 So, good afternoon. So, we will go on with the lecture. So, today we will talk about, so last time we finished talking about algebraic closure, but now we want to talk about splitting fields. So, so that is, you know, if I have a, if K is a field and F is a polynomial with coefficients in X, we are interested in a field where F splits into a product of linear factors. And we want to find somehow the smallest such field, smallest field over which F splits into linear factors. This will be called the splitting field. And we will see that it's uniquely determined up to isomorphism, up to K isomorphism. So, first we want to see that if we are given a polynomial, F and KX, we can find a field where F has a zero. So, a field extension where F, so the smallest field extensions in here. So, that we can find the field extension of K over which F has a zero. And we have in some sense seen this before, but let me state it. So, let F be again our polynomial in KX, and we want it to be irreducible. Then there exists a simple algebraic extension. So, maybe I call it F over K such that over F, maybe I call it L, but K such that over L, F has a zero. Such that F has a zero in F. And we can also say what the degree of the field extension is. And we have the degree of L over K is equal to the degree of F. Well, in some sense, so we just do again what we did before. We can use our F to construct this field. Namely, we will find that as this thing is irreducible, it's a maximal ideal in KX and we can take a quotient by this ideal generated by F by this maximal ideal. This will be a field and this will be the field we want. So, the only maybe funny thing is why F will have a zero over this field? So, proof. So, we want to use F to construct the field in the usual way. So, we know that F is irreducible that we had assumed. So, it means that F is a maximal ideal in the principal ideal domain KX. And so, we have used this many times. So, thus, if I put F to be the quotient, no, it was L, KX modeled by F, this is a field. And as we had seen before, we can identify K as the subfield of L consisting of the constant of the classes of the constant polynomials. So, it's the field, so L over K is a field extension. And so, now the claim is that our polynomial F has a zero in this field. Yeah, we have to somehow see what the zero is and I claim the zero is namely the class of the polynomial X. So, we have a, you know, we look at, so this is equivalence classes of polynomials, modulo this ideal. We take the equivalence class of the polynomial X and we claim that this equivalence class of the polynomial X is a zero of this polynomial. And this is actually essentially, by definition, we know that zero is equal to the class of F, because after all, we have divided by it. And if you write F equal to sum I equals zero to N, say, let's say N is the degree of F AI X to the I. Then, this means we have, this is sum I equals zero. To end this, taking the equivalence class, this is a ring homomorphism. So, we can write this as sum AI X to the I. And you have to remember that a constant polynomial is identified with a corresponding number in K. So, this is just sum I equals zero to N AI X to the I. And, you know, you see precisely that this says that the polynomial X is a zero of F. So, let me write this out. So, this is equal to F of X plus X is a zero of F. And clearly, so it's somehow we have done this in this very kind of total logical way. We have created this zero, this field extension, where this thing has zero kind of out of nowhere, just using the polynomial. So, and it is clear that L is equal to K, the class of X. L is equal to K, the class of X. Because every element in this quotient field is the equivalence class under this relation of a polynomial. And so, this means it's the same as the, you know, the polynomial in the equivalence class of X. Because for G, an element in KX, we have the class of G is equal to G of X by the same computation as here. So, finally, we want to see that the degree of this field extension is equal to the degree of F. So, the point is that F is irreducible. And by over KX, by dividing by the leading coefficient, we can assume F is monic. So, we have an irreducible polynomial which has a zero, an irreducible monic polynomial which has a zero in this extension. So, it means it's the minimal polynomial. So, thus, F is the minimal polynomial of X. And so, we know that the degree of the field extension by adding an element is equal to the degree of the minimal polynomial by some result that we had before. So, in this situation, so remark, you don't think I... So, in this situation where we just make the field extension in this way, we say we have formally adjoined a root of the polynomial. We say formally adjoined a root. The root is another word for a zero, in this case, of F. So, it's kind of... We say formally because we don't need another field in which this thing has a zero. We just create this other field by this construction. Okay. So, in particular, we find that... We have again our field F in kx polynomial. I think we don't... Off-degree n. Then... So, bigger than zero. Then, there exists a field extension L over k at most m and such that F has a zero in F. So, here, the previous result was... We find this when F is an irreducible polynomial. If F is not irreducible, we take any irreducible