Let be a field, and let be a finite extension of that field; in this post we will denote such a relationship by . We say that is a Galois extension of if the cardinality of the automorphism group of fixing is as large as it can be, namely the degree of the extension. In that case, we call the Galois group of over and denote it also by . The fundamental theorem of Galois theory then gives a one-to-one correspondence (also known as the Galois correspondence) between the intermediate extensions between and and the subgroups of :

Theorem 1 (Fundamental theorem of Galois theory) Let be a Galois extension of . (i) If is an intermediate field betwen and , then is a Galois extension of , and is a subgroup of .

is an intermediate field betwen and , then is a Galois extension of , and is a subgroup of . (ii) Conversely, if is a subgroup of , then there is a unique intermediate field such that ; namely is the set of elements of that are fixed by .

is a subgroup of , then there is a unique intermediate field such that ; namely is the set of elements of that are fixed by . (iii) If and , then if and only if is a subgroup of .

and , then if and only if is a subgroup of . (iv) If is an intermediate field between and , then is a Galois extension of if and only if is a normal subgroup of . In that case, is isomorphic to the quotient group .

Example 2 Let , and let be the degree Galois extension formed by adjoining a primitive root of unity (that is to say, is the cyclotomic field of order ). Then is isomorphic to the multiplicative cyclic group (the invertible elements of the ring ). Amongst the intermediate fields, one has the cyclotomic fields of the form where divides ; they are also Galois extensions, with isomorphic to and isomorphic to the elements of such that modulo . (There can also be other intermediate fields, corresponding to other subgroups of .)

Example 3 Let be the field of rational functions of one indeterminate with complex coefficients, and let be the field formed by adjoining an root to , thus . Then is a degree Galois extension of with Galois group isomorphic to (with an element corresponding to the field automorphism of that sends to ). The intermediate fields are of the form where divides ; they are also Galois extensions, with isomorphic to and isomorphic to the multiples of in .

There is an analogous Galois correspondence in the covering theory of manifolds. For simplicity we restrict attention to finite covers. If is a connected manifold and is a finite covering map of by another connected manifold , we denote this relationship by . (Later on we will change our function notations slightly and write in place of the more traditional , and similarly for the deck transformations below; more on this below the fold.) If , we can define to be the group of deck transformations: continuous maps which preserve the fibres of . We say that this covering map is a Galois cover if the cardinality of the group is as large as it can be. In that case we call the Galois group of over and denote it by .

Suppose is a finite cover of . An intermediate cover between and is a cover of by , such that , in such a way that the covering maps are compatible, in the sense that is the composition of and . This sort of compatibilty condition will be implicitly assumed whenever we chain together multiple instances of the notation. Two intermediate covers are equivalent if they cover each other, in a fashion compatible with all the other covering maps, thus and . We then have the analogous Galois correspondence:

Theorem 4 (Fundamental theorem of covering spaces) Let be a Galois covering. (i) If is an intermediate cover betwen and , then is a Galois extension of , and is a subgroup of .

is an intermediate cover betwen and , then is a Galois extension of , and is a subgroup of . (ii) Conversely, if is a subgroup of , then there is a intermediate cover , unique up to equivalence, such that .

is a subgroup of , then there is a intermediate cover , unique up to equivalence, such that . (iii) If and , then if and only if is a subgroup of .

and , then if and only if is a subgroup of . (iv) If , then is a Galois cover of if and only if is a normal subgroup of . In that case, is isomorphic to the quotient group .

Example 5 Let , and let be the -fold cover of with covering map . Then is a Galois cover of , and is isomorphic to the cyclic group . The intermediate covers are (up to equivalence) of the form with covering map where divides ; they are also Galois covers, with isomorphic to and isomorphic to the multiples of in .

Given the strong similarity between the two theorems, it is natural to ask if there is some more concrete connection between Galois theory and the theory of finite covers.

