 This lecture is part of an online graduate course on Galois theory, and will be about the algebraic closure of a field. So last lecture we saw that if we've got a field k and a polynomial p with coefficients in k, we can find a splitting field of p where this is generated by the roots of p and p factorizes into linear factors over L. We can obviously do the same thing with a finite number of polynomials p. We just multiply the polynomials together and take the splitting field of the product. Now what we can do is we can do the same for all polynomials in k of x. So what we're going to do is to extend k to a field k bar such that first of all, any polynomial in k of x factors into linear factors over the field k bar. And secondly we want k bar to be generated by roots of polynomials in with coefficients in k. So this is just like the definition of a splitting field except instead of doing it with one polynomial, we do it with all polynomials. And when we do that this field is called the algebraic closure of k. Now the construction of the algebraic closure is very similar to the construction of a splitting field except you do it infinitely many times. So the construction might be something like this. Let's first suppose the field k is countable. Then we all we do is we list the polynomials in k of x. So we might get p1, p2, p3. And there are only a countable number of polynomials so we can just list them all in a row. And then we take a field k equals k0. We extend it to k1, which is a splitting field of p1. And then we extend it to k2, which as you can probably guess is a splitting field for p2. And we carry on like this and we put a k bar to be the union. And you can easily see that that's just a field and contains a root of any polynomial in k of x. So any polynomial in k of x factorizes into linear factors. And what if k is not countable? Suppose k is uncountable. It's similar. But we well order the polynomials pi using the axiom of choice and then do something similar. Okay, well, if you don't know about well-ordering you probably don't care very much about uncountable fields either so we won't worry about it too much. But if you go to a set theory course, they tell you all about how to well-order things. And if you well order the polynomials, you can just sort of copy this proof and you get an uncountable well-ordered chain there. Anyway, a field is called algebraically closed. So field L is called algebraically closed if all polynomials in L of x have roots in L. In particular, that means all polynomials factorize into linear factors. And the algebraic closure of k that we've constructed is algebraically closed. Well, wait a minute. I seem to be kind of getting away with something because all I've proved is that any polynomial with coefficients in k has a root in k bar. But in order to claim that k bar is algebraically closed, I need to show that a polynomial with coefficients in k bar is algebraically closed. So I seem to be have slipped up a bit. So what we have shown, so we know that any polynomial in k of x has a root in k bar by construction. We want that any polynomial in k bar of x, and let me just put the bar in fluorescent pink so you notice it, has a root. Well, actually, this is not too difficult to show. If every polynomial in k of x has a root in k bar, then we can show that any polynomial in k bar of x has a root. So suppose that p of x is in k bar of x, and it might have coefficients might be say a naught, a one, a two, and so on. It might be x to the n plus a n minus one x to the n minus one plus a naught, where all the a i are in k bar. And now all we do is we look at the field of extensions. We take k and then we look at k with all these elements a naught, a one, a n minus one are joined. And then this is also contained in some field with k a naught up to a n minus one alpha, where alpha is going to be a root of the polynomial p. So we can just adjoin a root of p to k if we want. So this field k is certainly contained in k bar. And what we want to do is to show that we can also, if we want, pick alpha to be in k bar. Well that's quite easy because we notice that this extension here is finite because all the a i are algebraic over k. And this extension here is finite because alpha satisfies a polynomial with coefficients in these. So this extension here is finite. So alpha is algebraic over k. And what this means is that alpha is a root of a polynomial with coefficients in k. And we've shown that any such polynomial factors into linear factors in k bar. So we may as well assume that alpha is in k bar. So this shows that the algebraic closure of a field is algebraically closed which is really rather fortunate because it would be kind of stupid if it wasn't. So we can see that any field has, let's call this k, has an algebraic closure. So actually we only sort of really prove this for countable fields and mumbled something about set theory for uncountable ones but never mind. We can also sometimes ask about a weaker property or a weaker construction. Suppose instead of asking for all polynomials to have roots. Let's have the problem of find a field l with k contained in l so that all elements of l have a square root in l. So an algebraic closure would certainly have this property but what we want to construct is a smaller field that's only closed under taking square roots. And we would like the smallest possible field so we only want l to be generated by the operations of repeatedly taking square roots and field operations. Well we can construct a field like this. We could take k equals k0 and we can embed it in a bigger field k1 where