 So this lecture is part of an online Galois theory course and will be about Veder Bern's theorem, or at least one of Veder Bern's theorems because he proved several. So the one that we're going to talk about is that a finite division algebra, algebra is a field, in other words it's commutative. So let's just recall what these mean. So a division algebra is just a sometimes called a non-commutative field, which is rather bad terminology because all fields are commutative by definition. But what it means is you just drop the axiom about being commutative from a field. So the standard example that everybody knows is just the quaternions, H, which have a basis of one i, j, k over the real numbers. And the multiplication is going to be i squared equals j squared equals k squared equals i, j, k equals minus one as Hamiltonian famously wrote down on a bridge. Now if you've got a division algebra, so suppose d is a division algebra, we can look at the center of d and this is obviously a field. So d is a vector space over f and what we're, in general if you look at infinite dimensional division algebras over a field, it's rather hard to classify them. But you can ask the following question, given a field f, find all division algebras over f such that first of all they're finite dimensional and secondly f is equal to the whole of the center. If f isn't the whole of the center, you just have a bigger field and repeat with that. So we would like to classify all finite dimensional central division algebras over a field where central means the field is the whole of the center. And Wetterburn did this for finite dimensional division algebras over finite fields by showing that there really aren't any apart from the finite fields. So here's Wetterburn's theorem. Let's prove that any finite division algebra d is a field. And let's stop by doing the easy stuff. So we put f to be the center. So f is a field and we classified the fields. They have order q where q is some prime power. And d is a vector space of dimension n. So the order of d is equal to q to the n, rather obviously. And now what we do is we're going to, let's put g to be the non-zero elements of d. So the order of g is just q to the n minus 1 and g is a group. And now the main key idea of the proof or the first key idea is to look at the conjugacy classes of this group g. So let's just look at the conjugacy classes and we're going to count the number of elements in each conjugacy class. And what we do is we get the following formula. We have q to the n minus 1. So this is the order of g and it's equal to various terms. So the first term is going to be q minus 1. And this is just the order of the center. And the center is just the non-zero elements of the field f. And then we have to add a sum over various conjugacy classes. So this is going to be some conjugacy class. And then we want the number of elements of the conjugacy class. It's just going to be q to the n minus 1 over q to the ki minus 1. So I'll explain this in a moment. So this bit here is the size of the conjugacy class. And y is the size of the conjugacy class equal to this. Well, the size of the conjugacy class is equal to the order of g divided by the order of h, where h is the centralizer of some element of the conjugacy class. So we need to know what is the centralizer of an element of the division algebra? The centralizer in the whole of d is a subdivision algebra. So this subdivision algebra will have order q to the ki, because the subdivision algebra will have dimension ki over d. And what we want is not the whole division algebra, but the non-zero elements in the division algebra. So if we call this subdivision algebra e, then the order of e star is equal to q to the ki minus 1. So we should also note that ki divides n because d is a vector space over e. And e isn't actually a field, it's a division algebra, but you can still define vector spaces over division algebras, and we find d is a vector space over this division algebra, and therefore ki divides n. So now we want to obtain a contradiction from this, or at least we want to prove that n is equal to 1. And what we do, the key point is we look at these terms here, and we notice that all divisible by the cyclatomic polynomial phi n applied to q. So here we're using the fact that ki is less than n and ki divides n, which implies that this term here contains q minus zeta for any primitive n through zeta. And this, it's divisible by the cyclatomic polynomial. So this is divisible by phi n of q. So actually, Wetterburn's original proof didn't actually use cyclatomic polynomials and it had a rather more complicated argument. In fact, it had a slight flaw in it which was patched up by Dixon. This is actually quite common that the first person to prove something, who very often has some minor gaps or glitches in it, but this is usually not a big deal. I mean, the pioneering work is always a bit clumsy and full of errors. That doesn't mean the first person who did it shouldn't get the credit for it. Anyway, so we've now got the following equation. We know that phi n of q divides q minus 1. Well, this easily implies n equals 1. And that's because phi n of q is just a product of q minus various n-th roots of unity. And if you plot the roots of unity, so here's q and here's 1 and here's zeta. And you notice immediately that this distance q minus zeta is bigger than q minus 1 if zeta is not equal to 1. So if n is not equal to 1, phi n of q has absolute value bigger than q minus 1, so cannot divide it. So n equals 1, so the division algebra d is actually equal to its center f and is therefore commutative. So I'll just finish by making a few background comments about division algebras. There's actually a group called the Brouwer group of field f. The elements are isomorphism classes of finite dimensional division algebras with center equal to f. And it's not at all obvious that these form a group. I mean, what on earth is the group multiplication? Well, the multiplication, if you've got two division algebras d and e, you take that tensor product over the field f, and this is in general not a division algebra. However, it turns out to be a matrix algebra over some division algebra. And we define g to be the product of d and e in the Brouwer group. And you may look at this and think it's not actually even obvious that this product is associative, but that's not actually too difficult to check. And it's not at all obvious that this multiplication has an inverse. Well, the inverse is quite easy. The inverse of d is just d op, which is the same as d except you define x. You define a new multiplication x, y to be the old product but reversed. And then you can easily check that d tensed with d op over f is isomorphic to a matrix algebra over f where n is the dimension of d. And of course, the identity element is just the field f. Anyway, Brouwer showed that these form a group and this group turns out to be sort of quite important in class field theory. So Wetterburn proved that the Brouwer group of a finite field is trivial. That's just saying that all division algebras are fields. And the Brouwer group of the reals is a cyclic group of order two because how it's proved the only central division algebras over the reals are the real numbers of the quaternions. Of the complex numbers, the Brouwer group is trivial because the complex numbers are algebraically closed and it's quite easy to show that there are no division algebras over that. For other fields, the Brouwer group is often quite difficult to calculate. For instance, if you take a p-addict field, the Brouwer group turns out to be the rational numbers modulo the integers. If you take the field of rational numbers, the Brouwer group is even more complicated. It turns out to be a group of order two and then you take a sum overall primes of the Brouwer group of the p-addict fields and then that maps, there's a natural map from that to q over z and the Brouwer group of the rationals sits inside an exact sequence like this. So this is a sort of basic result of class field theory. You can calculate the Brouwer group of any field of algebraic numbers. And more generally, growth and dick even managed to extend the definition of Brouwer group to arbitrary commutative rings and even arbitrary schemes. Okay, the next lecture will be about norms and traces of fields.