In one direction, if the manifolds have an algebraic structure (or a complex structure), then one can relate covering spaces to field extensions by considering the field of rational functions (or meromorphic functions) on the space. For instance, if and is the coordinate on , one can consider the field of rational functions on ; the -fold cover with coordinate from Example 5 similarly has a field of rational functions. The covering relates the two coordinates by the relation , at which point one sees that the rational functions on are a degree extension of that of (formed by adjoining the root of unity to ). In this way we see that Example 5 is in fact closely related to Example 3.

Exercise 6 What happens if one uses meromorphic functions in place of rational functions in the above example? (To answer this question, I found it convenient to use a discrete Fourier transform associated to the multiplicative action of the roots of unity on to decompose the meromorphic functions on as a linear combination of functions invariant under this action, times a power of the coordinate for .)

I was curious however about the reverse direction. Starting with some field extensions , is it is possible to create manifold like spaces associated to these fields in such a fashion that (say) behaves like a “covering space” to with a group of deck transformations isomorphic to , so that the Galois correspondences agree? Also, given how the notion of a path (and associated concepts such as loops, monodromy and the fundamental group) play a prominent role in the theory of covering spaces, can spaces such as or also come with a notion of a path that is somehow compatible with the Galois correspondence?

The standard answer from modern algebraic geometry (as articulated for instance in this nice MathOverflow answer by Minhyong Kim) is to set equal to the spectrum of the field . As a set, the spectrum of a commutative ring is defined as the set of prime ideals of . Generally speaking, the map that maps a commutative ring to its spectrum tends to act like an inverse of the operation that maps a space to a ring of functions on that space. For instance, if one considers the commutative ring of regular functions on , then each point in gives rise to the prime ideal , and one can check that these are the only such prime ideals (other than the zero ideal ), giving an almost one-to-one correspondence between and . (The zero ideal corresponds instead to the generic point of .)

Of course, the spectrum of a field such as is just a point, as the zero ideal is the only prime ideal. Naively, it would then seem that there is not enough space inside such a point to support a rich enough structure of paths to recover the Galois theory of this field. In modern algebraic geometry, one addresses this issue by considering not just the set-theoretic elements of , but more general “base points” that map from some other (affine) scheme to (one could also consider non-affine base points of course). One has to rework many of the fundamentals of the subject to accommodate this “relative point of view“, for instance replacing the usual notion of topology with an étale topology, but once one does so one obtains a very satisfactory theory.

As an exercise, I set myself the task of trying to interpret Galois theory as an analogue of covering space theory in a more classical fashion, without explicit reference to more modern concepts such as schemes, spectra, or étale topology. After some experimentation, I found a reasonably satisfactory way to do so as follows. The space that one associates with in this classical perspective is not the single point , but instead the much larger space consisting of ring homomorphisms from to arbitrary integral domains ; informally, consists of all the “models” or “representations” of (in the spirit of this previous blog post). (There is a technical set-theoretic issue here because the class of integral domains is a proper class, so that will also be a proper class; I will completely ignore such technicalities in this post.) We view each such homomorphism as a single point in . The analogous notion of a path from one point to another is then a homomorphism of integral domains, such that is the composition of with . Note that every prime ideal in the spectrum of a commutative ring gives rise to a point in the space defined here, namely the quotient map to the ring , which is an integral domain because is prime. So one can think of as being a distinguished subset of ; alternatively, one can think of as a sort of “penumbra” surrounding . In particular, when is a field, defines a special point in , namely the identity homomorphism .

Below the fold I would like to record this interpretation of Galois theory, by first revisiting the theory of covering spaces using paths as the basic building block, and then adapting that theory to the theory of field extensions using the spaces indicated above. This is not too far from the usual scheme-theoretic way of phrasing the connection between the two topics (basically I have replaced étale-type points with more classical points ), but I had not seen it explicitly articulated before, so I am recording it here for my own benefit and for any other readers who may be interested.

— 1. Some notation on functions —

It will be convenient to adopt notation in which many of the basic hypotheses we use take the form of an associativity-type law. To this end, we will use two non-standard notations for functions (which is implicit in the usual notation for left and right group actions), sometimes referred to as reverse Polish and Polish notation (for left and right actions respectively).

