 This lecture is part of an online Galois theory course and will be about Kummer extensions. So in the previous lecture we showed that if you can solve a polynomial by radicals, then this is closely connected with the Galois group being solvable. At least if the characteristic is equal to zero, if the characteristic is equal to p, there are some additional complications. What we want to do is to think about the converse. So if the Galois group is solvable, does this mean we can solve it by radicals? Solve the polynomial by radicals. Well, if the Galois group is solvable, this means the Galois group can be split up, in some sense, into groups that are cyclic, and if we want we can split it up into groups that are cyclic of order n. So this suggests the following problem. Given an extension k contained in m, which is Galois, which is group, Galois group that is cyclic of order n, what can we say about m? In particular, is m of the form k with the nth root of something adjoined? Well, in general it's not, but it quite often is. So we're going to make the following two assumptions. When I say two, I mean three, but first of all, we're going to assume k contains all the nth roots of unity. Secondly, we're going to assume that the characteristic of k does not divide n. Again, if the characteristic does divide n, things are quite different. And the third minor one that we're going to assume is that n is prime. So this is not really necessary. So if n isn't prime, it just causes minor extra complications. If we drop one of these two assumptions, things get quite a bit more complicated. So this is just to simplify things a little bit. So here we've got k is contained in m, and we want to show under these assumptions that m is equal to k with the nth root of a adjoined for some a in k. And the problem, how to find a? Well, let's suppose the Galois group is generated by some element sigma. So sigma to the n is equal to one. And then we know that if a exists, the roots of x to the n equals a are going to be the nth root of a, fixed nth root of a, the nth root of a times zeta, the nth root of a times zeta squared and so on. Where zeta is a primitive nth root of one. We assume the characteristic doesn't divide n. So one zeta squared up to zeta to the n minus one of distinct as p does not divide n. So if p does divide n, as I said, we will get some extra complications from this. So the element of the Galois group must map the nth root of alpha to one of these other nth roots of alpha. And we may as well choose zeta so that it maps it to zeta times the nth root of alpha, say. And now what you need to do is we forget m is a field and think of it of m as a vector space over k and sigma as a linear transformation. So we're just forgetting about the multiplicative structure on m and we're forgetting that sigma preserves this multiplicative structure. And then you see that the nth root of a is an eigenvector of sigma with eigenvalue zeta. So if we know an element a, then we may be able to find it by taking an eigenvector and taking the nth power of that. So we look for eigenvectors of sigma on this vector space m. And again, we know sigma to the n is equal to one, so possible eigenvalues are just going to be one zeta to zeta squared and so on. And we have a sort of subtle problem here. Do enough eigenvectors exist? In other words, is sigma diagonalizable? Now in general, there's no particular reason why it should be diagonalizable. In this particular case, it is diagonalizable because we assumed that zeta is in k. So k contains all the nth roots and p does not divide, sorry, the characteristic of k does not divide n. So if these two assumptions don't exist, then we might not be able to diagonalize sigma. But in this case, we can diagonalize sigma. And let's see why sigma is diagonalizable. Well, let's just take n equals three, for example. And now suppose v is any vector of m. Then we can find eigenvalues by taking v plus sigma v plus sigma squared v. So this is now fixed by sigma. You see, we've just summed over the group g and applied all elements of g to v, and that's obviously fixed by sigma. Well, then we can also take v plus omega sigma v plus omega squared sigma squared v, where I'm writing omega for zeta because omega cubed equals one and I like writing cube roots of one as omega. And this has eigenvalue omega to the minus one, or possibly omega, since I always get a bit confused about this. And that's because we're summing over the group generated by omega times sigma. And we can also take v plus omega squared sigma v plus omega sigma squared v. And this is eigenvalue omega. And now if we take their sum, it's just three v. So v is equal to a third times the sum of these, the sum of three eigenvectors. And here we're assuming that the characteristic of the field is not equal to three. So this is one of the many places that we get problems with the characteristic of the field divides n. We wouldn't be able to show this. So any vector can be written as a sum of eigenvalues, so sigma is diagonalizable. In particular, we can find plenty of eigenvectors. Now let's pick an eigenvector with eigenvalue not equal to one. And we can do this because if everything had eigenvalue one, then everything would just be fixed by sigma, which isn't true because sigma is a non-trivial automorphism. So say sigma times omega is equal to sigma times w. That's a w, not an omega, is equal to zeta times w. I guess I should use a different letter. Let's say sigma of t is equal to zeta times t because w looks too much like omega. Then sigma of t to the n is equal to t to the n. So t to the n is fixed by the Galois group, so t to the n is in k. So m is equal to k root t, where we put t equals t to the n. So we found, we sort of shown the existence of some element t such that m is obtained by joining the nth root of t. The key point is just to look for eigenvalues of the Galois group. So let's work this out in an explicit example. So let's take the equation x cubed plus x squared minus 2x minus 1 equals 0. Now you remember this is the cubic polynomial whose roots, alpha 1 is 2 cosine 2 pi over 7, alpha 2 is 2 cosine 4 pi over 7, and alpha 3 equals 2 cosine of 6 pi over 7. And the reason we're using this example is that it's about the simplest example of an extension of the rationals with Galois group is cyclic of order 3. And it's generated by an automorphism that takes alpha 1 to alpha 2, alpha 2 to alpha 3, and alpha 