 This lecture is part of an online Galois theory course and will be about infinite Galois extensions. So far we've mostly been talking about finite extensions being Galois, but now we want to talk about more general algebraic extensions being Galois. So the typical example of this is the rational numbers contained in the algebraic closure of the rational numbers. The union of all algebraic number fields. And we will see this has a Galois group, which is of great interest in algebraic number theory. In fact, it's not too much of an exaggeration to say the whole of algebraic number theory is a study of this group. So first of all, what is an infinite Galois extension? Well, the following conditions are equivalent for an algebraic extension K contained in M. So first of all, it's separable and normal. Secondly, it's the splitting field of a set of separable polynomials. And thirdly, K is the fixed field Mg of some group G acting on M. Now, these three conditions are almost the same as the conditions for a finite extension to be Galois. The only difference is that instead of just fixing a single separable polynomial, we need a set of separable polynomials. Obviously, if you just have one separable polynomial, its splitting field is going to be a finite extension. One condition that we haven't got is the condition saying that the automorphism group of M fixing K is equal to the degree. And this doesn't really make a whole lot of sense if the degree is infinite. And for infinite extensions, these three conditions are all equivalent and we say that an extension satisfying them is a Galois extension. Proving their equivalent is a very easy consequence of the proof of finite extensions. The key point is that if you've got an algebraic extension K contained in M, then M is the union of extensions K contained K i for K i finite over K. So a lot of things can be just reduced to the case of finite extensions. And in particular, all these equivalences turn out to follow easily from the finite extensions. One thing we haven't talked about is the correspondence between subgroups of the Galois group and extensions between L between K and M. And in fact, it's not true that extensions L between K and M correspond to subgroups H of the Galois group G. There's one extra condition we need to add. It turns out the group G has a topology and subfields of M containing K only correspond to closed subgroups of G. So what we're going to do is we're going to show the Galois correspondence of infinite extensions except we've got this minor extra complication that we have to worry about the topology on G. So let's define the topology. This is sometimes called the Krull topology on the Galois group M over K. So as a group, this is just defined as in the finite case. It's the automorphisms of the field M that fix every single element of K. Now we've got K. We can think of it as being the extension K in M. What we do is we think of M as being the union of lots and lots of finite extensions. So all these extensions are going to be finite. In fact, we can even assume they're finite and Galois if M is a Galois extension. So the union is M. And let's think what an automorphism of M fixing all elements of K does. Well, if these extensions are Galois, any automorphism of any element of the Galois group M over K must act on each of these finite Galois extensions of K. So what we do is we get this collection of maps. So we have the Galois group of K over K. And mapping two, we have the Galois group of K1 over K and the Galois group of K2 over K and so on. So corresponding to this sort of lattice of finite Galois extensions of K, we get this lattice of finite Galois groups. So these are all finite. Now an element of the Galois group of M over K means we get an element of each of these groups and all these elements are compatible. So if we choose an element in this group, its image must be the element in this group we chose. So the Galois group of M over K is contained in the product of all the Galois groups of K I over K where K is contained in K I is contained in M and K I over K is finite and K I over K is Galois. And it's the subset of this consisting of all selections of these elements that are compatible with each other. So if you've done inverse limits, you know that this is called an inverse limit or a projective limit. So this Galois group is just an inverse limit of finite groups. And what we do is we give this product the product topology. So it's a product of finite sets. So the product topology is compact. And here we're taking this Galois group here and you can easily see that it's actually a closed subgroup. So this is actually also a compact group. It's a projective limit of finite groups and projective limits of finite groups are sometimes called profinite groups. So the classical example of a profinite group is the Galois group of an infinite extension. So now we've got to discuss this extra condition that subgroups of G should be closed if they correspond to fields. First of all, we notice that if alpha is an M, the subgroup of G, the Galois group of M over K, fixing alpha is closed. And that's because alpha is contained in L where K over L is finite. So by definition of the topology, anything fixing a particular element in this finite extension is a closed subgroup. So if K is contained in L is contained in M, subgroup fixing all elements of L is also closed. That's because it's the intersection of closed subgroups where you take the intersection over all alpha in L of the subgroup fixing that element. So the subgroup of the Galois group