Definition 7 (Polish and reverse Polish notation) Let be sets. A function acting on the right is a function in which the notation for evaluating at an element is denoted rather than the more standard . If are two functions acting on the right, we write for the composition that is more commonly denoted ; this way we have the associativity-type property for . Similarly, a function acting on the left is a function in which the notation for evaluating at an element is denoted rather than the more standard . If are two functions acting on the left, we write for the composition that is more commonly denoted ; this way we have the associativity type property for . We do not define a composition between a function acting on the right and a function acting on the left.

Remark 8 Functions acting on the left largely correspond to the traditional notion of a function, despite the reversed arrow in the notation ; functions acting on the right can be thought of as “antifunctions”, in which the composition law is reversed.

As a general rule, in this post covering maps and deck transformations will be functions acting on the left, while paths will be functions acting on the right.

When viewing the (left or right) automorphisms of a set as a group, we use juxtaposition rather than composition as the group law; thus, in the case of automorphisms acting on the right, the group structure we will use is the opposite of the usual composition group structure, whereas for automorphisms acting on the left the group structure agrees with the standard composition group structure. In particular, the group of right-automorphisms of a space is canonically identified with the opposite group of the group of left-automorphisms of a space . (Of course, the opposite of a group is in turn canonically identified with the group itself through the inversion operation , but we will refrain from using this further identification in this text as it can cause some additional confusion.)

As is customary, we will adopt the notational convention of omitting parentheses whenever there is an associativity-type law to prevent any ambiguity. For instance, if and , are functions acting on the right, we may write without any ambiguity thanks to (1).

— 2. Covers of manifolds —

We begin by revisiting the theory of covering spaces of manifolds. To align with the way we will be thinking about Galois theory, it will be convenient to assume the existence of a (connected) base manifold , such that all the other manifolds we will be considering are finite extensions of that base . Again, for simplicity (and to make the theory more closely resemble Galois theory) we will restrict attention to finite covers, though one could extend most of this discussion to infinite covers without much difficulty.

One can think of the base manifold as a category, in which the objects are the points of the manifold, and the morphisms are the paths in connecting one point to another , with the composition of two paths (connecting to ) with (connecting to ) being a path connecting with formed by concatenating the two paths together. We leave the composition undefined if the terminal point of does not agree with the starting point of . We will not distinguish a path from a reparameterisation of that path (in fact we will not mention parameterisations at all); in particular, we have the associativity law

We denote the space of all paths (up to reparameterisations) by . One should think of as being somewhat like a group, except that the multiplication law is not always defined. (More precisely, is not just a category, but is in fact a groupoid, although we will not explicitly use this fact here.)

To signify the fact that a path has starting point and terminal point , we write

If is not the starting point of , we leave undefined. Note this way that we obtain the associativity-type law

whenever one of the two sides is well-defined (which forces the other side to be well-defined also).

Now let be a finite cover with covering map ; here all covering maps are understood to act on the left. Given a point lying above a base point (thus ), and a path with starting point , we may lift up to a path in that starts at and ends at some point which we will denote by . This point will lie above in , thus we have the associativity-type law

whenever one of the two sides (and hence the other) is well-defined. It is easy to see that this defines an (right) action of in the sense that one has the associativity type law

whenever and are such that one of the two sides (and hence the other) is well-defined.

Note that if is connected and , then any other point can be connected to by a path in , which is the lift of some path in . Thus the action of is transitive in the sense that for every there exists such that .

If we have finite covers then the actions of on and are compatible in the sense that one has the associativity-type law

whenever one of the two sides (and hence the other) is well-defined. A deck transformation of (viewed as a cover of ) is a continuous map (acting on the left) such that

The space of deck transformations is clearly a group that acts on the left on . By working locally we see that deck transformations map lifts to lifts, thus one has the associativity-type law

whenever , , are such that one of the two sides (and hence the other) is well-defined. Conversely, one can check that any map that obeys the two properties (2), (3) is a deck transformation (even without assuming continuity of !). Hence one can in fact take (2), (3) to be the definition of a deck transformation; all the topological structure is now concealed within the path groupoid and its actions on and .

One consequence of (3) is that the action of this group is free: if is such that for some point , then we also have whenever makes sense, and hence by transitivity for all , so is the identity.

