 OK, so we had been talking about Galois groups. Recall that if l over k is a field extension, so then the Galois group of l over k is just the group of k automorphisms of l. Automorphism, such that phi restricted to k is the identity. So we had studied, we had kind of introduced these and studied it a little bit. And the last result that we proved was the following. So if we look at, so this was the following theorem, so if we have a Galois extension, then we have that if we look at the subset of all elements in l, which are fixed over under the whole Galois group, that this is equal to k. So k, so the set of all a in l, such that phi of a is equal to a for all phi in the Galois group of l over k, this thing is equal to k. So this we proved as the last thing. We will see that this is an important result for the development. So now we want to proceed. So here we have the Galois group of a field extension. Normally the field extensions that we are looking at is the splitting field of polynomial over k. And so therefore we will also say we have the Galois group of polynomial by meaning the Galois group of the splitting field of the polynomial over the ground field. So let f be polynomial provisions in k, non-constant polynomial. And so let l be the splitting field of f over k. Then the Galois group of f over k is the Galois group of l over k. I just tried this Galois f, although in principle also which field I'm considering over is important is defined to be the Galois group of l over k. And we won't be interested in studying such things. So we now want to say a little bit about this Galois group. So the first thing that we find is that this Galois group can be described as a subgroup of the group of permutations of the roots of f. We had some slightly similar result before we later compare them. But so we have the following result proposition. So we take f, a non-constant polynomial, and we let's say of degree n. So then we take r to be again the set of roots, the set of 0s over the splitting field of roots of f over the splitting field, which we call l. And then I say the first statement is that the Galois group of f, so the Galois group of f over k, is isomorphic to a subgroup of the symmetric group on the roots, so the permutations of the roots. And the order of the Galois group of f is a divisor. So divides the number n factorial, so n is the degree. And the second statement, which doesn't have much to do with it, is so if you assume that all the roots of f are simple, f has no multiple roots in its splitting root in l. So if all roots of f are simple, so we make this assumption, then we have then f is irreducible, if and only if. The Galois group acts transitively on the roots, transitively on the set r of the roots in l. OK? So let's first look at the first one. Remember before we had some result that for a simple algebraic extension, the Galois group of the field extension is isomorphic to a subgroup of the permutation group of the splitting of the zeros of the mineral polynomial of the element you join. This somehow looks similar, but it's not quite the same. So let's look at this. So by definition, after we have the Galois group of f, it's just the Galois group of l over k. And we want to see that it's isomorphic to such a subgroup. So first we have to see that this Galois group acts on these roots. So if, I mean, it's the same trick we have used already, half a dozen times. So if a is a root of l, so a 0 of l, then we can take f and phi is in the Galois group. Then we can take f of phi of a. This we had seen is the same as phi of f of a, because we had seen it actually on the coefficients of the polynomial. So this is phi of 0, so this is 0. So it follows that phi of a is also in r. So the Galois group sends roots of this polynomial to other roots of this polynomial. So it gives us an action of the Galois group on r. So thus, if you take the restriction to r, so phi from l to l maps to phi restricted to r, which is now meant from r to r, gives us a group homomorphism from the Galois group of l to the group of permutations of the zr group homomorphism from the Galois group of l over k, so Galois group of f to s of r, because here the group structure is by composition, here it is by composition, so it's the same thing. So we have such a thing, and now we want to say, we want to see that this group homomorphism is injective. So if phi, so maybe I call this u s, so phi is an element in the kernel of this restriction map, then this means that phi maps as the identity on the roots. So that means phi restricted to r is equal to the identity, then it is clear that phi is indeed the identity on l. Because we know that l is the splitting field of this polynomial, so it is obtained by taking k and adjoining to it all the roots. So we have that l is just k adjoint all the elements in r. So the elements in this l are just everything you can obtain by taking sums and products and quotients of elements in k and these roots. Now if phi is the identity on k and it's the identity on r, so it's the identity on this whole thing. So this means this is the identity. In other words, it means this