 Consider the very important polynomial x to the fifth minus one. Why is it so important? Well, the roots of this polynomial, we're reviewing this of course as a rational polynomial, the roots of the polynomial x to the fifth minus one are exactly the fifth roots of unity viewed as complex numbers of course. Now, this polynomial can be factored over the rationals into the prime factorization of x minus one and x to the fourth plus x cubed plus x squared plus x plus one. And which of course, the root of x minus one is clearly one. And so that accommodates the non-principle fifth roots of unity, a.k.a. one. Then which case that then leaves four principal roots of unity who would have to then be the roots of this polynomial x to the fourth plus x cubed plus x squared plus x plus one. And therefore this polynomial is the cyclictonic polynomial associated to the primitive fifth roots of unity. And so there are gonna be four roots of this polynomial. So let's think of it as a complex polynomial now for a moment. As a complex polynomial, f of x will factor as x minus, I'm gonna say zeta right here, for which zeta we're gonna take to be the complex number e to the two pi i over five. And then the other roots are gonna look like x minus zeta squared, x minus zeta cubed and x minus zeta to the fourth here. And this is again, we're viewing this as a complex polynomial in this situation. So it splits over the complex number field, of course, cause that's a algebraically closed field. These are roots of unity. So the important thing to remember about zeta here is that zeta to the fifth is equal to one. But it turns out that we also have this relationship here that if you take the sum of all of the roots of unity, if you take one plus zeta plus zeta squared plus zeta cubed plus zeta to the fourth, this is equal to zero. So the sum of the five fifth roots of unity does in fact equal zero. This is actually true for roots of unity in general because of this type of factorization going on here. Okay, so what we wanna do is compute the Galois group for the following field extension. So consider the field q join zeta, like so. Be aware that the four roots of this polynomial are zeta, zeta squared, zeta cubed, zeta fourth. So if you have zeta inside your field, you have every power of zeta, including zeta two, zeta three, zeta four. So joining one complex root of unity to q gives you the splitting field. It's gonna give you the splitting field for our cyclic atomic polynomial. And so we're viewing this, of course, over the rationals, like so. Let's say that e for simplicity is our field q join zeta, like so. And so we wanna compute the Galois group of e over q and then analyze the subfields by using the lattice of subgroups. That's the basic problem we want to do there. Compute this Galois group. So the first thing to note here is that if I take e over q, this is a degree four extension. And that's because this is just the same thing as q adjoin zeta over q, for which the minimal polynomial of zeta over q is four. And so the degree here is gonna be four. But q adjoin zeta is a Galois extension. Characters are zero, you're always separable, so that's not a big deal. And then our minimal polynomial splits inside of e right here. So this is, in fact, a splitting field that's a normal extension. This is a Galois extension here. So this tells us that the Galois group, Gal of e over q, and this example is equal to four. And so for simplicity, I'm just gonna call the Galois group g here, curly g, just to make it a little bit shorter to write whenever I have to write it in this situation. Let's put the whole thing in here. So as you're taking your notes that you do the same thing here. So we have this Galois group. And since it's order four, that means it has to be isomorphic to either the cyclic group of order four or the Klein four group. Now, we actually saw in a previous video, I gave you a Galois group of order four that turned out to be the Klein four group. That's not gonna be the case this time, where I'm actually gonna show you, and the reason I did this example, is that this time the Galois group is gonna be cyclic of order four. And let me show you how that is, all right? Because if in fact the Galois group is cyclic of order four, there has to be an automorphism of order four that it takes four iterations of the automorphism to produce the identity map. What map possibly could do that, all right? So I want you to note, I want you to note that two, if we view this as an element of Z five is in fact a primitive root. That means as we look at the exponents of two, mod five, we get basically the following situation. So we're gonna take our exponents here and we're gonna do then two to the n, mod five. So we'll start off with the zero power, that's just gonna get two to the zero, which of course is equal to one. We can do the first power, two to the first equals two, that's easy, two squared, that's gonna be a four, two cubed. Two cubed is gonna be equal to eight, which of course we wanna reduce that mod five, so that actually is a three. And then if you do the fourth power, you get two to the fourth, for which you could think of that as 16, but instead I'm gonna think of it more just as, well, I mean, I just 16 is perfectly fine because 16 reduces to be one, mod five, that works fine. I also think you could do it as just two times three, which gives you six, same basic thing. But note here that when it comes to the multiplicative structure, we're ignoring zero, so there's only four elements. So actually I knew this last one was gonna happen. So we get this chart right here that the four elements that are units inside of Z five, when we think of it