fixing all elements of some particular sub extension must be a closed subgroup. So that's the easy part of it. So what we want to do is to show that if we've got subfields K contained in L contained in M or sub extensions and closed subgroups H of the Galois group of M over K, then these correspond to each other. Let's just recall what the correspondence is. So first of all, if we've got a sub extension L, you remember this corresponds to the Galois group of all automorphisms of M fixing every element of L. And if we go back to fields, this corresponds to the subfield of M fixed by this. So we get Galois of M over L. And we notice L is obviously a subfield of this extension. And the problem is to show that it's the same. And to show these are equal, what we have to do is give an alpha in not in L. We need to find some sigma in the Galois group of M over L, not fixing alpha because that will show that L is actually equal to the things fixed by the Galois group. And what we do is we look at L contained in L alpha, and this will be a finite extension. Then we look at L contained in the splitting field of alpha of alpha over L. So that this is then a finite Galois extension. And by the theory of finite Galois extensions, we can find some sigma in the Galois group of this finite Galois extension, not fixing alpha. And then we can keep lifting it to larger and larger finite extensions. And if necessary, we probably have to use the Zorn's Lemma or something like that. So we can extend it to an automorphism of M over K. So the proof that L must actually equal the fixed field of this is, follows fairly easily from the finite case and a little bit of actual choice of Zorn's Lemma. So we've shown that if we start with a subfield and we go to a group and then we go back to a subfield, we get the same subfield that we started with. Notice that this didn't actually use anything about the subgroup being closed because if we go to a subgroup, that's automatically closed. So for this way around, we don't have to worry about it. Now we have to worry about the other way around. What happens if we take a group, go to a field and go back to a group. And for this we will have to use the fact that the group is closed. So here what we do is we take a subgroup H in the Galois group of M over K. And then we map H to a subfield M H. And then we map it to the Galois group of M over M H. And H is contained in this group here. And in general these are not the same in general. So these are not always the same. What we want to do is to show this group here is the closure of H in the Krull topology. So suppose sigma is not in the closure of H. Then we need to show that it acts non-trivially on M H. Which will be enough to show that if we start with a group H and go like that we get back to the closure. Well let's see what does it mean for sigma not to be in the closure of H? Well by definition of the topology it means we can find some finite extension L of K. So that sigma is not equal to any element of H on L. When I say finite extension I should have said a finite Galois extension. So that H acts by restriction on L. And by definition of the topology saying that an element isn't in the closure means we can find a finite extension. So sigma is not equal to anything like this. Now since K contains an L is now a finite Galois extension. We can apply the theory of finite Galois extensions. So some element of L H is not fixed by sigma. So we've found an element such that sigma acts non-trivially on M H. Because L H is certainly contained in M H. So the key point is that the topology on the Galois group just sort of quite formally reduces the correspondence in Galois theory. Just checking things for finite extensions. So the topology sort of encapsulates the idea that every element of M is contained in a finite extension of L. So now let's see some examples of infinite Galois extensions. First of all we can take any field K and take the separable closure of K. This is sort of like the algebraic closure except we only work with separable extensions. I mean you can take an algebraic closure and take the maximum separable extension of it. And this will be a Galois extension, usually an infinite one. And you remember I said there was this problem about the algebraic closure wasn't really unique. It's just that there are lots of algebraic closures which happen to be isomorphic. And of course the same thing is true for the separable closure. And you sometimes call the absolute Galois group of K to be the Galois group of K sep over K. And just as before there's not really any such thing as the absolute Galois group because this object here is a bit ambiguous. But the absolute Galois group is sort of unique up to isomorphism. So most of the time people don't worry about that. The next example let's just take K to be a finite field and try and figure out what is its absolute Galois group. Well we've got this chain of extensions of F so the finite field of order P is contained in Fp squared which is contained in Fp4. And it's contained in Fp cubed and these are both contained in Fp to the 6 and that's contained in Fp to the 9 and that's contained in Fp to the 5 and so on. So we get this huge lattice of extensions of finite fields. And you remember to get the Galois group of an infinite extension we take the inverse limit of the Galois groups of all the finite extensions. And we know what all these Galois groups of anything over the finite field is it's just cyclic. So what we do is we get these cyclic groups Z over 1Z and then we map Z over 2Z, Z over 3Z