In particular, if is a degree cover of (so that all fibres for have cardinality ), then the order of the group cannot exceed , since the orbit of any given point under this free action lies in a fibre of above and has the same cardinality as . If the order equals , we say that the cover is a Galois cover, use to denote the group , and conclude that the action of is transitive on fibres. Thus, if are such that , then there exists a unique such that .

We can now prove Theorem 4. Let be a degree cover. Let be an intermediate cover between and , and suppose that has degree over . A fibre of over has degree , and splits into fibres of over , each of which has cardinality (so in particular divides ). Pick one of these fibres, and pick a point in it. By the above discussion, there are precisely elements of such that lies in the same fibre over as :

Applying paths in on the right and using transitivity (and associativity), we conclude that , thus . As we have located elements of this group, this is in fact the entire group of automorphisms, so we conclude that is a Galois cover of and is a subgroup of . This proves part (i) of the theorem.

A similar argument shows that if then is a subgroup of . In particular, if and are equivalent then . Conversely, suppose that and are such that is a subgroup of . If are such that , then by the Galois nature of over we can find such that . But then , hence . Thus we can factor for some uniquely defined function . It is not hard to verify that is a finite covering map that is compatible with the existing covering maps, and hence . This proves (iii). Reversing the roles of and , we conclude in particular that if then and are equivalent. This gives the “unique up to equivalence” portion of (ii).

To finish the proof of (ii), let be a subgroup of , then we can define an equivalence relation on by declaring for every and . If we set to be the quotient space with the obvious quotient map , then one has whenever (which is equivalent to ). Thus we may factor

for some unique map . One may easily verify that are both covering maps, so that we have . To finish the proof of (Ii) we need to show that . Unpacking the definitions, our goal is to show that is the set of all such that for all . Clearly if then for all . Conversely, suppose that is such that for all . In particular, if one fixed a point , then for some . This implies that for all ; but the action of on is transitive, and hence , giving the claim.

Finally, we prove (iv). By parts (i), (ii), we may assume without loss of generality that , where if for some .

First suppose that is a normal subgroup of . Then implies whenever and . Thus the action of on descends to that of ; since the former is transitive on fibres above , so is the latter. Hence is a Galois extension of . Conversely, suppose that is Galois. Then for any and , there must exist such that . Since the right action of on is transitive, does not actually depend on , and is uniquely specified by . Replacing by for any , we have

but , hence

or equivalently . This implies that , thus is preserved by conjugation by arbitrary elements of . In other words, is a normal subgroup. The map described here can easily be verified to be a homomorphism from to with kernel , so the claim (iv) follows from the first isomorphism theorem.

Remark 9 Let be a Galois cover, let be a point in the base , and let be a point in the fibre . For any loop starting and ending at (so that , the point lies in the same fibre of as , hence by the Galois nature of the cover there is a unique such that It is easy to see that homotopic deformations of (preserving the initial and final point ) do not affect the value of . Thus, we have constructed a map from the fundamental group of at to the Galois group , such that As is connected, we see that this map is surjective. From (4) and the fact that that acts freely on ) we also see that the map is a homomorphism. (In many texts, the notation is set up so that this correspondence is an antihomomorphism rather than a homomorphism; we have avoided this by making the paths act on the right and the deck transformations act on the left, thus implicitly introducing an order reversal when identifying the two.) Thus we see that the Galois group is isomorphic to a quotient of the fundamental group . However, this isomorphism is not canonical, even if one fixes the base point , because it depends on ! If one replaces by another point in the same fibre for some , then the associated map , defined by has to be conjugate to in order to remain compatible with (4): Thus, if one does not specify the reference point in the covering space , the identification of with a quotient of is only unique up to conjugation. Thus the relationship between fundamental groups and Galois groups are a little subtle, requiring one to be aware of what base points have been selected, unless one is willing to just work up to conjugacy. (See also the above-mentioned post of Kim for some further discussion of this point.)

— 3. Extensions of fields —

We can now give an analogous treatment of the Galois correspondence for finite extensions of fields.

