 This lecture is part of an online Galois theory course and will be about Arten Schreyer extensions. So let me explain what these are. So the problem is as follows. Given an extension K contained in M, which is Galois, with Galois group cyclic, we want to describe what the extension M is. So last lecture we looked at the case where the Galois group is cyclic of order co-prime to the characteristic of K. This lecture we're going to look at the case when the Galois group is cyclic of order P and P is equal to the characteristic of the field K. So this makes the extensions completely different from what they were in the previous lecture. So the problem is to describe the extension M. And we first notice that M can't be described by radicals. The first problem is that if we take a radical extension of degree P, say we take K and we join the Pth root of some element A, this is inseparable because the equation X to the P minus A equals 0 has derivative 0, so all its roots are the same. So in particular it's not Galois. On the other hand, if you've got an extension K over M that's Galois with group Z over Pz, M cannot be obtained as a radical extension of A of K because as we've just pointed out radical extensions are inseparable, not Galois. So radical extensions of degree P and characteristic P have very little to do with Galois extensions. So if M is not given by taking a radical of something, we can ask what is M? Notice that such extensions are quite common. For example, if we take a finite field of characteristic P, this is contained in the field of order P to the P and this extension has degree P and is in characteristic P. So how are we going to construct M? Well, in the previous lecture we looked at eigenvectors and eigenvalues, so let's try the same. So we want to find eigenvalues and eigenvectors of sigma acting on M. Here sigma is going to be a generator of the Galois group of M over K. And M, well, we're going to forget that it's a field and just think of this as being a vector space over K of dimension P. So we just forget that M has a multiplicative structure. And let's try and find the eigenvalues. Well, sigma to the P is equal to 1, so if an eigenvalue is lambda, then we see lambda to the P equals 1. Well, this means lambda minus 1 to the P equals 0 because lambda to the P minus 1 is equal to that. So lambda equals 1 is the only eigenvalue. Well, let's find the eigenvectors. Well, the eigenvectors must satisfy sigma v equals lambda v, which is just v. So v is fixed by sigma. So the things fixed by sigma are just the elements of K. Well, this is a bit unfortunate because we were trying to construct the extension M of K by generating M by an eigenvector. But we've just found that every possible eigenvector is in K, so joining it to K is just going to produce K again. So what do we do? The solution is we use generalized eigenvectors. So these are the things that are very briefly mentioned in linear algebra courses and that no one pays any attention to because they look boring and technical. So I'll just recall the definition. So generalized eigenvector with eigenvalue lambda is something such that lambda sigma minus lambda to the Kv equals 0 for some K equals 1, 2, 3, and so on. So if K equals 1, this just says sigma minus lambda v equals 0, so v is then an eigenvector in the usual sense. For K equals 1, we get sigma minus lambda squared of v equals 0, and this is the case we're going to be most interested in. So you can also understand generalized eigenvectors by pushing a matrix into Jordan canonical form. So for example, sigma might look something like this with a lot of zeros, except on the diagonal and just above it. And if you apply this matrix to a vector with just one nonzero entry there, this is an eigenvalue. So sigma minus 1 of v is equal to 0. If you apply this matrix to something with two nonzero entries at the bottom, we get sigma minus 1 squared of v equals 0. So that's a generalized eigenvalue. And similarly, if we allow three entries, we get sigma minus 1 cubed of v equals 0. So you can see there's an increasing sequence of subspaces, which giving a filtration of the space m. Well, we want some sort of generalized eigenvalue. Well, the best case would be sigma minus 1 v equals 0, but this just implies v is in K, so that's no good. Well, the next best case is going to be sigma minus 1 squared of v is equal to 0. So let's think about what this means. So we can certainly find a vector with this property because the vector space is spanned by generalized eigenvectors. And if you've got a generalized eigenvector with sigma minus 1 to the nv equals 0, you can just keep on applying sigma minus 1 until we get this condition. So we can certainly find vectors like this. Well, what does this say? It says sigma minus 1 v is fixed by sigma. So it's in K because K is the things fixed by sigma. So this means sigma v is equal to v plus a for some a in K. Well, now we can just replace v by v over a and we find we get a vector with sigma v is equal to v plus 1. So this is what the extension m is generated by. We can find a vector v in m satisfying this condition here. Well, what we really want is a polynomial. So we want to find a polynomial with v as a root. Well, how do we do that? Well, we can do sigma of v to the p is equal to v plus 1 to the p, which is equal to v to the p plus 1. And now we see that sigma of v to the p minus v is equal to v to the p minus v. So v to the p minus v is fixed by sigma. So v to the p minus v is in the fixed field K. So we finally find our equation satisfied by v. v to the p equals v plus a for some a in K. And this is the equation that this equation is sort of we've now achieved our goal of finding an equation satisfied by v. So m is generated by a root of this equation here. So let's take a look at these equations in a bit more detail. So this is called the Arten-Schreyer polynomial x to the p minus x minus a. So we're going to join a root for it. Conversely, let's