 This lecture is part of an online course on Galois theory and will be about Hilbert's theorem 90. So I should first just say a bit about why it has such a funny name. The reason it's called Hilbert's theorem 90 is that it is Hilbert's theorem 90. More precisely, Hilbert wrote a famous report on the theory of algebraic numbers back in the 19th century. And one of the theorems in his report, you can see this is theorem 90, is exactly the theorem we're going to talk about. I don't know why this name has stuck, but anyway. So, incidentally, this theorem wasn't first proved by Hilbert. You can find more or less the same theorem in one of Comer's papers from many years before. So the original form of Hilbert's theorem says the following. And say the original form because there's another form due to eminertia, which is more general that we'll talk about later. Anyway, the original theorem says suppose K contained in M is cyclic. That means the Galois group is cyclic. Then, if the norm of alpha is equal to 1, this implies alpha is equal to beta over sigma beta for some beta. So here sigma is a generator of the Galois group. So sigma to the n is equal to 1, where n is equal to the degree of m over K. And alpha is any element of m, and n is just the norm. Well, so how are we going to find such a beta given that norm of alpha is equal to 1? Well, what we can do is we can take this equation and we can rewrite it a little bit. So this says that alpha times sigma of beta is equal to beta. And now what you do is you look at alpha times sigma and you think of this as being a linear transformation from m to m, where you think of m as being a vector space over K, of course. So sigma is an element of the Galois group acting, and alpha just means multiplication by alpha, but we're just composing them and thinking of them as being a single linear transformation. And what this says is that beta is a fixed point of alpha times sigma and also another key condition. We need beta is not equal to zero, otherwise you can't divide by sigma of beta. So we want to find a non-zero fixed point of this linear transformation, in other words, something with eigenvalue 1. Well, if we've got a group action, it's easy to find fixed points because we can just average over the group. So let's try and find a group action. Well, the thing to do is to work out the order of this linear transformation alpha times sigma. And in fact, it turns out to have order n, because you can see this as follows. If we take alpha sigma to the n, this is just equal to alpha, sigma, alpha, sigma, alpha, sigma, and so on. Now here, this sigma is acting on all these alphas and this sigma is acting on all the alphas to the right and so on. So if you work this out, it's alpha times sigma of alpha times sigma squared of alpha and so on, times sigma to the n minus 1 alpha times sigma to the n. Now the sigma to the n is acting on whatever vector you're acting on. Now let's look at what's going on. Here, this expression here is just norm of alpha, which is equal to 1. On the other hand, this is also equal to 1 because sigma to the n is equal to 1, sigma has ordered n. So alpha sigma to the n is just equal to 1. So alpha sigma generates a cyclic group of order, well, dividing n. And now we can find a fixed point. So to find a fixed point, all we do is we take beta to be theta plus alpha sigma theta plus alpha sigma squared theta and so on plus alpha sigma to the n minus 1 theta. And here we've just summed over all elements of the cyclic group generated by alpha sigma and applied them all to theta. And obviously, alpha sigma beta is equal to beta. Well, we haven't quite finished because there's a slight problem is beta none zero. So all we need to do is to find some value of theta such that this is none zero. And it seems kind of trivial, but if you try it a bit, it's not at all easy to show that some theta exists. It might be that this big linear transformation is identically zero. So we need to find some theta so that the sum is none zero. And how can we do this? Well, there's a very useful lemma called Arten's lemma, which can be used to show this. So this says suppose G is any group. And suppose sigma 1, sigma 2 and so on are characters of G. Well, what do I mean by character of G? Well, this means a homomorphism from group G to the multiplicative group of the field M. This is more or less the same as a character in representation theory, if you're wondering. Then Arten's lemma says that sigma 1, sigma 2 and so on are linearly independent over M. So what does linearly independent mean? Well, this means that if a1 times sigma 1c plus a2 times sigma 2c and so on is equal to naught for all c and some finite collection of these characters, then a1 equals a2 equals and so on equals zero. So there's no linear relation between distinct characters. So I should say these have to be distinct characters. Obviously, if sigma 1 is equal to sigma 2, we could just take a1 equals minus a2, which would be kind of rather silly. So let's show that there is no linear relation. Suppose there is a linear relation, choose the shortest possible relation if one exists. So we might have some relation a1, sigma 1 plus and so on plus ak sigma k equals zero. And since it's the shortest possible all ai's are none zero, so we can divide by a1, so we can assume a1 is equal to 1. And secondly, we can divide all the characters by the character sigma 1, so we can assume sigma 1 equals 1. So this just makes the notation a little bit simpler. So 1 just means the trivial character, so it's the character such that sigma 1g is equal to 1 for all g, which is obviously a character. And now what this relation means is that for all c, we have c plus a2 sigma 2c plus a3 sigma 3c and so on is equal to zero. So this is just a1 sigma 1c, which is equal to c