 So this lecture is part of an online course on Galois theory, and we will be talking about Arbeau's theorem. This is the very famous theorem, so general Quintic cannot be solved by radicals. So Quintic just means an equate polynomial of degree 5x to the 5 plus ax to the 4 plus bx cubed plus cx squared plus dx plus e. And solvable by radicals means you can't write down the solutions explicitly in terms of a, b, c, d and e using the field operations and taking nth roots of something where n is allowed to be anything at all. So a little bit later we're going to show that you can solve polynomials of degree 3 and 4 by radicals. Of course everybody knows how to do a degree 2 if you've got ax squared plus bx plus c equals nought, then x is minus b plus minus the square root of b squared minus 4ac over 2a. So the formulas for cubics and quartics are really rather messy to write out explicitly, but you can do them. What Arbeau done showed is that you can't do this for degree 5 equations. So what we're going to do is, well, first of all, we're going to work over a characteristic zero field such as the rationals. There are some additional complications for fields of characteristic greater than zero that I'll occasionally mention. So we'll usually be assuming have a default assumption that the field is characteristic zero unless I say otherwise. And what we're going to do is to show that if alpha can be expressed by radicals over a field k of characteristic zero, then alpha is contained in a finite solvable Galois extension of k. So solvable just means the Galois group is solvable. And we recall from the earlier group theory course that what this means is there's a chain of subgroups, 1 equals g0 contained in g1, contained in g2 and so on, contained in gn equals g, so that gi is a normal subgroup of gi plus 1 and gi plus 1 over gi is abelian. Actually, you can also assume that cyclic if you want, it doesn't really make much difference. So in other words, g can be kind of split up into abelian or even cyclic subgroups. So in order to show that a fifth degree polynomial cannot be solved by radicals, what you want to do is to find a polynomial of degree 5 whose Galois group is non-solvable. And there's one easy way to do this that we had before. You can take the rationals, take x1 up to x5 and take the subfield fixed by the symmetric group of order 5. So as we said earlier, this will be q e1 up to e5 where these are the elementary symmetric functions. So e1 is x1 plus x5 and so on. And then if we take the fifth degree polynomial x5 minus e1 x4 and so on, it has roots x1 up to x5. And since s5 is not solvable, this means that x1 up to x5 cannot be expressed in radicals using e1 up to e5. So the key point is that the symmetric group s5 is not solvable. You may remember from group theory that the symmetric groups s1 up to s4 are solvable, which is why we can solve polynomials degrees 1 to 4 by radicals, at least in characteristic zero. Well, okay, that shows there's, you can't write down a general formula that works simultaneously for all polynomials, but it doesn't quite rule out the possibility of doing it for any particular polynomial with integer coefficients. So some polynomials with integer coefficients you certainly can solve by radicals. For example, if we take x to the 5 minus 2, well, I can certainly solve this one by radicals and just take x as the fifth root of 2. And you could imagine that maybe by some sort of funny coincidence, every single polynomial that happened to have integer coefficients would have a solvable Galois group, even though the general polynomial doesn't. So we have to rule that out. So can we find a polynomial with integer coefficients or rational coefficients with non-solvable Galois group. Now, normally working out the Galois group of a polynomial of degree 5 is a real pain, but there's a rather easy trick for finding a few examples with Galois group S5, which is suppose that the polynomial f of x, let's let's work over the integers, is irreducible degree 5 and has exactly two non-real roots. It's easy to find examples of such polynomials, for instance x to the 5 minus 4x plus 2. You can easily check that its graph sort of looks something like that. It's got exactly three real roots. It's irreducible by Eisenstein. So why if it has these properties is it's Galois group S5. Well, the Galois group is contained in S5, which is just the permutations of all the roots. And its order is divisible by 5 because the polynomial is irreducible. So the first, when you add one root of it, you get an extension of degree 5 and you might get a higher extension by adding further roots. You might not. Who knows? Whatever, its order is divisible by 5, so g has a 5 cycle. Now there are exactly two non-real roots, so g contains a transposition. That's just an element that exchanges two roots and fixes the others. And this transposition is easy, it's just complex conjugation. Since there are exactly two non-real roots, complex conjugation flips those and fixes the others. Now we can see that any subgroup of S5 containing a transposition and a 5 cycle is the whole of S5. And that's easy to see. We may as well take the transposition as 1, 2. And what's the 5 cycle going to do? Well it's going to map 1 to something and something to something else. And what we can do is we can raise the 5 cycle to some power, so it maps 1 to 2. So a power of the 5 cycle is 1, well it maps 1 to 2 and then we may as well re-label all the other elements so that the 5 cycle is 1, 2, 3, 4, 5. So by renumbering