pick some field of characteristic pK and look at the fields generated by a root of this polynomial. So if alpha is a root of this polynomial, so are the elements alpha plus 1, alpha plus 2, up to alpha plus p minus 1. So these are the p roots. And now let's look at the Galois group of the splitting field. And suppose sigma is an element of the Galois group. Well, sigma of alpha must be alpha plus i for some i because these are the only other roots. And we see from this that we can think of sigma as being an element of z modulo pz because i is in z modulo pz. And you see that composing elements in the Galois group just corresponds to addition in z modulo pz. So the Galois group is a subgroup of z modulo pz. And there are only two possibilities. Either the Galois group is 1, in this case x to the p minus x minus a splits into linear factors, which isn't terribly interesting. Or the Galois group must be z modulo pz, in which case this forces x to the p minus x minus a to be irreducible. So we have this rather funny dichotomy. The polynomial x to the p minus x minus a is either irreducible or it splits completely into linear factors. There are no intermediate cases where it splits as a product of several factors of degree greater than one. So we can do an example of this. We might want to construct a finite field of order p to the power of p. Well, there's an almost canonical way of doing this. We can take fp x and quotient out by x to the p minus x minus 1, the art in Schreier polynomial. So you remember I said earlier that there was a bit of a problem with finite fields and that there doesn't really seem to be a best possible finite field. In order to construct a finite field, we have to pick an irreducible polynomial of some degree and there really doesn't seem to be a best way of doing that. Well, one case when there does seem to be a really canonical choice is just the fields of degree p, where p is the characteristic. By the way, I mentioned earlier in the section on finite fields that it wasn't really clear which of the irreducible polynomials x to the 4 plus x plus 1, x to the 4 plus x cubed plus 1 and x to the 4 plus x cubed plus x squared plus x plus 1 would be best to use. Well, as well as looking at art in Schreier extensions to the degree p, there's also a more general theory where you look at extensions to the degree p to the k and in this case you look at polynomials of the form x to the p to the n minus x minus a. So this would give this polynomial here as being the nicest irreducible polynomial for generating the field of order 16. That doesn't necessarily prove it wins because art in Schreier polynomials only work for extensions of degree of power of p and it's not at all clear that's the best thing to do if you're looking at extensions of arbitrary degree. There's also, you remember in Coma theory, we had a Coma pairing where we took the Galois group of k and joined all the nth roots of everything we could think of and had a pairing of this with k star over k star to the n and this mapped to the roots of 1. There's an analog of this for art in Schreier extensions so this might be the art in Schreier pairing. Here we take the Galois group of all degree p Galois extensions. So we take all the degree Galois extensions of k and kind of put them together in a big extension and there's a pairing between this and the additive group of k modulo the things generated by a to the p minus a. And what this pairing does is it takes an element of this Galois group and it takes an element a of this field here and the pairing what we do is that the pairing takes sigma and a to sigma of alpha minus alpha where alpha to the p minus alpha minus a is equal to zero. So to summarise in characteristic zero we we showed that the Galois group being solvable is sort of more or less the same as being solvable by radicals. Now in characteristic p greater than zero the analog is a little bit more complicated it turns out that the Galois group being solvable is equivalent to being solvable by radicals where we take the nth root of something where n is co-prime to the characteristic and art in Schreier extensions. So we somehow have to throw away taking p th roots because those give us inseparable extensions and we replace them by art in Schreier extensions. The proof of this is very similar to the proof for characteristic zero except that you have to use the fact that extensions with Galois groups cyclic of order p are now art in Schreier extensions not radical extensions. So I just finished by summarising that a comparison between Kummer extensions and art in Schreier extensions. So Kummer extensions are generated by an element where the Galois group acts as sigma alpha equals zeta alpha where zeta to the n is equal to one. Art in Schreier extensions generated by an element such that sigma of alpha is equal to alpha plus one. So in both cases you found an element where the Galois group acts in almost the simplest possible way either multiplying by a root of unity or adding one. You could hardly get any simpler than that. And this implies that alpha to the n is in K and this implies that alpha to the p minus alpha is in K. So the irreducible polynomial looks like x to the n minus a equals zero for some a whereas here we get x to the p minus x minus a equals zero. So art in Schreier extensions and Kummer extensions are quite similar except we sort of got these extra terms here. So the Kummer pairing is given by sigma a goes to sigma alpha over alpha where alpha to the n minus a equals zero here and here it's given by sigma alpha is sigma alpha minus alpha where alpha to the p minus alpha minus a equals zero. So here you can sort of see from this that Kummer extension is a sort of multiplicative theory and the art in Schreier extensions are a sort of additive theory. Okay, what we'll be doing next lecture is showing how to use Galois theory in order to write out the roots of cubic and quartic equations explicitly by radicals.