 Okay, so maybe we can start. So this is supposed to be the last lecture. So we were talking about solvability by radicals. So I hope you remember the definition I had wanted to relate this to solvability of groups. And so I had introduced what it means for a group to be solvable. And then I wanted to express this in terms of the commutator subgroup. So let me start again at that point. So we had the commutator subgroup. So we have if say G is a group and we have elements AB in G. We have the commutator, which is denoted like this. And it's just AB A to the minus 1, B to the minus 1. I hope it's just the correct order. Well, I decided to do the other round. It doesn't really matter, does it? And so then the commutator subgroup is the subgroup generated by all commutators. So it's denoted to G prime. And it is the subgroup generated of G generated by all commutators AB. Maybe I should just, by definition, the subgroup generated by something is the smallest subgroup containing all these elements. But you can also say differently, it consists of all products of such elements and inverses of them, which follow directly from the definition. So the subgroup of a group G generated by a subset finite or infinite m can be described as the set of all products A1 times times An, where n is some non-negative integer. So you take any length of product. And for all i, we have ai is an element of m. Or ai to the minus 1 is an element of m. So it's just all products of elements here and the inverses. You can see easily that this is the smallest subgroup which contains these elements of m. So we want to use that now, if I can find my page. Yeah, that's very sad. So anyway, so one second. Yeah, so that is interesting. So I seem to have, I don't think I want to improvise that. So maybe somebody can give me the notes so that I actually know what I'm doing. OK, so anyway, so now I want to show that. Let me see what I can do. Anyway, do it. So we have that. So this is the following lemma. So this lemma will allow us to decide whether a group is solvable in terms of the commutator subgroups as of iterating commutators. So g prime, the first is the g prime is a normal subgroup of g. And the second, the quotient g mod g prime is a billion. And the third is that if I have a normal subgroup of g, such that g mod low h is a billion, then h contains g prime. So these are the statements. So let's try to prove it. Well, first it is clear that if I take a b to the minus 1, this is just equal to bA. So it's also a commutator for all AB. So thus, according to this description, the commutator subgroup is just the set of all products of commutators. It just consists of products of some number of commutators. So g prime, the elements of g prime are products of commutators. Because by itself, they should be products of commutators in the inverses of them, but the inverse is this. Of a commutator is also a commutator. Now let's look if we compute, say, g to the minus 1 ABg for some element g and g. We can compute this. Well, this is g to the minus 1, a to the minus 1, b to the minus 1 ABg. And we can multiply by the identity here. So this is so g to the minus 1, a to the minus 1, g, g to the minus 1. So b to the minus 1, b. Let me just see whether I get the third. Yeah, I just put this in here. So I keep on going. So g to the minus 1, a, g, and g to the minus 1, b, g. And so if you look at this, if you take the inverse of this element, this will be this. Because you have to take the inverse of every factor and turn them around. And this gives you this. So this is nothing else. And you do this with all of them. This is nothing then. g to the minus 1, a, g, g to the minus 1, b, g. The commutator of those. So thus, if you have any, now let, so this will then imply that if you take any product of commutators and do a conjugation like this, you will again get a product of commutators. So this will show that it's a normal subgroup. So if I take, so let h equal to, say, a1, b1 times an, bn, element of g prime, then by this end g, an element in g, then by this statement, we get, if I take d to the minus 1 hg, this is equal to, first I can again put elements 1 in between. So I can write this as g to the minus 1, a1, b1, g times, and so on, g to the minus 1, an, bn, g. So in between, I've only introduced g to the minus 1, which is 1 at each factor. And this is equal to, according to what we just saw, g to the minus 1, a1, g, g to the minus 1, b1, g, and so on, g to the minus 1, an, g, g to the minus 1, bn, g. So this says nothing else that if I take any element in g prime, and I conjugate it by any element in g, I get an element in g prime. So this shows that this is a normal subgroup. So now for the second statement, so we want to show that this is a real, this is kind of simple. So let's take two elements in g. Then I want to see if I take, I want to