restriction is injective. And so we have this, therefore the garag of f is isomorphic to the image here, and so it is isomorphic to subgroup of S of r. It's isomorphic to its image in S of r, which is a subgroup. So a subgroup of S of r. And now it's clear that the number of elements in the garag group of f, it's the same as the number of elements in the subgroup of S of r. So therefore it has to divide the number of elements in S of r, which is, so this is the set of all permutations on this set r, and so this is the number of elements in r factorial, the number of elements in a symmetric group on so and so many elements is the number of elements factorial. And this number will divide n factorial because, so let me finish it because we know that r is smaller equal to n. The polynomial f has degree n, so it has at most n zeros. OK, so now the second part is a different thing. So we assume that, let's assume that f is irreducible, no. Let's assume that f that the, so we assume that the Galois group of f acts transitively on these roots, and that f is still reducible. So then as our statement is that, now let me, what? Maybe I can first, what is the problem? I didn't understand what you said. Maybe you're right, but I haven't understood the word. Yeah, well, so we want to show, but now I assume it's different so as a contra position, no. So we assume it acts transitively and f is irreducible, then we have to see that our assumptions are wrong. So we have to show that f has multiple roots. So I have, maybe I should have done it the other way around, but OK, so I do this direction. So this is, which direction do we assume? This is, will be this direction, no. So this then implies that if all the roots are simple, and the Galois group extensively, then f is irreducible. So let's have a look. So we take, so f is reducible, so we can write f. We can find two different factors of f. You can write g1 and g2 irreducible in kx, such that g1 times g2 divides f. So now we want to see what this means. So we know, we have assumed that the Galois group of L over k, so which is this Galois of f, extensively. Sorry, maybe just Galois of f extensively. And so I mean, I will write, I didn't write it here, so I do write L equal to the splitting field of f over k. So assume the Galois group of f, which is the Galois group of L over k, extensively on the roots, so on R. So R is still the set of roots of f in L. And we want to show that, so we, so let's see. Maybe as we are proving this, I will, maybe for a moment, leave it. Where am I? So let a1 be the root of g1 in L, and a2 of g2 in L. We know that f splits into an effector, so both g1 and g2 do, so we have this. So we have this. So then what do we have? So the Galois group, so then a1 and a2 are also roots of g of f. So as this group x, transitively, it follows there exists an element in the Galois group sending one to the other. So there exists, say, an element phi in the Galois group of L over k, which is Galois of f, with phi of a1 is equal to a2. But then we have the same story as above. I mean the same trick which we have used many times, like here. And we find that whatever, a1 is also 0 of g2, let's see. So if we take phi of g1 of a1, g1 of a1 is 0, so phi of it certainly is 0, but this is by the same token equal to g1 of phi of a1, which is a2. So we find that g1 of a2 is equal to 0. So as a g2 is irreducible with a2 as a 0, it follows that up to multiplying by a constant, it's the minimal polynomial of a2. So we can as well assume it is a minimal polynomial. So up to by a constant, g2 is the minimal polynomial of a2 over k. But we have found another one which vanishes there, so it follows that g2 divides g1. Now we can replace the, we can exchange the role of 1 and 2. And so we get that g1 divides g2. So if we assume, so that means, therefore, however, that g1 is equal to g2 up to multiplying by a constant. So we can assume that g1 is equal to g2. So that means that g1 squared divides f. fg1 times g2 divides f, so g1 squared divides f. But g1 is a polynomial of positive degree, so it will have some 0s. And if g1 squared divides f, all the 0s will be double 0s of f, will be multiple 0s of f. So thus, f has multiple 0s. And so this shows this direction. And now we want to prove the opposite direction, which I've done earlier. So now we want to show. So now we do assume that all the roots are simple as here. So we assume all roots are simple. And then we want to show that, so if we assume that f is irreducible and want to show that the Galois group is transitive. So if f is irreducible, so up to multiplying by a constant, it is the minimal polynomial of any of its 0s. I mean, it's irreducible. It has some, so f is reducible. And we take two elements, a and b elements in r. Then f is the minimal polynomial of a and of b over r. So now we can use, because it's an irreducible polynomial, which has this as a 0. So then we can use this old theorem of extension to field isomorphisms to splitting fields, which says thus there exists a unique, but if we don't care about unique, exists a k isomorphism, say Pc from kA to kB with Pc of A is equal to B. This was