 In this video, we're gonna compute the Galois group for the polynomial g of x equals x to the fourth minus two. And just to be clear, we're thinking this as a rational polynomial. The first thing we wanna mention is that this polynomial is in fact irreducible. By the rational roots test, the only possible roots of this polynomial would be plus or minus one, plus or minus two, which is very easy to check, they don't work. You would have to also go through an argument that this thing doesn't factor into a product of two irreducible quadratics. It's not too difficult, I mean, because you think of it as, oh, it factors as x squared plus ax plus b, and then x squared plus cx plus d, and you foil that out and compare the coefficients. You have a system of linear equations. You're gonna find out that system's inconsistent. You can do that. Although on this example, I feel like the easiest argument to show that this thing is irreducible is actually just to use Eisenstein's criteria. We like to use that one a lot, because after all, all of the coefficients, except for the leading term, are divisible by the rational prime two, but the constant term negative two is not divisible by four. So this is irreducible by Eisenstein's criterion. Now, in this situation, we very well know what the roots of this polynomial are in the complex number system, because after all, if you factor this as a difference of squares, you're gonna end up with x squared minus the square root of two, and then x squared plus the square root of two. And then, so this would give us the factorization of the polynomial, and Q would join the square root of two. But then if we go one more step, we could treat each of those as a difference of squares. This will factorize x minus the fourth root of two times x plus the fourth root of two. With the second one, x squared plus the square root of two, we have to treat these as different, this still has a difference of squares, so we'll use some imaginary numbers here. So we get x minus i times the fourth root of two, and then x plus i times the fourth root of two. And so this then gives us a factorization over the complex numbers, yes. But this will give us a factorization over the field Q would join the fourth root of two, which we definitely need the fourth root of two because that's one of the roots. But we want all of these other roots. This is just negative the fourth root of two. This is i times the fourth root of two, and this is time negative one times the fourth root of two. So notice if I adjoin i onto that field there, I can produce all of the roots of this polynomial. So our roots, just so we're clear, are gonna be plus or minus the fourth root of two and plus or minus i times the fourth root of two. Those are the four roots of this polynomial in the complex number system, for which clearly if I adjoin the fourth root of two, I'm gonna get plus or minus the fourth root of two. And if you also include i, those will produce these ones. So this, I should be saying this field adjoin x, right? This field adjoin x, that's what I meant to say there. And so that tells us that the polynomial does in fact split in this field, but is it the splitting field? It contains all four of the roots, right? Is there any smaller field that will contain all the roots? We'll have to investigate that one in just a second, but we can claim that in fact, this, I claim this is the splitting field and we'll show that is the case in just a moment. All right, so as we investigate this, the splitting field of whatever, the splitting field for this polynomial g lets for the moment call it e, okay? If we look at it over q, this will have, this degree will have the same order as the associated Galois group of e over q, for which we're interested in what that is. And so let's actually say that the Galois group here, Galois of e over q, just for convenience, I'll call it g for short, all right? So we're interested in computing this g, like so. Now, a very nice trick that we can do here is that if we look at just the fourth root of two here, this is a real number. And so I definitely can separate the fourth root of two from this other element i over here. And so I want you to consider the following factorization of the degree here. So the degree of e over q, we can factor this as e over q adjoined the fourth root of two, like so. And then you're gonna end up with q times the fourth root of two over q, like so. This second one is definitely equal to four because the fourth root of two is the root of an irreducible degree four polynomial so that degree is equal to four. So we know that four divides the order of this thing. I should also mention that because we're looking at the Galois group for a degree four polynomial, we can very naturally view g as a subgroup of s four. So we're looking for a subgroup of s four that's divisible by four, all right? That doesn't necessarily give us a lot of options. We just continue on from right there. You get the client four group, that's an option. You get d four, that's an option, the dihedral group. And then besides that, I feel like the only other option would be s four itself. A four, for example, I guess A four is divisible by four, yeah, I need to include that one. I felt like there was something else missing because A four has order 12, all right? So just right now, these are the four possible Galois groups you can get. And honestly, this is what happens in general. If you have an irreducible degree four polynomial, these are the four possible Galois groups. And so we wanna figure out which one this is. Now, if every root of g belongs to q adjoined the fourth or two, then that would mean the order of the group would be four