 This lecture is part of an online course on Galway theory, and will be about cyclotomic polynomials. So let's first explain what they are. Well, first of all we need to know what cyclotomic means. Well, cyclo just means circle, and atomic means cut. So the polynomial is where you cut up a circle. The circle you cut up is the unit circle in the complex numbers. And the way you cut a top is by choosing some roots of unity. For example, you might choose the eighth roots of unity, and you can see you cut the circle up into eight parts. And the eighth roots of unity are the roots of x to the 8 minus 1. However, the roots, the polynomial x to the 8 minus 1 is obviously reducible, because we can write it as, well, it's got the oneth root of 1, and it's got the other square root of 1, and it's got the two fourth roots of 1, and it's got the eighth roots of 1. So it factors as these four polynomials. Now these factors are the so-called cyclotomic polynomials. So this is phi 1 of x, this is phi 2 of x, this is phi 4 of x, and this is phi 8 of x. Where phi n of x is the polynomial with roots, the nth roots of 1 of order exactly n. So it's sort of pretty obvious that the nth roots of 1 are of order exactly d for some d dividing n, so this is just a product over d divides n of phi d of n. And from that you can check recursively that this polynomial has integer coefficients. And if you remember the Moebius inversion formula from number theory, this says that phi, that should be phi d of x. It says that phi n of x can be written as a product of d divides n of x dn minus 1 to the power of mu of n over d. I guess that should be a d. So let's work out a few of them just to see what they look like. So here's n, so n can be 1, 2, 3, 4, 5, 6. And the first few cyclotomic polynomials are, well, as obviously this only has 1 root 1, and then the only proper square root of 1 is minus 1, so here are the roots. For 3 we have x squared plus x plus 1, and the roots are omega and omega bar, where omega is minus a half plus root minus 3 over 2, of course. For we have x squared plus 1, and we have roots are i and minus i. For 5 we have x to the 4 plus x cubed plus x squared plus x plus 1, and the roots are fifth roots of 1, and I'm not going to write out explicitly, and for 6 we get x squared minus x plus 1, and the roots are a half plus or minus root minus 3 over 2, and so on. So it's not difficult to work out the first few. Let's try a slightly more complicated one. So for n equals 12, we would have to get x to the 12 minus 1, and then we throw out the sixth roots of 1, and we throw out the fourth roots of 1, but then we've thrown out the square roots of 1 twice over, so we should multiply by x squared minus 1. And if you work this out, it's just x to the 4 minus x squared plus 1. Or for 15, we would take x to the 15 minus 1, and then we have to divide by x to the 5 minus 1, x to the 3 minus 1, and we've thrown out 1 twice, so we have to add it back in, and this is a little bit more complicated to work out. It's x to the 8 minus x to the 7 plus x to the 5 minus x to the 4 plus x to the 3 minus x plus 1. And if you look at the few cyclotomic polynomials we've worked out, you see the coefficients usually seem to be 0 or 1 or minus 1. And there's a really good exercise, which is find the smallest n, so that phi n of x has a coefficient not equal to 0 or plus or minus 1. You need to go a surprisingly long way to find such an n. In fact, you are unlikely to find an n just by working out phi n of x for bigger and bigger values of n unless you're extraordinarily patient. What you have to do to solve this is you work out the cyclotomic polynomial for some small values of n and try and see some patterns that you can see what's going on. And if you notice a few patterns, you'll be able to figure out what the smallest value of n is. So a really fundamental property of phi n of x is that it's irreducible over the rationales. In general, it's not irreducible. For example, over finite fields it has a strong tendency to factorize into smaller things, but over the rationales it's irreducible. Now, for n prime, there's a well-known proof of this that everybody knows, which is if you change x to x plus 1, then we can find n of x becomes an Eisenstein polynomial, so we can apply the Eisenstein criterion, and I'm not going to go through this because you've probably seen it half a dozen times already. It also works for n prime power. Check that the Eisenstein criterion also works. But what we want to do is to prove for n composite, which takes a little bit more work, and for this we're going to use the Frobenius automorphisms that we mentioned last lecture. So first of all, we notice the Galois group of phi n of x is at least contained in the group of integers co-prime to n under multiplication. That's because if we take one primitive nth root of 1, then all nth roots of 1 are powers of it, so it's certainly a Galois extension, and any automorphism must map zeta to zeta to the i, and as before, we see the composition of the morphisms mapping zeta to the i and zeta to zeta to the j, map zeta to the i times j, so the Galois group is at least a subgroup of this. So we want to show that the Galois group is all of z over nz star, because this will show that the polynomial phi n of x is irreducible because the order of this is phi of n. This is Euler's phi function, it's nothing to do... Well, actually it is sort of to do with the cyclotomic polynomial, but it's the number of integers less than n co-prime to n, and it's also equal to the degree of phi n of x, which is presumably why they use this notation. So if we do that, we will have shown that phi n of x generates an extension of degree... that any root of this generates an extension of degree phi of n, so this would have to be irreducible. So we're trying to show the Galois group is all of this. And for this, we note that if p does not divide n, then phi n of x is separable, mod p. In other words, it's got no multiple roots, and that's because it divides x to the n minus one, oops, x to the n minus one, and the derivative of x to the n minus one is nx to the n minus one, and if p does not divide n, then we can divide out by this and see that the derivative is essentially x to the n minus one, and the highest common factor of x to the n minus one and nx to