its extension of field isomorphisms to splitting fields. And now we have that, so L is some extension of kA and of kB because a and b are both i and l. So it means, and l is the splitting field of f over k. So it's also the splitting field of f over kA, and it's also splitting field of f over kB. If we do it over intermediate field, it also still is. So l is also splitting field of f over kA and over kB. And now we had another statement that we can always extend an isomorphism between fields to an isomorphism between the splitting fields. We don't have any uniqueness, but we have existence. So by the extension of isomorphisms to splitting fields, there exists an isomorphism phi from l to itself with a because l, which is this given isomorphism on kA. But this then means that, for one thing, it's the identity on k because this one was, and it sends a to b. So phi is an element in the Galois group of l over k, so in the Galois group of f, with phi of a is equal to b. So this proves this. So you can see, I don't know precisely how difficult it is to follow this. Maybe it is a little bit difficult because we kind of always use these many arguments from before. But we have built up all this story. We somehow use everything we have built up so that makes it a little bit difficult because one has to remember everything we have done about this. And we tend to use the same results over and over again, but maybe it's not always clear which one is the best to use in a given situation. So I wanted to try to remove some confusion or maybe confuse you more, I hope not, by just saying again this thing which I just mentioned at the beginning of this proof, that when you have such a field extension, we had in two different situations we had that the Galois group was a subgroup of some symmetric group. And these are quite different, and in some sense, one is only interesting for theoretical purposes and the other one is more for practical purposes. So I mean, it's all, so L over k is a Galois extension, and so which is a splitting field of f of sample number f. Then we find that the Galois group of f, which is just the Galois group of L over k, is in two different ways a subgroup of a symmetric group. So the first one is what we just had. So if R is the set of roots of f in L, then we had seen, I've worked it out, that L is that the Galois group is actually subgroup, is isomorphic to a subgroup of the symmetric group on R. So is isomorphic to, and we can identify it with, is a subgroup of S of R, so of the permutations of the roots. And this is, if we look at the Galois group of f, this is how one, for concrete f, how one wants to study this Galois group after somehow, you study it as a group of permutations on the roots. So basically those permutations on the roots, which extend to isomorphisms of R, of the automorphisms of R. This is the Galois group. So this is how to study it for concrete f. So if I give you an exercise, compute the Galois group or whatever of f, then you would normally try to study it as a subgroup of the symmetric group on the zeros. We had seen something else, which has to do with the theorem of the primitive element. So we know that L over K is also a simple extension, because it's a normal extension. So it's simple, a simple algebraic extension, no normal and separated extension. So it means that we can write L equal to K of A for some element E and F. There is some primitive element. There's the theorem of the primitive element. So then we had seen under this situation, we can look at the minimal polynomial of A, so let G be the minimal polynomial of A over K. And then we had and maybe, so how could I want to say it? So let sigma be the set of roots of this polynomial G in F. Now the degree of the polynomial is equal to the degree of the field extension, which is equal to the number of elements in the Galois group. So as this is an irreducible polynomial with a zero and this is a Galois extension, we find that all the factors are different. So the number of elements in sigma is equal to the degree of the field extension, which is equal to the number of elements in the Galois group. So anyway, so let G, so let N be the degree of G, then the number of elements of sigma is equal to the number N because this cannot have multiple. So again, because this is a normal extension, this polynomial will split into linear factors over L. We have a normal extension of an irreducible polynomial, which has one zero, so it has to split. And it's also a Galois extension, so a separable extension. So therefore, it doesn't have multiple roots. We have an irreducible polynomial, then in the splitting field, it doesn't have multiple roots, so the number is N. And then the Galois group of F, which is just the Galois group of L over K, we had seen that this is isomorphic to a subgroup of the symmetric group on large N letters, which acts simply transitively on the set of root sigma. But these are two completely different descriptions. So here, if this polynomial has degree N, then these are N elements. And the degree of the splitting field could be in the most extreme phase, N factorial. And so then