3 to alpha 1 as we saw earlier. And what we want to do is solve this by taking cube roots. I mean in some sense we can solve it by taking seventh roots because if z to the 7 is equal to 1, then we saw that alpha 1 can be written as say zeta plus zeta to the minus 1. So by taking a seventh root of 1, we can solve this. But we don't want to solve it by taking seventh roots. We want to solve it by taking cube roots because the Galois group is order 3. So what we do is we find eigenvectors. And we find eigenvectors by just taking an element alpha 1 and we can sum it over the Galois group. So here we're going to find an eigenvector for the Galois group. Well, this isn't a terribly interesting eigenvector because the sum of a1, a2 and a3 is just given by minus the coefficient here. So this is minus 1, so that doesn't really get us very far. So let's try taking alpha 1 plus omega alpha 2 plus omega squared alpha 3. So you remember this is sigma alpha 1 and this is sigma squared alpha 1. Here omega is minus a half plus root 3 over 2i, so omega cube is equal to 1. So omega is the cube root of unity. And then we should of course take alpha 1 plus omega squared alpha 2 plus omega alpha 3. And the cubes of these are in the field cube of omega because they're fixed by this automorphism and so must be in the underlying field. Notice that we've sort of quietly adjoined omega to the rationals. So you can actually work out what they are. The cube of this one is 21 omega plus 14 and the cube of this one is 21 omega bar plus 14 where omega bar is minus a half minus that. Omega plus omega bar is equal to minus 1. So we can now write down what alpha 1 is. So alpha 1 is equal to 1 over 3 times the sum of these three vectors which is a third of minus 1 plus the cube root of 21 omega plus 14 plus the cube root of 21 omega bar plus 14. So here we have an expression for alpha 1 by taking cube roots of elements of various fields. It's really rather bizarre that we can do this because alpha 1 can be expressed as a seventh root of unity and it seems rather strange that you can express seventh roots of unity using cube roots of something else because in a seven is a completely different prime from three. So that's a summary of what we've done so far. If k contains m is Galois with group z over nz for n prime and so on So here we have k contains nth roots of unity. The characteristic of k does not divide n and we're going to assume n is prime. Then m is equal to k where we adjoin the nth root of a for some element a. Well we only wrote it out for n equals 3 but for other values of n we do something similar. So for example instead of taking the sum over 3 conjugates we would just take the sum over n conjugates so v plus sigma v plus sigma squared v plus sigma to the n minus 1 v and then we would take v plus zeta sigma v plus zeta squared sigma squared v and so on so you can see how it will go for other primes. Now we have the question when is k with the nth root of a adjoined the same as k with the nth root of b adjoined? Well we saw that the nth root of a is an eigenvector of sigma with some eigenvalue say zeta. Now the problem is the nth root of beta may have eigenvalue some power of zeta which is a bit of a nuisance. Well we can raise b to some power and this will change the eigenvalue to zeta if we choose the right power of b. So we're first going to change b so that these two things are the same eigenvalue. So let's assume the nth root of a and the nth root of b have the same eigenvalue. Well then the nth root of a divided by the nth root of b is fixed by sigma so it's in k. In other words a over b is the nth power of something in k. On the other hand if a over b is an nth power of something in k then it's obvious that these two fields are the same and that the nth roots are the same eigenvalue. So we see we get a sort of description of the possible extensions with Galois group z over nz as before we're just doing the case n equals prime. They correspond essentially they're classified by elements of the group k star over k star to the n where you take the multiplicative group and divide out by nth powers. In fact we actually get a sort of pairing called the Kummer pairing between two groups. What we do is we take the Galois group of k with all the nth powers of things are joined. So this is actually an infinite extension and we haven't actually done Galois groups of infinite extensions yet but never mind. And then on the other hand we can get the group k star over k star to the n and if we've got an element of this Galois group and an element of k star times k star to the n we can get a map to the group of nth roots of 1 and we do this as follows. So we pick an element of the Galois group of this big field and we pick an element a of k star over k star to the n and then we look at sigma of the nth root of a and this is going to be zeta to the i times the nth root of a for some value of i. So this is going to be the nth root of 1 we map it to. So in some sense it says that these two groups here are sort of dual. You remember there's a duality theorem for finite abelian groups where the dual of an abelian group of exponent n is the map from that group to the cyclic group of order n. And the group of nth roots of 1 is more or less the same as the cyclic group of order n except it's not quite. Well it's isomorphic as an abstract group but it has a different action of the Galois group of the nth roots of unity. So if you have some element of some Galois group actually non-triven on the nth roots of unity it will actually act differently on these two groups. So you've got to be a little bit careful. You shouldn't really identify nth roots of unity with the cyclic group of order n because you might get caught up by them having different actions. This is actually something called a tape twist. It's quite common in Galois theory to get sort of things that are in some sense twisted by roots of unity. So the action isn't trivial but gets sort of twisted by the action of a Galois group on roots of unity. So here these things are sort of dual except there's a sort of tape twist going on. Well I've been mentioning several times that things go wrong in characteristic p when p divides this exponent here. So next lecture we're going to be discussing the case of what happens when p in characteristic of the field divides the order of the cyclic Galois group and this will give us something called Artin Shrier extensions.