 This lecture is part of an online course in Galois theory and will be about transcendental extensions. So the transcendental extension is just an extension that isn't algebraic. So let's first quickly recall if we've got an algebraic extension we can sort of more or less understand it completely. First of all it can be split up into a separable extension and a purely inseparable extension. And separable extensions can be sort of more or less understood by turning them into Galois extensions and then we can apply Galois theory and mess around with Galois groups. Purely inseparable extensions will frankly the less said about them the better. So suppose we've got a transcendental extension K contained in M. What we want to do is to talk about how do we study this? Well what we do is we find a transcendence basis. So a transcendence basis is a maximal algebraically independent set. So the set might be x1, x2 and so on or something like that. So what does algebraically independence mean? Well algebraically independent means that there's no polynomial relation between these. So if some polynomial in xi, xj and so on is equal to zero this implies the polynomial itself must be zero. Now we want to show that a maximal transcendence basis actually exists. Well if you want to show that a maximal something or other exists in general you need to use Zorn's lemma and you can easily check that Zorn's lemma implies every field extension has a transcendence basis. So let's have a quick look at some examples. So suppose you've got any extension. We look at the field generated by a transcendence basis and then whatever's left over. And this extension here must be algebraic because if there was some element in M not algebraic over this then we can add it and get a bigger transcendence basis. On the other hand this extension here is purely transcendental. By this we mean that it's generated by a set of algebraically independent elements. So you can think of this field as being the field of rational functions over k in several variables. So it's not unique. This intermediate field isn't unique. For example if you have k contained in the field of rational functions in one variable over k then t is a transcendence basis for this rather obviously but so is t squared because k contains in k of t squared contains in k of t. Now this is an algebraic extension of degree two and now t squared is a transcendence basis. So there can be many different fields here such that this field is purely transcendental and this field here is algebraic. And you can see that even the degree of this extension here may vary. So I guess we should say we're really thinking of M as being the same as k of t in this case. A slightly less trivial example is if you take the field of rational functions on an elliptic curve. So if the elliptic curve is given by y squared equals say x cubed minus x then we take the ring of all polynomials in two variables and quotient that by y squared minus x cubed plus x and then we can take a field of quotients of this which would be k of x, y, except that y squared minus x cubed plus x is equal to zero. And in this case we can write k is contained in k of x and this is a purely transcendental extension and then this is contained in I can't think of quite a good notation for this field. You can't say it's the field of rational functions x and y modulo this ideal because this actually generates the whole field because it's a field. So here's the field of rational functions of an elliptic curve which I'll just write as elliptic curve but we could also have a different... Instead of using x as a generator for this field we could also take y as a generator for this field. So you can see this extension here is degree two because y has degree two over k of x and this extension here is degree three because x is degree three over k of y and so on. So again there's more than one possible purely transcendental extension you can find inside that. So another example, let's take the rational numbers contained in the complex numbers. Well here we have q contained in q with a transcendence basis for the complex numbers which is contained in the complex numbers and this transcendence basis needs an uncountable number of elements because if you write only a countable number then this field here would be countable and c is just the algebraic closure of this field so if there were a countable number of variables here then c would actually be countable. So here we have purely transcendental extension with an uncountable transcendence basis followed by taking the algebraic closure. So this sort of in some sense gives a view of what the complex numbers are it's the algebraic closure of the field of rational functions over q in an uncountable number of variables. Notice this implies that the complex number has an uncountable number of automorphisms because this field here has huge numbers of automorphisms for instance we can just permute all the basis vectors and that already gives us a huge uncountable group and then we can extend any automorphism of that to its algebraic closure you remember there isn't a unique way of doing that but there's at least one way of doing that. So the automorphisms of the complex numbers is absolutely huge so let's try and write some down. Well examples of automorphism as well as the identity and there's complex conjugation and you'll find it really hard to write down any others. In fact there aren't any other continuous automorphisms and writing down any other automorphism sort of involves the axiom of choice. People sometimes claim you can't write down explicit automorphisms in fact you can sort of. There are in fact models