factor of F. And according to this, this will have a zero in an extension of the degree at most... The degree of that irreducible factor is at most the degree of the polynomial. So, it's a trivial consequence. Now, we want to... So, this means that if we are given a polynomial, we can kind of adjoin roots. So, we first adjoin a root, and then we divide by x minus this root. So, if we have a polynomial of degree one less, it adjoin roots until this thing splits into linear factors. So, there will be a field extension where F splits into linear factors, and we take the smallest touch and call it the splitting field of F. So, definition. So, let F in kx be a polynomial of some degree. The finite extension, say, L over k is called a splitting field of F over k. If the following two properties hold... So, first, it should... F should split. So, F splits over k over L into linear factors, which, as you remember, means that we can write F equal to b times x minus a1 times and so on until x minus an, where the b and the ai are elements of L. And, as I said, the second statement is that L should be, in a suitable sense, minimal with this property. So, F does not split over any intermediate field of L over k into linear factors. So, second, F does not split over any intermediate field. Remember that it is a field which is a field extension of k, which is contained in L. So, now, we want to show that such splitting fields always exist, that it's kind of already intelligent and some simple properties how one can check that something is a splitting field. So, for simplicity, I now will assume that my polynomial is always monic. I can always achieve this by dividing by the leading coefficient. So, it somehow just simplifies my notation if I do that. So, let F in x be a monic polynomial. So, the first statement is if L over k is a field extension such that it's, in some sense, these are all kind of rather trivial statements, anyway, such that F splits over L into linear factors. So, that means we can write F equal to x minus a1 times x minus n. We assume still that the degree is n. Then, if I take the field extension, so the extension which lies inside L is bigger than k, which I obtained, but just joining the roots, I get a splitting field. So, if I take k a1 to an is a splitting field of F over k. In some sense, if you think of it, it's completely obvious, but anyway, we will prove it explicitly. So, the second statement is there exists a splitting field for our F. So, maybe you should again fix the degree here of degree n. So, there exists a splitting field of this polynomial. So, L over k of F of degree. So, the degree of the extension is at most L over k is smaller equal to n factorial. So, if n is the degree, then the degree of the splitting field is at most n factorial. And the third statement is again quite simple. So, if I have a splitting field, so L over k, the splitting field of F of F over k, and let say F in the intermediate field, so of this extension, so F lies between k and L. It's a subfield and L contains k as a subfield. Then L is also a splitting field of F over the field F. So, these are all very simple facts. It's kind of an exercise to see this. But I will still carry out the details. So, let's take our field extension here as given by one. So, let L over k be such a field extension, such that F splits over F. So, these A i are elements of F. So, we want to show that this is a splitting field. Clearly, F splits over k A 1 to A n into linear factors. Because, you know, after all the elements are here, no, we have to, you know, we have to, you know, that it splits here means, you know, that you can write this with coefficients in the field and the, you know, the coefficients are, we precisely have the A i, so this is certainly true. So, we have to see it's the splitting field, so that no smaller field that it cannot split over any smaller field. Well, in some sense, that's also obvious, but I just still carry it out. So, assume, so maybe I call this field here F. So, assume we have a, there's an intermediate field, we have M over k is an intermediate field. So, that means, I mean, so M is contained in F. Then you have to see that M must be equal to F. So, and such that our polynomial F splits M. So, that means we can write F equal to X minus C 1 times X minus C n, where the C i are elements in M. That means that it splits. But then obviously, if I take any of our, my A i's, so for all i, if I take F of A i, we know that this is equal to 0. And this I can write out as A i minus C 1 times A i minus C n in F, for instance. But if this is 0, we are in a field, so we have no 0 devices. This product can only be 0 if one of the factors is 0. So, that means that A i is equal to C j for some j. And this is true for all i. So, for all i, this element A i is equal to C j. So, in particular, it follows that A i is an element in M for all i. So, thus, I find that k A 1 to A n, which was our field F, is contained in M. Because this is the smallest field extension of k, which contains these elements. And M is a field extension, which contains them. So, it must be contained in that. Okay, so this shows number 2. And now, no, this shows number 1. And now number 2 is by some simple induction. So, how do we show number 2? So, we have our polynomial F of degree n. We have seen that we can find an extension