you would have this N would be some number up to N factorial, so it's a completely different group in which it plays. But we have seen that in some proofs, it's useful to have this viewpoint. But for complete cases, if it wants to study the Galois group of a given polynomial, we use this one. But it's just a bit, I always find a bit confusing that you have two different ways to view it as a subgroup of a, OK, and I should maybe say why this one is not useful in concrete cases. So assume you have this. You could say, given our F, we want to find, we want to study the Galois group via this description. But in order to do that, you have to find this primitive element, which is very difficult. And you have the field extension, which is the splitting field of this, but you have to find the primitive element of it, which is not very easy. If you have found the primitive element, you have to find the minimal polynomial of this primitive element. It's also very difficult. So it would be very difficult to use this description in concrete cases. You only know it exists, but to actually compute it would be difficult. OK, so maybe I can just look at one example for such a Galois group. So let's again maybe look at this polynomial. F is equal to x to the 4 plus 1. So as a polynomial with coefficients in Q, we had seen that if alpha is a root in the splitting field, we have that minus alpha 1 over alpha and minus 1 over alpha r roots. And so we have that Q alpha is the splitting field of F over Q. Now we want to determine the Galois group. So I claim that the Galois group of F over Q is isomorphic to z 2 times z 2. So there are two things. So we have that this is a simple algebraic extension. So we know that the degree of this extension, this polynomial is irreducible. Anyway, it's a simple algebraic extension. So the degree of the extension is equal to the degree of the polynomial. So as Q alpha Q is a simple algebraic extension, we have that the degree of Q alpha over Q is equal to the degree of x to the 4 minus 1, which is 4. This is assuming that we know that this is irreducible, but we can anyway know that if this was not irreducible, then it would be the 0 of some factor of it. So the degree would be smaller. But anyway, this polynomial is irreducible. And as this is a Galois extension, we know that the number of elements in the Galois group is equal to the degree of the extension this we had. So we have that the number of elements in the Galois group of F is equal to 1. So we know that this group has four elements. And now, in some sense, we can just try to find these elements. So this group is, maybe it was written like that, is Q alpha. So elements in it adjust. You take any expression with elements in Q. And you have alpha, some polynomial or some rational. In some sense, just some polynomial in alpha. So you have to do something. You have to find an automorphism of this thing, which is the identity on this, such that it gives you an automorphism of this field. So what you can do is, for instance, if you send alpha, so the map alpha to minus alpha is certainly a field automorphism. It's zero. So what I mean by this is that if you have any polynomial in alpha, so some i from 0 to n, a i alpha to the i, then I replace alpha by minus alpha. Then it sends this. So this element, f of alpha, is sent to f of minus alpha. And this is, if you take the sum and the product of these things, it's compatible with it. And the same if you take alpha goes to 1 over alpha. So you have to check that this map, so you take the identity on k, and you say you want to send alpha goes to minus 1 over alpha. So this defines in a unique way an isomorphism of this field k alpha, q alpha, to itself. And this is easy to see. So in this way, you get two automorphisms of our field q alpha, two k automorphisms. And you can see that they commute to 1 over alpha. So if you first send alpha to minus alpha, and then send it to 1 over it, you send alpha to minus 1 over alpha, and the same if you do the other round. So these two automorphisms commute. And they are, obviously, I'll call it, if you take the square of them, it's 0. So if you apply this twice, you get the identity. If you apply this twice, you get the identity. So these two involutions. So the group generated by these two elements is isomorphic to Z2 times Z2, which is isomorphic to Z2 times Z2. But we know that the Galway group of S has only four elements, so it means it's equal to Z2 times Z2. OK, so basically, and you can see that if you want to do other things, OK. And it's kind of clear that you don't, it's anyway kind of clear that there cannot be more elements. But that's the statement. So once you send alpha to minus alpha, it tells you what happens to all the other ones, obviously. And so you also see that there cannot be more than these four elements. OK, yes? Maybe? The Galway group should be isomorphic to S4, right? No. No? For a subgroup of S4? Yes, it's a subgroup of S4, yes. Just the permutations of the four roots. But this just means that S4 has very many subgroups. So we have a very small subgroup of this big group. So we see, in particular, that