 So this lecture is part of an online course on Galois theory, and will be about the fundamental theorem of algebra. So this says that the complex numbers is an algebraically closed field. In other words, every polynomial that's non-constant has a root. Calling it the fundamental theorem of algebra is not actually a terribly good name, because, first of all, it's not really a theorem about algebra, because you can't prove it or even construct the complex numbers purely algebraically. I mean, you have to sort of take the completion of the rationales to construct the reals and so on. And the other problem is it's not really particularly fundamental these days in algebra. I mean, you can do an entire course in algebra without even mentioning it if you want to. So it was probably conjectured in about the 17th century, although it was rather difficult for people to state it then, because the complex numbers hadn't really been properly defined or understood. You know, people used to talk about fictitious roots of polynomials, meaning negative roots, because they were sort of rather dubious about even about the negative real numbers, never mind complex numbers. Anyway, the first proofs of it seemed to be due to either Gauss or Argan, both around 1800. Gauss's proof was before Argan's proof, but it's not entirely clear that Gauss's proof was completely watertight. There may have been a small topological gap in it. But anyway, so we've already given two proofs or mentioned two proofs. First of all, we have a topological proof involving the winding number. So you remember we looked at a complex number z and made it go once around the origin, and then we looked at p of z for a polynomial and saw that p of z winds n times around the origin and were able to deduce from that that p has a root. The second proof comes from complex analysis. So, which may be almost the simplest proof, it says that if the polynomial p has no root, then 1 over p is bounded and holomorphic. So since it's got no roots, it doesn't have any poles, and it's easy to check it's bounded by looking at what happens for large values of z. So by Lueville's theorem, it is constant. So there we've got a proof that would be one line long if I wrote slightly smaller letters. So if we've got two perfectly good proofs, why do we need another proof? Well, it's a bit embarrassing to have something called the fundamental theorem of algebra only proved by using analysis or even worst apologies. So if something's called the fundamental theorem of algebra, you really wanted to have an algebraic proof just for the pride of your subject. Well, you can't really because as I mentioned, you can't even, I mean, the real numbers and the complex numbers just aren't really algebraic objects. But what you can do is you can try and reduce the amount of analysis you need to a minimum. So let's see what we need. So our assumptions are going to be as follows. So zero's assumption is that we have a finite field extension or contains C of fields of characteristic zero. And I'm assuming characteristic zero because I got a headache trying to work out whether or not this is what I say is true for positive characteristics. So I'm just going to assume characteristic zero because this makes it easier and is the only case I need. The first assumption is that any polynomial with real coefficients of odd degree has a root in the reals. And this is kind of obvious from the intermediate value theorem of introductory calculus or whatever because if say the leading coefficient is positive, then the polynomial will be negative for X, negative and large and positive for X positive and large. And somewhere in going from a negative number to a positive number, it must cross the axis. And because the real numbers are complete, it must have a zero. Notice that this part of the saying a polynomial of odd degree must have a root is not really algebraic. You need to sort of use completeness of the real numbers in order to prove the existence of a root. For instance, if you do this for the rational numbers, it's simply not true. A polynomial of odd degree over the rational doesn't necessarily have a rational root. The second thing we're going to assume is that C has no extensions of degree two. So this is equivalent to saying that every element of C has a square root because we assumed we're working characteristic zero. If we're working characteristic two, you have to be a little bit more careful about this equivalence, but we won't worry about that. So what we're going to do is just use these two assumptions and from now on, the proof will be purely algebraic. So the analysis we use is just the intermediate value theorem, which is really a very mild piece of analysis. So what we do is we look at R contains C and contain X and we're going to take X to be any finite normal extension of R containing C. Now to show C as algebraically closed, it's enough to show that X is equal to C because if C wasn't algebraically closed, we'd be able to find a non-trivial normal extension. Well, X is finite and normal and it's also several because we're in characteristic zero, so X over R is separable plus normal, so it's Galois. And let's put G to be the Galois group, so it's going to be some finite group. And now the idea is to translate our two conditions into conditions about the Galois group G. So we know subgroups of G correspond to extensions R contained L contained X between R and X. So our first assumption said that any polynomial of odd degree has a root. In other words, there are no extensions of R of odd degree apart from the trivial one. So this says that any subgroup of X of odd index is just X because an extension of odd degree of R corresponds to a subgroup of odd index. You remember the degree of the extension over R is actually the index of the corresponding subgroup, not the order of the subgroup. Now we want the other condition was that C has no extensions of degree two. So suppose H is the subgroup corresponding to C. So you remember by the Galois correspondence, this just means H is the Galois group of X over C. So the automorphism of X fixing all elements of C. Now extensions of C now correspond to subgroups of H. So H has no subgroups of index two because we said that C has no extensions of degree two. When I say odd index and X, that should be G I guess. I guess that should be G as well. So we've got two properties of a group. G has no subgroups of odd index and H and G has no subgroups of index two. And what we want to do is to show that H is trivial. Well this is easy. What we're going to do is just kill off this, prove this theorem by just quoting results from group two. So what we've done so far is we've taken a result about fields that C is algebraically closed and translated it into this purely group theoretical problem using Galois theory. Now we're going to use group theory to solve this. So what we do is we pick a seal of two subgroups of G. And now we notice this has odd index because seal of two subgroups always have odd index. So by one it is the whole of G because G has no subgroups of odd index other than G itself. So G has order two to the n for some n because any seal of two subgroups has order two to the n. So H has order two to the m for some n. So H is nilpotent. So any group of prime power order is nilpotent. So if H has order greater than one it has a subgroup of index two. Because by the theory of nilpotent groups any group of order two to the m for m greater than zero has a subgroup of index two and H can't have a subgroup of index two. So the order of H is equal to one so there are no extensions of the complex numbers. Okay that's enough about the complex numbers being algebraically closed.