of our field k of degree at most n, the degree of F, where it has a 0. So, by the previous corollary, so the one before this, there exists an extension, say, k 1 over k of degree at most n, such that F has a 0 in k 1. And now, we have the obvious thing. So now, we put the g equal to F. I mean, how do you call that as a 0? If it is 0, I call a 1 in k 1. So, I can take g equal to F divided by x minus a 1. So, this is now a polynomial. This is a polynomial in k 1 of x of degree n minus 1. So, by induction, we know that F splits over the field F, which is an extension of k 1 of degree at most n minus 1 factorial. We can move this by induction. And so, thus, we have that F. And obviously, we have the degree of F over k is equal to the degree of F over k 1 times the degree of F of k 1 over k. And we know this number is at most n 1 factorial. This number is at most n minus 1 factorial. This number is at most n. So, this number is smaller equal to n factorial. So, our polynomial F splits over an extension of degree at most n factorial. And then, we can again, as before, in part 1, find the splitting field of F over k as the subfield generated by the roots. So, if a 1 to an are the zeros of F in the field F, then we know that k a 1 to an is a splitting field of a polynomial F over k by part 1. And this is a subfield of F. So, the degree is at most, you know, so it is a, so it is more smaller than the degree here. In particular, the degree is still smaller than n factorial. Maybe I can still write down. And the degree of k a 1 to an over k is smaller equal to the degree of F over k, which was smaller or equal to n factorial. So, we are still fine. So, finally, we come to part 3. And that is really completely obvious. So, here, the claim is that if we have a splitting field over k and we have an intermediate field, then L is also splitting field over the intermediate field. But, you know, that's by definition. So, I just write it above here so that I don't have to write it out. So, for 3, F splits over k. So, F is an element in kx. So, also, in Fx. And F splits over k. And it does not split over any intermediate field between small k and, ah, let me see. So, L it was. Splits over this field L, if you remember. So, L over k is a splitting field and we take F. So, it's polynomial F splits over L. And it does not split over any intermediate field except for L between L and k. So, in particular, it does not split over any intermediate field between L and F because these are less fields except L. And so, that's it. So, therefore, by definition of the splitting field part 3 is trivial. So, now, let's look at a few examples which we will also look at again later. So, if F is a polynomial of degree D, if degree 2, do you want kx? So, an irreducible polynomial of degree 2. Then, and I have, and a is a 0 in an extension. Maybe I called b is a 0 in an extension of k. Then we have that splitting field. So, kb is a splitting field of F over k. I mean, that's obvious because if F has 1, 0, if I divide by x minus b, what the rest is a polynomial of degree 1 which automatically has a 0, so it splits. And, you know, I have a joint root. I mean, if I join the root, then I also have the other root. So, this is the splitting field. Let's look at a slightly more interesting example. So, we want to look at the splitting field of, say, x to the 4 plus 1 over q. It's also in the notes. So, we take any root. So, let alpha, say in complex numbers, be any 0 of this polynomial. So, you know, we have the c's are quite close. There will be a 0, but anyway, we have the 0. Then, we claim that q alpha is the splitting field of F over q. So, I mean, I will see in a moment, but this shows in particular we know that the q alpha, I mean, alpha is a 0. I mean, one can show that this polynomial is irreducible. Anyway, we will not, maybe. But, so this is the polynomial of degree 4. So, the extension here has at most degree 4. On the other hand, the theorem that we had before guarantees that there is a splitting field of degree at most 4 factorial, which would be much more. So, sometimes the degree of the splitting field is much smaller, even though this polynomial, as it turns out, is irreducible. And so, let's see why this is the splitting field. We have chosen a 0. So, I claim, so the claim is the splitting field. So, assume we have this alpha. So, then you see that if we take minus alpha, then this is also 0. Of course, if alpha to the 4 is equal to minus 1, then also minus alpha to the 4 is equal to minus 1 because we just multiply by minus 1 to the 4. And in the same way, we have 1 over alpha and minus 1 over alpha are also 0s of f in k alpha. So, we see we have 4 0s here, as many as the degree. So, it's split into linear factors. I mean, as long as we know that these are different. And they are also different because, well, if alpha is equal to minus alpha, it would follow that alpha is equal to 0. But this is impossible because alpha to the 4 is supposed to be equal to 1. And if alpha is equal to plus or minus 1 over alpha, then I can multiply by alpha. It follows that alpha squared is equal to plus minus 1. And therefore, alpha to the 4 will be equal to 1 and not equal to minus 1. So, it is not possible that these are equal. So, thus it follows. We have four different roots here, four different 0s. And therefore, we know that our