it can happen that the Galway group is relatively small. It could have, after all, 4 times 3 times 2 is 24. It could have had up to 24 elements, but it has only 4. OK, so now that's as much as I wanted to come to that. And now I want to kind of, I want to talk about the fundamental theorem of Galway theory. So we have already introduced all the players in this thing. So as I said, Galway theory is about understanding field extensions in terms of groups, in terms of the Galway groups. And so the fundamental theorem of Galway theory is that you have a Galway extension and you want to understand all the intermediate fields of this Galway extension in terms of all the subgroups of this Galway group. In fact, the claim is that there is a bijection between the intermediate fields and the subgroups, which kind of allows you to understand everything about these intermediate fields in terms of these subgroups. And so let me state this. So this is the fundamental theorem. So maybe we'll reintroduce this thing which I already had introduced, so definition. So if we take, again, L over K, a Galway extension, so let L over K be a Galway extension. And so for a subgroup, which I call H, in the Galway group, we want to consider the fixed field, which we had done in a special case before. So the fixed field, I don't want to call it, I mean call it the fixed field, is just fix of H, which is just the set of all elements in L, which are sent to itself by all elements in H. So the set of all A in L, such that phi of A is equal to A for all phi in H. So I hadn't said this before, but it's everything from the definition that this is a field. It's easy. Fix of H is a field. And in fact, a subfield of L, which contains K, so an intermediate field. Now it's clear. Now by definition, it's contained in L. It contains K because all elements in the Galway group fix all elements of K. And you can see if, for instance, A and B are fixed by phi, then A plus B will be, and so on. You just write it down. A times B will be, or the quotient will be, and certainly the element 1, the element 0, will be. So you find that fix of H is a field, intermediate field of this extension. So now we want to use this to our statement. And so you should also remark that we have proven, we have noted, so a previous theorem was that if we take the fixed field of the whole Galway group for a Galway extension, this is just equal to K. This was a previous theorem. And so now, and you can also have one more remark. So if we have an intermediate field of this extension, if F is an intermediate field L over K, we can look at the Galway group of L over F. So this, by definition, are all the automorphisms of L. To itself, which are the identity on K, on F. I mean, obviously, as K is a subfield of F, there anyway are the identity on K. So I can describe this as a set of all phi in the Galway group of L over K, such that phi restricted to F is the identity on F. And you can see that if you take the composition of such two, it will lie here again. So this is a subgroup of the Galway group of L over K. So what we have here is that if we're given a subgroup of the Galway group, we can associate to it in an intermediate field. And if we have an intermediate field, we can associate to it a subgroup of the Galway group. And the fundamental theorem of Galway theory says that these two operations give you inverse projections between the intermediate fields and the subgroups of the Galway group. So that there's one-one correspondence between the subgroups of the Galway group and the intermediate fields. So this is the following theorem. So as I said, fundamental theory of Galway theory, which says, so again, we have L over K, the Galway extension. So I write the Galway group of L over K just as G. So then I have the two mappings. So the main part is the following statement. We have the following maps, which are between, on the one side, the subgroups of the Galway group G. So with Galway group G. So the subgroups of the Galway group is in bijection with the intermediate fields. I don't think I have quite enough space here. Maybe I have to do it here. So I look at this thing. I have the, on the one side, I have the subgroups of G. On the other hand, I have the intermediate fields of this field extension. So the fields which lie between K and L. And I have now two maps, one in this direction and one in this direction. So the map in this direction is I send a subgroup H to the fixed field of H. And the map in the other direction is I send an intermediate field F to the Galway group of L over F. So these two maps are both bijections and they are inverse to each other. And you can also see that they are, if you want, I mean they're inclusion reversing. So that means if a field F prime is contained in F, then the Galway group of L over F prime will contain the Galway group of L over F. So it goes the other way around. So maybe something, it would seem more natural. I mean just by the, if one looks at it naively, that one takes the map that one associates to an intermediate field F, the Galway group of F over K. But that's not what it is. This is the thing that