 This lecture is part of an online Galois theory course, and we'll be about how to solve cubics and quartics using radicals. And we're going to do this using a knowledge of Galois theory. So quartic, by the way, means a degree four polynomial. Just as a warm-up, let's have a look at a quadratic. Everybody knows how to find the roots of AX squared plus BX plus C equals zero, let's try and do it using Galois theory. So we know that the root of this polynomial generates an extension of degree two, which whose Galois group is just cyclic of order two. So suppose the roots are alpha and beta. So if sigma is the non-zero element of the Galois group, then sigma alpha equals beta, and sigma beta equals alpha. If you remember, what we'd try and do to solve these is to find an eigenvalue of sigma. Of course, we're assuming we're working over a field that's not a characteristic two. So the eigenvalue is going to be minus one, and the corresponding eigenvector will be alpha minus beta. So we know from the work on Coma theory that alpha minus beta squared, oops, alpha minus beta squared, should be in the field we're working over. Now alpha minus beta squared can be worked out easily. It's a symmetric function of the roots, so you can express it in terms of the coordinates. It turns out to be B squared minus four AC over A squared, the usual discriminant of a quadratic equation of factors of A squared or whatever. We also know that A plus, sorry, alpha plus beta is equal to minus B over A, and from these two equations, you can easily work out alpha and beta, and you get the usual formula for a solution of a quadratic. So the key idea is you look at eigenvectors and eigenvalues of certain elements in the Galois group. So now let's look at cubics, and I'm going to take x cubed plus Bx plus C equals zero. If you're not working over a field of characteristic three, then you can get rid of the coefficient of x squared just by adding something to x. The history of the solution of the cubic is a little bit murky. It appears to be solved by ferro and possibly independently by Tata Aglia, but it's difficult to know exactly what happened because both of these guys kept their work secret. You see, apparently in those days in Italy, the way you got a mathematical job was by challenging some professor to a mathematical duel where you would both give each other problems, and if you lost the duel, you lost your job. So if you could solve a cubic, this gave you an incredibly powerful weapon because you could just pose a lot of cubic equations to your opponent and win any duel. So ferro kept all his work secrets and Tata Aglia apparently told his work to Cardano but swore Cardano to secrecy and then got rather annoyed when Cardano later published a solution which Cardano claimed was ferro's solution or whatever. So it was all rather complicated and everybody ended up in bad terms with everybody else. Anyway, we're going to do it using Galois theory, which of course they didn't have in those days. In fact, they didn't really even have negative numbers. They were really struggling with... This was back in the days when people were still trying to set up algebra and didn't really quite have the concept of the real numbers. So anyway, let's take the roots to be alpha, beta and gamma and the Galois group will be a subgroup of S3, but let's take it to be S3. And S3 has a composition series. We take the identity group and then we have the alternative group A3 and we have the group S3. And this is a group that's cyclic of order 3 and A3 is a normal subgroup of S3 and has the quotient group is order 2. Now the fixed point fields we get from this give us a tower of fields. So we have k is contained in the summer field which is contained in k, alpha, beta and gamma. And you know from Galois theory that what we should do is first work out this mysterious field here by taking the root of something and then take a cube root of something to get to this field. We assume that k contains all the roots of unity we need which are square roots which is automatic and cube roots. So we want to work out what this field is. It's generated by some element fixed by A3 and we want it to be an eigenvector of A3 over S3. So we want S3 over A3 to act on it as minus 1 or rather we want the non-trivial element to act on it as minus 1. We want to think of some polynomial alpha, beta and gamma that's invariant under a cyclic permutation of them but changes sign if you swap alpha and beta. And the obvious thing to try is alpha minus beta, beta minus gamma, gamma minus alpha. So this is more or less the famous discriminant or possibly the square root of the discriminant of the polynomial. So let's call it delta and so we know delta squared is now a symmetric function of alpha, beta and gamma. So it's a polynomial in the coefficients and in fact you can work out the polynomial it turns out to be minus 4b cubed minus 27c squared which you can get after doing a little bit of algebra. So that sort of goes from S3 to A3 but this field here is k of delta which you can take by taking a square root of this expression here. Now to go from here to here we need to take a cube root and we remember how to do this. What we do is we look at the three roots alpha, beta and gamma and we add them up and the sum of them is actually 0 because we had x cubed plus 0x squared plus bx plus c equals 0 and now we take the element y which is going to be alpha plus omega squared beta plus omega gamma and z which is alpha plus omega beta plus omega squared gamma and we know that these are eigenvalues so these eigenvectors of the element sigma with sigma cubed equals 1 so sigma is just the permutation taking alpha to beta to gamma to alpha or possibly the other way round. Now we know from general theory of cummer extensions that y cubed and z cubed must be in k of delta