In this setting, we will take the base space to be the category of integral domains (with morphisms now interpreted as “paths”). Thus, points in this space are integral domains, and the paths in this space are ring homomorphisms from one integral domain to another, which we will think of as functions acting on the right, in particular the composition of and will be denoted (rather than the more traditional ). As before, we use the notation to denote the assertion that is a ring homomorphism from to , so in particular we have the associativity type law

whenever either side is well-defined. As previously mentioned, we will ignore all set-theoretic issues caused by the fact that is a proper class rather than a set, and similarly for the covers of introduced below. We let denote the collection of all ring homomorphisms connecting points in . This is analogous to the space of collections of paths in the base manifold in the covering space context. Oen caveat however is that while paths in manifolds always have inverses, the same is not true in , because not every ring homomorphism between integral domains is invertible. This complicates the situation somewhat compared to the covering space case, but it turns out that the additional level of complication is manageable.

Given a field , we define an associated space , whose points are ring homomorphisms to some ring , where we view these homomorphism as acting on the right. There is an obvious map (viewed as a function acting on the left) that maps a ring homomorphism to its codomain . If , are such that and , then the composition is a ring homomorphism from to and is thus also an element of . Thus acts on the right on in the sense that

whenever either side is well-defined, and the action is compatible with the base in the sense that

whenever either side is well-defined.

Remark 10 One can define for any commutative ring in the same fashion. If one takes to be the integers , then is in one-to-one correspondence with , since for every integral domain there is a unique ring homomorphism from to . Thus we are really thinking of all our spaces as being covers of . In the usual scheme-theoretic language, one views the schemes as lying above the base scheme .

Remark 11 The characteristic of the field naturally restricts the base points that actually lies above. For instance, if has characteristic zero, then points of only lie above those base points for which the unit of has infinite order; similarly, if has a positive characteristic , then points of only lie above base points for which the unit of has order . This helps explain why fields of different characteristic seem to inhabit “different worlds” – they lie above disjoint (and disconnected) portions of the base space ! However, we will not exploit the field characteristic in this post.

As mentioned in the introduction, each space has a distinguished point , which is the identity map on and is the representative in this formalism of . This point makes is “directionally connected” in the following sense: every other point in is connected to by a path in the sense that . Indeed, one just takes to be precisely the same homomorphism as . (More generally, in every commutative ring , every point of will be connected via a path from a point in , namely the kernel of which is necessarily a prime ideal since is an integral domain. I like to think of as a sort of “penumbra” lurking around the much smaller set .) However, the connectedness does not flow the other way: not every point in is connected back to by a path, because paths need not be invertible. (For instance, if is an embedding of into a larger field , there is no way to map back into by a ring homomorphism while preserving , as homomorphisms of fields have to be injective.)

Suppose we have a finite extension of fields, then the inclusion map , viewed as a function acting on the right, is a ring homomorphism. As such, it also can be viewed as a map from to : any point of , when composed with , gives rise to a point of . We relabel this composition map as (now viewed as a function acting on the left), thus

for all . By construction we see that

and similarly for any nested sequence of fields one has

Now let us look at the fibres of above . We need the following basic result, which relates the degree of the field extension to the degree of the cover :

Proposition 12 Let be a degree extension of a field . Then every fibre in of a point in has cardinality at most .

Proof: Unpacking all the definitions, we are asking to show that every ring homomorphism into an integral domain has at most extensions to a ring homomorphism .

We induct on . If the claim is trivial, so suppose and the claim has already been proven for smaller values of . Then is larger than , so we may find an element that lies in but not . Let be the degree of , thus , is a degree extension of and is a degree extension of (in particular, divides ). By induction hypothesis, every ring homomorphism on has at most extensions to , so it suffices to show that has at most extensions to .

Let be the minimal polynomial of with coefficients in , thus is monic with degree , and . Any extension of to must obey the law , where is the monic polynomial of degree with coefficients in formed by applying to each coefficient of . As is an integral domain, has at most roots. Thus there are at most possible values of ; since is completely determined by and , we obtain the claim.

