 OK, so maybe I will start. So we were talking last time about the fundamental theorem of Galois theory. I will restate it. So the statement is a bijection. So if let L over K be a Galois extension, then we have bijections between the Galois extension, say, with Galois group G. Well, maybe I can just start like that. So we have bijections between the subgroups of the Galois group and the intermediate fields. And so we have these bijections which are inverse to each other. To a subgroup, we associate the fixed field. So these are all elements in L which are fixed by so sent to themselves by all elements in H. And to an intermediate field, we associate the subgroup of the Galois group, which is the Galois group of L over this intermediate field. So these are mutually inverse bijections. So if we start with any subgroup of the Galois group, take the fixed field, and then the Galois group of L over the fixed field, this is just H. And if you do the other round, if you start with the field, take the Galois group of L over the field, and then the fixed field of that, this will be the original field, F. So now, in some sense, the next thing is still, in a suitable sense, part of the principle or the main, the fundamental theme of Galois theory. Namely, one can ask ourselves, how can we, what is the condition on the subgroup so that this intermediate field, F, is also Galois over K? Or one could also ask, so which means what's the condition for F over K to be a normal extension? In the same way, you can ask yourself, special subgroups that you can have are normal subgroups. You can ask yourself, what kind of intermediate fields correspond to normal subgroups? The point is that the result is that these two notions of normal are the same, which is not a coincidence because they actually, I think, normal subgroups are called normal subgroups because the group theory, essentially, was born out of Galois theory. And so normal subgroups are subgroups for which the intermediate field is normal. So let me state that. So we want to show, again, in the situation that L over K is a Galois extension. And so at F, say, in an intermediate field, then we want to show that F over K is a normal extension if and only if the Galois group of L over F is a normal subgroup. So let's start doing this. So first, we introduce a notation. So if we have our intermediate field, so let F be an intermediate field of our Galois extension L over F, L over K, and let alpha be an element in the Galois group, then we call alpha of F what? It's just alpha of F. I mean, alpha by itself is an automorphism of L to itself. So alpha of F is the image of this automorphism. So this is just a set of all alpha of A with A in F. And you can check immediately that alpha of F is an intermediate field. So you just have to see that it's a field. So it's kind of clear. This alpha will send 1 to 1 and 0 to 0 because it's a field homomorphism. It sends the sum to the sum of the product to the product. So therefore, it follows and to the quotient to the quotient. So therefore, it follows that alpha of F is a field just because alpha is a ring homomorphism. Or OK. And so we have the following statement, lemma. So first we have a simple statement which we use in order to prove this. So we take again F as such an intermediate field of a Galois extension. Do we need that? Yeah, L over K. And we take some element in the Galois group. Then we want to say what is the Galois group of alpha of F, of L over alpha of F. So the claim is then the Galois group of L over alpha of F is equal to we take the Galois group of L over F and we basically conjugate it with alpha. So we take alpha to the minus 1, the Galois group alpha. And we see that's almost direct from the definition. Or it is directly from the definition. We just check what it means. So we take an element in the Galois group, say, of this. So we take an element in the Galois group of L over K. We have to ask ourselves what is the condition on it so that it's an element here. So then phi is an element in the Galois group of L over alpha of F, if and only if it sends every element of alpha of F to itself. So I could say it equivalently, this is if and only if I take phi of alpha of A is equal to alpha of A for all A in F. This is the same as saying that it sends every element in alpha of F to itself. Well, obviously, now this just means you multiply here by alpha is an automorphism. You can compose with the inverse of it. So this is equivalent to saying that alpha to the minus 1 phi alpha, if I applied this is of A is equal to A for all A in F. So this means that alpha to the minus 1 phi alpha is in the Galois group of L over F. That's what it means. And so conversely, this means if now we conjugate, we multiply with alpha on this side with alpha to the minus 1 on the other side, we find that phi is in. So this is equivalent to saying that phi is an element in alpha to the alpha Galois of L over F, alpha to the minus 1. So that's very simple. It's just the definition. Now we want to actually use this to prove this statement. So we have, again, the same situation. So L over F over K Galois extension, F an intermediate field. So then the following are equivalent. So first, that F over K is normal. The second is that if I apply alpha to F, this is equal to F for all alpha in the Galois group of L over K. So I'm not saying that every element of F is mapped to itself. It just means that every element of F is