 OK, maybe I will just proceed. So we were talking about normal extensions. So a normal extension was an algebraic extension such that every irreducible, so just recall, so that means a normal extension. This is an algebraic extension such that every irreducible polynomial, so if I call this L over k, such that every irreducible polynomial f with coefficients in the smaller field, which has 0 in the bigger field, splits into linear factors over L, splits over L. This is somewhat strange definition, if it doesn't look very likely to happen. But we had this theorem that said that, so the extension, a finite extension L over k is normal if and only if L is the splitting field of a polynomial with coefficients in k, if and only if L is a splitting field of a polynomial in kx. So in some sense, it says a strange thing. So if there's one polynomial in kx of which is the splitting field, so which splits, then all polynomials, which are irreducible in kx and have a 0 in L, will already split and also the converse. So we had started to prove. So obviously, it's kind of clear that the difficult direction is this one. And this we had already shown. And it was a somewhat tricky proof. And now the other direction is quite simple. It's kind of the trivial direction. So we take a finite normal extension. So we have to show that L is a splitting field of some polynomial. So we can write equal to the k, adjoint some elements a1. So we know that the finite algebraic extensions are the finitely generated algebraic extensions. So there are finitely many elements ai in L, such that L can be written like this. So where the ai in L are algebraic over k. So let fi be the minimal polynomial of this element ai over k. Then this minimal polynomial has obviously a 0 in L, namely the ai. And so it splits because this is a normal extension. So fi has a 0 in the ai in L. Thus, ai fi splits over L into linear factors. And now we put f to be the product of the fi of L. So then, obviously, it splits. And obviously, and clearly, k a1 to an is the splitting field or splitting field of f over k. Because the polynomial splits over it. And it is obtained from k by adjoining some of the roots of this polynomial. We had seen that if we have a field over which this thing splits, and it's just obtained by adding some of the roots, then it's the splitting field. Our roots are 0s of f. And so we have, in this kind of stupid way, we have obtained that L is indeed the splitting field of a polynomial. We just take any set of elements which generate the field extension. And we take the product of their minimal polynomial. OK. So that was that. Now I want to very briefly introduce the characteristic of a field. So this, for the moment, is all I say about normal extensions. Now I talk about the characteristic of a field. It's just this thing that if you have an element n in z bigger than 0, bigger than 0, and we have k field, then we can, we have written n times 1. So 1 is the element 1 in the field to be 1 plus 1 n times. So we had introduced this for any ring, this notation, for any element a. So now it can happen. Sometimes we would have, for some fields, we have n times 1 is equal to 0 in k for some n, a positive integer. For instance, we have seen, so if p is a prime, prime number in the integers, then we had seen that we had this set zp, which as a field we also call fp. So these are just the numbers modulo p. So as elements, these are elements 0, 1 to p minus 1. You add them by adding them and taking the rest. So we had seen that this is a field. And if we take n times 1, so p times 1, this is what we had denoted like this. So we take p and take the rest of division of p by p, which is equal to 0. So if we add 1 p times, we get 0 in this field. So we see that in this case there is some such number. On the other hand, obviously in the rational numbers, we have n times 1 is never equal to 0 for n bigger than 0. So we want to give a name to this phenomenon. So the smallest number, the smallest positive integer for which n times 1 is equal to 0 in the field is called the characteristic of the field. And we say the characteristic is 0 if there is no such number. So definition, let k be a field. So the characteristic of k is the smallest positive integer with n times 1, which means 1 plus 1 is equal to 0 if such an n exists. Otherwise, it is equal to 0. It is defined to be 0. And we denote the characteristic by characteristic of k. So we have just seen that the characteristic of the rational numbers is equal to 0. This will also be 2, for instance, for the real numbers. And the characteristic of the finite field, fp, well, it's easy to see that this will be equal to p. Obviously, for no smaller number, positive integer than p, we will get 0. All the other ones are different. So now we want to make a few remarks about this point of remark. First, the characteristic of a field is either 0 or prime number. So