one has to have. The other one, the other correspondence wouldn't, just doesn't work. It is kind of, just want to remind you that it is maybe the opposite of what one might think if one is naive. So this is the first statement. The other two statements are, so this is the main result and the other two results are very simple. So it says if I take the degree of L, so maybe I should state this explicitly. So these are what, that the inverse to each other obviously means that if I take the fixed field of the Galway group of L over F, this is equal to F for all intermediate fields. So this is end. So this means that if you start with a field here, you send it to the Galway group and then back, you get back the field. And the other inclusion is that if you take the Galway group of L over the fixed field of a subgroup H, this is H for all subgroups, H of the Galway group, which I had denoted by G. So these two statements are equivalent to what I've said. And as I said, the rest of the statement is a little bit of bookkeeping. So just say that I have that the degree of L over the fixed field of a subgroup H will be equal to the number of elements in H. And basically equivalently, if I take the degree of the fixed field of H over K, this will be equal to the index of G in H, of H in G. We'll see that later. That's basically, these two and three are basically trivial, but we will see then. And then the third one is similar on the other side. If I take L, so F is always intermediate field. H is always a subgroup of the Galway group. So the degree of L over the intermediate field is obviously equal to the number of elements in the Galway group of L over F. And if I take the degree of F over K, this is equal to index in G of the Galway group of L over F. As we will see, two and three are basically obvious, but now we have to, I'm concerned, proving part one. So we want to see that this is true. So we have to prove these two statements, because that's equivalent to the statement of one. So and it turns out that one of them we have already essentially proven. Let's see, why? So let F be an intermediate field, our extension L over K. So then I say L over F is a Galway extension. So let's see, I mean, we follow from the things we know, but maybe should review it. So L over K is a Galway extension. So L is a splitting field of some polynomial with coefficients in Kx. L is splitting field of some F in Kx. So then L is also the splitting field of the same F over this intermediate field F. You have seen this a few times. So L is also a splitting field of F over the intermediate field, large F. So we know that an extension is normal, if and only if, I mean, finite extension is normal, if and only if it's a splitting field of some polynomial in the ground field. So it follows that the extension is normal. And now we have to show that it's separable than it is a Galway extension. For this, you have to see the story about the multiple roots. So let A, no, let say, now say that G the minimal polynomial of some element A, A in F over K. So the minimal polynomial over K, you see. And let say H be the minimal polynomial over F of the same A. So we know that the extension, so that the extension L over K is separable. So that means this minimal polynomial G has no multiple roots in its splitting field. As L over K is separable, G has no multiple roots in splitting field. But now we take a minimal polynomial of the same element over F. We have certainly G of A is equal to 0. So H must divide G. So if it divides it, so once you split it into linear factors, the linear factors will be a subset of the linear factors of G. So if there are no multiple factors for G, there will be no multiple factors for F. And it's for H. Thus H has no multiple roots. If you remember to be a separate extension, you need that if you have an irreducible polynomial, which has a 0, so an irreducible polynomial in the smaller field, which has a 0 in the bigger field, then it has no multiple roots in its splitting field. That's the same as saying, because an irreducible polynomial is up to multiplying by a constant the same as a minimal polynomial of some element, so if you take a minimal polynomial of any element in F, then it has no multiple roots in its splitting field. And that's what we have proved. So thus, we have that L over F is separable. And so it's a separate normal, so thus it's a Galois extension. So we see that this is, if we have an intermediate field, we see that the big field over the intermediate field is always a Galois extension. And then we can use this theorem that we had before. If we take the fixed field, if you have a Galois extension and take the fixed field under the whole Galois group, then this is the smaller field of the extension. So by the theorem of last time, we have that if we take the fixed field of the Galois group of L over F, this is equal to F. So we had before, we had any Galois extension L over K, and the fixed field was this. But now as we find that for an intermediate field, this thing is also a Galois extension. We can apply it to the intermediate field and get this. And so we basically get for free