mapped to an element of F. Because otherwise, this would be the Galois group of L over F. But this is not what I'm saying. And the third one is that the Galois group of L over F is a normal subgroup of the big Galois group. OK, so we want to see this. So this is also a useful remark that this is equivalent to F over K being normal. So we have to prove all these implications from 1 to 2. So we assume that F over K is normal. And let we take an element A and F. And we have to show, I mean, just to remember that we have to show. And we, what is it? We have to show. And so maybe we also let alpha be an element in the Galois group of L over K. And then we have to show that alpha of A is also an element of F. Then we have proven for every element A in F and every element in the Galois group of L over K, alpha of A is in F. OK, so let's see. So we take F be the minimal polynomial of A over K. As F is normal over K, and this small F is a polynomial, an irreducible polynomial in Kx with a 0 over F, it follows that F splits over the field F. So I could also say this all roots of say F in L lie already in. So we have now, if we take our element alpha in the Galois group of L over K, then as usual, we have that if we take F of alpha of A is equal to alpha of F of A is equal to 0. We have used this many times. So alpha of A is also, alpha of A is a 0, is a root of F. And so we know that all roots of F in L lie in the field F. So it follows that alpha of A is an element in F. And that was all. But we used the fact that we have a normal extension. Well, actually, we have to show a little bit more. We wanted to show that these two are equal. Now we have shown only that alpha of F is contained in F. So what I have said, Rotea is not completely correct. So we have to show this, but we have also shown the other. So thus, we have that alpha of F is contained in F. Now we can apply the same argument for the automorphism alpha to the minus 1. So doing the same with alpha replaced by alpha to the minus 1, we get alpha to the minus 1 of F is contained in F, which is equivalent to, but it implies that F is contained in alpha F. So F is equal to alpha F. So this shows 2. Now we want to go from 2 to 3. So we have this identity, which we're going to use in a moment, but whatever. Maybe I'll leave it for one second. 2 to 3 want to show that the Galois group of L over F is a normal subgroup of this. So we take alpha and element in the Galois group of L over K. Then we know that if we take alpha, the Galois group of L over F, alpha to the minus 1 by what is written here, this is equal to the Galois group of L over alpha of F. So by assumption, we assume that here 2 is 2, so that alpha F is equal to F. So by assumption, this is equal to the Galois group of L over F by assumption, because alpha F is equal to F. So the Galois group of L over alpha F is equal to the Galois group of L over F. So by assumption. So this says we have the subgroup such that for every element in the big Galois group, if I conjugate by it, I get back the same subgroup. So this means precisely this is a normal subgroup. So now we come to the last bit. So we want to go from 3 to 1. So we assume that the Galois group of L over F is a normal subgroup. And you have to show that the extension is normal 3 to 1. So we assume the Galois group of L over F is a normal subgroup of the Galois group of L over K. So we know, therefore, if we use again this lemma, that if we take the Galois group of L over K over F, this is a we basically just reverse this. This is so for all alpha and element in the Galois group of L over K, we have that if we the Galois group of L over F is the same as if we conjugate by this element. And we had seen that always this is equal to Galois group of L over alpha of F. Now, I haven't stressed it before. But if you remember, the fundamental theorem of Galois theory says that you have this bijection. So the map which associates to an intermediate field, the Galois group of L over the intermediate field, is a bijection. So that means if these two Galois groups are equal for F and for alpha F, it means that F is equal to alpha F. The map which sends an intermediate field to the Galois group of L over the intermediate field is a bijection to the subgroups. So by the fundamental theorem of Galois theory, it follows that F is equal to alpha F. So in some sense, we actually have just reversed this step. But now we have to reserve the reverse the other step. So now we take again. So now let's take let A be an element in F. And we take, say, F the minimal polynomial of this thing. So let's see. So what do we want to show now? And let F be an irreducible polynomial. F in kx be irreducible with F of A is equal to 0. So we take an irreducible polynomial with a 0. So actually, I don't have to do like that. So let F be an irreducible polynomial in kx with a 0 A in F. Then that the field extension F over k is normal means that F already splits over the field large F into linear factors lie in. So we know that as L over k is a normal extension, we know that this polynomial F splits over L into linear factor. So we have to show that for any element B in L, for any root B of this polynomial in L, that it lies already in F. Roots B of F in L lie in F. So as I said, as I just said, as L over k is a Galois extension, is in particular normal, we know that under this assumption, in this case, we know that F splits over L into linear factors. And thus, if