compute a, g prime, times b, g prime. So these are the, I take the product of these cosets. We know that this is ab, g prime. But we can squeeze in between here an element in g prime and get the same thing. So we can take the commutator of a and b. So this is equal to ba, ab, g prime. But as ab lies in g prime, this is the same as ba, g prime. So this means, and obviously this is the same as b times g prime times a times g prime. So we see this quotient is commutative, is abial. And for the third statement, it's kind of similar. So I don't remember precisely how. So we assume, so we want to assume that the quotient by h. So h is a normal subgroup and g mod h is abial. So we have that a h times b h is equal to, so we know, we assume that this is equal to b h times a h. This is just ab h and this is equal to ba h. So but ab h is equal to ba times ab, if I'm not mistaken. And we know, thus we have, ba h is equal to ba abh. But that says nothing else that ab is an h. And so we see that h contains all commutators of elements in g. And so, therefore, h contains the commutator subgroup. OK, so this is all quite simple. I think with this I can go back to my things. So now we want to see. So that should allow us. So you can now kind of, once you have that, you have a g prime. You can kind of iterate the process of taking the commutator. So you can take the commutator of the commutator, the commutator of the commutator of the commutator. Each time you get a smaller group, a smaller subgroup, such that the quotient by that smaller subgroup is a billion. So you can somehow, so a normal subgroup, so that the quotient by that is a billion, which is somehow precisely what we want to get a solvable group. There you have a chain of subgroups, such that each one is normal in the previous one, such that each quotient of one by the next is a billion. So you get this here automatically. If you start with g, take g prime, then the commutator of g prime, and so on. The only question that you have to ask yourself is whether in the end you can end up with one, because this is what you need for it. So let's set this up. So we have the definition. So g is still a group. And we write also g0 for g, and g1 for g prime. And inductively, we write g n plus 1. It's kind of like taking the derivative equal to gn prime. So we start with g, we take the commutators of g, then we take the commutators of the commutators, and so on. And so we get somehow a chain of groups. And now this will give us a criterion for a group to be solvable in terms of commutators. So lemma, g is solvable if and only if there exists an n such that if I have kind of the nth commutator, this is equal to 1. So I kind of have sketched some sense one direction. So let us see. So we assume that the nth commutator, so this is this, that this is equal to 1. Then we have a chain of subgroups. So we have a chain, so we have g is equal to g0, which contains g1, which are the commutators, which contains g2, and so on, until gn, which is 1. So we have such a chain. And we know that according to the lemma I just wiped out, we have each one is a normal subgroup in the previous, and the quotient is a b. And for all i, we have gi is a normal subgroup in gi plus 1. Maybe I write gi, well, whatever, in gi minus 1. And the quotient gi minus 1 divided by gi was write it wrong. This quotient is a b. And this is precisely the condition, I mean the definition of g being solvable, that you have such a chain. Thus g is solved. So this is in some sense the easy direction, but it's not a big difference to the other. The complement, the other direction would be that if we have a chain with some subgroups, we can find a chain where these are given by commutators. So conversely, we assume that g is solvable. So then we have this chain of subgroups. Thus then there's a chain in gi. So that g is equal to the 0th one, which contains h1, and so on, until hn, which is equal to 1. Such that each one is normal in the previous one. The quotient is a b. So it's precisely what I had here, such that hi is normal in hi minus 1. And hi minus 1 divided by hi is a b for all i. Now we have to see that we can somehow relate, get the special one in terms of the commutators. So basically we have to show that we have to find that there is some commutator, which is equal to 1. But the lemma said that if I have any subgroup in a group such that it is any normal subgroup in a group such that the quotient is a bn, then it contains the commutator. But as you see, hi minus 1, hi is a bn. We have hi contains the commutator of hi minus 1. This was the last part of the statement for all i. So thus, if I kind of do this inductively, we have that h1 contains g prime, then h2 contains h1 prime, which is g2. And we can do this inductively, hi contains h so inductively. So by induction, we have that this contains hi minus 1 