polynomial f must just factor as a product of these linear factors. So, thus we have that x to the 4 plus 1 is equal to x minus alpha times x plus alpha x minus 1 over alpha x plus 1 over alpha. In particular, our polynomial splits into linear factors over k. So, thus f, so x to the 4 plus 1 splits over k alpha into linear factors. And as we have obtained this field extension by joining a 0 of this polynomial, we know that this is the splitting field because it means that joining this one element is the same as joining all four of them. And therefore, k alpha is the splitting field. So, it was q. I don't know whether here it's already has been, we have worked over q. And therefore, q alpha is the splitting field, a splitting field of f over q. So, sometimes it's enough to just add 1, 0 in order to get all of them. Finally, let's look at one case where this is not true. This is also, so this is the splitting field, say, of the polynomial x to the 3rd minus 2 over q. So, obviously, we can see that over C, we can write this polynomial, I mean, as you might have learned in high school as x times x minus 3rd root of 2. So, we take the 3rd root of 2 in the complex numbers in the usual way, x minus C e to the 2 pi i divided by 3 third root of 2 and x minus e to the 2 pi i. So, twice 2 pi i divided by 3 third root of 2. Okay. So, we know that these three numbers are zeros of this polynomial because if I take, I mean, third root of 2 is, you know, just the real number with the property that it's third power is 2. And if I have e to the 2 pi i divided by 3 and take it to the power 3, it gives me 1. And so, these three are therefore the roots and therefore this is the product. So, it splits into linear factors like this. So, therefore, we know that q third root of 2 and e to the 2 pi i divided by 3. So, if I join these two elements, this is a splitting field of this polynomial over q because it's easy to see that if we have this element and this element, then obviously we have these three elements in the field. And conversely, if we have an extension of q which contains these three elements, then we can just divide this by this and get this. And so, the field extension given by joining these two elements is the same as the field extension given by joining these three elements. And so, if we join these three, we know it's the splitting field and so, this is the splitting field. Okay? So, let's see. And we can see. So, it is clear that the degree of q third root. So, by the isothine criterion with the number 2, this is an irreducible polynomial. So, if we join any root, the degree of the field extension is 3. It's equal to the degree of the polynomial. So, we know that the degree of the field extension q third root of 2 over q is equal to 3. And on the other hand, we also know, so, this number e to the 2 pi i divided by 3 is certainly not in q a joint third root of 2 because this is a real number and this is not a real number. It's not an element in q third root of 2 because it is not in r. On the other hand, if we divide this polynomial by the polynomial, this x minus third root of 2, we get the polynomial of degree 2. So, the degree of this field extension must be 2. So, q, if we take this, so, q third root of 2, the degree of this field extension is 2. And so, altogether, the degree of the field extension of the degree of the splitting field over q is 6. So, this is equal to 6, which is 3 factorial. So, in this case, we actually do find that the degree of the splitting field is equal to the factorial of the degree of the polynomial. So, that was these splitting fields. I have these examples. And now, we want to, again, talk about extensions of field isomorphisms. So, remember, not very long ago, we were talking about extensions of field isomorphisms to simple algebraic extensions. So, if you have an isomorphism of two fields and we have a simple algebraic extension on one side, and on the other side, given by polynomials which are related by this isomorphism, then we found that the field isomorphism can be extended to the extension, and, in fact, can be extended uniquely when we require that a given root of the minimal polynomial is sent to a given root of the minimal polynomial on the other side. And here, we want to prove something similar. We want to, except for this uniqueness, we want to show that if we have a field extension, if we have a field isomorphism between two fields and we have a polynomial, and the polynomial which is the image of it under this field isomorphism, and we look at the corresponding splitting fields, then the isomorphism between the original fields extends to an isomorphism of the splitting fields. But we do not claim to have the uniqueness, and that actually is the thing which makes Galois theory non-trivial. So, let me state it. So, extension of field isomorphisms to splitting fields. So, we will always be interested in these field isomorphisms because finally, we will want to study field extensions and so, L over K in terms of the K isomorphisms of L. So, we are somehow interested in how field isomorphisms extend to bigger fields. And so, let us do this for the splitting fields. So, the theorem is the following. So, let say more phi from K to K prime be a field isomorphism, and we take let F be a polynomial in Kx, maybe polynomial. And now remember that we had this, so