we have this, it also splits over F. If its roots in L lie in F, it also splits over F. So we take a root. So let B in L be another root. So F is an irreducible polynomial in kx. So up to multiplying by a constant, it is the minimal polynomial of A, and therefore also the minimal polynomial of B. So A, so if I have a monic irreducible polynomial which has a 0, it's the minimal polynomial of that, but I always can divide by the leading term. So thus, F is the minimal polynomial of A and of B over k. But then we have this statement about if we have a Galois extension and we have two polynomials, two elements in the bigger field which have the same minimal polynomial, there is an element in the Galois group which sends one to the other. Thus, there exists an element, say, phi in the Galois group of L over k with phi of A is equal to B. But we have that for every element in the Galois group, F is equal to alpha for, or in this case, F is equal to phi of F. So thus, our element B is an element in phi of F which is equal to F. And so we have found that this root already lies in F, which means that we do have a normal extension. So thus, F is normal. OK, so we see that the intermediate fields for which we have a normal extension correspond precisely to the normal subgroups. In this case, we can also easily compute the Galois group of F over k. It's just the quotient of the Galois group of L over k by the Galois group of L over F. When we can form the quotient, because this now is a normal subgroup, so we have a quotient group. And the quotient will be the Galois group of the intermediate field over k. So corollary. So again, let L over k be a Galois extension. And F intermediate field, intermediate field, intermediate field such that if I take F over k, this is normal. This is a normal field extension. Then, not very surprisingly, we have that the Galois group of F over k is equal to the quotient of the Galois group. If you want isomorphic to the quotient of the Galois group of L over k by the Galois group of L over F. So we can only write down this quotient, because F over k is normal, because this ensures that this would be a normal subgroup. So we have a quotient group. And OK. So that's, I mean, this is quite simple. We just have to remember more or less the definition. So if we have an element alpha in the Galois group of L over k, then we know that if we restrict that it sends F to itself, then alpha of F is equal to F, as we had seen. So therefore, if we restrict alpha to F, it becomes an automorphism of F. So thus, if I take alpha restricted to F, this is a field automorphism from F to F. And it will be an element in the Galois group of F over k. It's a field isomorphism from F to F, because obviously it's inverse alpha to the minus, is the restriction of alpha to the minus 1. And it's the identity on k, because original alpha was. So we have a, thus, we can take any element in the Galois group of L over k and restrict to F. This obviously is a group homomorphism. So the restriction map, so from R from the Galois group of L over k to the Galois group of F over k, which sends alpha to alpha restricted to F, is obviously a group homomorphism. After all, the group structure is by composition. And what is the kernel? So an element alpha lies in the kernel. If alpha restricted to F is the identity, so it sends every element of F to itself. But this is precisely the definition of the alphas, which are the identity on F, is precisely the Galois group of L over F. So with kernel, Galois group of L over F. So we have a group homomorphism whose kernel is this. So then we know that the image of this group homomorphism is isomorphic to the source divided by the kernel. So thus, so the image of R, which is a subgroup of the Galois group of F over k, is isomorphic to the Galois group of L over k divided by the Galois group of L over F. So then to show our claim, we have to show that R is surjective. So why would it be surjective? Well, we don't really have to care. We know in order to see it's surjective, we know this is isomorphic to the image. It's maybe enough to say that this has as many elements as this. So you want to see or that this one, the Galois group of F or k. So if we take this quotient, the Galois group of L over k divided by the Galois group of L over F, so these are both Galois extensions. So for Galois extension, we know that the degree of the field extension is equal to the number of elements in the Galois group. So I divide the number of elements. I could say if I take the number of elements in this quotient, this is the quotient of the number of elements. And this is equal to L over k divided by L over F. But now by the what was the degree theorem, we have that L over k is L over F times F over k. So it follows that this is equal to F over k by the degree theorem. And as F over k is a Galois extension, it follows that this is equal to the number of elements in the Galois group of F over k. And so we have an injective map from this quotient to the Galois group of F over k. And this quotient has as many elements of this. So it's a projection. That's the isomorphic to Galois group of L over k divided by the Galois group of L over F. So these are all rather simple consequences of this principle theorem. So let me study one. So I hope it's kind of clear. Anyway, we have seen. So this, anyway, is a simple corollary. But we can, in a simple way, describe the Galois group of such