the characteristic of k is 0 or a prime number. And the second one, which is even easier, is that if l over k is a field extension, then the characteristic of l is equal to the characteristic of k. So it's invariant undertaking field exception. So this is quite simple. So we take i to be the set of all integers such that n times 1 is equal to 0 in k. So if I add 1 n times in k, I get 0. So this is an ideal. It's easy to see. So i is an ideal in z. Now if we add two of these, we all take the difference. It would still be the same by the distributive law. It will still get 0. And if you multiply by an integer, if this product was 0 and you multiply an integer, it will still be 0. So it's an ideal. And note that it's easy to check that by the distributive law, we have if we take n times 1 multiplied. So this means n times 1 means you add 1 to itself n times. This is here the multiplication in k here, m times 1. And this will, it's easy to check that this is the same as you take the product in z and add 1 times. It's an easy exercise. Follows from the distributive law, OK? You can check this. That this will hold, but just by this standard properties of multiplication addition. So thus it follows. So thus if n times m times 1 is equal to 0 in k, then it means that this product is equal to 0 in k. And here we're in a field. So the product of two elements in the field can only be 0 if one of them is equal to 0. So then it follows that n times 1 is equal to 0 or m times 1 is equal to 0. So that means, therefore, that this ideal i is a prime ideal. i in z is a prime ideal. And it's easy to see that we know all the prime ideals in the integers. So the integers are principal ideal domain. So every ideal is generated by one element. If this element is plus minus prime number, then we get a prime ideal. And also if the element is 0. And if the element is not a prime number, so it's a product of two primes, we don't get a prime ideal. So it's a prime ideal. And this, so we know that, therefore, i is equal to the 0 ideal or i is equal to the ideal generated by p, where p is a prime number. So let me repeat it. Again, now I explained it. So z is a principal ideal domain. So every ideal is the ideal generated by one number. And we can take this number to be non-negative because an element and a number and minus the number generate the same ideal. So if this is a number which is a product of, which is not a prime number, a positive number which is not a prime number, so n times m, you can see that by formula like this, you find that the ideal is not a prime ideal. So however, therefore, the only thing that rests is that the number is 0 or prime number. And then it is obviously prime ideal. So it means that we have this. And this precisely says, so in this case, the characteristic is 0. In this case, the characteristic is p. So you can also see that there's some justification for saying characteristic 0 because really means this ideal is generated by 0. And here, the ideal is generated by p. So the number 0 also comes here, whereas in the original definition, it is if there's no number. But in this case, the number is 0. And the second statement is trivial. So what do we have here? If we have a field extension, the element 1 in L is the same as the element 1 in k. So 1 element L is the same element as 1 element k. So therefore, if you add it, n times, you stay in k. So therefore, this n times 1 is the same element in L and k. So the condition that this element is 0 is the same whether you're in L or in k because it's the same element you're talking about. After all, k is a subset of L. So we had talked the last time a little bit this time about normal extensions. I'd said that there are two crucial properties that we want field extensions to have in order to be somehow nice. Namely, they should be normal and separable. And if they are normal and separable, we call them Galois extensions. And these are the field extensions we want to study. And so I have introduced what normal extensions are. And so now I have to introduce what separable extensions are. Again, it comes with a somewhat crazy definition like the normal extensions did. But in this case, the situation is in some sense even better. Under reasonable assumptions, all field extensions are separable so that you don't even have to worry about anything at all, at least in characteristics 0. So let's see. Separable extensions. So separable extensions, the condition has to do with multiple roots. So if you have a polynomial f in kx, it will split over some extension field. And it will have a number of 0s. It can happen that some of these 0s coincide. So if you write it as a product of linear factors, then there is x minus a1 times x minus a2 and so on, x minus aN, it might be that a1 is equal to a2. So it's x minus a1 squared plus and so on. So in some sense, the 0, a1 occurs twice. And so we will see that when this happens