Example 13 Let and be the fields from Example 3. The inclusion map is a point in . The fibre consists of points , with being the field automorphism of that sends to (in particular, is the distinguished point of ). In contrast, the distinguished point of has empty fibre , since there is no ring homomorphism from to that fixes .

Remark 14 If two points in are connected by an invertible path , thus and , then it is easy to see that induces a one-to-one correspondence between the fibre and the fibre ; in particular, the two have the same cardinality. However, in contrast to the situation with covering spaces, not all paths are invertible, and so the cardinality of a fibre can vary with the base , as can already be seen in the preceding example. However the situation is better for Galois extensions, as we shall shortly see, in which all non-empty fibres are isomorphic.

We continue analysing a field extension , viewing as analogous to a covering space for . By analogy with covering spaces, we could define a “deck transformation” of over to be a map (acting on the left) such that

and such that

whenever , are such that one side (and hence the other) of the above equation is well-defined. Let us see what a deck transformation actually is. If we apply to the distinguished point of , we obtain a ring homomorphism . From (5) we have

which on chasing definitions means that is the identity on . Thus is a -linear map; it also preserves , so it has trivial kernel and is thus injective. As is finite dimensional over , it is also surjective; hence is in fact a field automorphism of that fixes . The action of on any other point of can then be computed by writing where is the path connecting to and using (6) to conclude that . Conversely, given any field automorphism that fixes , one can generate a deck transformation by setting for all ring homomorphisms ; by chasing definitions one verifies that this is a deck transformation. Thus we have a one-to-one correspondence between deck transformations and field automorphisms of fixing . (This correspondence feels reminiscent of the Yoneda lemma, though I was not able to find a direct link between the two.)

Exercise 15 Show that the correspondence between deck transformations and field automorphisms given above is a group isomorphism. (When functions are expressed in more traditional language, this correspondence is an antiisomorphism rather than an isomorphism. We are able to avoid this reversal of order by making deck transformations act on the left and field automorphisms act on the right.)

One consequence of the above correspondence is that the space of deck transformations of over is in one-to-one correspondence with the group of field automorphisms of fixing . If is an intermediate field, then certainly any deck transformation of over is also a deck transformation of over , so is a subgroup of .

From the above discussion, we see that a deck transformation of over is completely determined by its action on the distinguished point . In particular, the action of is free on the distinguished point of in the sense that the points for are all distinct. As they also lie in the same fibre of above , we conclude from Proposition 12 that the order of cannot exceed the degree of the extension:

As stated in the introduction, we call a Galois extension of if we in fact have equality

and then we rename as . This is canonically equivalent to the usual Galois group over fields. Thus, if is a degree Galois extension of , then implies that the fibre containing above must have cardinality exactly , with acting freely and transitively on this fibre. In fact the same is true for every non-empty fibre:

Lemma 16 Let be a degree Galois extension of , and let be a point in . Then the fibre containing above has cardinality exactly , and acts freely and transitively on this fibre. In other words, whenever are such that , then there is a unique such that .

Proof: As is a field and is an integral domain, the ring homomorphism must be injective (it maps invertible elements to invertible elements, and hence maps non-zero elements to non-zero elements). Each Galois group element corresponds to a different field automorphism of , hence the homomorphisms , are all distinct. This produces distinct elements of the fibre of ; by Proposition 12 this is the entire fibre. The claim follows.

Thanks to this lemma, we see that behaves like a degree covering map on some subset of (the space of ring homomorphisms for which is “large enough” that there is at least one extension of to ).

We can now prove part (i) of Theorem 1, in analogy with the covering space theory argument:

Proposition 17 If is a Galois extension of , and is an intermediate field, then is a Galois extension of .

Note that we have already demonstrated that is a subgroup of , so this proposition completes the proof of Theorem 1(i).

Proof: Let denote the degree of over , and the degree of over , thus is a degree extension of . Consider a non-empty fibre of above . By Lemma 16, this fibre consists of points. It also can be decomposed into fibres of above , indexed by a fibre of above . By Lemma 16, the latter fibre consists of points, while by Proposition 12, the former fibres all have cardinality at most . Hence all the former fibres must have cardinality exactly . In particular, the fibre of above has cardinality . This fibre consists of those ring homomorphisms that fix . As discussed previously, these ring homomorphisms must be field automorphisms, and thus generate elements of (or . Thus has cardinality at least , and hence by (7) it has cardinality exactly . Thus is a Galois extension of as required.