intermediate fields if it's normal. So let me look at one example, which we had before. I mean to show in a simple case how one can analyze these field extensions. So we had this particularly simple example of polynomial F equal to x to the 4 plus 1, which is a polynomial with coefficients in Q. And we had seen that if alpha is a root of this, then it splits into linear factors. So if alpha is a root of F, then over Q. Alpha, we have, as you have seen a few times, that F can be written as x minus alpha times x plus alpha times x minus 1 over alpha times x plus 1 over alpha. And we had also seen that the Galois group is isomorphic to z times z2 times z2. So the Galois group of F, I could say. So the Galois group of Q alpha over Q, we had seen was isomorphic to z2 times z2. And the elements could be described as, so obviously you have the identity. So we call this the identity of the identity. Then we have the map that sends alpha to minus alpha. So the elements are, so the map alpha is sent to minus alpha, which I call. How do I call it? Mu. I mean, you can see if you send alpha to minus alpha, there's a unique way how this becomes a Q automorphism of this field. You know precisely where to send the other roots. And you can have, say, delta, which sends alpha to 1 over alpha. And then you can take, say, the composition of these two. So this would be delta, which sends alpha to minus 1 over alpha. So in particular, this thing, this z2 times z2, we can see what its subgroups are. And then we want to see what the corresponding intermediate fields are. And now I have a little bit of room. So we have, if we take here, we have here the diagram of subgroups. So we have, say, 1. So the neutral element, which always means it's contained. So this is contained in subgroups generated by mu, in a subgroup generated by delta, and in the subgroup generated by mu delta. And they all just have the element 1 in common. And then we have the whole group, which I may be called. So the curve group of F. So you have this diagram of fields. And we know there's a bijection of this with the corresponding fixed fields. So the fixed field of the whole Galois group is obviously just Q by the principle theorem of Galois theory. So if you send, you have to think a little bit, but if you send alpha to minus alpha, you can certainly see that alpha squared will be invariant under this action. And it's easy to see that alpha squared is, in this case, the alpha to the squared will be minus 1. So it's not an element. So alpha to the 4 would be minus 1. So alpha squared is not an element of Q. So you do have an extension of at least of degree 2. So we can look here to this. So we find that the fixed field of this mu certainly contains this. On the other hand, we see that this field is not equal. So we had Q was the fixed field of the whole thing. So the fixed field of this certainly contains Q alpha squared. Because we have just seen that alpha squared is invariant. And it's easy to see that the degree of this Q alpha squared over Q is indeed 2. You can certainly see that Q alpha squared is not equal to Q. And then the degree must be 2 because there's no more room. Obviously, the 1, you have Q alpha. And then here, if you take the delta, you send alpha to 1 over alpha. So something which certainly will be invariant under this is Q of alpha plus 1 over alpha. And again, you can see, you can check that this will not be an element of Q. So therefore, this will indeed be the fixed field. And in the same way for the other one. So anyway, I do not do the details. But this would be the corresponding diagram of fields. Maybe one should look at more interesting. What's the question? Yes? So it's supposed to correspond in the same order. So maybe for the moment, what time is it? So I now want to go a little bit back to the classical problems that people were looking at, which had motivated looking at this stuff. So this is about solutions of polynomial equations by radicals. So you have a polynomial in Qx with coefficients in Q. And you want to find the formula for a 0 of that polynomial just in terms of writing a formula with somehow in terms of the coefficient of the polynomial and then plus, minus, and taking some roots. So in the 17th, 18th, and some extent 19th century, this was studied quite a lot. So in particular, the first instance of this are the cubic polynomials. So I want to briefly review what happens there. So let me just write it down. So a big classical problem was the solution or was the solution polynomial equations by radicals. So you want to find a 0 for a polynomial in, so find a 0 for a polynomial in Qx, so in, say, Qx normally, in terms of where you just do the normal operations using, only you can say, plus, minus, and somehow the nth root. And this was first, so I expect, so in school, you have studied the case of polynomials of degree 2, OK? And I can briefly do this from school, you know? So if you have a polynomial, so the formula for quadratic polynomial, so this is, so we have f equal to x squared plus px plus q. Obviously, we can normalize it to be harmonic by dividing by the leading coefficient. And then the formula is the usual one. So we have the solutions are, so what is it? Minus p halves plus or minus square root of p divided by 2 squared minus q, OK? So this everybody knows. So incidentally, I was told that, so in Germany, when you want to become a professor, you have to give a talk on some advanced