prime. And by induction, this will be equal to the prime of the i minus 1 commutator. So if I take the last step, we have that hn contains 1 we know is equal to hn. But by what we have said here, it contains gn. So that means that gn is equal to 1 because it's a subgroup. So therefore, we see the statement. So if there is such a chain, there is also this chain. And we find that we can decide whether the group is solvable just in terms of iterating taking the commutator. And the main reason why we prove this here is because it shows that if we have a subjective homomorphism from a solvable group to another group, then that will also be solvable. Let g be a solvable group. And say phi from g to h be a subjective group homomorphism. Then h is also solvable. And this now with this statement is quite easy because essentially by definition, the image of the commutator subgroup of g will be the commutator subgroup of h. And so by iterating this, we find that the lemma applies. So let's see. So if we take any elements a, b in g, then if I take phi of the commutator, as it's a group homomorphism, it sends the commutator to the commutator. It's just a to the minus 1, b to the minus 1, so on. So it's compatible with these products. So this is equal. So the commutator of phi of a and phi of b. And note, so thus we see that the image of the commutators of g lies inside the commutators of h. But we also know that the map is surjective. So every element is phi of a for some a here. So therefore, every commutator is the image of the corresponding commutator there. So we find that h prime is equal to the image of the commutators of g. And now, obviously, the higher commutators are just obtained by doing it again. The map will still be surjective. So we find, and therefore, we have hn is equal to phi of gn for all n. And so therefore, if g is known, so thus, or maybe I call this i. Now we have this n for which, so thus, if g is solvable, there exists nn such that the nth commutator of g is equal to 1. And therefore, if I apply phi to it, it sends 1 to 1. So thus, hn is equal to phi of gn is equal to phi of 1 equal to 1. Yeah, I have assumed surjective. I maybe should have written it out. Surjective group homomorphism. Otherwise, it obviously wouldn't make sense. But maybe I shouldn't have abbreviated it because it's an important assumption. So I write it out. So surjective group homomorphism. OK, so this is the statement. And now we want to go back to the solvability by radicals. So we know now enough about solvable groups to be able to proceed. And we want to go towards proving the relation between the Galois group being solvable and the polynomial being solvable by radicals. So first, we have to deal with roots of unity. So somehow, we want to somehow restrict our attention to be in extensions. And this we can do in our situation by making sure that we have all the roots of unity available that we want. The root of unity is just an element in the bigger field such that some power of it is equal to 1. So let's see. So definition. Let we take some positive integer, positive. And let L be a field. So an element, zeta in L, is called root of unity if some power of it is equal to 1. So if there exists an nth root of unity, if zeta to the n, so if I multiply it in terms of itself, this is equal to 1 in the field. And it's a primitive nth root of unity if this n is the smallest power for which you get 1. If zeta of n is equal to 1 and zeta of i is not different from 1 for all i, which lie strictly between 0 and n. So it means that n is the smallest positive integer for which you get 1. Obviously, 1 is always an nth root of unity for all n, but it's usually not a primitive nth root of unity. And in the complex numbers, you have, for instance, if you take e to the 2 pi i divided by n, this will be a primitive nth root of unity as an example. So now we want to see that if we have a field and we adjoin to it the primitive nth root of unity, we get in this way a Galois extension with a b and Galois group. So lemma let n be a positive integer. And we take a field, I don't know. Should we OK? Any characteristics here? I don't remember. Yeah, maybe I'm just looking for security of characteristics here. And so let zeta be the primitive nth root of unity in an extension field L of k in an extension field of k. Then if I just take the extension generated by this one element k x e over k, this is a Galois extension and the Galois group is a b. That's quite simple. So we have to see that the Galois extension you have to see, as we are in characteristics 0, it's enough to see that it's a splitting field of something. Because then it means it's normal. And in characteristics 0, normal and Galois is the same because separable is no condition. So the claim is that if I look at this field k xe, we find that the polynomial say x to the n minus 1 splits