and it would have to be z four in that situation. I guess, excuse me, v four. I guess there is the, man, my list is incomplete. There's a lot of options here, isn't there? There's also the possibility you have a cyclic group of order four. Like for example, when we did the irreducible cyclic tonic polynomial for fifth roots of unity, that was an irreducible degree four polynomial and the group turned out to be z four in that situation. Ooh, so be careful about that one. So if this thing contained all of the roots of g, then in fact, we would have that the degree of the extension would be four and this would be one of the groups in that situation. I claim that's not gonna be the case. We're actually gonna be bigger than that. And this is because, looky here, we need some imaginary numbers because the splitting field, if it's not this, right? Be aware that the splitting field, e, we know for a fact it does equal q a join, the fourth root of two, we have i times the fourth root of two, right? We have negative the fourth root of two, which is already in there and then negative i fourth root of two is already in there. So in fact, we can get away with this, we can get away with this field right here. This is in fact the splitting field because this contains, we've only adjoined roots of the polynomial but we can algebraically just construct all of the roots of the polynomial. So this is in fact the splitting field. I want you to know though that the element i is equal to i fourth root of two over the fourth root of two, which is an element of this field e. So we do get in fact that with these observations here, we do get that e is a subfield of q a join of the fourth root of two and i, right? Because this contains the fourth root of two, it contains i, so it's, I think I put my any calling in the wrong direction there. Let me try it out again. Whoops, that's not my eraser. There we go. So we do get this. So i is contained inside of here. The fourth root of two is contained inside of there. So the splitting field does contain it. But on the other hand, our polynomial splits over the field q a join the fourth root of two and i. And since splitting fields are minimal extensions that make the polynomial like split, we see that in fact, these two things are one of the same things. So we can represent the splitting field like this or like this, whichever one we prefer, okay? And so in particular, this, our splitting field is not q a join the fourth root of two because this right here is a real field. It's contained inside of, let me draw a little here. This is contained inside of the real numbers. This is a real field, but e is not contained inside of the real numbers. So as such, this extension cannot equal one. That is there's something that you gain by going bigger. So how big is it? Well, if you would join just i like we did here, be aware that over the rational numbers, the minimal polynomial of i is of course, x squared plus one. It's a quadratic polynomial. And so when you look at q a join the fourth root of two, does this polynomial split or not? If it splits, that actually means i already belongs to a real field, which is impossible. So that tells us that this field extension is in fact two. So now that we have got both pieces, this tells us that our field extension has degree two times four, which is equal to eight. And while we had a little bit of confusion beforehand on what are the possible Galois groups we can have, I can definitely tell us that the Galois group in this example is going to be D four. How do I know that? Well, when you look at S four, its order is 28. 28 factors as eight times three. So since our group G is a subgroup of S four, S four up to isomorphism has only one group of order eight because eight is the order of a Seeloff subgroup, which Seeloff subgroups are always isomorphic to each other. For S four, of course, you can have a couple of different Seeloff subgroups. We don't have to go through that but they're all gonna be isomorphic to D four. And so the fact that we have a order eight subgroup of S four that tells us by Seeloff theory it's gotta be D four. And so that then answers the first question, what is the Galois group of this polynomial X to the fourth minus two? What I wanna do next is then consider this Galois correspondence because D four is a group we have studied extensively in the past here. So I'm gonna scooch this up a little bit more here. And what I'm gonna do is quickly draw the Hase diagram for D four, but I'm gonna intentionally draw it upside down. And the reason I do that is because the Galois correspondence throws things it switches the order of things. So the big becomes the small and the small becomes the big. So I'm gonna put D four at the bottom and put one at the top. And so then the group, the Hase diagrams can have the following structure. You have a cyclic group of order four and this is generated by the four cycle R. You're gonna have two other groups of order four. You have a Klein four group. This one would be generated by say R squared and S. You have another Klein four group V four which is generated by R squared and RS, like so. Likewise, there's going to be, there's lots of groups of order two of importance is the subgroup generated by R squared which is the center of D four. In fact, it's a subgroup of all of these four groups of order four. But then there are other ones of course there's gonna be a cyclic group of order two generated by S. There's gonna be a cyclic group of order two generated by R squared S. Those are contained inside of this group and of course contain the trivial subgroup. We also have a cyclic subgroup generated by RS and we have a cyclic subgroup generated by