topic. And then afterwards, people will decide on how good the lecture was. And so somebody apparently gives some very difficult lecture. And so after the lecture, there's people are supposed to ask questions. And so nobody could ask a reasonable question, because they hadn't understood the lecture. And so they asked him, can you tell us something about the solution of quadratic polynomials? And he couldn't remember the formula. And so he didn't become professor. So it's important to know this formula. OK, so now let's look at the case of the, so in the 70s centuries, maybe in the 16s, the formula was found for cubic polynomials. So we have the formula of Cardano and Tertaglia, I think, found a formula for cubic polynomials. Also that is quite funny. So in those days, mathematics was some kind of spectator sports. So you would have these two people who would kind of challenge each other. There would be some public thing. And everybody would give to the other one a cubic equation to solve. And then they would kind of make a duel. So if somebody did something that the other one couldn't solve, then they would kind of earn their money with these duels. So it's kind of, and so they found this formula, but they kept it secret so that they could win duels with it. And I think Cardano published it, which Tertaglia was very angry about. Anyway, but now let's look at it. So we will just read your life this formula. It's quite simple. So we look at the cubic polynomial that f, so say like this, we would call it a2x, plus a1x plus a0, b cubic polynomial. So anyway, we can just, so we have this cubic polynomial. And we want to find the solution. So first one makes a variable transformation, which in some sense is the basis of solution also for the quadratic one. We want to get rid of this coefficient. That's very easy. So we make the variable change x. We replace by x minus a2 divided by 3. So we compute f of x minus a2 divided by 3. If you do this, we see that this coefficient vanishes. And obviously, if you find a 0 of the polynomial, so if, say, alpha is a 0 of f of x minus a2 divided by 3, then not very surprisingly, alpha plus a2 divided by 3 is a 0 of f. So we can certainly make this transformation and get rid of the quadratic term. So with this, I can assume that f is of the form x to the 3 plus what we around we wanted it, px plus q. So we get rid of the quadratic term. So now we want to find the solution. And now comes the trick, which is very strange indeed. I don't know why. So I mean, this is not really Galois theory. I just review this result. Come the trick, we put x to u minus v. So we write this polynomial now as a, you know, we put u minus v instead of x. Nobody can hinder us from doing that, but why this should help is completely unclear. But let's just do it. So if you make f of u minus v, well, you can compute it. Maybe you don't detail. So you will get this. I think you will be able to verify this. Unless, obviously, one has to be able to distinguish the u from the v here. Otherwise, it's difficult. OK. So we have this equation. And now, so now comes somehow a miracle. So now there comes some kind of, we go on with the trick. So we want to find, you know, u and v. So we want to find the u minus v so that this is 0. So one way, the easiest way, so certainly the easiest way how this can happen is if this one minus this one plus q is 0 and also this is 0. So a sufficient condition for f of u minus v equal to 0 is these two and 3u v minus p is equal to 0. I mean, obviously, we do not know whether we can find the solution for these two equations. We have made our problem more difficult. But if we do find the solution for these two equations, we have a solution for our original problem. But now each of these equations is much simpler than the original one, obviously. So first, the second one we can obviously solve. So is it? So the second equation gives what is it that v is equal to p divided by 3u. And we put this into the first. And, well, so if I'm not mistaken, you get 3 to the 3. I mean, there's some cancellation occurs, which, well, not so much. So 3 to the 3u to the 6 minus p to the 3 plus 3 to the 3u to the 3q is equal to 0. So I claim that this is what you get. So we kind of multiply, obviously, with the correct power. So by itself, v is something by u. But we multiply out with the corresponding power of u so that it gets polynomial. And now what you see here is that some miracle has occurred. Because this is a quadratic polynomial in u to the 3rd. So we can use our wonderful old thing from school. So we put y equal to u to the 3rd. And then we find the equation. Then the equation becomes, if I'm not mistaken, y squared plus qy minus p divided by 3 to the 3 is equal to 0. And now everybody knows how to, so we have divided by the correct powers of q and so on to get this. And so now we can use the quadratic formula that we have just, that's written here. And we find that y is equal to, now p and q have changed their role. This is now, so minus q halves, because q is what we force p. Plus or minus, but we're taking it, just the plus part, q halves squared minus no plus p divided by 3 to the 3rd. So this is a solution for the y. Now we just have to remember that y was equal to u to the 3rd. So u is the 3rd root of that of this. And we had put v equal to p divided by 3u. So v, which one do I want? No, no, I don't think I, yeah, this however I didn't want that. So yeah, I want to