into linear factors. Why is that? Well, we find n different roots by taking all the powers of this and we see that they are all different. So if I take the set zeta to the r for 0 smaller equal to r, smaller equal to r, n minus 1. So these are certainly roots of unity. So r0s of x to the n minus 1. Because if I take zeta to the r to the n, this is zeta to the n to the r. So it's 1 to the r, which is 1. And I also claim they are all different, all distinct. So if so say zeta to the i is equal to zeta for the j, for i and j smaller than n, we can divide by this. So I have zeta of maybe I do it like this. I mean, it doesn't matter, obviously. It's zeta to the j minus i. If I divide by this, it's equal to 1. But we know as zeta is a primitive n to the unity, the smallest possible power. I mean, there's no power of zeta, which is smaller than n and bigger than 0, which gives you 1. And so it follows j minus i must be equal to 0. So in other words, i is equal to j. So that means all these roots are distinct. And so therefore, we have found n0s for our polynomial x to the n minus 1. And it therefore splits into linear factors x minus zeta times x minus. So x minus 1 times x minus zeta times x minus zeta squared and so on. So that's x to the n minus 1 splits. So as we have obtained this field extension by adding a 0 of this polynomial, it follows that, since by adding all the 0s, it follows that k zeta is the splitting field of this polynomial. By definition, if I have a field over which it splits, and if there's no intermediate field over which it splits, then it's a splitting field. And obviously, if it's obtained by adding a 0, then it must be the splitting field. OK. So this is the splitting field, so it's a Galois extension. Because we had proven that the field extension is normal, if and only if the bigger field is the splitting field of the polynomial with coefficients in the smaller field. And in character 6-0, normal and Galois is the same. So now we want to see that this is an abelian. That this is an abelian. So let G be the Galois group of k zeta over k, which is the same as the Galois group. OK. So we want to show it's abelian. So let's take two elements. So in note, so we know that the Galois group, maybe I'll just say, so if sigma is an element in G, then zeta of sigma must be a 0 of this polynomial. In fact, we had seen that the Galois group acts as a subgroup of the permutations of the 0s of the polynomial, of which it's a splitting field. And in fact, and if so in fact, so thus zeta of psi will be equal to psi to the j for some j. And we know, so we know it's a subgroup so that the restriction of the Galois group to just the permutations of the roots of this polynomial is an injective group homomorphism. So if two elements in the Galois group do the same thing to the roots of unity, then they are the same. So as the restriction, so I can just say G is a subgroup of the permutations of the set 1 zeta to the n minus 1. And thus, if for elements sigma and tau in G, we have sigma of zeta is equal to tau of zeta, then obviously it follows because it's a group homomorphism that sigma of zeta to the i is equal to tau of zeta to the i for all i, then sigma is equal to tau as elements in the Galois group. And so let's just see what they do. So let's take two elements. So let sigma and tau be elements of G. So then sigma of zeta will be equal to some power zeta to the i, and tau of zeta will be equal to some power zeta to the j, because that's what it is. And so we want to see that the two different products are the same. So if I take, say, sigma tau of zeta, this is sigma of zeta to the j, sigma is a field isomorphism. So it takes the power to the power. So this is sigma of zeta to the j. And this was, so this is equal to zeta to the i. The j is equal to zeta to the i plus j. And obviously, if you do it the other round, the same way, tau sigma of zeta, you can obviously see this will be just a place here i by j. And in the end, I get zeta to the, say, i plus j or j plus i. So they are equal? What? Plus. Let me see. Yeah, you're right. It doesn't really matter, but I think you're right. Zeta, is it correct? Yeah, yeah, yeah, yeah, yeah, yeah, sorry. Anyway, but it doesn't, as both multiplication and addition is commutative, it doesn't matter. But you're right that it must be times. OK, but anyway, the same argument works. And so it follows that thus, the group is commutative. Or I'll be OK. And now, so we want to finally show that solvability by radicals has something to do with the group being, with the Galois group being solvable. So we always have to somehow find some part of some Galois group which is Ibn, so now we want to, we have seen if you want to add the primitive root of unity, the Galois group for that is Ibn. And then the other statement is that if we do have a field which has enough roots