by our assumptions. And now we just multiply it by b. Well what's b? Well we pick b with sigma 2b not equal to b, which we can do because sigma 2 is not equal to sigma 1. So we just multiply this by b, we get bc plus b a2 sigma 2c and so on is equal to zero. And then we can also change c to b times c, so we get bc plus a2 sigma 2bc plus and so on is equal to zero. And now we use the fact that all the sigmas are characters, so we get bc plus a2 sigma 2b sigma 2c plus a3 sigma 3b sigma 3c, so on is equal to zero. And now what we do is we just subtract this relation and this relation. And the factors of bc cancel and we find b minus sigma 2b times a2 sigma 2c plus various other things is equal to zero. And now we notice that this term here is none zero because a2 is not equal to zero because we had a shortest possible relation. And b minus sigma 2b is none zero because sigma 2 is not equal to 1 and we picked it. And what we have here is a shorter relation. And this contradicts our assumption that we'd already chose the shorter relation so no relation exists. So in other words characters are linearly independent. Now let's go back and use this to show that we can find the theta such that beta doesn't exist. Actually before that we should just have a sort of warning. We're going to take g to be the multiplicative group of m, so a character sigma i is just a homomorphism from m star to m star. And let's put a warning in fluorescent pinks that everybody notices it. Warning, if sigma i is in the Galois group then sigma i is a character. But there are two possible ways of multiplying elements sigma i. We can take a product of characters, so if we've got any two characters which are homomorphisms from m to m we can obviously just multiply them together and we get sigma i, sigma j of x. This is the product of characters is sigma i of x times sigma j of x. So this is a sort of point wise multiplication. And the product of character is not equal to the composition of elements of the Galois group. So composition of Galois group, instead of doing point wise multiplication we actually compose two maps here. So this would say sigma i, sigma j, x is equal to sigma i, sigma j of x. So you see that this notation sigma i, sigma j is actually ambiguous. We can either be talking about point wise multiplication or about composition and these are completely different operations. For example, this thing here is abelian and this can be non-abelian because the Galois group might not be abelian but this multiplication is also abelian. And you also notice that this product here might not be in the Galois group. If you take two elements of the Galois group and multiply them by this operation it's a character but there's absolutely no reason why it should be an element of the Galois group. So that's a rather confusing thing about characters. It's sort of slightly ambiguous what you mean if you're multiplying two characters because there are two ways of multiplying them. Well anyway, let's go back to finish the proof of Hilbert's theorem 90. So we have characters 1, sigma, sigma squared, up to sigma to the n minus 1. And these are linearly independent by Artein. In particular, 1 plus alpha times sigma plus alpha squared sigma squared and so on is not identically zero otherwise this would give a linear dependence relation with coefficients 1 alpha, alpha squared and so on. So if we apply this to theta it means theta plus alpha sigma theta plus alpha squared sigma squared theta and so on is not equal to naught for sum theta in M. And this is what we wanted to prove because this is the element beta. So we found an element beta that's fixed by alpha sigma and is none zero. So this finishes the proof of Hilbert's theorem 90. Funny thing that the hardest part of this proof is proving the apparently trivial fact that some rather general element isn't always zero. So let's do an example to show that Hilbert's theorem 90 really proves a result about cummer extensions as a special case. So I suppose K contained in M has Galois group, say Z over PZ. And suppose K contains a primitive p-th root of 1 denoted by zeta. And let's see what Hilbert's theorem 90 says about it. Well we notice that norm of zeta is equal to zeta to the P which is equal to 1 because it's fixed by sigma. So by Hilbert 90 we know that zeta is equal to beta over sigma beta for sum element beta. Well this means that sigma of beta is equal to zeta minus 1 of beta which is exactly the formula we use to show that beta is a p-th root of unity. So beta not equal to zero. So sigma of beta to the P is equal to beta to the P. So beta to the P is in K. Beta is not in K because of this condition here. So M is equal to K with the p-th root of beta adjoined. So we recover the theorem about Kummer extensions in that if K contains a p-th root of unity, then M is generated by p-th root of something. So Hilbert's theorem 90 is a sort of generalization of this fact to the case when the field doesn't contain a p-th root of unity. We can't show that it's generated by the p-th root of something but we can still show that it has this property that if something is norm 1 then it's given by whatever that formula was. Just before going on to Nerta's generalization of Hilbert's theorem 90, let's just have another application of Arting's lemma. We can show that the trace is not identically zero for Galois extensions K contained in M. Well we've already shown this for arbitrary separable extensions so proving it for Galois extensions is just a special case of that. But if you remember the proof for arbitrary extensions used a lot of rather messy looking calculations with determinants and so on. Using Arting's lemma it's completely trivial because we notice that trace of theta is equal to sigma 1 of theta plus sigma 2 of theta and so on plus sigma N