for our field extensions, for polynomials which were irreducible before and then we extend it, then they get multiple roots, that this causes problems. And we don't want these problems to be there. And so we will exclude it by definition. So we want to exclude certain ways how multiple roots can occur. And for this, we make the definition of a separable extension. So let me make the definition. So first I have to want to define again what a multiple root is. So let f be a polynomial to coefficient in some field k. And l, an extension of k over which f splits. We have f is equal to say b times x minus a1 times x minus aN. So then I can write here m1 to mN. So we can write this product of linear factors. They might repeat themselves. So where the mi are some positive integers and the ai are some elements of l. So then as we know, the ai are the zeros of f, but we also call them the roots of f. And an element, so if the number m1 is equal to 1, we call a, so if mi, if this power here is equal to 1, we call ai be a simple root. And if it is bigger, then we call it a multiple root of order this number mi, order mi. OK, so this is the definition. And now we want to see what a separable extension is in terms of these multiple roots. The definition looks formally somewhat similar to how we did for the normal extensions. So the definition, an algebraic field extension, again, l over k is called separable if the following holds. If every irreducible polynomial, again with coefficients in the smaller field, which has a 0 in l, does not have multiple roots. And in fact, it does not have multiple roots anywhere. That means it doesn't have multiple roots in its splitting field, OK? So it can be written as a product like this, where the powers are all 1 if you make the field big enough so that all the 0s exist. So this is the definition. As you see, it's a bit involved. But anyway, so the statement is that if you have an irreducible polynomial which has a 0 in this field, then it does not anywhere have multiple roots. So in order to make, so, and then just to complete it, so if the field k is such that every algebraic field extension of it is separable, we call the field perfect. Every, so if for given k, all algebraic field extensions are separable, k is called perfect, OK? So now this is somewhat, so we have this, the definition is in terms of these multiple roots. So in order to do anything with this, we have to have a criterion for the existence of multiple roots. We have to have a way to check whether a polynomial has multiple roots somewhere or not. And this is something that I expect you know about from calculus or high school or whatever, that a polynomial has a multiple roots. So a is a multiple 0 of a polynomial if it's a 0 of the polynomial and of the derivative. So you can easily just check by, and so the same we can do here where we define the derivative in the formal way by just using the formula from calculus. So we, where is this? So we have a criterion for multiple roots, which will be in terms of the derivative. So let me formally define the derivative of polynomial. So obviously not by calculus, but by just formally writing the formula via any field, not in the real numbers. So definition, so let f be a polynomial. So I can write it as sum i equals 0 to n, ai x to the i be a polynomial with coefficients in k. So the derivative of f is defined by the formula that one would prove in calculus if this was a polynomial with real coefficients is f prime. Well, everybody knows what one gets. Sum i equals 1 to n, i times ai times x to the i minus 1. That's what you learn. So and this is obviously also a polynomial. So this is just our definition. And it's easy to check as an exercise that the rules that one knows for the derivative for functions also hold here for these polynomials just by doing the computation. Namely, so if a, b are some elements in k and f and g are polynomials, then I have that if I take a f plus b g and take its derivative, this is a f prime plus b g prime. I mean, I expect everybody of you can prove that from this formula. And also maybe I should also say if a is a constant, then the derivative is 0. This is even more obvious if you look at the formula. And we also have the, what's it called, the Leibniz rule, which one has to prove somehow by induction or something. I mean, anyway, it's an easy exercise. So if I take f times g prime, this is f times g prime plus f prime times g. I think everybody of you would be able to use this definition to prove these statements quite easily. So now we want to also prove this criterion that one has in calculus that in multiple root you have when it's a 0 of f and of f prime. So lemma, so let f in polynomial in kx, a non-constant polynomial, and that l over k be a splitting field of k, actually any field over which it splits anyway, of k. Then a in l is a multiple root of f, f and only f. f of a is equal