 So we just proved in our previous video, we proved if e over f is a finite Galois extension, then we get that the order of the Galois group of e over f is actually equal to the degree of the extension e over f like so. And we're going to use this result to actually compute the Galois groups of every finite field that you possibly could think of. So imagine that we have some finite fields e over f. Now f does not necessarily have to be the prime field zp. It could be but it doesn't have to be. And so we're going to prove that the Galois extension of e over f is in fact always cyclic and it's going to be the cyclic group of order k where k is the degree of the extension in that situation here. So in order to apply this theorem, we have to argue that this is a Galois extension. Why is it always a normal extension? Why is it always a separable extension? Right? Which we've discussed these previously. We've proven that every finite field is a splitting field over zp and therefore it will be a splitting field for the same polynomial x to the p to the k minus x because this is a polynomial over f as well. And so you get that since e is a normal extension over zp, it'll be a normal extension over f. And likewise, it's a separable extension over zp so it'll be a separable extension over f as well. So that's to say that whenever you have a situation like the following, f is contained inside of e which is contained inside of k. If k is Galois over f, it's going to likewise be Galois over intermediate field as well. And so for finite fields, we've already proven that essentially every finite field is going to be Galois extension, a finite Galois extension of zp. So it doesn't matter which finite fields we have here. We have a Galois extension in that situation. All right, so we want to prove that this thing is cyclic. Since we know this is a Galois extension, we know the order of this group is going to be k, but why is it cyclic? We need to produce a generator and this is going to be the so-called Frobenius map. The Frobenius map is going to be the map which sends every element of e called alpha to the pth power, alpha to the p here. This is going to be a field automorphism because of freshman exponentiation. Since we're working characteristic p, we have that if you take alpha plus beta to the pth power, this is equal to alpha to the p plus beta to the p. So it preserves addition. It will also naturally preserve multiplication since multiplication is commutative in the situation. Notice that sigma here is going to preserve the subfield zp. That is, if you take sigma of a, this is a to the p, which is the same thing as a as a consequence of fair mouse little theorem. So this Frobenius map sigma belongs to the Galois group of e over zp. Now that's not the Galois group over e over f necessarily, but notice it's going to be in there. And so we know from a previous result here that there's a primitive root inside of e. There is some element alpha, so at every element of e, some element alpha here, so they take any power of alpha, you'll eventually get all of the elements in e. So alpha is a primitive root in e over zp like so. So when you look at this extension, we get that the extension of e over zp, let's call that degree n. All right, I just need to give it a name there. So let's call it degree n right here. So that would be saying that like, of course, e is the finite field of order p to the n. Not a big deal. Now, since alpha is a primitive root and since the field has order p to the n, it turns out that the order of alpha, when we think of the group e star there is going to be p to the n minus one. That's a little of a typo right there should be p to the n minus one. Sorry about that. So in particular, if you then raise alpha to the power p to the n, you're going to get back just alpha itself. So alpha to the p to the n is going to give you alpha. But alpha to the p to the n, that is just sigma to the nth power of alpha. So if you apply sigma n times to alpha, you're going to produce back this right here. And so alpha is fixed by sigma to the n. Now, as every element in e is algebraically generated by alpha here, this actually tells us that sigma to the n, like so, is the identity map. Okay, because alpha is fixed and so therefore everything is fixed. So if we think of the order of this automorphism, sigma, it's going to be n. So wait a second, n was the degree of this extension e over zp, like so. And so this tells us that, again, by the result we proved previously that the Galois group e over zp, it has order n because of this identity. We then produced an element which has order n. So this group has to be cyclic. The Galois group e over zp has to be cyclic. Of course, the Galois group of e over f is a subgroup of e over zp because fixing f is more restrictive than fixing zp. And so if you're the subgroup of a cyclic group, then you have to also be cyclic. And so it'll be generated by some power of sigma as well. So there's some power of the Frobenius map that'll generate this Galois group. And this is what happens for every Galois group over finite fields. Some power of the Frobenius map will generate it. And so we have now completely classified Galois groups over finite fields. So Galois groups over infinite fields is a little bit more complicated. We're going to focus on, of course, the case where we have characteristics zero, infinite fields with prime characteristics go beyond the scope of this course. So take a look at the next video to learn more about the fundamental theorem of Galois, which is sort of like the climax of this entire lecture series. So I do hope you check it out. If you learned anything about Galois groups in this video, please like this video, like all the videos, in fact. Subscribe to the channel to see more videos like this in the future. And please post any questions in the comments below. If you have any, I'll be glad to answer them as soon as I can.