R cubed S. They're contained in that and they each contain the trivial subgroup. So while I'm not necessarily the best drawing of this Hase diagram, I can tell you that this is in fact the Hase diagram for D four. And so the question then comes down to what are, what's the correspondence? I drew this Hase diagram upside down because that's how I want to look at the lattice as in terms of fields. So the big correspondence of the fall because Galois correspondences are order reversing. So that means the smallest field coincides to the largest group. So you're gonna have q on the bottom and then you're gonna have q adjoin the fourth root of two comma i on the top which is equivalent of course to q adjoin the fourth root of two and i times the fourth root of two. You can get it with that as well, whichever representation you want. I forgot to draw one line that looks like right here. Okay. And she mentioned that all of these all of these degrees are degree two, degree two, degree two, degree two, all of these. I mean, after all d four is a two group. So these extensions have to be degree two. P groups always have those extensions there. All right, so we have the obvious ones taken care of, what more can we do? Well, we might need to specify exactly what are these maps? Like can we make sense out of what R and SR? So R is gonna be a four cycle. If the four cycle is gonna be constructed by permuting the four roots of our polynomial g of x. And so for our representation, we can take R to be the fourth root of two, maps to i times the fourth root of two, comma negative fourth root of two, and then comma negative i fourth root of two, like so. So if you think about that just for a moment, R is gonna be this four cycle. It's gonna permute cyclically, there are four roots. So it goes from a real one to the imaginary one to the negative one to the other imaginary one and then back again to the positive one. And so that's what our permutation's gonna do. R cubed would do similar, but work through this thing backwards. What's gonna be fixed by this operation here? Notice that we send a real to an imaginary, purely imaginary, then back to a real, then back to a purely imaginary. What does it do to i after all? So if I take R of i, what's gonna happen there? Well, the way that this is explained, it has to do with the roots. i's not actually a root of this thing, but I can still make some sense out of it, because after all, like we mentioned earlier, i can be written as i times the fourth root of two divided by the fourth root of two, for which as this is a field operation, we can, a field automorphism, we can apply this R to the numerator and denominator separately. i root two is going to go to negative root two and then at the bottom, we have the fourth root of two, that's gonna go to i times the fourth root of two. The roots of two are gonna cancel, so we get negative one over i, for which that simplifies as a complex number just to i. So i is actually left fixed by R, and in fact, the fixed field associated to this automorphism group R is exactly that. This is going to be the field q of joint i. After all, we know that the fixed field associated to R must be a degree two extension of d four. q i does exactly that. So the fixed field in the situation is going to be q i. All right, that was quite kind of helpful. What about the other generator of the dihedral group? So we're using the fact that the dihedral group d four can be written as R comma S, right? What about the other maps? Well, R squared is relevant to bring up here. What does R squared do? R squared is gonna swap these two. So we end up with the fourth root of two, minus the fourth root of two, and then these two are gonna get swapped. So you get i fourth root of two and then negative i fourth root of two. What does it do to i? Well, since R fixes i, R squared will also fix i. So when we look at these things, we'll come back to this one in just a moment because after all, we're gonna be very interested in what's fixed by this right here. We can say some things. Actually, what we'll talk about right now, we can look at some things here because notice that i is fixed by R, so it's gonna be fixed by R squared, but there's gotta be something else. There's gotta be, there's a bi-quadratic extension going on here. So a second square root, what else is gonna get fixed by this? Well, I want you to note that if you look at R of the square root of two, the square root of two is actually just the fourth root of two times the fourth root of two. And so when you apply i to that, you're gonna get i, excuse me, if you apply R to that, you get i fourth root of two and then you're gonna get i fourth root of two. So when you put that together, you're gonna negative the square root of two. So R does move the square root of two to its conjugate. Well, then if we square things, this actually, you'll switch back. So actually the square root of two is fixed by this map as well. And so that then gives us that bi-quadratic extension we are looking for. This fixed field coincides with the square root of two comma i, okay? And so then there's gonna be two other subfields other than q adjoin i. And I can predict exactly what they're gonna be here. One of them is gonna be q adjoin the square root of two. The other one is gonna be q adjoin i times the square root of two. These are both quadratic extensions that live inside of this field. After all, we talked about bi-quadratic extensions in a previous video. So these fields have to coincide in some regard. Now, because we have a specific meaning for R squared and S and these things, we should be more precise. Up to symmetry doesn't matter too much, but let's figure out which