to 0, and f prime of a is equal to 0. So obviously, in order to be a root, we must have that f of a is equal to 0. So let's just see. Let r be the order of the root a. So r could also be 0, but anyway. So then we know that f, then we can write f equal to x minus a to the r times g, where g is a polynomial in l of x, which does not vanish at a. That's how you find it. If g would be 0 at a, it is divisible by x minus a, one more power. So the order is precise, obtained in such a way. f can be written as x minus a to the r times g, with g of a is different from 0. And now let's take the derivative of this equation. So then, if I take f prime, this is equal by this Leibniz rule, and another rule that you can also easily check. If you take x minus, these are all the things that you know from high school. Anyway, so if you take the derivative of this, then it follows from the definition that, again, the usual rule applies. Namely, if you take the derivative of this, this is x minus a to the r. So this is r times x minus a to the r minus 1 times g. So you take the derivative. So this is the derivative of this, and then times g plus the derivative of the other side. And so you see that this part is divisible by x minus 1, x minus a to the r, and this one is precisely visible by x minus a to the r minus 1. So you find that the whole sum, the highest power of x minus a, which divides it, is r minus 1. If I put in this part, if I divide this by, I can divide this by x minus a to the r minus 1. So I get r times g. And if I apply this to a, I get something non-zero. And here I get x minus a times g prime. If I apply this to a, I get 0. So this is how I get it. And so we see that r is bigger than 1, if and only if f prime has a 0. OK. So now we want to, so this is all a little bit kind of obvious. I mean, you know that this is the case. Now we want to come to some consequences which are a little bit, I mean, if one doesn't now applies this in the context of our field extension, we find that something slightly non-obvious happens. I mean, if you are, so that it almost looks important, that it seems pretty clear that normally a field extension must be separable. Because this one can see further in the following theorem. Let f in kx be irreducible. Then f has no multiple roots in its splitting field, if and only if. If I take the derivative of f, this is not the 0 polynomial. So it's not equal to 0, the 0 polynomial. So that's a rather big step. Here we said a multiple root is if you have a 0 of f and the derivative. Now we say in this situation where we start with an irreducible polynomial and we ask about the multiple roots in the splitting field, this can only happen if the derivative is the 0 polynomial. Now, so maybe I should say, normally you would think if you are working over the real numbers, if you take the derivative of polynomial, it can only be 0 if the polynomial is a constant. But this is not quite clear to here, because if the characteristic of the polynomial is not 0, it can happen also polynomial of positive degree that the derivative is 0. But on the other hand, if the characteristic is 0, it cannot happen. So this theorem will immediately apply that every field of characteristic 0 is perfect. So that in particular, this condition of separability is not usually such a strong one. So let's prove it. It's not so difficult. So we take a splitting field of f. Clearly, if our derivative is equal to 0, so if f prime is the 0 polynomial, then all roots, then by what we have seen here, all roots of f in L are multiple roots. And as we have a polynomial which is irreducible in particular of positive degree, it has some roots in its splitting field. And so it has some multiple roots. So this is the trivial direction. So this is obvious, because every root will be multiple root if f is equal to 0. f prime is equal to 0. Now, f is identically equal to 0. What? It's equal to 0. But 0 as a polynomial, not that f prime of a is equal to 0, but the derivative is the 0 polynomial. So it has multiple roots if and only if the derivative is the 0 polynomial. Now, what? Oh, sorry, sorry, sorry. Yeah. OK. I didn't look what I had written. Obviously, you are right. So which way around do I want to formulate it? Yeah. So in fact, I mean, I was sure of what I meant, but I wrote precisely the opposite. So obviously, this is if and only if it is not identically equal to 0. So sorry. Obviously, the rest of the other round would be nonsense. Thank you. OK. So equivalently, it has multiple roots if and only if this is identically equal to 0. So one of the two has to be negated. OK. So now let's do the non-travel direction. So assume we have a multiple root in L. So that means we have that f of a is equal to f prime of a is equal to 0. But that means no f is irreducible. And so up to multiplying by non-zero constant, it means that f is the minimal polynomial of a. So as f is irreducible