half of the main theorem. And so now we have to deal with the other half. And that's a bit more difficult. So now let H be a subgroup of this Galois group. So G, which was the Galois group of L over K. And we have to show this second statement. So we want to show if I take the Galois group of L over the fixed field of H, this is equal to H. Then we have shown part one. So maybe write for simplicity write F for this fixed field. So now we want to again apply this wonderful theorem of the primitive element. So I don't need the statement anymore because what I need to prove is written here. So by the term of the primitive element, we can write L is equal to K of A for some element A in F. Now we write down a crazy polynomial, which actually will finally turn out to be the minimal polynomial of A over F. But let's see. So let's take polynomial F. This is defined as follows. It's a product over all elements in this group H. So that means just a fun factor for primitive H. X minus H of A. So we take all elements in the subgroup of the Galois group and we take a product over X minus this element applied to our root A. So this is a polynomial of degree, the number of elements in H. So this is by itself a polynomial in L of X of degree, number of elements in H. So we have these complicated polynomials. Now we want to say something about it. So the first thing is I want to claim that all roots of F are distinct. So I mean, no. So this large H is a subgroup of the Galois group. So if it makes you happy, I can write it phi. Then it would be phi of A. So that's an automorphism of L over K. They apply this to A. So H was, if you look at it, H is a subgroup of the Galois group. So I apply this automorphism of L to A, which is an element in L. And I take the product over all elements in this finite group and I get this polynomial of degree, the number of elements in it. So I first want to say that all roots of F in L are distinct. So the point is, this again, what we had before, so we have this simple algebraic extension K of A. Then we know that for any to read to 2 zeros of this polynomial, there exists a unique element, a unique K isomorphism, which sends one root to the next. So in particular, so this means that if phi of A is equal to C of A for two elements here, then the difference sends, so if phi of A is equal to C of A for phi C in H, then it means that phi composed with psi to the minus 1 of A is equal to A. So that means you have an element in the Galois group which sends A to A. But we know that for any two roots, for any two elements in two roots of this polynomial, there's a unique element in the Galois group which sends this element to another one. So there's a unique one which sends A to A, namely the identity. So thus phi is equal to psi for the minus 1. So it means that all these elements here are different. And so these are obviously the roots of the polynomial. And so it's like this. This polynomial has no multiple roots. All these are different elements. Yeah, yes. Yeah, you know, well, that we will see. F is a fixed coefficient. Yeah, yeah, but A is just an element of L. Just about F is equal to L. Yeah, yeah, yeah. That's correct. But in the moment that doesn't contradict anything, I say it's true that if A would be in F, then phi of A would be equal to A. But that, you know, obviously what it says is that unless L is equal to F, this cannot happen, which is also clear because A is supposed to be a primitive element. No, the field extension is supposed to generate the whole of L. So if A would lie in F, it could not possibly generate the whole field. So that's not a contradiction. OK, but you're right. If A would be in F, which cannot be, then I would be in trouble. Yes, that's what I meant. Thanks. OK, so they are all distinct. And so let's go on. Now we can write sum i equals 0 to whatever number of elements in H, bi x to the i. So we can write it as polynomial like that. And in power i, we have that the bi are elements in F. It's a polynomial in L of x. Now we want to show that actually this is a polynomial with coefficients in F. So to show F is an element in F of x, that is bi is an element in L for all i. And so let's see. So we want to take another element. So basically for this, we have to show that this polynomial is invariant under this Galois group. That the coefficients of the polynomial are invariant under the Galois group. And that, if you think of it, is almost obvious. So let's say C in element in H. So I can apply this thing, which I had, I think, denoted by C star F. This was defined to be the sum i from 0 to H. And we apply this element C in the Galois group to the coefficients. So C of bi x to the i. Now if you think of it, you can also do this by applying, if you look at this expression. This is a ring homomorphism. You obtain this by applying this psi to the zeros here. If you multiply it out, you will get precisely the coefficients. So this is equal to the product over all phi in H, as before, x minus psi of phi of A. So we can write like this. But if you look at it like this, it is clear that that is the same as F. Because if psi composed with phi is an