just remember that we had this. So, phi gives us an isomorphism from Kx to K prime of X. Now, we have this map which I called phi star from Kx. Maybe I put it not in the theorem, put it here below. So, we have phi star from Kx to K prime of X, which sends a polynomial sum ai X to the i to some phi of ai X to the i. We used this before when we had a similar theorem for simple algebraic extensions. And I put there for F prime, so it is not derivative, notation F prime equal to phi star of F. So, let K be the splitting field, be a splitting field of F over K. So, large K, or maybe I don't call it K. So, let's say L be a splitting field of F over K. And let L prime be a splitting field of F prime over K prime. Then there is an extension, then there is an isomorphism from L to L prime which extends phi. So, then there exists an isomorphism, large phi from L to L prime with phi restricted to K is equal to small phi. By definition, L is a splitting field of F over K. So, K is a subfield of L. And so, it makes sense. And so, in particular, the cases which particularly we are interested in would be, for instance, that K is equal to K prime and phi is the identity. So, if K and K prime are two splitting fields of our polynomial F over K, so then this statement says, so maybe call it again L prime, then there exists a K isomorphism, phi from L to L prime. So, this is just a special case here where we have taken K equal to K prime and phi equal to the identity. K isomorphism is just an isomorphism of L between these fields which is the identity on K. So, this means that two splitting fields of the same polynomial of F in Kx are K isomorphic. So, the splitting field in particular is not really unique, but it's unique after isomorphism. Okay, so let's see how you want to prove that. So, somehow, we have before proven a similar statement for simple algebraic extensions. So, the good strategy might be to somehow use that here, to somehow use the fact of simple algebraic extensions to prove the theorem here, and in fact we can do that. So, we make an induction over the degree of L over K. So, the case, so if the degree of L over K is equal to 1, then it follows that L is equal to K, no, let me know. And so, that would mean here that F splits over K into linear factors. But this phi is an isomorphism, so if something splits into linear factors and apply an isomorphism, it still splits into linear factors. Also, F prime splits over K prime. And so, we have just K is equal to, so L is equal to K, L prime is equal to K prime, and phi is our isomorphism, so the statement holds. So, phi is an extension of itself. So, in this case, the start of the induction is clear. So, now we have to do the induction step, and for this, we want to use the result for simple algebraic extensions. So, we now assume if L over K is bigger than 1, so then it means our polynomial F does not split into linear factors, so it contains an irreducible factor of degree bigger than 1 as an irreducible factor g in Kx of degree bigger than 1. No, because if it would factor into linear factors, after all, then it would split. So, we will join a root of this. So, we look at what happens if we just take 1, 0, join one of the zeros of g. We get an intermediate field extension. So, we put let say g prime to be again applying this isomorphism. This polynomial we get from g, and this g prime is also irreducible because under isomorphism irreducible polynomials are sent to irreducible polynomials. So, over L, this F splits into linear factor, so it has all the possible zeros, so also g splits into linear factors. So, in particular, there is a zero of g in L. So, let A, what was the field, in L be a zero of g. Now, you know, and the same we can apply here. So, let A prime in L prime be a zero of g prime. So, we have, you know, we know that all zeros, so that this splits into linear factors over L because F does, and it's a factor of it. So, we have this. Now, we can look at the extension. So, thus, we have kA over k is a simple algebraic extension and the same way k prime of A prime over k is a simple algebraic extension. So, now, and we are precisely in the situation of our previous theorem for extension for field isomorphisms for simple algebraic extensions. So, there exists an extension of our F from k to k prime to a map from kA to k prime of A which sends A to A prime. So, we can look at this by previous theorem about extension of field isomorphisms to simple algebraic extensions. We have that there exists. Anyway, so, I just hope you can remember. An isomorphism may be phi tilde from kA to k prime of A prime which is actually unique, but we don't care. Isomorphism, which is an extension of phi, so with phi tilde restricted to k is equal to phi and phi tilde of A is equal to A prime. That is such a thing. And now, you know, we are kind of done which might not see. I mean, so, it is clear that if we look at the extension L over kA times kA over k is equal to the degree of L over k. So, it follows as this number was bigger than 1 is equal to the degree of g. So, equal to the degree of g bigger than 1. It follows that the degree of L over kA is smaller than the degree of L over k. So, and we also know, we just proved it in the previous corollary, that L is also the splitting field, is a splitting field of our polynomial f over kA. Because we know that