to multiplying. Yes? Ha ha ha ha ha ha. Yeah. OK. Yeah, so basically, all the signs were the other round. But anyway, what? No, but in the theorem now, I thought what we did it. It has no multiple roots if and only f prime is not identically zero. I think now it's correct. Yeah. But obviously, I could have said it has multiple roots if and only if that would have been here. I have twice said no. Maybe that was too complicated for me. And so f is irreducible up to multiplying. And so it follows that up to multiplying by constant in k, so to alpha in k, we have that f is the minimal polynomial of a over k. And f prime, the minimal polynomial, it means that this f generates the ideal of all polynomials which vanish at a. So it means that f prime also is this in this ideal. So thus, f prime is in the ideal generated by f. That means f divides f prime. But this now is rather unlikely because if f prime is not zero, it follows that the degree of f prime is smaller than the degree of f just by the definition that I wiped out of the derivative. You can see that the degree of the derivative is one less than the degree of f. So if f prime is not identically zero, then f prime, the degree of f prime, is smaller than the degree of f. But we know that when we multiply two polynomials, the degree of the product is a sum. So this is impossible. If something is divisible by f, then its degree must be at least bigger equal to f. So this is a contradiction. So the only way out is that f prime is equal to zero. So it's not particularly difficult. And so as I said, it follows as a corollary every field k of characteristic zero is perfect. And this is because, as I said, so if the characteristic, this is because if the characteristic of k is equal to zero, then if f is a polynomial in kx, which is not constant, then f prime is not equal to zero. Just from the definition, if f is equal to sum a i x to the i, and we can assume, so a i is in k, we can assume that a n is different from zero by only taking the then f prime is equal sum i equal zero, equal one to n i times a i x to the i minus one. And now the leading term is n times a n. Because if you can easily see if the characteristic not zero, then for any element in k, once you, the statement for one, that n times one is equal to zero, if n times one is never equal to zero, then n times a is never equal to zero for non-zero element in k. Maybe I should have proven that. Anyway, so thus f prime is not zero. So on the other hand, for instance, if you take the polynomial x to the p minus whatever, p squared minus x to the p. So if I take the derivative of this thing, then in fp, this will be zero. So in, so if I view this as a polynomial as of fp of x, then the derivative of this, or maybe whatever, and 2p, so this is the element in fp, which is p times 2t, p times one, but p times one is equal to zero. So this is zero plus zero is equal to zero. So certainly can happen in characteristic p that the derivative of a polynomial is zero, even if the polynomial is not constant. OK. So this is, and so we see, I mean, something which we will not prove, which is also true, is that, however, also every finite field is perfect. And you have just seen that this criterion somehow obviously fails, but one can prove that also every finite field is perfect. It's not in the notes. But here, we only prove that every field of characteristic zero is perfect. But there are some fields of characteristic p, which are not finite, which will not be perfect, but we will not talk about that. OK. So now we want to prove a real theorem. So we have, in some sense, one of the main reasons why we have introduced this separable extensions is that we can prove the following theorem, which is the theorem of the primitive element. So it says that if we have a finite separable field extension, then it is simple algebraic extension. So remember that in the beginning, we studied simple algebraic extensions, which are extensions by just adding one algebraic element. And we had been able to completely describe them. And so now we find that this description holds, in some sense, for instance, in characteristic zero, for all field extensions. Because if you have a finite, for all finite field extensions, because if you have a finite several field extension, it is a simple algebraic extension. So if we have a field extension, which is obtained by adding a finite number of algebraic elements, then we can just find one element so that the field extension is given by adding just this one element. Obviously not necessarily one of the elements we added, but some combination of them. And so I will try to prove that. So this is a theorem of the primitive element. So statement is, let L over K be a finite separable extension. So for instance, in characteristic zero, it could be just a finite extension, finite field extension. Then it is a simple extension. So that means there