[SOUND] Our next subject is the solvability of equations by radicals. So we consider K of characteristics 0 and embed it into some algebraic closure, K bar. A definition, a finite extension. E of K is solvable by radicals. If there exists the elements alpha 1 and so on are generating. E and such that alpha i to the power ni belongs to K of alpha 1 and so on, alpha i- 1 for some natural numbers and i. In other words E is obtained by adjourning nis roots of some ais. For example one can take q of the cubic root of 2 + 3 of square root of 7, something like this. But one can also take several expressions of this form. Cubic root of 2 + 3 square root of 7. And then the 5th root of 4 + 5 square root of 11, okay. Then we will have the square root of 7 and the square root of 11 as alpha 1 and alpha 2. And then alpha 3 will be under this cubic root. And alpha 4 will be the 5th root. So another definition. Let P be a polynomial over K. We say P is solvable by radicals. If this [INAUDIBLE] field of P is contained in some extension. Solvable radicals if there exists an E solvable by radicals. And containing all roots of P. Well maybe it's better to say in this case that the equation be equal to 0 is solvable by radicals. So more precisely, It would say that the equation, P equal to 0 is solvable, by radicals. But to make the definitions shorter, we shall say simply that the polynomial P is solvable by radicals. So let me list a few elementary properties. Clearly a composite extension of solvable extensions is solvable. Secondly, if we have, L an extension of K solvable by radicals. By definition, such an extension is finite. Then there exists a Galois. Finite Galois extension. Containing. Well, let's give it a name. E containing L, and also solvable by radicals. This is something we have already seen. It suffices to check for E. The composite extension of all images of L in K bar. Indeed, take the composite of all images of L in K bar. Now, you can also view it as the images of l by the Galois group of K bar over K. Or this is the same as images of L by the Galois group of K bar over K. Well, I should better use the plural of those are the images of L and K bar are the same. As the images of L by the Gal group of K bar over K. Let me now talk about solvable groups. This shall be a brief reminder since this is not a course on group theory, you are supposed to know some group theory already. So I somehow I presume that you are familiar with this definition but I will recall the definition of basic properties. So let G be a group. Which is called solvable, If it has a filtration. A filtration is the following thing. We have G = to G0. Contains G one contains. Gr- 1 contains Gr, which is trivial. So a Gir subgroups and we shall require that those are normal subgroups. Gi is a normal subgroup. Of Gi- 1. So we can form a quotient. And we require the quotient to be abelian. So Gi- 1 / Gi is abelian. So for example, the group of permutations on three elements is solvable. Since S3 contains A3, contains the trivial group. [COUGH] A3 is a normal sub group. The quotient A3 by A3 has only two elements in particular it must be cyclic of order 2. And A3 is cyclical for order of 3. S3 itself is not abelian, but it is solvable. It has a filtration which is just written here with a abelian factors. Abelian quotients, okay? S4 is also solvable. Indeed S4 contains A4. Even permutations on four elements contains a group which I shall denote by K, contains e. I will explain what K is. Here K is the subgroup of double transpositions. So it consists of, Identity, then (12)(34), (13)(24), and (14)(23). So subgroup, Of double transpositions. A double transposition is a product of two transpositions with distinct support, right, which permute the distinct elements, okay. So now, let me rewrite this sequence. I have S4 contains A4 contains K contains e. So A4, the subgroup of even permutations is a normal subgroup of, of S4. The quotient is cyclical for the two. Then K is a normal sub group in A4. Of course, a conjugate of a double transposition is always a double transposition and the quotient, Is of order 3, so it is also cyclic. K is not cyclic, but it is abelian. K is abelian. Isomorphic to The product of two cyclic groups of order 2. So it follows that S4 is solvable. [MUSIC]