if I have a polynomial with coefficients in the smallest field and I have an intermediate field between the smallest field and the splitting field, then the splitting field is also the splitting field over the intermediate field. That was basically for trivial reasons. So, therefore, we can apply the induction hypothesis to this extension, this extension and the corresponding extension here. So, by induction, replacing k by k of A and k prime by k prime of A prime, we have by induction there is a field isomorphism phi from L to L prime, whose restriction to the smaller field, which in this case is kA, is the given map which we had called phi tilde. So, with phi restricted to kA is equal to phi tilde. Phi tilde was written in a different way. But once we have this, we can also further, I mean, restrict it further to k. So, in particular, phi restricted to k is equal to phi because the restriction of phi tilde to k was phi prime. So, here we have constructed our extension of field isomorphism by this induction, so by somehow reducing it to the case of a simple algebraic extension. But if you look at the proof, you can somehow see that we don't really know much about how we explicitly can describe this isomorphism. In particular, we have no uniqueness statement. So, we do this inductive proof and it's not really clear how many. So, remark, we have no idea how many extensions phi from L to L prime of phi exist. So, whether this is unique or whether there are many, so we don't really know. And in fact, the answer in general is also it depends. Depending on the field extension, sometimes there's only one extension, sometimes there are very many. And in some sense, that is the thing which makes the study of field extensions interesting because one can describe, okay. So much for this, how much time? Not very much. So as a, now we kind of, so we are slowly approaching Galois theory. So, we will introduce two properties of field extensions, two nice properties. One property is that the field extension is normal. And another property is that the field extension is separable. So, and we'll see in a moment what they are. But anyway, field extension is a Galois extension and we will want to study most Galois extensions if it's both normal and separable. So, we want to introduce these two properties that we need in order to be able to study field extensions well. One by one and so we start with normal extensions. So, I maybe just say it again. Field extension will be called, so maybe an algebraic field, whatever, called a Galois extension if it satisfies two properties, namely if it is normal and separable. So, we need to introduce these two properties and then we can study these Galois extensions. And we will find that for Galois extensions we can describe what happens very precisely in terms of some group which is called the Galois group. And so, therefore we want to introduce these things. So, we introduce first in what a normal extension is. So, the definition is a bit strange maybe. It actually looks rather strange in the first moment. So, an algebraic extension, say L over K, so it's supposed to be an algebraic extension if it's all the elements are algebraic, is called normal. If every irreducible polynomial, so f which coefficients in the smaller field which has a zero in the bigger field already splits into linear factors. So, one thing this word normal we will find out later is somehow related also to normal subgroups. So, there is a group associated to this extension which is the Galois group and normal extensions correspond to normal subgroups. So, there is some relation between the two words normal but we are not far away from that now. But if you look at this definition it looks like a pretty bad one because how are you ever going to check that? You know, if you have this field extension you take any irreducible polynomial, you know, there will be very, very many which has a zero in L and then it already is supposed to split. You know, we don't even, you know, it doesn't even seem to be obvious whether there are any normal extensions at all. But in fact, it's not at all so difficult. Well, first give me an example of a non-normal extension which we have already seen. So, example, if we take q, the third root of 2 over q this is not normal. And because we have seen that, you know, in the example before that obviously the third root of 2 is a zero of x to the third minus 2 which is an irreducible polynomial over q in q third root of 2. But the polynomial does not split there, does not split into linear factors over this field because we have seen that, you know, in the complex numbers there are two other roots and they are complex numbers. We can certainly never obtain a complex number which is not real in this field which is the subfield of the real numbers. Okay, so we know that there are some field extensions which are not normal, but now we want to see that it's very easy. I mean, there's an easy criterion for a field extension to be normal. Namely, a finite algebra extension will be normal if and only if the bigger field is the splitting field of a polynomial from the smaller field. So, theorem, finite extension, so finite field extension L over k is normal if and only if k L is the splitting field of a polynomial F in k x. That's somewhat surprising because, you know, it's a very