and then we get Z over 4Z and here we get Z over 6Z and so on. So what we have to do for an element of the Galois group is to pick an element of every one of these cyclic groups in such a way that all these picks are compatible with each other in the obvious sense. And this is the inverse limit of all finite quotients of Z and it's called the profinite completion of Z sometimes denoted by Z hat or Z bar or something like that. In fact you can do this profinite completion operation for any group whatsoever you take all its finite quotients and take the inverse limit of those. And that gives you a, in some sense it's the universal profinite group generated by a group. Well what does the profinite completion of Z look like? Well you recall the Chinese remainder theorem says that Z over MNZ is equal to Z over MZ times Z over NZ for MN equal 1. And what this means is that the profinite completion of Z kind of splits up into, because we can take the inverse limit of Z over 2Z goes to Z over 4Z and so on. And then we take the product of Z over 3Z goes to Z over 9Z and so on and then we do the same thing for 5 and so on. So if we take all these inverse limits and multiply them together the Chinese remainder theorem says that that's the profinite completion of Z. Well if you've done p-addict numbers you recognise this is just the 2-addict numbers and this is the 3-addict numbers and this is the 5-addict numbers and so on. So the profinite completion of Z which is the absolute Galois group of a finite field is just the product of the p-addict numbers for all primes p. And in particular, slightly surprising, this profinite completion actually has zero divisors because it's a product of various rings whereas the individual p-addict integers don't have zero divisors. So that's what the absolute Galois group of a finite field looks like. It's just a product of all the p-addict numbers. Now let's do an example of the group generated by all roots of unity. So we're going to take q contained in the maximal cyclotomic field of q. So this is generated by all roots of 1. And as I said earlier, in some sense the fields generated by roots of 1 are the easiest extensions to study. So this is the sort of maximal easy extension of the rationals. As soon as you go beyond this, your things start to get more complicated. Let's work out what its Galois group is. Well, as before, we first start by writing the cyclotomic field as being a union of finite extension. So we get q, well, this is contained in q with the fourth roots of unity, and it's contained in q with the eighth roots of unity, and it's contained in q where we add in the third roots of unity, and these are contained in q where we add in the twelfth roots of unity, and so on. So we get this big lattice of various fields. And we know what the Galois group of q contained in, if we're drawing all the nth roots of unity, the Galois group is, it's not quite z over nz, it's now the multiplicative group of z over nz. So the Galois group of this is, it looks a bit like the Galois group of an absolute Galois group of a finite field, except we take the units of the integers mod n instead of the integers mod n. So we get z over 1z, that maps to z over 2z star, z over 3z star, and z over 6z star, and so on. And just as before, we can apply the Chinese remainder theorem, and we find this group is split as a product. So we take z over 2z star, maps to z over 4z star, and so on, and then we multiply that with z over 3z, the units of 3 over 3z, and the units of 3 over 9z, and so on, and do the same thing for 5. So we see that this is equal to the units of the two-addict integers times the units of the three-addict integers times the units of the five-addict integers, and so on. So what do the units of the p-addict integers look like? Well, it's just sort of like asking what does z over p to the nz look like? And we recall that if p is odd, this is z over pz, sorry, z over, this is isomorphic to z over p minus 1z, coming from z over pz times z over p to the n minus 1z. And for p equals 2, it's a little bit different, it's z over 2z times z over p to the n minus 2z. There's a general property of Galois theory or number theory over the integers or rational numbers. The prime two always, always, always goes wrong. There are numerous theorems, like the Kronecker-Weber theorem about the maximal abelian extension of the rational or Sheffarovich's theorem where every solvable group is a Galois group over the rational. Pretty much any theorem, the first time it was proved, the person who proved it got the prime two wrong, there was some mistake that needed to be corrected. So, I don't know, two always goes wrong. This is the first time it goes wrong, when you try and work out the units of z modulo p to the nz2 which starts behaving badly. In particular, this group is cyclic and this group is non-cyclic. So, we find the units of zp star for p odd is isomorphic to z over p minus 1z times the p-addict integers. p2 star is isomorphic to z modulo 2z times the two-addict integers. So, the Galois group of the cyclotomic extension of the rational is a product of all the p-addict numbers together with a product of various finite groups, finite cyclic groups of order p minus 1 or 2. In particular, the rational numbers have a lot of zp extensions. I'll just mention these briefly. There's a very beautiful subject in number theory called Iwasawa theory which is basically a study of zp extensions. A zp extension is an extension of algebraic number fields with Galois group zp, the p-addict integers for some prime p. So, we