of unity, then adding an nth root of something will give us something with a, will give us a field extension with an Ibn Galois group. So this is the next, what I here call theorem. I don't know whether this is reasonable theorem. So we take a positive integer and let k, a field that contains a primitive nth root of all unity. And we take any element in k and we want to take its nth root. So let L be the splitting field of x to the n minus a over k. Then first, we have that L is equal to k of alpha, where alpha is any root of this polynomial x to the n minus a in L. And furthermore, so this is the first statement, the second statement is that the Galois group of this extension is AB of L over k is AB. OK, so let's see. It's actually the proof is rather similar to the previous one. So we take our root. So we have alpha is a root of it. So alpha, the root of x to the n minus a in L. Then if I take and let zeta be a primitive nth root of unity, then I can form alpha, zeta alpha, until zeta to the n minus 1 alpha. Obviously, are all zeros of x to the n minus a. Because if I take this to the power n, I get zeta of n times a alpha to the n, zeta of n is equal to 1. So these are zeros of x to the n minus a in L, or actually in k of alpha. And obviously, now I want to claim they are again all distinct. So again, assume that zeta of i alpha is equal to zeta of j alpha for 0 smaller equal to i smaller than j smaller than n. Then I can anyway, alpha is not 0, because then a 0 of this polynomial and a was not 0. So then it follows that zeta of i is equal to zeta of j. And we have seen that this implies that i is equal to j. So thus, we have found n zeros of this polynomial of degree n. They are all distinct. So we find that this polynomial splits. And as we have obtained it by just adding one root, we find that this k of alpha is the splitting field. x to the n minus a splits over k alpha. And therefore, by the same argument as above, we find that it is the splitting field. And L is equal to k of alpha. And why is this group a b n? This is again a very similar argument as above. So if I take two elements sigma and tau in the Galois group of k alpha over k, then again, the elements in the Galois group act as permutations on these things that they are determined by how they act as permutations. The homomorphism, the restriction homomorphism from the Galois group to the group of permutations of the roots is an injective group homomorphism. So we only have to see what they do on the roots. So we have that sigma of, and again, so then sigma, if I take sigma of theta to the i alpha and same for tau, this is equal to theta to the i sigma of alpha for any i because this is an element in the ground field. And so it just pulls out. So therefore, the elements in the Galois group are just determined by what they do to alpha. So thus, if sigma, so thus, elements in the Galois group of k alpha over k are determined by sigma of alpha. So now let's look at, and it's clear that sigma and sigma of alpha will be equal to alpha theta to the k, well, to the l times alpha because it must be another root for some l. So if we now have these two elements sigma and tau, we want to see they commute. So they have to see that if I compute them in one way or in the other, I get the same thing. Then the argument is the same as before. So if I take, so assume this, and tau alpha equal to theta of s, maybe I call this s and I call this t, alpha for some s and some t, then say sigma tau of alpha will be equal to sigma of theta to the t times alpha. And now this element in the ground field k, so it just pulls out. This is theta to the t times sigma of alpha. And this was theta to the s times alpha. So this is equal to theta to the, in this case, I was right. In this case, s plus t times alpha. And in the same way, if you compose them the other way, you get it also. If I take tau sigma times alpha, it will be by the same argument, also theta to the say t plus s times alpha. So we see that the group is commutative. Sigma tau is equal to tau sigma. And the group is commutative. OK, so now we want to come to the last main theorem. I think I will not be able to. OK, so the theorem is the theorem of Galois, which kind of, decides for us in terms of group theory whether a polynomial is solvable by radicals or not. So let k be a field of characteristic 0. And let f be a polynomial with coefficients in k. So then f is solvable by radicals, if and only if the Galois group of f, which means the Galois group of f over k, is solvable. So this was a big result then, because you will see it shows that not all polynomials of degree bigger equal to 5 are solvable by radicals. So now we will only prove one direction. So we will prove this direction, that if f is solvable by radicals, then the Galois group is solvable. The converse direction is not more difficult, but