with k1 you would join all square roots of elements of k0. So this is just like what we did for an algebraic closure exception. Instead of taking all polynomials we're only taking polynomials of the form x squared minus a0. Now it's certainly not true that k1 is necessarily closed under taking square roots but that doesn't matter. We can extend it to a bigger field where we adjoin square roots of elements of k1 and you can continue this an infinite number of times and let's put l to be the union of all these fields. Then you can see that l is closed undertaking square roots and secondly l is generated from k by the usual field operations plus minus times divide and taking square roots. Now you see if we were really lazy we could have used a similar argument for the algebraic closure. So we could have said well k1 is going to be generated by all roots of polynomials with coefficients in k0 and then k2 is going to be generated by all roots of polynomials with coefficients in k1 and we sort of continue and construct the algebraic closure to be l. And that would save having to prove that k1 is algebraically closed. In fact we showed that k1 is in fact equal to k2 but if we were too lazy to do that we could just use this argument instead which would be kind of a bit sloppy. So this is a fairly common construction mathematics. If you want to construct an algebraic object closed under some operation you can quite often get it by just iterating a countable number of times and taking the union. By the way if you wonder why anybody should care about fields closed under taking square roots that's related to the old problem of doing things like squaring a circle or trisecting an angle because the numbers you can construct using a ruler and compass turn out to be exactly the numbers you can construct from the rational numbers using the field operations and taking square roots. So a number can be constructed by ruler and compass if and only if it's contained in the field you get like this by starting with the rationals and taking its closure under square roots. These are sometimes called Euclidean numbers. So let's have some examples of algebraically closed fields. Well the example that everybody knows is you take k to be the field of real numbers and then the algebraic closure is the field of complex numbers. So this is sometimes called the fundamental theorem of algebra which says that the complex numbers is algebraically closed. In fact this is a particularly easy example because we only need to take a degree 2 extension of r whereas in general the algebraic closure of a field would be a huge infinite extension. And how do we prove complex numbers are algebraically closed? Well we'll be giving an algebraic proof of it later on in the course and if you've been to a complex analysis course you've probably seen a proof of it using Louisville's theorem. What I'm going to give now is a very short topological proof. So I suppose p of x equals x to the n plus a n minus 1 x to the n minus 1 and so on plus a naught is a polynomial with coefficients in the complex numbers. We want to show that p has a root and what you notice is that if x is large say x is equal to r sufficiently large then this bit here x to the n is bigger than this bit here so it has bigger absolute value than a n minus 1 x to the n minus 1 plus a naught and so on. In fact you can assume it's much bigger and that's pretty much all we need to know to show if it's algebraically closed. So let's draw a kind of picture of the complex plane and here I'm going to draw so this is going to be so this plane I'm going to draw x and now I'm going to let x go around a big circle of radius r and I ask what what happens to the values of the polynomial p of x? Well as I said x to the n you remember from complex analysis goes around the origin n times but it's perturbed slightly by this but since this is very small the perturbation isn't going to be very big so if x is doing this p of x will sort of wiggle a bit and it will sort of go around the origin n times so this is what p of x does and now imagine making this circle smaller and smaller so instead of looking at radius r I might look at some smaller radius and then what's going to happen well it's going to go I don't know this bit will no longer be smaller than this bit so we can't really assume it goes goes roughly around n times and if I make it smaller still so I just get a yellow spot there then what happens to p of x is it will just look like this actually that's not terribly legible let me do it in green instead so I'll put a green spot there so and now let's look at what happens so here this pink circle goes around n times anti-clockwise where n is equal to the degree of the polynomial p on the other hand this yellow or possibly green spot goes around the origin zero times so as you imagine contracting this circle here its image starts off going around the origin n times and ends up going around the origin zero times and you know if you sort of imagine this this thing going around as being a loop of string or something what we've got is this loop of string kind of going around the origin three times or n times like that and you're trying to move it around so it ends up going around the origin nought times but if you imagine the origin as a sort of post there it's sort of caught on the post and you you can't move it to going around the origin nought times without at some intermediate point passing through the origin so there must be some intermediate circle between this big circle in the origin such that if your if x goes around that circle it must actually pass through