as a ring, the four units, because after all Z five is a field, one, two, four, and three here, they can all be generated as powers of two. So two is in fact this primitive root. And so what we can do is we can construct an automorphism from E to E, which be aware that the basis of E is determined by Zeta, right? An arbitrary element of E looks like some number plus some number plus Zeta. I mean, we can write this out here. A plus B Zeta plus C Zeta square plus D Zeta cubed because one Zeta Zeta squared Zeta cubed is a basis. So this is an arbitrary element of E. I'm gonna take the automorphism that's induced by the relationship that Zeta maps to Zeta squared, okay? So that means this element here would transform into, well, I'll say it like this. This element here is sigma will transform into A plus B Zeta squared. What is it gonna do to Zeta squared? Well, you're gonna, it's an automorphism. So you're gonna square it. So this would give you C Zeta to the fourth. Here, you're gonna get D times Zeta to the sixth, right, which of course is just Zeta, like so. You'll notice that I have a Zeta to the one, two, four. I don't have a three here. What can actually be useful is that while this is a basis, sometimes it might be better to define it using a spanning set. So I'm actually gonna throw in E times Zeta to the fourth. Again, this is not an independent set anymore, but it is still a spanning set. In that situation, you're gonna get E times Zeta to the fourth is gonna be Zeta to the eighth, which is Zeta three. So that's where that happens. So be aware, this map is gonna permute around these Zetas, right? They're in a different order now, but they're all there. And so that's if you do Zeta once. If you do Zeta twice, Zeta squared is going to go to Zeta to the fourth, like we observed. And if you do Zeta again, you're gonna get Zeta three. And if you do it one more time, you end up with back Zeta. And since, if we know what Zeta goes, then we know where everything goes. And so this does in fact show us that Sigma is a field automorphism of order four, like so, that it takes four iterations for this in order to produce the identity map. And so this is then going to establish that the Galois group is cyclic of order four. So I want us to think about that for a second then. If we were to draw the Galois group, we have Z four on the top. There's gonna be the trivial group on the bottom. And I want you to think of, what are the corresponding fields, right? Well, what is the field that is fixed by every automorphism here? Well, that is going to be the rational numbers, which is at the very bottom of our picture right here. Now, which field is going to be fixed by only the identity? That's gonna be the whole field, Q a join Zeta, like so. Now, we know that our Galois group is generated by this element Sigma. So it's just the squaring map, we squared the Zetas. Because it's a cyclic group of order four, there should be an intermediate field Z two. These are both degree two extensions in that situation. There's gonna be a cyclic group of order two that sits in between them, which since the Galois group is generated by Sigma, the subgroup of order two must be generated by Sigma squared. And so it begs the question, what does Sigma squared do to these maps, okay? So we saw what Sigma did over here. What if we do Sigma twice? Well, Sigma squared does nothing to the rational number. So A is left alone. Let me, I'm gonna erase these exponents here. I mean, powers of Zeta have to map to powers of Zeta, but then what happens here is that Sigma, it went to Sigma squared, then it went to Sigma to the fourth because we're just squaring twice. So then of course, if you square twice here, you square two, you get four, you do that again, you get eight, which is the same thing as three. Here, if you square three, you're going to get six, which is one, you square it again, you're gonna get a two. And then notice what happens here with the fourth power. If you square it once, you're gonna get eight, which is the same thing as three. And if you square it again, you're gonna get six, which is the same thing as one. So in that case, you get the following. So this is what Sigma squared is gonna do. And so I want you to compare the coefficients, all right? With regard to the rational part, we didn't do anything to it. So A can be whatever it wants. A just has to be some rational number. No worries there whatsoever. So you have full freedom there. But I want you to look at B. Sigma and Sigma and Sigma to the, excuse me, Zeta and Zeta to the fourth both have a coefficient of B. But notice here that Zeta to the fourth and Zeta one have the same coefficient of E. So this tells us that B has to equal E and E has to equal B. So there is this correspondence there. So that's perfectly fine. So as long as B equals E, that seems okay. And then we look at this. Sigma squared maps to Sigma, excuse me, Zeta squared goes to Zeta cubed. And Zeta cubed goes to Zeta squared. So when you compare these things, here's the Zeta squared. They have to have the same coefficient if this was a fixed element. And this has to be the same element if it's a fixed element. So we end up with C equals D, like so. And so this fixed field that we've constructed here, so E sub Sigma squared, we're required. We're taking a span where one, nothing happened to one here. We have that B and E have to be the same thing. So Zeta and Zeta to the fourth have to have the same coefficient. So that would be the linear combinations of thing involving Zeta plus Zeta to the fourth. And then also we have Zeta squared and Zeta cubed have to likewise have the same coefficient like so. And so this field is the