we would have to prepare a little bit more, I mean, make a few more preparations. Somehow you have to show, in order to find these radicals, you actually have to find that you have to split up your Galois extension into pieces where the Galois group is cyclic, not just a bian, and you then somehow have to show that if such a group is solvable, then you can have such a chain of subgroups so that the quotient is not just a bian, but also cyclic. But we don't have the time to do that, so I will just do this the other direction, which is anyway maybe the thing people were mostly interested in. So now we assume that this thing is solvable by radicals, and we have to somehow find this chain of subgroups, which shows that the Galois group is solvable. So let L be the splitting field of f over k, and then the Galois group of f is just the Galois group of L over k, and we have to show that this group is solvable. So let's see how we can attempt to do this. So first, so we know that f is solvable by radicals. So there exists such a chain of subfields, which such that it splits over the last one, where each time it's obtained by joining just a radical. So there exists a sequence. So our field k, I also call k1, which is contained in k2, which is k of an element alpha 2. And we keep on going. The last one is k is equal to kn minus 1 of alpha n. Such that first, we have that for all i, there exists an element, say ei in ki, do you want it this way? Say in ki minus 1, and element ri, the positive integer, such that we have that alpha i, that ai is equal to alpha i, to the ri. This just means that we have obtained it by always taking, extracting some ri through it. Here the k2 is obtained from k1 by taking the r2 through it of an element a2, and so on. So we just, by extracting root, we get extension, such that first this thing splits, so f splits over kn into linear factors. And we had seen, although we didn't prove it, but I said it was a somewhat difficult exercise, we can assume that this kn over k is a Galois extension. So we have to use this to somehow find out something about the Galois group of our extension, l over k. We knew that it's all. So we know that if we extract such a root, this will be an abelian. The Galois group of this extension will be an abelian extension. So the Galois group will be abelian. If we have all the roots of unity that we need, so we therefore will, as a first step, join these roots of unity. So note that as l is the splitting field of f over k and f splits over kn, we find that l is a subfield of kn. So now we want to have all these roots of unity. So let m be the smallest common multiple of all the r i. So we have here r 2, r 3, and so on until r n. These are the power, so the root we extract, the power of the root we extract. Take the smallest common multiple of those and let zeta be a primitive mth root of unity in some extension of k. So then obviously, and we put f, so another field, equal to k, zeta, we join this primitive mth root of unity. So note that if we take zeta to the m divided by r i for any i, these are i's, this is a primitive r ith root of unity. Namely, if we take this to the power r i, obviously, you get 1 because it's the same as zeta to the m. And if you take any smaller power of this thing, you will not reach until m, and so you will not get 1 because zeta was a primitive nth root of unity. So if we join this thing, we have primitive ith root of unity for all i. So we want to kind of change our field extension by replacing in some sense k by f, so that we don't have to worry about these roots of unity. Primitive of unity for all the i, that we have here, i from, and now we make a new chain. So we had this chain of fields like this, a sequence of, and we want to instead add this f to it. So we make a new chain, so we put f0 equal to k, f1 equal to f, and we put fi equal to fi minus 1 of alpha i for all i bigger equal to 1. So basically, we just have, we make one step more here. We replace here k by f and make the corresponding thing larger. And the k still we have in the beginning, we just shift them by 1. So we have this somewhat longer change of, so we have this. So this we have, it does have a chain. So f0, which is k, is contained in f1, which is f, contained in f2, and so on. Contained in fn. So note that here, we have this alpha i is the i th root of an element ai in the previous field. It's an element ai in here, the previous field, but we just have a joint, the f. So we have, therefore, we have a joint, a radical, so the r i th root of, so we have that as fi is equal to fi minus 1 of alpha i. And alpha i to the r i is equal to ai in fi minus 1. We had this previous lemma, which said that under this condition, this extension fi over fi minus 1 is a Galois extension, and the Galois group is a bigger. Thus, by previous theorem, we have that fi over fi minus 