of set theory where you can write down explicit automorphisms of the complex numbers. The way you do that is by taking something like the Goedel's universe of constructible sets and Goedel showed that in constructible sets you can sort of write down an explicit choice function and doing that you can write down some explicit automorphisms of the complex numbers but they're really horrible you wouldn't want to actually use them for any serious purpose. So in practice there's no sensible way to write down an automorphism of the complex numbers other than these two. However there are huge numbers of them at least if you're not some sort of intuitionist. So an obvious question is do any two transcendence bases have the same size? So the answer is yes. I'm not going to prove this in general. I'm just going to prove for finite transcendence bases. So for finite transcendence bases you do it like this. Suppose you've got two transcendence bases. So you've got kx1 up to xm contained in something and you've got ky1 up to yn contained in your bigger field m. And what we want to do is to show that m is equal to n and for that what you do is you let's suppose we can pick some yi not equal to any xi. And then we're going to add this yi to the x's. So we get x1 to xm. And now since x1 up to xm is maximal there's some polynomial pyi x1 to xm equal nought with p0 equal 0. But then what we can do is some xi occurs in this non-trivially so we throw out some xj that appears in this with non-zero coefficients. So then we get a new transcendence basis consisting of x1 up to xj minus 1 xj plus 1 up to xm yi. And so we've replaced one of the x's by a y and you can keep on doing this. You can just replace xi's by yi's until you've replaced all the xi's by y's and that shows the number of x's at most the number of y's and by reversing the argument you can show the number of y's is equal to the number of x's. So this implies m is equal to n. So you can sort of draw a little picture of this if you like. So if I have all the x's as x1, x2 up to xm, orange and I have the y's in green. What you do is you find one of the y's that's not in the x's so you make it orange and then you find one of the x's that's dependent on that say x1 and you cross it out. So you've got a new orange set and then you continue doing it again, you add another of the y's and you cross off another of the x's and so on and so you just sort of turn all the green points orange one by one. You notice that involves, that assumes the bases are finite and for infinite bases we have to repeat infinitely many times and this may be a problem because the infinitely many times you have to repeat may actually be uncountable and to do this you need to use sort of variations of the action of choice. Anyway, I'm not going to bother with that because if you're talking about the difference between different sorts of uncountable objects you're not doing algebra, you're doing set theory and this isn't a set theory course. So any two transcendence bases have the same number of elements and the number of elements of a transcendence basis is called the transcendence degree. I really wish they'd thought of a word shorter than transcendental for this because it's getting real pain writing it out. You may notice that the proof that any two transcendence bases have the same number of elements is sort of formally rather similar to the proof that any two bases of a vector space have the same number of elements. In fact, these are both examples of something called a matroid. So I'll just have a very brief introduction to matroids to show that these two things are the same. Matroid consists roughly of a set. Now I'm going to cheat a little bit because most definitions of matroids assume that this set is finite but I want to apply this to vector spaces or field extensions or whatever. And it has some special subsets called bases and there must be at least one base and they also have this exchange property. Let's say these are special finite subsets because I don't want to get into the theory of infinite matroids. Actually I'm not even sure there is a theory of infinite matroids. So these have the exchange property that if A and B are bases and A is in A but not in B then we can find B that's in B but not in A so that A without A but together with B is also a base. And you notice this is exactly the property we were using to show that any two transcendence bases have the same number of elements. And you can apply the same proof to show that if you've got a matroid then any two bases of matroid have the same number of elements. So let's have some examples of matroids. So we can take the set to be a vector space and the bases are sets that are linearly independent and that span the vector space V. Or we can have a field extension K contains an M and here the bases are going to be algebraically independent sets and that generate M as a field over K. Let me give you a third example just to show that matroids don't always look like things over vector space as the set could be a graph. And the bases are going to be some sets of edges that have no loops but contain all vertices. So you can think of this as being a spanning forest which is sort of like a spanning tree except it might be disconnected if this graph is disconnected. And you can check that any two spanning forests of a graph have the same number of elements and the proof is essentially just the same as before it just uses this exchange property. So in other words bases of a vector space and transcendence bases of an extension and spanning forests of a graph are all examples of matroids. Now we'll come on to some applications of the transcendence