element in H, and if I fix my psi, then these are in the same way, all elements in H, as if I just take the phi. Because multiplying with psi is a projection of H. In the group G, if I multiply by one element of the group, this is a projection of the group to itself. So this thing is equal to F. And this says nothing else. So we know that for all elements in the Galois group, for all elements in H, we have that if I apply this element to the coefficients of the polynomial, they are sent to itself. So it means these coefficients lie in the fixed field of H. So in F. So thus, F is a polynomial with coefficients in X. And obviously, we have F of A. F of A is equal to 0. So if I take the minimal polynomial of A over F, over this field F, this will divide the polynomial F. So thus, the minimal polynomial of, so which I call maybe G, of A over the intermediate field F divides the polynomial small f. So we know that L, so if I take K of A, this is equal to L. But F is contained in L. So therefore, this is also equal to F of A. Because if I just is the smallest field which contains A and F. But we know that L contains. So it's clear that these are equal. So therefore, if I take the Galois group of L over F. So obviously, the number of elements of this is equal. We have a Galois extension. We know for a Galois extension, the number of elements in the Galois group is equal to the degree of the field extension. And we have a simple algebra extension. The minimal polynomial G of A over F divides F. So thus, we have that the degree of G is smaller equal to the degree of F, which by our design was equal to the number of elements in H. So this L over F, it's a simple algebra extension. So K of A over F. The degree of the field extension is equal to the degree of the minimal polynomial, which is the degree of G. We have just seen that this is smaller equal to H. But if you think of it, this does it because clearly, what is this Galois group? This is the Galois group of L over the fixed field of H. F was the fixed field of H. So this is equal to the Galois group of L over the fixed field of H. Obviously, any element in H lies here because these are supposed to be the automorphisms of L, which are the identity on H, which are the identity on the fixed field of H. But any element of H has this property. So this certainly contains H. So thus, we have a group which contains H, but has at most as many elements as H. So it's equal to H. And this was the second thing we had to show. Incidentally, this also shows, if you think of it, that we actually have proven, as these numbers now have to be equal, we have also proven that this degree is equal. These two are equal. So it means F was actually proven that the minimal polynomial of our element A was this F. But I mean we don't need it, but that's what comes out. So this was the main part of the proof. So we have proven this bijection between the subgroups and the, and now I will very briefly, the subgroups that need to be revealed. So I'll very briefly do these, which are essentially trivial. We just have to remember our definitions. So maybe I write again, so we do two. So I have left it here so that we can see it. So let again F equal to the fixed field of H. So we have seen that L over F is a Galois extension. It's true for any intermediate field. So thus we know that the degree of L over F, so this, is equal to the number of elements in the Galois group. That's for Galois extension, we have this. So F was fixed of H, which we have seen to be this group was equal to H. So this is this thing. And then, obviously, this thing is just dividing. So note that if I take L over F, so just the degree theorem, times F over K is equal to the degree of L over K, which is, you know, this is also a Galois extension. The Galois group is G, which is equal to the number of elements. So the, so, so the degree of the Galois extension is equal to the number of elements in the Galois group. And L over K was equal to G. And so if you put this together, we get, we just divide by this, and we get, get the second state. And number three is, you know, as, you know, similarly straightforward, maybe even, even simpler. So again, the first formula says that standard, you know, we know that as L over F is Galois, so it follows that the degree of L over F is equal to the number of elements in the Galois group. This was a theorem. And then, again, we have the same formula. In order to get the rest, we have again that L over F times F over K is equal to L over K. I hope it's right. It's equal to G. And so you can just, again, I hope that's correct. Yeah, you can again just divide. Okay. So maybe, so these are just trivial. But so the real difficulty is the main thing. But, you know, it's still useful to, to have always these formulas, you know, it's just basically the degree theorem that we use forwards and backwards. But what, what, what, I mean this was supposed to be like this. Okay, maybe I will stop here and kind of, I, yeah, if you, I have to be in the meeting very soon, but if you want, I can also stay here until the last moment if, but I think somebody will go on strike. So, okay.