one is which. What is S after all? S, there are some options here, but I'm gonna take S actually to be complex conjugation. So we're gonna send i to negative i and negative i backwards. In particular, the fourth root of two is left fixed by that, okay? So if we take, let's look at the element which is fixed by R squared and it's fixed by i, excuse me, it's fixed by S. Now, R squared is fixed, it fixes the square root of two and since the square root of two is a real number, it's also fixed by complex conjugation. So that tells us that the square root of two is fixed by both R squared and S. Q adjoin the square root of two is a degree two extension, so that's the one that fits the bill there. So then by process of elimination, this one's gotta be Q adjoin i times the square root of two. But I want you to check that for a moment. R squared, it fixes i, it fixes the square root of two, so that's an option. What does R S do in this situation? That one can be a little bit more difficult but I'm going to squeeze this over here for a moment. What would R S look like? So if we write this as a permutation, R S is gonna look like the fourth root of two is going to swap with i times the fourth root of two. It's going to move around i and negative i and then lastly, it's also gonna move around negative the fourth root of two. Let me scooch over a little bit more and negative i times the fourth root of square root of two. Now it takes a little bit of effort to see those things but be aware that's what R S is going to do. And so we should probably also include R cubed S by a similar calculation. You can verify that the fourth root of two is going to swap with negative i times the fourth root of two. We're gonna get that, of course, i and negative i swap places. If we know that, we know everything else but I'll be complete here. And we also get that negative the fourth root of two is going to swap with i times the fourth root of two. And so that gives us the complete picture in that situation. And so what's fixed by all of these elements? R squared, like I said, it's going to fix i and the square root of two. R S, notice what happens here, is that if we look at i, it's i will swap with negative i and the square root of two. So if we do this thing twice, right? Much like we saw before, S doesn't do anything to the square root of two. So if you do this to the square root of two, R will send it to its conjugate. So you end up with negative square root of two. That gives you a double negative so you do end up with i root two. So in fact, this is the correct fixed field there, degree two extension. All right, coming back to our picture. All right, let's fill in the missing pieces left. Well, this is gonna be a lot easier given the information we have already. We now need a degree four extension and this is gonna coincide to those things fixed by complex conjugation. So this has to be a real field and in fact it's gonna be the maximal real field, which is going to be the, Q would join the fourth root of two, okay? What is then fixed by R squared S? Again, this has to be a degree four extension and this is going to be Q would join i times the fourth root of two. Notice I need a degree four extension, which I know this is a degree four extension because it's a root of the same polynomial of G here. So it's degree four, but it has to also contain Q join the square root of two as a subgroup. So this is the only other one that does that, but you can also verify that R squared S is going to fix this element and exactly that. S does nothing to the fourth root of two, R squared sends the fourth root of two to its negative. S will send i to negative i and R squared S will do nothing to i. So the net effect is a double negative. So this is the fixed element. So that side was a lot easier. Let's then come back to this side. I'm gonna show RS and R cubed again because this is the hardest part and this is where the Galois theory really is helpful because we can pull from the groups exactly the element here. What's gonna be left fixed by RS? Believe it or not, this is gonna be the element Q join one plus i times the fourth root of two. Now this is not a root of the polynomial, but this is an element that lives inside of our splitting field clearly enough and be aware that this does contain i squared of two. And you can mostly see that by taking the square of this thing. If you take, for example, one plus i times the fourth root of two and you square it, be aware that you're gonna square one plus i which is going to give you two i in that situation. You square the fourth root of two, you're gonna get the square root of two. You can divide by two and thus you obtain i root two. So this seems like an appropriate candidate right there. Also, we should mention where did I even come up with this element in the first place? Well, I looked at this cycle right here. If you add the cycle together, you end up with the fourth root of two plus i times the fourth root of two which of course, if you factor, you get exactly this element right here. That's what I was going for. And so that's a fixed element, but it's also, it's not in any proper subfield. It's an extension in that regard. Looking at r cubed, you can do a similar thing. You're gonna get a q adjoined, one minus i times the fourth root of two, like so. And then I'm gonna zoom out here. We can now look at our lattice here of subfields. By analyzing the subgroups of D4, we now have completely determined all of the subfields of the field. Q adjoined the fourth root of two and i. And it's pretty, pretty awesome how we were able to use the Galois group to completely determine all of these subfields and how they're related to each other.