Corollary 18 If , then .

Proof: By definition, every element of is fixed by , thus and hence

On the other hand, by Proposition 17 we have . Finally, from (7) we have

Therefore all inequalities must be equalities, and as claimed.

Now we prove Theorem 1(ii). Let be a subgroup of , which we can also identify with a subgroup of . Let denote the set of elements of that are fixed by ; this is clearly an intermediate field between and :

On the one hand, every element of can be identified with an element of . By Proposition 17, has order exactly equal to . Thus

To prove the reverse inequality, we use an argument of Artin. We first need a simple lemma:

Lemma 19 (Producing an invariant vector) Let the notation be as above. Let be the -fold Cartesian product of (viewed as a vector space over ), and let act componentwise on : Let be an -invariant subspace of (i.e., is a vector space over such that whenever and ). If contains a non-zero element of , then it also contains a non-zero element of .

Proof: Let be a non-zero element of with a minimal number of non-zero entries. By relabeling, we may assume that is non-zero; by dividing by , we may normalise . If all the other coefficients lie in then we are done; by relabeling we may thus assume without loss of generality that . By definition of , we may thus find such that . The vector is then a non-zero element of with fewer non-zero entries than , contradicting the minimality of , and the claim follows.

Remark 20 This argument is reminiscent of the abstract ergodic theorem from this previous blog post, except that now we minimise the norm rather than the norm, since the latter is unavailable over arbitrary fields . While this lemma is simple, it seems to be the one aspect of the Galois correspondence for field extensions that does not seem to have an obvious analogue in the covering space setting; on the other hand, I do not know how to prove the fundamental theorem of Galois theory without this lemma or something like it (though presumably the original arguments of Galois proceed differently).

Corollary 21 We have .

Proof: Suppose for contradiction that , thus we may find elements of that are linearly independent over . We form a vector out of these elements, and consider the orbit of this vector. Let be the space of vectors orthogonal to this orbit, where we use the usual dot product

in . This space has codimension at most in and thus has a non-zero vector since . Clearly is -invariant. By Lemma 19, contains a non-zero vector whose entries lie in . By construction, is orthogonal to , thus there is a non-trivial linear relation amongst the with coefficients in , contradicting linear independence.

We thus have , giving the existence part of Theorem 1(ii); the uniqueness part is immediate from Corollary 18.

Theorem 1(iii) is also immediate from Corollary 18, so we turn to Theorem 1(iv). Let . First suppose that is a normal subgroup of . If and , then fixes , thus fixes . Thus , hence by Corollary 17 we have . Thus the action of descends to that of ; equivalently, every deck transformation of above descends to a deck transformation of above . Since the fibre of above has cardinality , and splits into fibres of cardinality above indexed by a fibre of above that has cardinality at most by Proposition 12, and , we conclude that the latter fibre must have cardinality exactly . As acts transitively on the fibre of , it must also act transitively on this fibre of . Thus there at least distinct deck transformations of above , and so is a Galois extension of thanks to (7), and every deck transformation of above arises from a deck transformation of above . Thus we have constructed a surjective map from to which can easily be verified to be a homomorphism. The kernel of this homomorphism consists of all deck transformations of above that preserve the fibres of above ; applying the deck transformation to the distinguished point , one sees in particular that the field automorphism of associated to this transformation preserves , and hence this transformation lies in . Conversely every deck transformation in descends to the identity transformation on . Thus the kernel of the homomorphism is precisely , and the isomorphism of with then follows from the first isomorphism theorem.

Conversely, suppose that is a Galois extension of . Let and . The points , of lie above the same point of ; from Lemma 16 there thus exists such that

On the other hand, as is the distinguished point of , there exists such that . Applying on the right to the above equation, we thus have

But lie above the same point of , thus , and hence

By Lemma 16 again, this implies that for some . As connects to every other point in by a path in , we conclude that , thus . In other words, is a normal subgroup of .