3)^2 + 3 (x^3 + 3 x^2 y - 9 xy^2 - 3y^3)^2$$
which gives us our infinite family of solutions. Now we have to prove that infinitely many of these cubics are irreducible and that infinitely many of their splitting fields are distinct which seems a bit painful; at this point this seems to me like more work than using cyclotomic fields.
For degree $4$ things are easier because we can piggyback off of the degree $2$ case: we can consider the biquadratic extensions $\mathbb{Q}(\sqrt{a}, \sqrt{b})$ where $\gcd(a, b) = 1$.
For degree $5$ we want to exhibit infinitely many quintics with Galois group $C_5$ with nonisomorphic splitting fields, which without cyclotomic extensions seems hard again. In general things will be hard in prime degree $p$ since the Galois group must be $C_p$ and we have no intermediate fields to piggyback off of. For large $p$, and by "large" here I mean $p \ge 5$, I don't think it's clear a priori that there exists even one such extension.