the n minus one is obviously equal to one if n is not zero. That means zero in the field, so p does not divide n. So if p doesn't divide n, then it's co-prime to its derivative, so all its roots are distinct mod p. So factor phi n of x mod p, and we get that it's equal to product g1 of x, g2 of x and so on. So you remember we get a Frobenius automorphism of the field generated by a root for each factor gI, and this Frobenius automorphism is rather confusingly called phi p, which shouldn't be confused with the cyclotomic polynomial. We're rather overusing the Greek letter phi, I'm afraid. And what does phi p do? Phi p of zeta is zeta to the p in mod p, sort of more or less by definition, the Frobenius automorphism raises everything to the p from our mod p. However, all roots of unity are distinct mod p. We just showed this because the polynomial is separable. So phi p of zeta must be zeta to the p in the field generated by roots of unity. So phi p of zeta equals zeta to the p in q of zeta. So we found an automorphism of q of zeta, which raises each nth root of unity to its pth power. So the Galois group, which is contained in z modulo nz star, contains p as an element of z over nz star, whenever p does not divide n. But these elements generate z over nz star rather obviously. So the Galois group is the whole of z over nz star. So the cyclotomic polynomial phi n of x is irreducible. As we saw last lecture, it's quite often reducible modulo p, even if n equals 2. Now we'll give a couple of applications of this. So first of all, we're going to give an application to Dirichlet's theorem, which says that if a and n are co-prime, there are infinitely many primes congruent to a mod n, I guess I probably ought to have n greater than 0. And we're going to do the special case when a equals 1, which has a quite elementary proof. And we can see this as follows. Suppose p divides phi n of k, where k is some integer, and suppose p is co-prime to n. Well, if p is co-prime to n, this means the nth roots of 1 mod p are all distinct. It means there are n of them. So the map from nth roots of 1 in the cyclotomic field to nth roots of 1 mod p is injective. So p divides phi n of k implies k is a root of phi n mod p. So k has order exactly n in z over pz. That's because the roots of phi n mod p are the roots of unity that have ordered exactly n, because the map from nth roots of unity to nth roots mod p is injective. Well, that means k has order n in the group of z over pz star. That's order exactly n, of course. And this group has ordered p minus 1. So this group has an element of order exactly n, so n divides p minus 1. So we've shown that if p divides some value of this nth cyclotomic polynomial to co-primes n, then it has to be 1 modulo n. So this gives us a way of constructing lots of primes that are 1 modulo n. So suppose we have found p1 up to whatever that are 1 mod n. Now we pick p dividing phi n of p1 up to all the others. So we just take a product of all the primes that we found. Well, we notice that phi n is 1 plus something, 1 plus x times something. So this is 1 modulo p1, p2, and so on. In particular, p is not equal to p1, p2, and so on. So p is a new prime. Sorry, I forgot to put an n in here. So this is 1 mod p1, p2, and n. So p is not equal to any of these, and p does not divide n. So p is a new prime. p does not divide n, p is not equal to p1, p2, and so on. And p is congruent to 1 mod n because it divides phi n of this. So we can generate an infinite number of primes that are congruent to 1 modulo any positive integer. So as far as I know, it's a completely open problem to find a similar elementary proof of Dirichlet's theorem for p congruent to a modulo n. I mean, we have Dirichlet's original proof using L series and group representation theory, which sort of irritates people sometimes because it uses analysis in order to prove a result in number theory. And that's the sort of idea that results in number theory should be proved using just pure number theory and using analysis as cheating. There are proofs of this which don't use analysis. I mean, Selberg found one as a variation of his elementary proof of the prime number theorem, but they are incredibly difficult and far harder than Dirichlet's theorem. There seems to be no easy proof of this except for the general case. But the case of primes 1 modulo n is easy if you know what a cyclotomic polynomial is. So to finish, we're going to prove a special case of the inverse problem. So the inverse problem for Galois theory is, given a finite group, find an extension of the rationals with that as Galois group. So suppose g is abelian, so we're going to find an extension q contained k with Galois group g. I should have said g is going to be finite, of course. So as we pointed out, finding an extension with any given finite group is Galois group is easy if you don't specify what the base field is. But if you insist the base field should be the rationals, then it gets much harder. Anyway, over the rationals we can do this using the irreducibility of the cyclotomic polynomial and using the special case of Dirichlet's theorem. So we write g as a product of groups. Let's write it as a product of cyclit groups of orders n1, n2, n3, and so on. And now we're going to pick primes p1, p2, p3, and so on, all distinct with p i as common to 1 modulo n i, which we can do by the special case of Dirichlet's theorem that we just proved. And now we look at the cyclotomic field generated by p1, p2, and so on, roots of unity. So if we look at the field generated by roots of unity of order p1, p2, and so on, then we know the Galois group of this because the corresponding cyclotomic polynomial is irreducible. So the Galois group is z over p1z star times z over p2z star, and so on. Because you remember we chose p1 and p2 and so on to be distinct. And now this contains the group z over n1z because n1 divides p1 minus 1, and this group contains z over n2z and so on, which is equal to our group g. So we found a cyclotomic field whose, actually, sorry, we don't want the Galois group containing it, we want the Galois group mapping onto this. And then that will give us corresponding to this, we get an extension of the rationals with this as a Galois group. So the next lecture will be giving some other applications of cyclotomic polynomials.