 So in the earlier videos of lecture 33 introduced Gaoua groups and we did some examples. And I kept on pointing out the following observation that the order of the Gaoua group was always equal to the degree of the field extension. Now, it turns out that this is true in general, but there's an important assumption we have to put into play here. Now, first of all, we do want these to be finite extensions because what does it mean for the infinite case, do, do, do, do? You can extend it to the infinite case as well, but that's gonna go beyond the scope of this lecture series. So we're gonna focus on finite there, but it also requires that it be a Gaoua extension. All right, so E as an extension of F needs to be a normal extension, which means that it is the splitting field for some collection of polynomials, but also it needs to be a separable extension. That is that polynomial collection, that split, you can't have any repeated roots in that side of that collection. That's what we need, that's what we need for E. It's a separable normal extension, hence a Gaoua extension. When you have a Gaoua extension, then it turns out that the order of the Gaoua group is equal to the degree of the extension. And while this one is typically a lot easier to compute than this one, this helps us compute Gaoua groups for finite extensions, which is very, very important. And the proof goes in the following way. So suppose, since it's a Gaoua extension, E over F is gonna be a finite, finite Gaoua extension, normal extension. So E is gonna be the splitting field for some single polynomial, okay? Because in general, to be a normal extension, there should be some collection of polynomials. But as we're a finite extension, that means at worst case scenario, we attach a finite number of roots to the field F to form the field E. And so if you had an infinite collection of polynomials, then, well, I mean, maybe they overlap on the roots or whatever, but we can only add a finite number of roots over these polynomials to E to form E, as we attach them to F to form E. And as such, there's only finitely many roots in general. We can just take the minimal polynomials of those elements over F, take their least common multiple, that's our polynomial F right here. So because it's a finite nominal extension, without the loss of generality, we can suppose that E is a splitting field for a single polynomial F of X, whose coefficients come from the field F. And we can also suppose that since it's a separable extension, because that's part of Galois definition there, that F has no repeated roots. Again, that's what it means to be a separable extension. We can assume that F has no repeated roots. Now we're gonna proceed by induction to prove this result. So the base case is gonna be pretty simple. What are we inducting on? We're inducting on the degree of the extension E over F. So the base case is when the extension E over F is degree one, that only would happen if F and E are exactly the same polynomial, excuse me, the same field. And the Galois group is all of the automorphisms of E that fix F. So if we fix F and E is equal to F, then the only field automorphism available to us is the identity map, because we have to fix F, but F is everything, so we fix everything to the identity map. So the Galois group contains only one element in that situation, because your Galois group of F over F is just the identity map, which we're gonna denote that as one. And therefore the order of the Galois group of F over F is one, which was equal to the degree. And so that gives us the base case, that's pretty trivial. So then for our induction hypothesis, assume that the formula holds for all Galois extensions of degree K, such that K is anywhere from zero up to N. And I guess I said zero here. Well, this is degree zero extension B, that should be a one. We're gonna go from anywhere from one up to N, but not necessarily including N right here. And consider the situation for which the extension E over F is exactly degree N, okay? So let P of X be an irreducible factor of F over F of X, because to be a splitting field, E doesn't necessarily have to be a splitting field of an irreducible polynomial. We do require that F of X doesn't have any repeated roots though, since it's a separable extension. So let P of X be an irreducible factor of F of X, which F splits over E. And suppose that P of X has degree F, okay? Now since F, F of X splits over E, each of its factors must also split over E. So P of X splits over E as well. And so all of the roots of P of X are included inside of E. Let alpha be a root of P of X. And consider the chain of extensions E over F adjoined alpha and F adjoined alpha over F in that situation, okay? So in this situation, I want you to consider this extension first, okay? Because P of X is a polynomial, an irreducible polynomial of degree R. So that tells us that the field extension F adjoined alpha over F is a degree R extension, as we've seen previously. And so we also have that degrees factor. And so if we take the field extension, it's degree E over F, this will factor as E over F adjoined alpha. And you times that by F adjoined alpha over F. And so we can solve for the degree of E over F adjoined alpha, because you're just gonna get the degree of E over F divided by the degree of F adjoined alpha over F here. This is really the part where the finite extension comes into play here. This cardinal arithmetic would be a little bit more complicated in the infinite case, which we can handle it, but again, for the sake of our electric course, we're not gonna go there. So we don't have to worry about it right now, but you can repair this proofs to work with infinite Galois extensions as well. But again, Galois extensions in infinite case get a little bit more complicated. You have to introduce the Zyrinski topology to make things hunky-dory to get the fundamental theorem of Galois. So again, we're not gonna worry about that. We'll just worry about finite Galois extensions. Now by assumption, E over F is degree N and by construction here, F adjoined alpha over F is gonna be degree R, like so. So our inductive hypothesis is going to apply to this field extension because