exists an element A in L, such that L can be obtained by K by just adding this one element A. So we might have constructed L by adding many elements to K, but we can find one which is sufficiently complicated to give us the whole of L if we add it. OK, so let's see. So this is, we want to prove that. So we will make a simplifying assumption, which is not necessary, but which simplifies the proof. So I don't prove it quite in this generality. So for simplicity, we assume that K is infinite. So this statement proves also if K is a finite field, but for one thing it's anyway somewhat special case. And secondly, the proof that I give will not work for a finite field. One needs a different proof. This proof is simpler for infinite fields. So for instance, if K is something like Q, then the proof would be OK. So for instance, if the characteristic of K is equal to 0, then it's clear that K is infinite. And if the characteristic is P, then we still want it to be infinite. So anyway, as we will see, the proof is somewhat difficult, a little bit with trick. So let's see how we do it. So we have that. So we start. So we know that L over K is a finite extension. So it follows it's a finite algebraic extension. Because if it's finite, it must be algebraic. So we have that we know that we can write L equal to K. We join some elements A1 to An with A i and L. Now we know that the finite extensions are the finitely generated algebraic extensions. So if we have a finite extension, we can find finitely many elements so that if we join them to K, we get L. And we'll have to show that we can somehow change these elements so that in the end, we only need one. We do this so we want to, anyway. So we will do our proof by induction on N. So we want to prove that we can do it with one element. We want to prove this by induction on N. So we have L, which is equal to K A1 to An. So by induction on the number of generators, we have that. So by the induction, so if N is equal to 1, we are done. We have nothing to prove. So by induction, we have K A1 to An minus 1 is equal to K of A for some element A in K A1 to An minus 1. We know if the induction hypothesis, which we can use here, is that if we have N minus 1, if we join N minus 1 elements, we can do this by just adjoining 1. So thus, L, which is K A1 to An, is K A comma An. So which means that by using induction, we can assume that the number N that we have here is equal to 2. So we can reduce to the case that we join just two elements. Thus, we can assume L is equal to, say, K AB. And to show that exists an element C in K AB, such that this element K AB is equal to K of C. So using induction, we have reduced to the case that we have only two, the joint, only two elements. And actually, what we will show, we will show something slightly stronger, which will, for instance, not hold in the case of finite field. So this is where the proof is different. So in fact, we can choose C equal to A plus Z times B for suitable element Z in K. So that means we can write, we can find a linear combination of these two with coefficients in K such that this thing is generated by this linear combination alone. And now we have to somehow find this linear combination. So now we go to some field where, let me see. So let first let F be the minimal polynomial of A over K, and G be the minimal polynomial of B over K. And now we choose ourselves a field where both F and G split into linear factors. We choose ourselves a field extension of KC where both F and G are product of linear factors. So we can, for instance, first take the splitting field of F over K of C, not K of C. We don't have C yet over K of A. So let me write it a bit better. F over L be a field extension such that both F and G, so L is this KAB after all, so both F and G split over into linear factors over F. So as I said, we can do this by taking first the splitting field of F over L. And then, so this will be some field F prime. And then we take the splitting field of G over F prime, and this is our F. Over this field, both F and G will split into linear factors. So now we want to write, so we write down the roots, so zeros of F and G over this field F, large F. So X1 until Xn be the roots of, say, the polynomial F over the field large F. And one of the roots is this A. We take the first one to be A. And we also take the roots of G over the field large F, call them Y1 to Ym. And the first one, we take B. We know that it splits into linear factors. One of the zeros of it is B, so we can take this to be this one. So we have all these roots. And now is the place where we use that our field extension is separable. The fact that the field extension is separable means that as F and G were minimal polynomials, they were irreducible to begin with. And they have a 0 here. So it means all their zeros are different. So the Ai, so A is different from all the Xi, and B is different from all the Yi, except