can now see that there are several examples of zp extensions because if we look at the Galois group of the cyclotomic field over q, this has a lot of maps on to zp because it's got zp as a factor. So, for each prime, there is at least one zp extension of q. In fact, the Chronic-Ava theorem turns out to imply that the rational numbers have only one zp extension which is given by this construction. So, what this means is that k equals k0 is contained in k1, it's contained in k2 and so on, and the union is equal to l. And this extension is Galois group z modulo pz. This extension is Galois group z modulo p squared z and so on. And it turns out there's a really funny obstruction to constructing zp extensions. You might think these are easy to construct. Let's just construct an extension with Galois group z modulo pz and then extend it to an extent with Galois group z modulo p squared z and so on. This is a sort of extension problem about Galois fields. So, suppose we've got groups one goes to A, goes to B, goes to C, goes to one. So, we've got an exact sequence of groups. And suppose we have found an extension of the rationals, q contained in l, with Galois group of l over q equals C. And so, can we extend to q contained in l contained in m so that the Galois group of m over q is equal to B? And this seems a very plausible, slight extension of the conjecture that every group is a Galois group. For instance, an obvious way to try and prove that every solvable group is a Galois group, which is Shafarevich's theorem, is to sort of do induction on the length of a group. So, you take a solvable group and you can factor it like this with an abelian group here with a smaller solvable group here. And then we find a Galois extension with this group as its solvable group and try to extend it to an extension with a whole of B as its Galois group. And although this is a plausible way of doing things, it just doesn't work. Rather surprising thing is that for the rationals there are some fairly simple obstructions to doing this. And I'll just give one example. Let's just take the simplest possible extension. So we take nought goes to z modulo 2z, goes to z modulo 4z, goes to z modulo 2z, goes to 0. And now I'm going to choose a Galois extension of the rationals with Galois group z modulo 2z. And we're just going to take qi over q, which obviously is Galois group of order 2. And then there is no extension q contained in qi contained in L with the Galois group of L over q cyclic of order 4. There are plenty of extensions of the rationals with cyclic group of order 4, I mean we constructed some of that earlier, but you can't get them by first going to the extension qi of order 2 and then trying to extend that. So let's see why that doesn't work. So suppose q is contained in qi is contained in L with the Galois group of L over q z modulo 4z generated by some element sigma. And we know that this bit has Galois group sigma squared of order 2. And what we do is we can pick A in L so that sigma squared A is equal to minus A. That's because sigma squared is an endomorphism of L of order 2 and since we're in character 6-0 we can split it up to eigenspaces with eigenvectors 1 and minus 1. So this is just an eigenvector of sigma, sorry of sigma squared. Now this implies that sigma A over A is fixed by sigma squared as you can see. So sigma A over A must be of the form B plus C i in q of i because anything fixed by sigma squared is in q of i. And if we apply sigma we find sigma squared A over sigma A is equal to B minus C i because sigma of i is minus i. And now if we multiply these together so we take the product we find that sigma squared A over A is equal to B squared plus C squared but sigma squared A over A is equal to minus 1. So minus 1 is the sum of two squares which is not possible. So we can't have a cyclic extension of the rationals of order 4 that contains q of i as its quadratic subfield. So in fact there are quite a lot of obstructions like this to extending Galois groups over the rationals. So this raises the question of what does the Galois group of q bar over q look like? Well as I said this is a completely open problem. We don't even know how to solve the inverse Galois problem which says that every finite group appears as a quotient of this. Well we can split it up like this. So we have 1 and this contains the Galois group of q bar over the maximum cyclotomic extension of q which is contained in the Galois group of q bar over q which maps to the Galois group of the cyclotomic extension of q over q goes to 1. So we have an exact sequence like this and this group is known. So we worked it out earlier. It's just the units of the profinite completion of the integers. So that's very well understood. And we can ask what is this group? And it can't be a free profinite group which as you remember is the profinite completion of a free group because if it was then we could always solve the extension problem because you can always lift free groups. However Schaffer-H asked is this a profinite completion of a free group? So the example we gave showing we can't lift q of i to a cyclic extension shows that this group here is definitely not the profinite completion of a free group but as far as we know this might be which would give a sort of reasonable description of what the Galois group with the algebraic closure of q looks like. So in some sense Schaffer-H's conjecture says all the problems about lifting extensions come from the cyclotomic part of the Galois group of the rationals. Okay that's all about infinite Galois extensions.