 This lecture is part of an online Ganwar theory course, and will be about primitive elements. So what is a primitive element? Well, suppose we've got a finite extension of fields. Then m is generated by a finite number of elements. So it's equal to k alpha 1, alpha 2, up to alpha n, for some alpha i. And we've seen we can do a lot of arguments by induction, because we first prove something for k, and then we prove it for k alpha 1, and then we prove it for k alpha 1, alpha 2, and so on. And this works, but it's a bit messy and tiresome, and it would really be much neater if m was generated by just one element of m. So we can ask, is m equal to k alpha for some alpha? And if so, alpha is called a primitive element. So I don't know why they use the term primitive element. A more sensible thing would be to call it a generator, but anyway, that's the terminology. So, and the answer is yes, if the extension m over k is separable, and sometimes no, if not. So what we're going to do is we're going to prove that there's a primitive element whenever the extension is separable, and then give an example to show it fails in general. So we first observe that if m over k is separable and finite, we're assuming all extensions are finite for the moment, then there's only a finite number of intermediate extensions between l and m. And this follows fairly easily, because what we do is we just extend m to a larger normal extension of k, let's say normal and separable. So we could, for example, just take the irreducible polynomials defining a set of generations of m and n could be their splitting field. And then we look at the Galois group, which is the Galois group of n over k. And this is finite, of course. And now we notice that subgroups of g correspond to extensions between k and n. And of course g is finite, so it has only a finite number of subgroups, so they're only a finite number of extensions between k and n. And then, of course, they're only a finite number between k and m, because m is contained in n. And now the idea is we want to choose alpha not in any extension l with k contained in l, contained in m, and l not equal to m, of course. Because if it's not in any of these finite number of extensions, then the field it generates can't be any of the proper subextensions, so it must be the whole of m. Well, all we've got to do is to prove that m is not the union of all these extensions. And this usually follows easily from the following lemma. Suppose k is an infinite field and m a finite dimensional vector space, then m is not the union of a finite number of proper subspaces. So proper means they're not equal to m. And we can prove this by induction on the number of subspaces. So if we've got subspaces v1 up to vk, what we do is we sort of draw v1, v2 up to vk minus 1. And then let's draw vk in a different color. And what we do is we find an element not in v1 up to vk minus 1. And we don't have to assume it's in vk, but if it's not in vk, we're done, so we may as well assume that. And then we know vk is not the whole of v, so we pick some element not in vk. And as before, we're finished unless it's in one of these vi's, although we don't really have to assume that. And then what we do is we draw the line through these two points. Oops, I seem to have missed that point, so let me make this point a bit bigger. And we notice this line is not contained in any vi. So it has at most one point in common with any of these subspaces v1 up to vk. Well, it's got an infinite number of points on it because our field is infinite, so it must have an infinite number of points not in any vk. So the line contains points not in any vi. So over infinite fields, a vector space can't be the union of a finite number of proper subspaces. And of course, this immediately implies that if k is infinite, and if the extension m over k is separable, this means m is equal to k of alpha for some alpha. Because we've shown there are only a finite number of intermediate extensions because the field extension is separable and their union can't be m because k is infinite. Well, fine. What if k is finite? So it's a finite field. Well, there are several ways of doing this. One is to show that we look at the multiplicative group of m. And this is a cyclic group. So any generator of this cyclic group generates the field m. I mean it generates the multiplicative group of m under multiplication, so it certainly generates the whole field and is primitive. So the only thing we need to recall is why is this cyclic? Well, the reason why this is cyclic is that any finite subgroup of m star for any field m is cyclic. This is really nothing to do with m being finite. And the reason for this is that if we call this finite subgroup g, g has at most n solutions to g to the n equals 1. And that's because these are just the roots of an nth degree polynomial and there are at most n roots of that polynomial because we're working over a field. And this condition here for a finite abelian group implies g is cyclic. And there are several ways of proving that. There's a sort of quick and dirty one which is just a quote, the fact that any finite abelian group is a product of cyclic groups and then you can check that if this product is not cyclic, it has more than p elements of order p for some prime p. So there are several ways of doing that. So this proof is a little bit strange because the proof of the existence of a primitive element for finite fields k is quite different from the proof of existence for infinite fields k. I don't really know of a neat proof that works for both finite and infinite fields. So let's just have a quick example. Suppose for example we have k contained in, well the simplest example of a field with more than one generator is we might just take k, the field over k generated by say the square root of 2 and the square root of 3. And then we know the subfields from Galois theory are k, k root 2, k root 3 and k root 6. And of course the whole field M itself. So all we need to do to find a primitive element is to find an element not in any of these fields here. So if we look at a plus b root 2 plus c root 3 plus d root 6, well in order for it not to be in any of these three subfields, we must have at least two of b, c, d are none zero. Sorry, when I said k here I should have said this should be the rational numbers. These are all the rational numbers. So if at least two of b, c and d are none zero then this element is primitive. Well so far we've shown that separable finite extensions all have primitive elements. What about inseparable extensions? And the basic rule of inseparable extensions is their counter examples to almost everything. Well if we have an inseparable extension of degree p for p the characteristic, that's certainly generated by one element because the only subfields are going to be k and m. So the degree had better be at least p squared. So let's take a field of k of characteristic p greater than zero and we're going to look at the field generated by rational functions of two variables t and u. And inside this we can find a field, also a field of rational functions in two variables where big t is t to the u and sorry t to the p and big u is u to the p. And now we notice, so this field is big k and this field is big m. And now we notice that if a is an m, a to the p is in k. And that's because we've got the Frobenius mapping by a equals a to the p. This is a homomorphism of fields and it maps m into k because it maps a set of generators of m into k and since there's a homomorphism it must map the whole of m into k. So if a is an m, a is a root of x to the p minus b was not for b in k. So it generates a field of degree p. However, we notice that this extension has degree p squared because both t and t has degree p over the field and then u adds an additional factor of p. So there are no primitive elements. We also notice that the extension k, t, u contained in k, t, u is finite degree p squared but has an infinite number of sub extensions. So you remember for separable extensions we proved there only a finite number of intermediate extensions. We see from this that this is not actually true for all field extensions. Okay in the next few lectures we're going to be discussing fields with cyclic Galois group and fields generated by taking a root of some element and this will lead to the famous theorem that you can solve polynomials by radicals if the polynomial has degree at most four but not in general if it has degree five or more.