degree. So the first application is the definition of the dimension of an algebraic variety. So an algebraic variety is going to be some subset of affine space which is just a fancy way of saying an n-dimensional vector space defined by some polynomial equations. For example we might have an elliptic curve y squared equals x cubed minus x which will be some sort of elliptic curve like that. And we want to define its dimension. It's obvious what the dimension of this is. It's 1. However it's quite tricky to come up with an algebraic definition of dimension. And one of the oldest and simplest is as follows. We look at the function field which is all rational functions on the algebraic set. So that's just the field of quotients of the field of regular functions. And dimension is then defined to be the transcendence dimension of the function field as an extension of whatever field we're working over of course. This works fine for dimensions of algebraic varieties. It's not used much in modern algebraic geometry because you really want to define dimension of schemes which are more general in algebraic varieties. And this definition just goes horribly wrong for most schemes. For example suppose I take the ring of Laurent series over a field. Now this really ought to have dimension 1 because it's just adding one variable. But if you look at its transcendence degree it turns out to be infinite because there are lots of power series that are algebraically independent. So this definition is a rather old definition of algebraic variety used in the first half of the 20th century but algebraic geometers don't use it very much anymore. So for another application of transcendence degree we can try and classify algebraically closed fields. So here we point out the complex numbers can be obtained as the algebraic closure of a purely transcendental extension of Q. And we can do the same thing for any algebraically closed field F. What we do is we take the prime field F0, so this is the smallest field contained in something and it's either rationals or it's the integers mod P for some P. And maybe I shouldn't have used F0 for that, it's called the prime field G. So we have G is contained in F and we can take a transcendence basis for F so we get G containing X1 up to whatever it is and then this is an algebraic extension, F is just the algebraic closure of GX1 up to X whatever it was. So we see from this that there are two invariants of the algebraically closed field F. The first is it's characteristic which can be 0, 2, 3, 5 or something and the second is it's transcendence dimension over the prime field G which can be any cardinal number. So we've only actually proved this is an invariant if the cardinal number is finite and sort of waved our hands a bit for infinite cardinal numbers. So this is one of the rare cases when you can actually give a fairly simple classification of all models of some sort of first order theory. By the way, notice that we can ask the number and we can ask what is the number of algebraically closed fields of cardinality and P and sorry of characteristic P and cardinality K for some cardinal number. So K has to be an infinite cardinal number a finite field isn't algebraically closed and we see from this classification that if K is equal to the smallest cardinal LF0 then there are a countable number of fields because the transcendence degree can be equal to 0, 1, 2 or LF0. On the other hand, if the cardinality is some higher number then the transcendence degree just has to be LF alpha and there's only one example. Now this actually sort of gives an example of a categorical theory. Well, so categorical theories were originally defined by Veblen as being theories for which there is only one model and people very soon notice that the Skollum-Lohenheim theorem implies that the first order theories can't be categorical unless they've only got a finite models. However, you can define a theory to be categorical in a certain cardinality. So the theory of algebraically closed fields is categorical in any uncountable cardinality but it's not categorical for countable cardinalities. This actually gives a nice example of a theorem in model theory called Morley's theorem which says that if we have a countable first-order theory, so countable means it's only got a countable number of axioms and first order means it's in first-order logic which I'm not going to explain in detail, if we have a countable first-order theory it says if it is categorical in some uncountable cardinal, it is categorical in all uncountable cardinal. Uncountable cardinals. And you can see from the example of algebraically closed fields that this fails if we allow countable cardinals. So, um, algebraically closed fields are categorical uncountable cardinals because there's only one model up to isomorphism but they have lots and lots of different models in some countable cardinality. Morley's theorem was the beginning of a huge explosion of model theory. It's one of the most important and basic theorems of model theory. Rather sadly, Morley died just a few weeks before I made this video. Anyway, um, the fact that the theory of algebraically closed fields is categorical in some cardinality. So, if it's categorical in cardinality k, this implies the theory is complete. Um, it means you can decide the truth of any statement about um, algebraically closed fields of given characteristic, um, from the axioms, um, that's any first-order statement. Um, you may be a little bit puzzled about this if you've heard about Goedel's incompleteness theorem because Goedel's incompleteness theorem says that mathematical theories are incomplete. And here I'm claiming the