the origin the the number of times you go around the origin is sometimes called the winding number so what we're saying is the winding number starts off being the degree of the polynomial here and goes down to zero there what i said sorry the winding number of the image goes down from n to zero so at some point it must change and the only way it can change is if if the curve passes through the origin so what we have shown in fact is that anything of the form x to the n plus f of x has zero if f of x is less than x to the n for x having absolute value r for some fixed value of r that's because if f of x is less than x to the n then x to the n plus f of x will have winding number n because it's this is sort of forced to stay roughly in the same direction as x to the n so it's the same winding number as x to the n um so this doesn't need to be a polynomial at all it can be any continuous function um so um so that gives the complex numbers as a fairly natural example of an algebraically closed field so let's have some natural examples of algebraically closed fields um there are lots of examples you can get just by taking the algebraic closure of a field but the problem is is not very natural they're rather hard to write down explicitly i mean our construction said you know you had to write down this infinite sequence of polynomials and keep adding roots and you know actually doing that would be a horrendous bookkeeping operation so let's see some natural examples well first of all we have the complex numbers secondly we've got things like the algebraic closure of q bar inside the complex numbers that so these are the so-called algebraic numbers which are the elements of c that are algebraic over the rational numbers and um since the complex numbers are algebraically closed it's fairly easy to check that all the elements that are algebraic over the rational numbers are also algebraically closed so this is called the field of algebraic numbers and there's an entire subject called algebraic number theory devoted to studying just this field um apart from this it's really quite hard to write down natural examples the the only really natural one i know of is is the field of preser series um these were actually invented by isaac newton but there are too many things named after newton so some of newton's inventions are named after other things so you remember the lorant series or rather formal lorant series are just the power series or formal power series over the complex numbers except you're allowed to have a finite number of inverse powers so for example we might have x to the minus 3 e to the x would be a lorant series x to the minus 3 plus x to the minus 2 plus x to the minus 1 over 2 factorial and so on um now preser series are a slight variation of this so what we do is we take the union over all n of greater than or equal to one of lorant series in x to the 1 over n so we take x to the 1 over n and then we're allowed to um invert so it's so you might have a series like x to the minus a half plus 2 plus 3 x to the half plus 4 x and so on that would be a typical preser series and newton more or less showed that preser series are algebraically closed i mean he didn't actually state that because algebraic closed algebraically closed fields hadn't been defined and wouldn't be defined for several hundred years but in spite of the fact that they hadn't been defined newton still more or less managed to show that the the field of preser series is algebraically closed he sort of proved a theorem equivalent to that um well we showed existence of algebraic closures um now we should discuss uniqueness of algebraic closures and here we run into the same sort of rather funny problem that we had for um splitting fields so any two algebraic closures of k are isomorphic and the proof is much the same as the splitting fields except you have to repeat things an infinite number of times instead of a finite number of times so i'm not going to bother doing that but the trouble is this isomorphism is not unique and in general there's no way of making it sort of canonical and this runs into a it's not really a serious problem but it's a sort of rather irritating bookkeeping problem in that you can't really speak of the algebraic closure of a field because you know if two people come up with algebraic closures of the field they may be isomorphic but there may be no good way to agree on what the isomorphism between them is um for instance you get a problem like this if you're a category theorist and you try and make um suppose you take a field and you ask is taking the algebraic closure is this a functor so you remember from category theory that to make it a functor you've got to say if you've if you've got a field k and a map from k to l and you take an algebraic closure of l is there a natural map from the algebraic closure of k to the algebraic closure of l and if you think about it a bit it's really not easy to to make this into a functor because you know which which isomorphism you're going to take and I mean I think if you work very hard and using the use the action of choice a lot you might be able to sort of push through something but it's really rather a mess and there's a sort of analogous very similar problem to this in algebraic topology so what I want to explain is that is the problem of uniqueness of the algebraic closure is related to the problem of defining the fundamental group of a topological space x so um you know if you've got a topological space say a surface with genus 2 surface then you can define a fundamental group whose elements are roughly loops started on