for 1 and 1 here. So as L over K is a separable extension, we find that all the roots, so find the Xi, are pairwise distinct. So I should say, I take the roots here, I count them with multiplicity. So if I would have a multiple root, I write it four times. If A is a root of order 4, then so it means what I really mean that I can write. So I maybe haven't so with multiplicity. So maybe I will, as it doesn't matter so much, I will just write it one more carefully. So that means that F is equal to X minus X1 times X minus Xn. That means that it's the roots. And G is equal to X minus Y1 times X minus Yn. And then the statement is that if I write it like this, then all the Xi are distinct among each other, because that means you have no multiple roots and all the Yi are distinct, Ym. So as L over K is separable, we have that all the Xi are pairwise distinct and all the Yi are pairwise distinct. Because as I said, F was an irreducible polynomial over KX, and so the roots over the splitting field are all simple, because we have several extensions. So until now, we don't seem to have done very much. And now I want to find this Z. We'll do this as follows. We use that they are all distinct. So we can write, so let me first write it down. So thus, for E, first I write down, then we check it, 1 to n. And J from 2 to m, we have that ZiJ equal to Xi minus A divided by B minus Yj is the unique element field F, such that I write like this, A plus ZiJ times B is equal to Xi plus ZiJ times Yj. Now this looks a bit difficult to adjust first, but in some sense, it is completely obvious. If you look at this formula, if I want to solve this, A plus ZiJB equal to this, so I can bring something on the other side. So this is equal if and only if A, maybe I can write X. So I put this to the other side. So Xi minus A, if I subtract the A, is equal to ZiJ times B minus ZiJ times Yj. Just have this equation. I bring things on the other side. So this statement is equivalent to that, that too. I think also, which means that ZiJ times B minus Yj, so it's equivalent to this being equal to Xi minus A. So we know that all the roots of all these Yj's are distinct from each other. Y1 is equal to B. So all the Yj with j bigger than 1, this difference is not equal to 0, so I can divide by it. So this is equivalent because we know that B minus Yj is not 0 for j different from 2. And we have assumed there that j is different from 1. This is equivalent to ZiJ, as I said, is equal to Xi minus A divided by B minus Yj. So I've now just checked the statement which I had made. This element ZiJ is the unique element of F, which such that this equation is satisfied. I've given you a proof. So now we use this assumption that our field k is supposed to be infinite. So we have for all pairs i and j, we have one such element ZiJ in F. So altogether, these are at most m minus 1 times n elements. So our field k has infinitely many elements. So it certainly has an element which is not one of these. So let Zi in k be different from all the ZiJ. So as I said, i from 1 to n, j from 2 to n. So we can just find an element. This is possible as k is infinite. And now the claim is, so now we put. So thus if we take A plus Z times B, then this is different from Xi plus ZiJ for all i from 1 to n and all j from 2 to m. This is how we have chosen. So now the claim is that this is enough to make sure that if I put C equal to A plus Z times B that this will be equal, will generate the same field extension. So let C be equal to A plus Z times B. And the claim is that kC is equal to k of AB, which was our field F. Now obviously, this inclusion kC is contained in kAB is clear because C is an element in kAB. After all, it's just a linear combination of A and B. So what we have to prove is the other inclusion. And so for that, it is enough to show that both A and B are contained in kC. So we will now want to show that B is contained in kC. Then for A it's trivial by the formula. So I want to see, show B is an element in kC. I mean, we do this again with trick. So we write down some crazy polynomial. We take a polynomial H with coefficients in kC. So let B be defined by, so we put H of X equal to F of C minus Z times A. That's X. So you replace, you know, you have a polynomial, some AI X to the I, and you place the X by this. So you get some AI listing to the I. And you can see that this would be a polynomial with coefficients in kC. And now what can we see? So one thing that we find, what is the value of H at B? While we just put it, this is F of C minus Z times B. Now, C was this one, so if I subtract Z times B from it, I get A. And F was a minimal polynomial of A, this is equal to 0. So we find that B is a 0 of H. So thus, it follows. So X minus B is a divisor of H. And we want to use this fact that it's a divisor to show that H, that this B actually is in kC. So we know, obviously we also know that X minus B is a divisor of