of theta. And this is not not for all theta because if it was it would be an obvious linear relation between the elements sigma 1 up to sigma N of the Galois group. So trace of theta is none zero for some theta. Notice incidentally this immediately implies the bilinear form is none degenerate because if A is not equal to zero then the inner product of A and theta over A is the trace of theta which is none zero. So for Galois extensions Arting's lemma gives a very short proof of the fact that the bilinear form is is none degenerate. So now we're going to go on to Nerter's form of Hilbert's theorem 90 and life is really unfair because not only was Hilbert's original theorem 90 not really due to Hilbert but Nerter's version of Hilbert's theorem 90 is always called Hilbert's theorem 90 and everybody forgets about Nerter but whatever. So what Nerter's theorem says is suppose K contained in M is any finite Galois extension. So instead of taking just a cyclic extension we're going to allow any Galois extension. And now it says that any one co-cycle is a co-boundary. So what is a one co-cycle? Well a one co-cycle is a map from the Galois group to M star. Strictly speaking this is a one co-cycle with coefficients in M star. And it's usually noted by sigma maps to an element alpha subscript sigma and I sometimes accidentally write my sigma so it's hard to tell their subscript so watch out for that. And it has to satisfy the following rather bizarre looking condition. Alpha sigma tau is equal to alpha sigma times sigma of alpha tau. And I'll explain where this weird looking condition comes from a little bit later. By the way this is the condition when you're taking a one co-cycle with values in a multiplicative group. If this were an additive group then you'd have alpha sigma tau is equal to alpha sigma plus sigma alpha tau which people sometimes use. And what is a one co-boundary? Well a one co-boundary is something where alpha sigma is equal to beta minus sigma beta or beta divided by sigma beta depending on whether you're an additive or multiplicative. Here beta is a fixed element of I guess it would be M star in this multiplicative case. Let me put this in brackets because we're not really going to use it. And the first co-homology group of the Galois group with coefficients in M star is just a group of one co-cycles divided by one co-boundaries. So this is just like an algebraic topology where a first co-homology group is some collection of co-cycles over co-boundaries except usually co-cycles are a little bit more intuitive than this weird expression here. So Nertis theorem says that this group is the trivial group. In case you're wondering, yes there are co-homology groups for higher dimensions. For instance the second co-homology group of the Galois group with coefficients in M star turns out to be the Brouwer group that I mentioned earlier. But we won't be talking more about this now. So we should try and explain. Well I've given you the definition of a one co-cycle but I haven't really explained what it is. So I should say what on earth is a one co-cycle? So I mean what is the motivation for it? Not what is this rather incomprehensible definition? Well let's look at the following expression. I take alpha sigma times sigma and I'm going to multiply it by alpha tau times tau. And I'm going to think of both of these as being linear transformations on M. Sigma and alpha sigma are both linear transformations because this is an automorphism of M and we can think of this as multiplication by the element alpha sigma. And if I work at what this is, it's alpha sigma times sigma alpha tau times sigma tau applied to something in M. And by the one co-cycle condition this is equal to alpha sigma tau times sigma tau. So in other words the map taking sigma goes to alpha sigma dot sigma is a sort of action of the Galois group on M star. Except it's a bit of a funny action because it's an action on M star as a set. It doesn't necessarily preserve the group action. So the action of the Galois group on M preserves the group action. But this is a sort of twisted action which doesn't preserve the multiplication. However it still has an action of the Galois group on the set M. So that's what a co-cycle is. You can think of it as being a way of twisting the action of the Galois group on M so it's still an action although no longer preserving the group operation on M. And now we can ask what is the condition for being a co-boundary? Well if alpha sigma is equal to beta over sigma beta this just means that alpha sigma times sigma beta is equal to beta. In other words beta is fixed by this funny twisted action. So if the action preserved the group action of M star then the fact that beta was fixed would not be terribly interesting. You could just take beta to be the identity element. But what this says is that if this action is a co-boundary then this funny twisted action still has a fixed point even though it might not be the identity element of M. Anyway with this interpretation of a one co-cycle Nurtis theorem becomes almost trivial to prove. If you don't think of this as being a linear transformation it's really hard to prove Nurtis theorem. So here's a proof of Nurtis version of Hilbert's theorem 90. Maybe it should be called Nurtis theorem 90 or something. So suppose alpha sigma is a one co-cycle. Then we get action of the Galois group on the set M star. This follows just by mapping sigma to the element alpha sigma times sigma. And we want a fixed point. And to get a fixed point we just take the average over the group. So we take an element beta which is sum over all elements of sigma of alpha sigma times sigma of some theta. So beta is always fixed by the Galois group. Well wait a moment we just said that this funny twisted action didn't preserve the group operation. And