the degree here is gonna be less than N because R is greater than one in that situation. So by our inductive hypothesis here, we are going to get, which I mean, I guess it's certainly as possible that P of X could be a linear factor in that situation. Maybe we'll come back to that one in a second because if you have a linear factor, you can remove that from F and so we can assume that none of the, we can assume without the loss of generality that none of the roots of F of X belong to F. Otherwise we'd be in this case as we were above. So yeah, we can assume that R is a least degree too larger. So N over R is gonna be strictly less than N. So our inductive hypothesis applies in that situation. Since E is also the splitting field of F of X over F adjoined alpha, right? If so E makes F of X split over F, so if you take an intermediate field, it's gonna still split and this is a minimal extension. So E is a splitting field of F of X over F adjoined alpha and the degree of this extension is less than N, we can use the inductive hypothesis to tell us that the Galois group of E over F adjoined alpha is equal to the degree of E over F adjoined alpha, which is gonna equal N over R in that situation. Let beta be another root of this polynomial P of X. There's gotta at least be two because P of X is not a linear polynomial. And then consider the map tau which is gonna send F of alpha, the field to the field F adjoined beta and this is gonna be the unique isomorphism that fixes the base field F, but then sends alpha to beta. Such a map always exists, okay? There could actually be lots of maps that potentially do that but there's at least one that'll do it. I mean, I should say that in this situation there's only one map that'll send alpha to beta because if we think of this as a vector space, your basis is going to be just one alpha, alpha squared, alpha cubed all the way up to whatever the degree needs to be, right? Continuing on here, this is a span of the powers of alpha. And since this is supposed to be a field isomorphism, it's gonna preserve multiplication, preserve exponents. So if you know what alpha does, you know what it does to every element of the field. So there is one, there's only one map tau we have in mind right here, okay? Now, of course, since F adjoined beta can be embedded inside of the algebraic closure without the loss of generality, we can actually think of tau actually as a map from alpha beta to the algebraic closure of F. And since alpha adjoining to F is an algebraic extension by the extension theorem, the isomorphism extension theorem from previous work, we can actually extend this to some map, an injective map from F bar to F bar which then forces it to be a field isomorphism in that case. It's not necessarily a unique extension. There could be potentially other isomorphisms but there's at least one. There does exist some map tau bar which is a isomorphism of the algebraic closure. The nice thing about splitting field these normal extensions, if you restrict a field isomorphism on the algebraic closure to some splitting field below that restricts to an automorphism there. Cause after all, every automorphism is going to permute conjugates. And E turns out just to be, since it's this normal extension, E is just gonna look like some alpha one, some alpha two, some alpha three up to whatever alpha K, excuse me, E. I did not write that correctly. Let me try that again. We get that E is equal to F adjoin all of these elements and tau bar is just gonna permute these things around. In which case that means the image of F adjoin all these conjugates will be F adjoin all the conjugates still. So if you take tau bar of E here this is gonna give you back E. So the restriction of this map to E gives you an automorphism on the field E. So this is a very common trick we do. We can take a map, we can extend to the algebraic closure, we can restrict it to a splitting field. The fact that it's a splitting field is very necessary here because as automorphisms permute conjugates, if you don't have all of the conjugates, then you might actually exit your field. And then what guarantees you have all the conjugates, you're a splitting field. That's the whole point of a normal extension there. So you can restrict automorphisms of the algebraic closure to splitting fields. It's a very, very common trick here. And so this gives us some map tau that's gonna do exactly this. Alpha goes to beta. There's gotta be some map that sends alpha to beta. Okay? And so we'll call that map tau bar. It belongs to the algebraic closure, excuse me, it belongs to the Galois group of E over F. Since P of X splits over E, there actually are our choices on what beta could be. So this map, this original map tau was dependent upon our choice of alpha, excuse me, a choice of beta. We can map, we can construct an automorphism that sends any element to any of its conjugates. And as there are our choices for those conjugates, they are gonna give us our different possibilities for this map tau. And so the algebraic, excuse me, the Galois group here contains at least our many elements. This Galois group E over F in that situation, because you have these our choices for the original map tau from before. Now, consider, and this right here is where the separability comes into play because I know P of X doesn't have any repeated roots because F of X didn't have any repeated roots. So I have actually our distinct conjugates to alpha. So I get exactly our choices for tau here, at least our choices, right? Because like I said, when you extend it to the algebraic closure, there potentially could be other possibilities going on there. It doesn't really matter which one you choose though, when you restrict it down to E, you're always gonna get the exact same map here, which we can call that tau. So we have our different choices for tau. We also had this map sigma from before, excuse me. We don't have sigma yet, let's introduce it now. Let sigma be a map that belongs to E over F, the joint alpha, right? So this is a subset of Galois of E over F. In fact, this is a subgroup, in fact. And I want you then to of course, recognize that if I take sigma times tau, this belongs