the 5 roots we can arrange that the transposition and the 5 cycle are like those. But then conjugates of 1, 2 under 5 cycle. Well conjugates just mean you re-label these elements according to the cycle. So its conjugates are going to be 1, 2, 2, 3, 3, 4, 4, 5 and 5, 1 but we don't really care about that. And now it's well known and easy to check that these generate the symmetric group S5. Details of that are in the course on group theory if you want to look them up. So we found an explicit polynomial over the rationals that can't be solved by radicals. By the way if you're wondering about this condition that there are exactly two complex roots it actually is essential. So we can ask what if there are zero or four complex roots. Well in these cases an irreducible 50 degree polynomial might indeed be solvable by radicals. For example x to the 5 minus 2 has four complex roots and it's irreducible and it's obviously solvable by radicals. In fact we can take a quick look at what goes on because this will be an example of what we're going to do later. So how do we find the splitting field of this? Well we do it in two steps. First of all we add in the fifth roots of 1 and there are there are four primitive fifth roots. So you remember x to the 5 minus 1 over x minus 1 is x to the 4 plus x cubed plus x squared plus x plus 1 which is irreducible. So if we write this out in the complex plane the four we're going to add the four primitive fifth roots of 1 like that. So we get an extension q contains in q zeta where zeta is say one of these primitive fifth roots of 1. And now we add in fifth root of 2. So we get q contains q zeta contains q the fifth root of 2 together with zeta. And now this extension is normal and for that matter Galois. Notice if we just add in the fifth root of 2 we wouldn't get a normal extension. We first need to add in the fifth root of unity. And now we can sort of see what the Galois group is because the Galois group of this we saw earlier is just the little abelian group of order 4 which consists of the units of the integers mod 5 whereas this bit of the Galois group well we can multiply the fifth root of 2 by any fifth root of unity. So the Galois group of this bit is z over 5z. So all together we've got a group of order 20 and you can check it's actually none abelian because this group z modulo 5z star acts non-trovian z modulo 5z. By the way remember when you go to a Galois group the groups are sort of upside down to the field. So if you have the Galois group the z over 5z is a normal subgroup of the Galois group and the group divided by z over 5z is this group of order 4 z over 5z star. So it's very confusing because the here we have a subfield but the subfield doesn't correspond to this subgroup it corresponds to a quotient group and in this particular case this quotient group also happens to be a subgroup but in general it won't be. So the point of this example is that we're getting a solvable extension but we get in two steps. We first add in roots of unity then we add in the radical that we're really interested in. So that's an example with four complex roots. What about an example with zero complex roots? Well this doesn't have to be non-solvable either. For example we could take cosine of 2 pi over 11 which is equal to zeta plus zeta to minus 1 where zeta is a primitive 11th root of 1. So why are we taking 11? Well see if I can draw 11 things correctly so we get 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4. Okay so here we've got 11 roots of unity and one of them is the number one which we're not really interested in and then we can project the others onto the real line and as usual we get these numbers cosine of 2 pi over 11 cosine of 4 pi over 11 and so on and all together we get 5 of these which is why it shows 11 because 11 minus 1 over 2 is 5 and we can ask what is the Galois group of the fields generated by these? Well we sort of did this for cosine of 2 pi over 7 and it's pretty similar here so the Galois group of q of zeta over q is just c over 11z star which is cyclic of order 10 and the Galois group of cosine of 2 pi over 11 modulo q will be a quotient group of this where we quotient out by the element plus or minus 1 so this is isomorphic to a cyclic group of order 5. So what we've got here is a polynomial with it's an irreducible polynomial with five real roots which are these numbers cosine of 2 pi over 11 4 pi over 11 and so on and its Galois group is definitely solvable in fact it's even cyclic of order 5 so we really did need to assume there were exactly two complex roots if there are any other number of complex roots that doesn't force it to be non-solvable although of course it might be. Well so that's given some examples of fifth-degree polynomials with non-solvable Galois groups but now we better explain why if an equation is solvable by radicals we get a solvable Galois extension so what I want to do is to sketch this implication. So suppose something is solvable by radicals we're going to construct the Galois extension in several steps so first of all we add in all roots of unity we need and so we're going to take our field k which might be the rationals and add in a root of unity of some high order it's going to be a root of unity of order n where n is big enough so that so that every radical we're taking is a radical of order dividing n and we can ask what is the so we're going to take this to be a splitting field of x the n minus one for some large n and remember we're working characteristic zero so this is actually