 In this video, we are going to compute the lattice of subfields for the field e, which is which for us in this video is going to be q, the rational numbers are joined the square root of three and the square root of five. Now, notice that e is in fact the splitting field for the polynomial p of x, which is x squared minus three, and x squared minus five. Now, of course, if we view this as a rational polynomial, we get that x squared minus three and x squared minus five are, in fact, irreducible polynomials. They're quadratic, so we can check by the rational roots theorem, the only possible roots would be plus or minus three and plus or minus five, respectively, that doesn't work. You could also use Einstein's criterion to show, since three and five are prime numbers as integers, that this polynomial, these polynomials are irreducible. Okay. Now, of course, p of x, if we think of it as a polynomial over e, then these polynomials actually factor, you're going to get x minus the square root of three, you're going to get x plus the square root of three. You'll get x minus the square root of five, and then you're going to get x plus the square root of five, like so. And so we see that this polynomial does in fact split over the presumed splitting field e in that situation. And so yeah, that's what we're going to get here. We get four possible roots for this polynomial. We're going to get plus or minus the square root of three, and we're going to get plus or minus the square root of five. And so if you view these as complex numbers, right, the complex field is in fact an algebraically closed field by the fundamental theorem algebra. So every rational polynomial will split inside of c. And so oftentimes we go to the complex numbers to try to understand what the roots of these polynomials are. And so this polynomial p has four roots. There's the square root of three, negative square root of three, square root of five, and negative square root of five. Notice that our field e contains exactly those four elements, right? Because if you contain the square root of three, then you're going to contain negative square root of three. It's added inverse. And if you have the square root of five, then you're going to contain the negative square root of five. So we see in fact that e is the splitting field for p, because this is the minimal extension of where we joined roots of p to q, and we got this. So we're going to join all four of the roots, because as the roots come in these additive inverse pairs, once you took one from each, you got the other. And so I want to mention that because of the factorization of p right here into two irreducibles, you'll notice that these two numbers here, the square root of three and negative square root of three, these are conjugates of each other because they have the same minimal polynomial because that's x squared minus three. And likewise, plus or minus the square root of five, these are likewise conjugates of each other, because again, they have the same minimal polynomial, x squared minus five. So there is this relationship between the two. The two are very much close to each other in that regard. So to make conversation a little bit easier, we're going to introduce some other notation here. So we're going to introduce the field f one, and this is going to be the field q adjoin the square root of three, where likewise going to introduce the field f two, which is going to be q adjoin the square root of five. And so I want to analyze these fields for a moment. So if you look at f one, there exists an automorphism sigma, which goes from f one to f one, which be aware that f one as a set as a vector space it looks like the span, which we're going to take rational coefficients here, as there's lots of different fields that we might want to specify which vector space we're talking about. This is a q span here. This is going to be the span of elements of the form one plus the square root of three. Since the minimal polynomial, the square root of three is x squared minus three, we see that f one is going to be a degree to field extension of q. So f one over q. So it's going to be a degree to extension. It's a quadratic extension at situation. So coming back to sigma here, sigma is the map that is determined by the relationship. So sigma will, sigma will take the element a plus b square root of three, and it maps it to a minus b times the square root of three. So in a previous algebra class, like college algebra, intermediate algebra, so like math 1050 math 1010 at SUU, you would call this the conjugation map, right, you're replacing the number a plus b square root of three with its conjugate. Right, this notion of conjugate is trying to generalize that concept we're replacing the number with its conjugate. Essentially, this is the map that sends the square root of three to the negative square root of three. There is an onomorphism there. I want us now to return to the field e. We could actually represent the field e as f two adjoin the square root of three, because after all, f two is q adjoin the square root of five. So if you take the square root of five and then you would join the square root of three, that's exactly what the field e was to begin with. Right, so e can be written as an extension of f two. So in particular, we have that f two sits in between e and of course q, like so, and you can accomplish this by joining the square root of three. I want you to be aware that this map sigma, which we're viewing it as an f one map, automorphism, this can actually be extended to map on e. Okay, because in that situation, if you think of it not as if you think of e as a rational span, your basis is going to be four elements. But if you think of this as an f two span, this is exactly the same thing. And so e over f two is still going to be a quadratic extension, and we can do this exact same map as before. So for now, instead of thinking of a and b as rational numbers, we think of these as elements of f two. Alright, and so sigma, we very much can think of this map from e to e. But with this perspective, note that yes, sigma is going to send the square root of three to its conjugate the square root of three. But it's also when you restrict it to f two. This just gives you the identity map on f two. So in particular, sigma, we can identify with a map inside of the Galois group e over f two, it fixes f two. So in particular, this is a subgroup of the Galois group of e over f, which of course is q in the situation. So this map, which it fixes the square root of five, but conjugates the square root of three. So be aware that our map is looking something like the following. If we take an arbitrary element, we'll say that a plus b square root of five, like so. And then you have a c plus d square root of five times the square root of three. This map is going to send it's going to lead and so for each of these numbers a b c d. These are in fact rational numbers. This map is going to send the first number which belongs to f two, it's just going to leave it alone. And then with the second one, you're going to take the negation there. And so you get c plus d times the square root of five times the square root of three. So that's what that thing ends up doing in that situation. All right, I want you to note that this map is order two, right? If we take Sigma squared, you get back the identity on the identity on e, by the way. So this was the identity earlier on f two. But if you do Sigma squared, I believe the identity on everything, everything