take the other one. I take the other equation. So I have u to the 3rd plus, minus v to the 3rd is equal, plus q is equal to 0. So we find that v is equal to the 3rd root of y plus q. And if we put, now we can, we know that our x, that we want to find the solution of this u minus v. And we can just put everything together. And we find that this will be equal to the 3rd root of minus q halves, so this what's written here. So the square root of q halves squared plus p 3rd to the 3rd. Minus the thing which you obtain by this. So we have to add q. And there's a minus q of it becomes plus q halves. So 3rd root of q halves plus square root of q halves squared plus p divided by 3rd, OK? So this is the solution. And we have made this ansatz which was not directly justified. But as it leads to a solution, it's fine. And so this is Cardano's formula. So this is not really your algebra, it's just a trick. But still, this is this old solution. And there's a, so that's as much as I wanted to say about this. It's a similar, more complicated formula which also works for polynomials of degree 4, which I certainly am not going to write down. And so people were trying to do it for higher degree, for degree 5 and higher. And somehow they didn't succeed. I think maybe there were some people who thought they succeeded, but then it was not true. And so people tried to understand or tried to wanted to figure out whether at all it was possible. And so this became, then, the problem. So it's a general question of solvability by radicals. So there's a similar formula for degree 4, but nobody found one. So quite a lot of groups here. And then Cardano 3 was somehow developed in order to understand this problem, to see whether there will be a formula, or for which polynomials you can find a formula in terms of radicals for the roots. So it was first proven by Abel that there is no general formula which does it for all polynomials. And now instead, so I will not deal with that now. So Abel, so I think maybe he was 18 or something, that there is no formula, no such formula for degree 5, or bigger, I think. But now the question, but so then the question kind of that post itself is for which polynomials is their formula? How do you decide whether you can express the roots of polynomial in terms of radicals and taking in terms of square roots and tires? And this is what Galois did, and famously wrote down in the night before the dual, if you know there's this kind of. So now let's try to formulate the question. So the question is under what conditions? So when can one express the roots of a polynomial f, kx, say in terms, so maybe normally the old question was obviously was normally with coefficient q in terms of radicals. So now we want to write this in a more formal way. So let me kind of make this question a bit more precise so that we actually have a way to talk about it properly. So we don't restrict our attention to q. We take k, say, a field of characteristic 0, for the rest of the section, let k be a field of characteristic 0. And we want to make precise what we mean by the fact that you can express the roots of polynomial in terms of radicals. So definition, the field extension, say, I can take large k over small k, is called a radical extension. Well, it's obtained by just joining roots of elements in k, so we want more roots to it. So extension, if there is a chain k, the large field is somehow the last one in the chain. It contains another field, m minus 1, and so on. And in the end, the 0th one is k, such that each step is obtained by adding some kth root of an element in the previous field. So such that l i plus 1 can be written as the field extension of l generated by one element, say, b i. And so for all i, with b i is 0 of the polynomial x to the n i minus a i for some element a i in l i. So the next field is always obtained from the previous one by adding some n i root of an element of the smaller field. So we have this field extension obtained by only step-by-step adding radicals. So thus, b i is an n i root of a i, as you can see. And so we call a polynomial f with coefficients in the small field is called solvable radicals. Well, if there is a radical extension over which it splits, say k over k, small k, such that f splits over the larger field. Yeah, but actually, this is not quite what I do. Radical extension, which is also Galois, such that f splits over k into linear factors. And obviously, this condition, so if one forgets about the story that this should be Galois extension, this is obviously the reasonable formulation. We want to find the field extension, which is just obtained by adding really n-th roots of elements step-by-step, such that f splits. Now, for technical reasons, we also assume we also want to impose that this is a Galois extension. But in fact, it is true that if there exists such an extension for which this thing is Galois, or which k over k is Galois, if there exists such an extension over which it splits, then there also exists a Galois radical extension over which it splits. But we don't have time to prove it. It is actually an exercise in the notes, but I don't think a particularly easy one. So exercise, so the condition that k over k is Galois can be removed. If there exists a radical extension over which it splits, then there is another radical extension, which is also Galois over which it splits. Namely, if there is a radical extension such that f splits over k, then we