1 is Galois extension, and the Galois group is a bigger. Now, you have to keep in, so furthermore, and obviously, ln is contained in fn, because again, our polynomial ln was the, l was the splitting field of our polynomial f over k, and it's still, obviously, the polynomial also splits, if it splits over kn, it certainly splits over fn, because only a bigger field. So and we know, so kn is this splitting field, so we know that kn over k is a Galois extension. Thus, we have that kn is the splitting field, or splitting field of some polynomial g with coefficients in x, because we know Galois extensions are always splitting fields in characteristic series in equivalent, and but on the other hand, it's clear that fn then is the splitting field of g times x to the m minus 1 over k, because clearly, this polynomial splits into linear factors, because you know, we have all these primitive roots of, we have the primitive roots of unity, so this one splits and this one splits, you know, because this is just the, you know, we can, you know, after all, fn can be just described as kn, a joint, a primitive mth root of unity, so this also splits, and it's the smallest such field, because if, if this polynomial splits, then the field must contain a primitive mth root of unity, and it must contain all the roots of g, and so it must contain this, so fn is the splitting field, and so therefore, as this is the polynomial with coefficients in k, we have fn over k is a Galois extension, so see, it's kind of okay, and so if you think of what you've done, we have kind of proven something rather similar to what we want to prove, let me put this down, so fn over k, which is f0, is a Galois extension, and so if we take any intermediate field, then, so if you have a splitting, so fn over k is a splitting field of some polynomial, so then if we take any intermediate field between fn and k, and if we take fn over this intermediate field, this is also the splitting field of the same polynomial, so therefore, it is also a Galois extension, so thus fn over fi is a Galois extension for all i from 0 to n, if you want, and we have that f1 and therefore, all fi contain all primitive i-throups of unity for all these i's here, so therefore, this extension, this over this is a Galois extension, and the Galois group is a b, which is equal to fi minus 1 over fi over fi is a Galois extension, and the Galois group is a b, but now we have to remember that, okay, so we also know that the first one, that f1 over f0 is a Galois extension, because here we have just, with a b in Galois group, because we had seen if we have a, if we join some primitive root of unity, the corresponding Galois extension is, we put an extension as Galois and with a b in Galois group, so now we use the fundamental, the second part of the fundamental theorem of Galois theory, which said that, you know, the fundamental theorem of Galois theory said that if you have the intermediate field, fields correspond to the subgroups of the Galois group, and if the extension for this intermediate field is a Galois extension, then this subgroup is a normal subgroup. So here we have all these Galois extensions, so it means that we get normal subgroups. So by second part of fundamental theorem, Galois theory, we know that as these fi over fi minus 1 is always a Galois extension, it follows that, it follows that, just see, so we have that fn over fi is a Galois extension. So that the Galois group of fn over fi is a normal subgroup, so we have, we have that, so we have the statement that this here is a Galois extension. So then if one kind of does it, then the statement is that if I take the large one, fn over fi is a normal subgroup of the Galois group of fn over, which way around does it go, in this case fi minus 1, this is a smaller field. Now we have that the intermediate fields of fn between fn and f0 correspond to the subgroups of this Galois group, and such a subgroup is a normal subgroup if and only if, and the subgroup that it corresponds to is always this, fn over the Galois group of fn over fi, and this group is a normal subgroup if and only if the, this fi, no I made a mistake, yeah, if and only if this is a Galois extension, let me just say it again. So I have to make one more step, you know, that's, I'm not sure I'll prove this, this may be something I have slightly different, okay, so and we know that if we take the quotient, the Galois group of fn over fi minus 1 divided by the Galois group of fn over fi, this will be equal to the Galois group of fi modulo fi minus 1. Now this is how we have this, and so we see that, but we know that this Galois group is, so we have seen that this is abelian, no, so this is and so, but Galois group of fi over fi minus 1 is abelian, so thus if we look at the chain of Galois groups like this, so thus if I take the Galois group, we'll see fn over