and we can do some algebra and we find y cubed is equal to minus 27c over 2 plus 3 root 3i over 2 delta except that should be a minus sign. So we can work out y and z by taking the cube roots of these and then we can find the root alpha as being y plus c over 3 so that allows you to solve a cubic. So we can give an explicit example just to check this all works. I mean for instance if you take x cubed plus x plus 1 equals 0 we find the minus 4b cubed minus 27c squared is equal to minus 31 so the discriminant delta is the square root of minus 31 and then we find that y is equal to the cube root of let's try and get it right minus 27 over 2 minus 3 root 3 i times root minus 31 over 2 and z is equal to the cube root of minus 27 over 2 minus 3 root 3i plus I guess root minus 31 over 2 and you find y is about minus 0.304 and z is about 0.99 and we find from this that the root alpha is about minus 0.68 and you can actually do this all on a calculator and check that this really is a root of the equation. So that does cubics. Oddly enough degree 4 polynomials are almost easier than cubics or rather the jump in difficulty from quadratics to cubics seems to be more than the jump in difficulty from cubics to quartics. So quartics were solved by this guy Ferrari who was apparently a student of Cardano who shouldn't be confused with Ferro who solved the cubic. I think Ferro and Ferrari both actually mean Smith or something like that in Italian so this is the word for iron. So for degree 4 we have a polynomial which looks like x to the 4 plus ax squared plus bx plus c equals 0 and you remember from Galois theory we have to look at the group s4 and it contains the following chain of subgroups. We have the trivial subgroup and this is contained in the subgroup z over 2z times z over 2z which is contained in a4 of order 12 which is contained in s4 and this quotient is z modulo 2z. This quotient is z modulo 3z and this quotient is z modulo 2z squared. So if you look at the corresponding fields k is contained in something is contained in something is contained in our field k alpha beta gamma delta I guess there are now four roots. So this says that we should first take the square root of something then take a cube root of something then take a bunch of square roots and then we're finished. So let's see how to do this. Well we need to know what this group is. Well this group of order 4 is generated by, consists of the identity together with the products of 2, 2 transpositions. So we want to find generators for this field here and so we want to find eigenvectors for this group. So we want to find eigenvectors for the group z over 2z times z over 2z and that means we want to find vectors that have eigenvalue either 1 or minus 1 under this element and either 1 or minus 1 under that element and that's quite easy to do. We can take the elements alpha 1 plus alpha 2 minus alpha 3 minus alpha 4 alpha 1 minus alpha 2 plus alpha 3 minus alpha 4 alpha 1 minus alpha 2 minus alpha 3 plus alpha 4. So you see this under the elements 1, 2, 3, 4 this is eigenvalue plus 1 minus 1 minus 1 and under 1, 3, 2, 4 the eigenvalues are minus 1, 1, minus 1 and I guess we could also add the element alpha 1 plus alpha 2 plus alpha 3 plus alpha 4 which is eigenvalues plus 1 and plus 1 but in the special case this would actually be zero so we don't need to worry about it. Anyway you see if we can work out these four expressions we can find alpha 1 and alpha 2 and so on just by linear algebra and alpha 1 is going to be the average of these you just sum these up and divide them by 4 for example so we want to work out what these are well now you just notice that the squares alpha 1 plus alpha 2 minus alpha 3 minus alpha 4 squared and alpha 1 minus alpha 2 plus alpha 3 minus alpha 4 squared and alpha 1 minus alpha 2 minus alpha 3 plus alpha 4 are now going to be permuted by S4 so they are three roots of a cubic with coefficients in K and I'm not going to write out this cubic explicitly because it takes a fair amount of rather messy algebra so what you do is you find this cubic you solve the cubic you can then take the square roots of the three roots of the cubic and do some linear algebra to find the roots of the degree 4 polynomial so you can see reducing degree 4 to degree 3 is actually sort of almost easier than solving a cubic of course you've got to solve a cubic as well so you can see what's going on here that this group here is really S3 which is the Galois group of a cubic so we were sort of going from there getting to a cubic and then the cubic is this bit of the Galois group it's essentially because S3 happens to be a quotient of S4 well you might ask what does the solution actually look like I found a copy of the solution on Wikipedia that I will just show you it sort of starts off like this and then it goes on and on and on and on and on and on and on so you can see why I did not try and write it out explicitly by hand so what's going on is that for degree 4 we know that S4 acts on a set of three elements so we can sort of reduce solving a quartic to solving a cubic and that's because S4 has a subgroup because S4 factors through S3 so it's got a subgroup of index 6 which is group Z2 squared if we try and do the same thing with S5 well S5 is order 120 and we can find a subgroup of order 20 which is the Frobenius group so this means that S5 acts on a set of six elements and when people try to extend the solution of a quartic to the solution of a quintic they discovered that whereas for a quartic you could reduce the equation to one of degree less when you try to solve a quintic you end up having to solve a degree 6 equation which only made things worse of course Arbel and Gower later discovered the reason for this is that the group S5 is not solvable so there's no way to find some clever way of solving it by radicals