can make k a little bit larger, so it becomes Galois, but it's still radical. Then there exists a radical Galois extension, say f over k, such that f contains k as a subset. And that obviously does it because clearly our polynomial f, as it splits over k, it splits over the field f. But it's not so difficult, but we don't have the time to prove that now. So we will put this additional assumption. And so what we want to prove is that the solvability of polynomials by radicals has something to do with the property of groups, namely solvability of the group. So that the theorem, which is due to essentially to Galois, is that the polynomial can be solved by radicals, if and only if the Galois group is solvable. But first I have to tell you what it is. So whether a solvable by radicals has to do of the Galois group. So the Galois group of f, which by definition is the Galois group of the splitting field of f, namely, again, this is called solvability. And we want to introduce this now. And as usual, group theory in its beginning was mostly motivated from studying field extensions, mostly by studying polynomials. So the notion of solvable for a group comes from the notion of solvable by radical for a polynomial. Maybe now solvable groups are more important than the question of whether polynomials are solvable by radicals, but it comes from there. So let me define what this is. So a group G is called solvable. It's some kind of generalization of Ibn, so that you can make successive quotients so that they have a chain of subgroups so that all the quotients are of normal subgroups, that all the intermediate quotients are a billion. So if there's a chain of subgroups, so we have that G, so to speak, the 0th one, contains a subgroup G1, contains G2, and so on. Such that these GI are all subgroups. The last one is the trivial subgroup, such that the following hold. So the GI are themselves just subgroups, but we have that GI plus 1 is a normal subgroup in GI. GI plus 1 is a normal subgroup in GI. And so if it is a normal subgroup, we can talk about the quotient group, and the quotient group is a billion. So the group G itself doesn't have to be a billion, but you can somehow kind of filter it so that all the intermediate pieces are a billion. So for instance, we have that, so as an example, obviously a billion groups are solvable. That's not so exciting, but maybe the simplest non-commutative group which is solvable would be, so S3 is solvable, can be used in several more or less equivalent ways. So we have, in this particular case, we have the group of two cycles, so which is generated by the cycle 1, 2, 3. Is a normal subgroup. I think you had seen that or something, but anyway it's very easy. So just 1, 2, 3, 1, 2, 3 squared, 1, 2, 3 to the third. This is a normal subgroup, and you can easily see the quotient S3 divided by this normal subgroup. This normal subgroup has three elements, so this must be a group with two elements. There's only one such group, so this is isomorphic to Z2. You can also see the same thing in a different way. We had introduced the sign of a permutation. So this sign is a map in this case from S3 to the group 1 minus 1 with multiplication, which is the same as Z2 as a group, and the kernel is precisely the even permutations, and the even permutations is precisely this normal subgroup. So this again shows that this is a normal subgroup, and the kernel is isomorphic to Z2. So we see that this is a solvable group. In general, we want some kind of way to decide whether a group is solvable or not. So we want to have some kind of way to decide whether a group is solvable. And this is basically by finding a canonical way of getting this chain. So there's some kind of procedure of producing this chain. And so if this procedure works, it's solvable. If the procedure does not work, it's not solvable. And this has to do with the commutator subgroup. So, finishing. And then I think I have to go before we get into trouble. So we take G a group, and if you have two elements, A, B, and G, we can form the commutator. This is A, B. It's just another way to write A, B minus B, A. So it's just a commutator of these elements. Clearly, the commutator will be 0 if and only if A and B commute. Let's say that. So A, B is equal to 0 if and only if. Well, A, B equal to B, A. That's not very deep. Except that it's not 0. We take our group and multiply it. Have it multiplicativity. So it's 1. And this is not minus. So I mixed here. So I denote my group multiplicatively because I don't assume it's commutative. So it's A, B times. And I maybe have also decided what I wanted in this order. I decided the other round. A to the minus 1, B to the minus 1, A, B. So I just have denoted the group with multiplication, not with addition. Obviously cannot mix it. And then the commutator is 1 is equivalent to this. And then the commutator subgroup of G is the subgroup generated by our commutators. So the commutator subgroup is usually called G prime. Sometimes called the right group. G of G is the subgroup generated by all commutators A, B, I with A and B elements in G. And clearly, the group is a billion if and only if the commutator group is equal to just 1. So G is a billion if and only if G prime is equal to 1. Because that precisely says that all the commutators are 1. OK, so I think my time is up. There was already some nervous person trying to come in. So we meet on Tuesday. OK, anyway, we meet next week.