f0, which is k, contains the Galois group of fn over f1 and so on, the Galois group of fn over fn minus 1, maybe I write it with less room, Galois group fn over fn minus 1, this is a chain of subgroups and then the next one, the last one would be the Galois group of fn over fn, which is equal to 1, so just the identity, so we find a chain of subgroups, such that the last one is 1, such that the quotient here, this divided by this, so this is, if I take two consecutive ones, this, the next one is always a normal subgroup and the quotient is abelian, so this chain demonstrates that the Galois group of fn over k is solvable, so this chain shows that the Galois group of fn over k is solvable, now in the moment that doesn't look so great because we need to show that the Galois group of l over k, so not this fn that we have somehow constructed, so we have to somehow relate the two, but that's not so difficult because we have that l is contained, you know, if n is a subfield and so the Galois group of l over k will be equal and again, this is again, and it's a l over k is a normal extension, so it follows that the Galois group of, is equal to the Galois group of fn over k, again by this kind of second part of the fundamental theorem of Galois theorem divided by the Galois group of fn over l, we find this and so we see that the Galois group of l over k is a quotient of this group, so in particular we have a surjective formomorphism, just a kind of natural projection from this group to this group with kernel this, so thus there is a surjective formomorphism from the Galois group of fn over k to the Galois group of l over k, now and then we had this result that if a group is solvable and we have a surjective formomorphism to another group then also that group is solvable, thus also Galois group of l over k is solvable and this shows the part of the theorem that we wanted to show, I'm already over time so I do not, I wanted to give some example so I would maybe just say it in words, so or maybe I will just write down a couple of statements so that we kind of see that this actually can help you to show that some polynomials are not solvable by radicals, so the first statement is that I mean it's a theorem that the symmetric group in n letters is solvable if and only if n is smaller equal to 5, to 4, so if n is bigger than 4 it's not solvable, so if we have a polynomial whose Galois group is sn for n bigger than bigger equal to 5 then this polynomial is not solvable by radicals and then there's the following theorem which is the proof is in the notes, obviously I will not prove it, so let p be a prime number and let say f be an irreducible polynomial in qx such that over the complex numbers it has p minus 1 real roots, p minus 2 real roots, 0s in r so to speak it will have p 0s over the complex numbers and p minus 2 of these are actually real then the Galois group of f over q is the symmetric group in p letters, so therefore if p is at least 5 and we have such a polynomial then it will not be solvable, not solvable by radicals. In the notes I give an example where one can check by a elementary calculus that it's not solvable, I mean that this condition is fulfilled, so if we take an example if we take the polynomial f equal to 2x to the 5 minus 10x plus 5 it's a polynomial of degree 5 and the claim is that this has precisely 3 real roots and it's irreducible by the Eisenstein criterion with the number with the prime 5 as you can easily see and the claim is that it has precise 3 real roots and you can basically see this if you do some kind of you know you know what you do in calculus in school you take some function you discuss the your maxima minima and so on so you find that the function looks like this so it has a minima here at so here's the point 1 so at 1, minus 3 it has a minimum at this is 0 this is minus 1 so at here at yeah x obviously that's a misprint and you find here it has a maximum so you will see you have a minimum here a maximum here then it must somehow have a zero in between by the intermediate value theorem and as though the coefficient of x to the 5 is positive if you let x go to infinity obviously it goes to infinity and so it must go like this it must have another zero here and for the same reason it goes to minus infinity here so if you just apply the intermediate value theorem and you find you see that there must be three at least three real roots and you can find that these are the only extremer values by applying the usual things you apply calculus and so you see that it has precisely these zeroes by this kind of curve discussion and therefore this polynomial is not solved by radicals okay I'm sorry I went over time but I wanted at least to finish seeing something so next I mean there's no next time but if you agree there will be at some time soon some some tutorial and maybe you stop here