theory of algebraically closed fields is complete. Well, if you look at Goedel's theorem a bit more closely, it says that a mathematical theory is incomplete if it can encode the natural numbers and it turns out the theory of algebraically closed fields can't actually encode the natural numbers. Um, I mean the natural numbers are contained in any algebraically closed field but you can't actually define them using the language of first-order fields. Anyway, if you want to find out more about that you need to check into a course on mathematical logic. I'm getting a little bit sidetracked. Anyway, let's get back to algebra. Um, so for Galois extension, if we looked at the Galois group, so I suppose we've got a transcendental extension, we can ask what about um, the group of all automorphisms of m over k. I'm going to denote this as the Galois group although I don't think it's really called the Galois group if it's a transcendental extension because Galois theory doesn't apply. And we can ask, do subgroups of this group have anything to do with extensions between k and m? Well, quite a lot. Um, so finite subgroups of the Galois group of m over k correspond one-to-one to extensions k contained in l contained in m with l contained in m Galois and finite. Um, in fact this statement is a more or less trivial consequence of Galois theory for finite extensions. I mean I said if you've got a finite group acting on an extension then the fixed points form a Galois extension and so on. So, so as long as you're only looking at extensions of finite co-dimension, you do indeed get a Galois correspondence. Um, for example, let's just look at the extension of the complex numbers contained in the complex numbers with one transcendental element. So you can think of this as being meromorphic functions on the Riemann sphere c union infinity, which you can also think of as being the projective plane, sorry the projective line, if you are an algebraic geometry. And if you've been to a Caughton complex analysis you know all the automorphisms of this extension. The automorphisms are just that tau to a tau plus b over c tau plus d. And this is the group pgl2 over c, where this is the projective general linear group. So that means you take all matrices a, b, c, d with a, d minus b, c not equal to 0 under multiplication and you're quotient out by the diagonal matrices. And this is a three-dimensional group of the complex numbers and it's very nice and well understood. And we can find all its finite subgroups. This was done by Felix Klein probably by other people. Incidentally if you want to know about this he's got a very nice book on this called The Icosahedron where he explains all this. Anyway the finite subgroups well up to isomorphism they're all cyclic or dihedral or they're the alternating group A4 or the symmetric group S4 or the alternating group A5 so these are orders 12, 24 and 60. And you may notice that these groups are the same as the finite groups of rotation in three-dimensional space. And that's not a coincidence because I mentioned that we have the Riemann sphere so well the Riemann sphere is a sphere so you can take any group of rotations of a sphere such as the icosahedral group that certainly acts on a sphere and this will also act on the Riemann sphere so we get each of these finite groups acts as a group of automorphisms on K of X. So for example we can find the complex numbers is contained in some field L which is contained in C of X where this quotient has degree 60 and is isomorphic to the icosahedral group A5 so you can indeed construct sub-extensions of transcendental extensions using a sort of variation of Galois theory. However in general we don't get a one-to-one correspondence and things can get very weird. So let's have a look at rational functions in two variables so this is transcendence degree two now for one variable what we were doing is we were looking at the projective line over the complex numbers which is automorphism group PGL2 of C and we saw this acted on the field of rational functions in one variable because this is just rational functions on the projective line in two variables we can do the same thing we can take the projective plane which is actually on my PGL3 of C that's 3 by 3 matrices modulo-diagonal matrices and this easily gives us automorphisms of the field of rational functions in two variables and if you think this is going to be the full automorphism group by analogy with this case then you are completely wrong because the automorphism group of CXY is much bigger than PGL3C it certainly contains PGL3 as a subgroup and how much bigger well this is actually infinite dimensional and it's quite easy to see it's infinite dimensional let me write down an infinite dimensional space of automorphisms we can map X to X and map Y to any given polynomial in X plus Y and the inverse of this maps X to X and it maps Y to Y minus this polynomial in X and the space of polynomials is infinite dimensional so already we've got an infinite dimensional abelian subgroup of this group and then we can there's nothing special about X and Y it's a trick with any two generators so this group is huge it's called the Cremona group and algebraic geometers have been studying it for well over a century and still haven't got it completely figured out we have at least figured out the finite subgroups rather Dolgachev and this Kovsky I hope this spells his name right found all the finite subgroups this was actually quite recent only about 10 years ago and this involves a fair amount of rather complicated algebraic geometry as far as I know the describing finite subgroups of larger purely transcendental extensions is wide open