G, because G was the minimal polynomial of B, so B is a 0 of it. So we find that this is a common divisor. We want to show that X minus B is a greatest common divisor of G and H, so say in F of X. So whatever field they both split. And when we have done this, we will see that this will prove it. This is actually quite simple, because G splits over F into linear factors. So if you have a greatest common divisor, it must be a product of these linear factors up to multiplying by a constant. So a divisor of G is, so up to multiplying by a constant, a product of these linear factors, of some of these linear factors. Remember that these linear factors were X minus YI from I from 1 to M. So the greatest common divisor must be a product of some of these. But such a thing can only be a factor of H if these YIs are 0s of H. But X minus YI, maybe I had denoted them by J, the index just to keep it as before, is a 0 of H, if and only if H of YJ is equal to 0. So but what is H of YJ? I mean, you have, after all, defined H in this way. So anyway, I wiped out the definition of Z, but I expect you. So H of YJ, by our definition, is equal to F of C minus Z. And we had precisely defined C, this Z in such a way that, so which means maybe I write it out as F of plus Z minus Z times YJ, is that correct? So we had defined our Z precisely in such a way that this can never be equal to a 0 of F. So by definition of Z, we have that this thing, A plus ZB minus Z times YJ, is different from XI for all 0s, XI of F in the field F. So that means this is not 0. So this will never be 0. So that means none of these factors divides our H. So the only factor, the only kind of factor of G, which also divides H, is this one, X minus B. And this means it is the greatest common divisor, is a monic, I mean, it's monic, as you can see, greatest common divisor H and G. Now you have to see, now if you remember, we had this statement. If we have, after doing the Euclidean algorithm, as a consequence of this Euclidean algorithm, I proved that if you have two polynomials with coefficients in a field and you take their greatest common divisor, their monic greatest common divisor with coefficients in a bigger field, then in fact the polynomial will lie in the smaller field. So in this case, we have so as H and G are both in KC of X, because in fact G, after all, is a polynomial with coefficients in K, so it's even in K of X. It follows that their greatest common divisor, if required to be monic, has coefficients also in KC of X, is in KC of X, if we reply this to be monic. So it means that X minus B is a polynomial in KC of X. And now obviously that means just that B is an element in KC. And then obviously we are done because C is equal to A plus ZB. So C is equal to A plus ZB. So in other words, A is equal to, now this is very difficult, ZB minus C. So this is an element in K. This we now know is an element in KC. And this is also an element in KC. So it follows the whole thing is an element in KC. So you find both A and B lie in KC. So plus KAB is contained in KC. So that's, anyway, that proves it. It's a bit difficult to understand what somehow the idea of the proof is. I mean, it's somehow, also if somebody would ask you to prove this theorem, you would find that rather difficult to achieve and so would I. So somehow, anyway, the trick is maybe the story that you want to see that this X minus B is a greatest common divisor of these two polynomials. And this will, then once you have that by some rather trivial fact, it follows that B is in KC. And in order to find that this is the greatest common divisor, you have to somehow achieve that, you know, it must, you want to make it the greatest common divisor of two polynomials in KC. You have to make this kind of strange trick of kind of cooking up this H in order to make the greatest common divisor. And then you want to choose an element here with the property that you have, that they have precisely one zero together, namely B. And for this, you have to choose this Z appropriately in this way. So it's a, you know, it's several tricks and one, you know, it's not the kind of natural proof you are given the problem. There's an obvious thing you want to do and you do it. It's just, you know, once somebody has done it, one can understand it, but one still doesn't quite understand why one would think about it this way. Anyway, it's true. And we have proven it. Okay. Ah, the time is up. So I now wanted to start something else, but then I won't. Okay. So that was as much as I wanted to say now about separable extensions. So next time we will first briefly talk about finite fields. So we'll classify all finite fields up to isomorphism. And then we will go on with a study of studying field extensions. We will really go into Galois theory. We'll introduce the Galois groups and then kind of slowly come to try to start to the proof of the main theorem of Galois theory. Okay.