field for which, again, Zeta, Zeta fourth have the same coefficient. Zeta squared and Zeta cubed have the same coefficient. These are the things fixed by Sigma squared for which I want you to note that if I take Zeta plus Zeta to the fourth and you square that, that's gonna equal, when you get a Zeta times Zeta, which is a Zeta squared, you're gonna get a Zeta times Zeta to the fourth, which is Zeta to the fifth, that happens again. And then you're gonna have a Zeta to the fourth times Zeta to the fourth. Well, as we've mentioned that one already, that's just gonna give you a Zeta three. And so when you put these things together, notice Zeta to the fifth is just, that's just one. So you end up with two plus Zeta squared plus Zeta cubed, for which is, this is the side of a field, which will contain the rational, but I have a two, I can subtract that off. So be aware that if you have a Zeta plus Zeta to the fourth, you also have a Zeta squared plus Zeta cubed. And so in summary here, what we get is that our fixed field is in fact, q adjoined Zeta plus Zeta to the fourth. All right, and so that then is the field that sits in between here, q adjoined Zeta plus Zeta to the fourth. Like so, these are degree two extensions, right? This extension corresponds to this one right here and this one coincides with this one right here. Again, this example, because the lattice is top down, bottom up symmetric with each other, it can sometimes confuse the Galois correspondence, which then flips the lattice upside down. So that's the thing is this extension coincides with this extension and this extension coincides with this one right here. But the point is, since we know the Galois group, we found a subgroup and this subgroup, we then were able to construct it, since we had the subgroup, we were able to construct its corresponding fixed field and then we got that this field is a subgroup. We found it using the group structure. We might have not realized this was here before. We found it using the group structure. And since this is a normal subgroup of Z4, that tells you that this is in fact a normal extension. That is it's a splitting field for some polynomial, which it's degree two, so that's not too surprising, but still we can discover the structure of the subfields using the structure of the subgroup, that's fantastic. Now this example does generalize to the so-called cyclic tonic extensions. Cyclo, make sure I spell this right. Cyclo tonic extensions for which for a cyclo tonic extension, you are adjoining a primitive intrude of unity and it generalizes this process. Now in general, this thing could be much more complicated. In this example, we saw the Galois group was cyclic. In general, the Galois group could be, well, it could be much larger. It doesn't have to be cyclic. It will always be a billion. That's a very nice result. And in fact, there is a very impressive result in algebraic number theory that basically says that every abelian extension, so we say that a field extension is abelian if the corresponding Galois group is an abelian group, we can, one can actually prove using some very impressive mathematics that are beyond the scope of this course. You can show that every abelian extension can be embedded inside of a cyclo tonic extension. And so in some respect, these cyclo tonic extensions generalize all the possible abelian groups. Excuse me, all the abelian extensions you can do is they're very, very important. Now for every cyclo tonic extension, it always looks like Q adjoint zeta because if you get one root of unity, you get all, well, if you have one primitive root of unity, you get all of the intrudes of unity by taking powers. And since primitive intrudes of unity are conjugated to each other, if you get one, you get all of them. So adjoining one primitive root of unity gives you all of them. It's always gonna be a splitting field, so that's very nice as well. And it always will contain a degree two, a unique degree two extension, well, maybe not unique, but there is always a degree two extension that coincides with this one, where in general, you don't think of it as zeta plus zeta to the fourth, you think of it as zeta plus zeta to the negative one, like so. And this field actually can also be expressed in the following way. It's q adjoined zeta, and then you intersect this with the real numbers, because after all, we're adjoining complex roots of unity. Well, what if we look at the subfield of q zeta that coincides with only the real numbers, the intersections there? And so this is actually a very common trick we do in general, not just for cyclotomic extensions, but in general, as we're adjoining complex roots to our rational field, we often look at what is the field intersect the real numbers? Well, if it was a real extension, you don't restrict anything. But in this case, since we are adding non-real complex roots, this will give us a degree to extension. And so looking at the maximal real subfield will be a very useful thing when we explore Gaoua groups in the next lecture. It was very useful in this case as we looked at the fifth cyclotomic extension. So thanks for watching these videos. I know they kind of ran on the longer side, but these Gaoua group computations can be very complex and a lot of details we had to cover. If you did learn anything about computing Gaoua groups or the fundamental theorem of Gaoua, please like these videos, subscribe to the channel and see more videos like this in the future. And of course, if you have any questions, please post them in the comments below. I'll be glad to answer them. We'll do some more calculations of more complicated Gaoua groups in the next lecture, lecture 35.