to the Galois group of E over F, because sigma, it doesn't just fix F, it fixes F of joint alpha, but in particular, it fixes F. So since this one fixes F and this one fixes F, the product will fix F, and so it belongs to here, okay? Now, consider what are the possibilities here? So if you allow sigma to be an arbitrary element of this subgroup, and you have these elements tau, which there's our choices for tau, depending on the r conjugates of alpha there, I want you to think of how many possible, how many possible automorphisms do we have here? Well, this group right here has order in divided by r, and tau has r options. And so this then accounts for in, and this counts for in automorphisms in the group, the Galois group of E over F. Notice that in of course is the degree of this, all right? So what we've now done is we have shown that the order of this group is at least in this situation. But on the other hand, if you take two elements, pi and phi that belong to this Galois group E over F, I want you to be aware that the following statements are equivalent. That pi belongs to the coset associated to F, associated to phi right here, because after all, the Galois group of E over F of joint alpha is a subgroup. So we can talk about its left coset. Look at the left coset associated to the map phi, right? Pi belongs to this coset, if and only if the quotient pi inverse phi belongs to this subgroup. But if it belongs to the subgroup, that means it fixes E over joint alpha. Now, if the quotient fixes E over joint alpha, what that means, that happens if and only if the restriction of pi and phi to the subfield E over joint alpha, that means that the functions do the exact same thing. These functions don't have to fix E over joint alpha, but because their quotient fixes E over joint alpha, that means they act identically on the elements of E over joint alpha. Now, because of Lagrange's theorem, it turns out if you take some subgroup, in this case we can think of this as the subgroup H, then every element belongs to some coset of H right here. So in particular, you have this thing going on that the group, the Galois group of E over F, it can be partitioned using the cosets. So every element belongs to one of the cosets, okay? And that's where this factorization comes into play right here, that if you fix the representatives of the cosets, then we're gonna have the order of the subgroup, many options in that coset, and we can add up all of those cosets together. So what are the possibilities? Who represents the same coset? Well, this is what we're saying right down here, that two automorphisms represent the same coset exactly when their restriction to F over joint alpha is the same. So basically you can ignore the F over joint alpha part. And so we've now counted all of those possibilities here because this is then the elements of the subgroup H and there are gonna be R possibilities for what you can do here because what you do to F over joint alpha determines these possibilities here. And so the different options we have for R are all the possibilities we do to alpha. So what you do to alpha determines which coset you belong to. If you do nothing to alpha, then turns out you belong to H here. If you send alpha to beta, well, you belong to that coset. If you send alpha to gamma, you belong to that coset. That's what we've now spelled out here and there are options there. And so by Lagrange's theorem, when you multiply these two things together, that gives you everything in the group because the cosets partition it. And therefore we've counted every element, thus establishing the function by induction, right? I wanna now summarize what great result we've just established here. Suppose that we have an extension of fields. So F sits inside of K, which sits inside of E. And suppose that E over F is a Galois extension, okay? So previously we have shown that degrees of extensions are multiplicative. So if we take the extension E over F, we can factor this through the intermediate field K, E over K times K over E, like so. Now because of the previous theorem we just proved here, we get that the order of the Galois group, Gal of E over F, this is equal to the degree of the extension like so. Now because E is a Galois extension over F, we also get that E is a Galois extension of K, right? Because this is a splitting field over a polynomial set with no repeated roots. Those coefficients of polynomial come from F, which means their coefficients are in K as well. So we do get that the E is Galois over the intermediate subfield there. And so we also are going to get that this field extensions degree is equal to the associated Galois group. So we're gonna get the order of the Galois group of E over K, like so. But then by Lagrange's theorem, we get that if you take a subgroup, because this is always a subgroup of that group, if you take the order of the subgroup, then times it by its index, for which I'm gonna struggle to fit that in there. So we take the index of Gal of E over F, colon Gal of E over K. So the product of the order of the group with the index of the group gives you the whole group. The Lagrange's theorem tells us that. Well, since we have equality between the product and one of the factors, then it's gotta be true that these factors are equal to each other as well. So we can infer this very important relation for Galois extensions that the extension K over F, the degree is equal to the index of the Galois groups E over F over E over K. And this actually justifies why we used index notation to describe the degrees of field extensions, because there is this correspondence between the degree of the extension of a field with the index of a group, at least in the case of Galois extensions. Now this result we proved when you do have a Galois extension, it turns out that you get the other, if you don't have a Galois extension, you still get that the degree is gonna get upper bound of the Galois group. You can't have more automorphisms than the degree of the extension, but there are cases where you don't have it, since you don't have a Galois extension, you don't have enough automorphisms, and this inequality can be strict in that situation.