suddenly we're taking an average over the Galois group which implies it does preserve the group operation. So what's going on? Well this twisted action sigma goes to alpha sigma dot sigma does not preserve the multiplication on M but does preserve the addition. In fact these elements are just linear transformations so that they preserve the additive structure of M so it's perfectly okay to take an average over the group using the additive structure. So we found an element fixed by the Galois group and as before we need beta is not equal to zero and we can choose theta so beta is not equal to zero by Artin's lemma that we had before. So just as in the cyclic case there must be some theta such that beta is none zero otherwise we would have a linear dependence between all the elements of the Galois group. So next we'll just check that the Noter's non-Abelian form of theorem 90. So let's show that Noter's theorem 90 implies Hilbert's theorem 90. So let's suppose K contains an M is cyclic and suppose that norm of alpha is equal to one so we want to show that alpha is equal to beta over sigma beta where sigma generates the Galois group. In order to apply notice then we need a one-coast cycle where we get a one-coast cycle as follows we put alpha one equals one remember we've got to have an element alpha something for any something in the Galois group and let's put alpha sigma equals alpha. Now the co-cycle condition implies that alpha sigma squared must be alpha times sigma of alpha and it implies that alpha sigma cubed must be alpha times sigma of alpha times sigma squared of alpha and similarly you can go on and you can check alpha of sigma to the N must be alpha times sigma alpha times sigma squared alpha up to all the way times sigma to the N minus one of alpha times I guess that's it and now sigma to the N is equal to one so these two elements have to be equal in order for us to get a one-coast cycle well they're equal as norm of alpha equals one so norm of alpha implies that this is a one-coast cycle so saying alpha is norm one is really just the one-coast cycle condition so this means that alpha sigma to the I is equal to beta over sigma to the I beta for some beta not equal to zero and if we take I equals one we just find alpha is equal to beta over sigma beta which is the conclusion of Hilbert's theorem 90 so Hilbert's theorem really is a special case of Nurtur's theorem next we can just say a little bit more about what a one-coast cycle is first of all one-coast cycles sort of classify things that become same over M so for example suppose we take R contained in C and we want to classify Lie algebras over the reals so it doesn't really matter whether you're classifying Lie algebras or some other algebraic structure I'm just saying Lie algebras because that's a common thing people like to classify now simple Lie algebras over the complex numbers have been classified by Carton and Killing and you might want to classify simple Lie algebras over the reals well if you've got a simple Lie algebra over the reals you can tensor it with the complex numbers and you get a simple real Lie algebra over the complex numbers so the question is if you've got a Lie algebra over the complex numbers how many Lie algebras over the reals give rise to it? for example if you've done a Lie algebra course you know you can take the Lie algebras SL2 and SU2 and these are different over the reals but if you tensor them with the complex numbers they both become isomorphic so the general problem is suppose we've got some two algebraic structures they might be Lie algebras they might be something else and suppose if you tensor them over M they become isomorphic so suppose there's an isomorphism I between them then we want to sort of find you know if we've got one possible V we want to find all other possible Ws that become isomorphic to it over M so that we can classify whatever these structures are and now we've got an action of the Galois group on both of these because it acts on M so we can put a sigma equals i sigma i to the minus one sigma to the minus one so a sigma is going to act on this M see we'll apply sigma to the minus one then i to the minus one then sigma then i and similarly we're going to get a sigma tau which is going to be i sigma tau i to the minus one tau to the minus one sigma to the minus one and we get sigma a tau which is going to be sigma and then we put a tau i tau i to the minus one tau to the minus one sigma to the minus one so the action of sigma on one of these things is just going to be given by conjugation and then we can check that a sigma tau is in fact equal to a sigma times sigma a tau which is the one co-cycle condition. So in other words, an isomorphism between if we take two different things over a field, an isomorphism between them over the extension field is just a one co-cycle. And it's not terribly difficult to guess that a one it's going to be two one co-cycles are going to differ by one co-boundary if and only if these two things were isomorphic over the field k you first thought of. So in other words, the first co-homology group of one co-cycles modulo co-boundaries classifies things over k if you know the classification over this bigger field m. So one co-cycles tend to turn up quite a lot when you're trying to classify things over fields that are not algebraically closed. For instance, if you're trying to classify elliptic curves, elliptic curves are easy to classify over the complex numbers, but then number theorists want to classify elliptic curves over say the rational numbers. And this starts producing lots of one co-cycles with values in the automorphism group and elliptic curve and so on. Okay, so that's all about Hilbert's theorem 90. Next lecture we'll probably be doing something about infinite Galois extensions.