a separable polynomial. If you work in characteristic p one of the complications is that this polynomial is no longer separable in general. Anyway the Galois group the elements of the Galois group all take zeta to some power of i sorry to some power of zeta where i is in z modulo nz star. Now it's not necessarily true that all elements of z modulo nz turn up in the Galois group that depends on the field k we're going to show what they do for the rationals but in general they don't so and as usual the zeta to the i times if we compose the automorphism taking zeta to the i and zeta to the j we get zeta to the ij so the Galois group is a subgroup of z over nz star in particular it's abelian and therefore solvable. As I said it may not be the whole of this group it's sometimes less but it doesn't really matter if it's smaller than this because it's still abelian and that's all we really care about. Now once we've got once we've added in all the roots of unity the next step is to so let's put k1 equals k with our root of unity then let's put k2 equals k1 together with some nth root of an element alpha in in k1 and let's put k3 to be k2 where we apply some other root sorry this doesn't have to be an nth root it could be some other root and so we built up a tower of fields like this except that's not quite right well we saw the problem with this earlier the problem is that kn might not be so kn might not be normal just recall the very simple example of this you might take q contains in q root 2 contains in q with square root of the square root of 2 and then this extension here is not normal well that's quite easy to fix the reason is this that this isn't normal is we only took the square root of this number and we forgot to take the square root of all its conjugate so what we should really do is take q square root of the square root of 2 and then we take the square root of the square root of minus 2 so these are the conjugates of the square root of 2 so when we're building up these fields we shouldn't take k2 to be k1 where we we take a root of alpha 1 we should take it k1 with nth root of alpha 1 and the nth root of all conjugates of alpha 1 and similarly we shouldn't take k3 to just take a root of alpha 2 but we should take it to be k2 with some root of alpha 2 and all conjugates under the galva group and and if we remember to put in all the conjugates this ensures that each of these fields is going to be normal and we also need to know what does the galva group of this look like well what's the galva group of let's take a field l nth root of alpha over l where l contains the nth roots of 1 and we're assuming it contains exactly n nth roots of 1 so it's not in doesn't have characters dividing n or anything like that then the galva group consists of elements taking the nth root of alpha to the nth root of alpha times zeta to the k for some k in z modulo nz because any nth root of alpha must be the nth root of alpha we first thought of times times some power of our primitive nth root of unity and you can see the composition of these automorphisms corresponds to addition in z modulo nz so the galva group of this is a subgroup of z modulo nz as before it doesn't necessarily have to be the whole of z modulo nz but again this doesn't matter it's a subgroup of a cyclic group so in particular this group here is abelian so every time we add an nth root we get a normal extension with a cyclic galva group so what happens is we can get a chain of fields we get various fields like this up to k whatever it is such that each of these groups is a normal extension with galva group abelian so it's either contained in z modulo nz star or it's contained in z modulo nz for some where n is the largest possible where n is something big enough so it includes all possible radicals we're taking so but you remember by including radicals of all conjugates we could also arrange that this extension here was normal and therefore galva because we're working in characteristic zero so we've got a chain of fields such that each field is a is an abelian extension of the one earlier now if we look at the corresponding galva groups i guess that shouldn't be an m now each of these is um so each of these groups is a normal subgroup of the earlier one and the quotient is abelian so it's one of these groups here so the galva group with the whole extension we've got is therefore solvable because we can break it up into abelian groups so this completes the sketch of arbor's theorem that fifth degree polynomials are not solvable by radicals we can ask the converse if the galva group is solvable can we represent elements in the field so i suppose we've got k in m can we represent elements of m using radicals and the answer to this turns out to be no in general there are actually some problems in characteristic p that we're going to discuss but it turns out to be true in characteristic zero and this is what we want to talk about next you see that if the extension is solvable we can split it up into a lot of extensions each of which has a galva group that is cyclic of order p so we have the following problem suppose we have a galva extension k over m with galva group a cyclic group of prime order um using p for that is probably bad since p isn't necessarily the characteristic but anyway then we can say what can we say about the extension l over k and this question is going to be the topic of the next lecture so we want to describe the simplest sorts of galva extensions which are those whose galva group is cyclic of prime order