this surface so so a typical loop might go around there and then background there and if you've got two loops you can sort of try and compose them and you can try and make that into a group well you've got to be a little bit careful because if you've got two loops it's not quite clear how you compose them what you've really got to choose is choose a base point and instead of looking at arbitrary loops you should just look at loops that start at this base point and go round and then if you've got a loop that starts at that base point and if you've got another loop that starts at that base point and goes around there say then you can compose them and get a loop that goes like that so you've um by it's not too difficult to make to make these loops into a group if they're taken up to home a topic and the problem is so you've got this fundamental group is usually denoted by pi 1 so you might say well you've got a fundamental group pi 1 of x well it seems to depend on the choice of base point well it doesn't really because if you pick another base point then um for any loop here um you can turn it into a loop at this base point by going along a fixed line between these base points then round that loop and then round there and from this it's not too difficult to see that pi 1 of x doesn't depend up to isomorphism on the choice of base point however if you try doing that you run into some rather nasty problems that the problem is that the isomorphism between these groups for these two different choices of base point depends on the line you've chosen between these two points so we've got the same problem that we had for algebraic closures of fields what we should do is not define the fundamental group of a space x but define the fundamental group of x with respect to a fixed base point x zero and this map here is not a functor if you've got a map between topological spaces you don't get a well-defined map between this whatever it is however you do get a well-defined map for these things so this is a functor and now i want to show that same sort of problem happens for algebraic closure so if you've got a field k we can um embedded in an algebraically closed field k bar and we can define an absolute Galois group Galois group and i'm going to just define this to be the automorphism group of k bar that fixes all elements of k so it's just a group of symmetries of k bar and you should think of the absolute Galois group of a field k as being analogous to the fundamental group of a topological space x in fact as i mentioned a few lectures ago growth and dick had a had a formalism in which it more or less is the fundamental group of something if you're willing to use et al topology and the problem is that the fundamental group of a space x isn't a functor if you don't pick a base point and similar the absolute Galois group of a field k is well it's not really a functor in other words if you've got a map from k to a space k one and you embed this in an algebraic closure k and embed this in an algebraic closure k bar then if you look at the automorphism group of k bar there's no natural map from this automorphism group to this automorphism group well what you can do however is you can define an absolute Galois group if you pick an algebraic closure k bar now picking an algebraic closure k bar turns out to be very similar to picking a base point x zero so if we pick a base point x zero then we have a nice fundamental group which which is doesn't depend on all these funny ambiguities similarly if you want to define the absolute Galois group of a field we really have to pick an algebraic closure even though all algebraic closures are isomorphic if we don't pick one we run into nasty problems about this thing not being a functor so most of the time it's harmless to talk about the algebraic closure of a field and you don't need to worry about this ambiguity but sometimes you do by the way if you do need to worry about this ambiguity that there's a construction called a groupoid that i just quickly mentioned that sort of gives a neat solution to this problem so groupoid is a category such that all elements such that sorry all morphisms are isomorphisms and now in these two cases we can construct a fundamental groupoid that doesn't depend on a base point and we can construct an absolute Galois groupoid that doesn't depend on the algebraic closure and we do this as follows the category here has objects points of the space x and the morphisms from a point x naught to x one are the same as homotopy classes of paths from x to x one sorry from x zero to x one in the space when i say paths i mean homotopy classes of paths but there's not enough space to write out homotopy here now this groupoid i'm going to define a category whose points are algebraic closures k bar of k so i said there's really no canonical choice of an algebraic closure although they're all isomorphic so you just take all of them and the morphisms sorry that's not points it's objects and the morphisms from one algebraic closure k to another algebraic closure k are just going to be isomorphisms of fields from the first one to the second so if you're bothered by this lack of uniqueness of the algebraic closure you need to go and find out what a groupoid is and work with the absolute Galois groupoid rather than absolute Galois group but as i said most of the time you can sort of get away with the absolute Galois group and not worry about this this this rather complicated nonsense okay next lecture we're going to give another application of the splitting field of a field which is to construct the finite fields or galois fields