different... Now, here we say every irreducible polynomial splits and here we basically are requiring that one polynomial splits but, you know, we take the smallest field where this one polynomial splits. So, that doesn't seem to be very likely to be true, but still, that's our theorem. So, let me see how much time we have. So, we can still try. So, obviously, the difficult part is this... If it is a splitting field, we have to show it's normal. So, this is really the thing which is unexpected. So, therefore, it must require some kind of idea. And here we will use these things about extension of field isomorphisms in a somewhat unexpected way. So, we take a splitting field of some polynomial F with coefficients in the smaller field k. So, now we have to take any irreducible polynomial and see that it splits, which has a zero, and have to see that it splits. So, let G, your polynomial with coefficients in k x, be irreducible with a zero alpha in F. Now, we have to see that all the zeros that it has in any extension of L actually already lie in F. So, we have to show G splits over L into linear factors. So, let now better be another zero of G in some extension of L. We know that we can divide by x minus alpha, and then this will have a zero over some extension. So, let better be another zero of F over an extension of L. So, we have to show that better actually lies in L. Because, you know, by induction we can, you know, so that means all the zeros over any extension lie in L. So, it splits over L into linear factors. So, then result F splits over L. So, let's see. So, now our polynomial G is irreducible, and alpha and better are two zeros. So, we have a simple algebraic extension, you know, k of alpha and k of better. And so, there is an isomorphism between them. So, since G is irreducible, we have that, say, k alpha over k and k better over k are simple algebraic extensions with the same minimal polynomial, same minimal polynomial. So, alpha and better have the same minimal polynomial, which is G. And so, by the theorem that we had for extension of field isomorphisms, we find that there is an isomorphism from k alpha to k better. So, maybe call it phi from k alpha to k better, which sends alpha to better. I don't know what I actually need anyway. Phi of alpha is equal to better. Okay, so this is the first thing. But we also have to use the story again. The other story with the field, extension of field isomorphisms for splitting fields. So, L is a splitting field of G over k. So, it's also the splitting field of G over k alpha because that's an intermediate field. So, L is a splitting field of G over k. And k alpha is intermediate field. So, L is a splitting field of G over k alpha. Now, not G. You know, you should pay attention, no? F, okay? And on the other hand, if I take L of better, so if you join better to L, it's already in L, it doesn't become bigger, but otherwise it's different, is a splitting field of F over k of better. Because, you know, certainly it splits, F splits over L of better because already splits over L. And the smallest field which contains k better and or which it splits is this one because, you know, the condition that contains k better means precisely that it contains better. So, this is the splitting field of this. But now, you know, there comes somehow the trick because, you know, this somehow means we can extend this isomorphism here to an isomorphism between these two which means that they have the same size. So, by the extension of field isomorphisms to the splitting field, we have that there is an isomorphism of fields phi from L to L better with phi restricted to k of alpha is equal to our map phi. But, you know, now this is really, so somehow that looks a bit like cheating because now we have all of, you know, all of a sudden these two fields have the same size. And now, obviously, we are done. We just have to do it. So, we have this. So, in particular, we have that phi restricted to k is equal to the identity. So, we have an isomorphism from this field to this field which is the identity on this. So, if it's an isomorphism between fields, it's in particular an isomorphism of k vector spaces. So, you know, it's an isomorphism which is the identity on k, so it's an isomorphism of k vector spaces, of k vector spaces. So, this means these two fields have the same dimension. So, L and k, so we have that L over k is equal to L of beta over k. So, if you use the degree theorem, it follows that L, maybe I can wipe it out, by the degree theorem we have that L of beta over L is equal to L of beta over k divided by L over k, which is 1. So, this means these two fields are equal. You know, we know that this is a field extension and the degree is 1, so it follows that L is equal to L of beta and that means beta is in L. Okay, so this is a rather surprising proof, so we can also see that it was quite useful to prove this statement about the extension of field isomorphisms because it gives you kind of a nice handle to do this. So, this is only one half of the proof, but so the trivial direction we will do next time. I mean, the almost trivial direction we'll do next time and we don't want to rush it now. So, I expect we see each other on Wednesday. So, are there any questions or comments? Am I too fast? A little bit. Okay, so I maybe can try to slow down. Okay, yeah, thank you.