is left fixed in that situation. If you do it twice. Now, if you do it only once though, because after all now we're very interested in what's the subgroup generated by Sigma here. It will contain just the identity in Sigma. So as a group, this thing is isomorphic to the cyclic group of water two. I'm interested in what is the fixed field associated to this, this map Sigma right here because notice when you look at this expansion right here, we have this a plus b times the square root of five. We have negative c squared of three minus d times the square root of three times the square root of five, for which you can put those together and just think of it as a square root of 15, like so. If this equals the original number a plus b squared of five plus c times the square root of three plus d times the square root of 15 here, you'll notice because of the difference of signs here that since these things are linearly dependent, we have to have that c equals negative c, which implies that c equals zero. And so you see the same thing on the square root of 15 right here as well. We get that c equals d equals zero, if this is in fact a fixed element. But with a, you just get a equals a, so a could be anything b could be anything. So I want us to note here that if we look at the subfield of e that is fixed by Sigma, we get exactly f2. So the way we've constructed this Sigma conjugates the square root of three but it leaves the square root of five alone. Therefore the fixed field is in fact going to be f2, which was q adjoined the square root of five. All right, so that gives us one of the automorphisms in our Galois group. I want you to be aware that by a similar reasoning, we have some map tau, which is a map from e to e. And for what's it going to do, it's going to, it's going to send the square root of five to negative square root of five like so. And it's also going to leave tau when you restrict it to f1. Now it's going to give you the identity in that situation, the identity on f1. So in particular, if you take something like tau of a plus b times the square root of five plus c times the square root of three plus d times the square root of 15, like so. What this map is going to do, it's going to negate the square roots of 15. And so you end up with, in this situation, a minus b times the square root of five plus c times the square root of three. That's the square root of 15. It's the square root of five times the square root of three. So that square root of five gets replaced with a negative square root of five. So it also will conjugate the square root of 15, like so. Now, if this is a fixed element, if this is a fixed element, make a comment there about that. If this is a fixed element, then this should equal a plus b times the square root of five plus c times the square root of three plus d times the square root of five. These are different signs, their coefficients would have to be 0, b, and d are both 0, and so then you have no restriction on the choice for a coefficient a, no restriction on the coefficient for c. So in that situation, we see in fact that the fixed field associated to the subgroup of tau is going to be f1, which is q adjoin the square root of 3. Again, just like with sigma here, notice that tau has order two as an on a morphism. If you do it twice, you get back, you get back the identity right there. All right, so with that said, let's make a few other observations here. Let me move on to a new page here. What do we know? So we have that the Galois group of e over q, it clearly contains the identity. It contains this element sigma, it contains tau as well. I'm not sure I wrote them in the different order, right? Then there's potentially some other things going on inside of that. We'll come back to that in just a second. I claim that we know the order of this Galois group. The order of the Galois group is in fact equal to, well, since it is a Galois extension, this is going to equal e over q, the degree of that extension. Well, how big is that? Picking one of your favorite of the intermediate fields. So we'll take e over, say, f1 here, and then f1 over q. We can factor in that way because, like we already observed, these fields f1, f2 sit in the middle. So you have e, it contains f1, it contains q. We could also done f2 would make much of a difference. Now, f1 here, of course, is q enjoying the square root of three. And the middle of a polynomial we mentioned was x squared minus three. So this degree is two because we had joined a quadratic element to it. Now, I claim that when it comes to this one right here, this extension is the same thing as the degree of f2 over q. And the reason I say that is because e, and we'll leave that on the screen here, the reason I say we can do that is because in this situation, e is actually equal to the field extension f1 enjoying the square root of five. We did it the other way around before, but this one's also true. So e can be formed by extending f1 by enjoying the square root of five. And the square root of five does still have as its minimal polynomial x squared minus five. Its minimal polynomial over q is x squared minus five, which means that its minimal polynomial over f1 has to divide x squared minus five. As this is a quadratic polynomial, if it factors at all, it has to split. And so that would mean that f1 contains the roots of x squared minus five. That is f1 would have to already contain the square root of five or the negative square root of five, which it doesn't have that. You can argue that in fact that f1 does not contain any square roots of five whatsoever. So this is a proper extension. So it's going to have to be degree two as well for the same basic reasoning. And so when you put this together, the Galois extension has degree four, which means the Galois group has order four. Now, how many groups have order four in general here, right? If we take our Galois group of e over q, I'm going to call it just g for short. I hope that's okay right now. I'll do curly g right here. There are basically two options. This group is either the cyclic group of order four or it's the Klein four group. That is z2 cross z2. So those two options. And so it turns out with the information I have already, I can determine which of these two groups we have. Because notice, clearly we have the identity, but our Galois group contains two automorphisms of order two. Which of these groups can do that? Well, the Klein four group has three elements of order two, but the cyclic group though has one unique element of order two. Besides the identity, other two elements of the cyclic group are elements of order four. And so it turns out that our Galois group can't be cyclic because we have too many elements of order two. And as such, it turns out that this means that the Galois group has to be the Klein four group in this situation. So that's what we have here. Our Galois group is in fact equal to the Klein four group. And that tells us that the fourth element we haven't accounted for yet has to be the product of the other two elements of order two. So our Galois group is going to be one tau sigma and sigma tau. So what does sigma tau do though, right? If we look at that sigma tau or tau sigma, it turns out while functions in general don't commute, these ones will. If I take an arbitrary element a plus b root five plus c root three plus d root 15. Let's apply these two. Tau is going to conjugate the square root of five, thus also the square root of 15. So this is going to give us a minus b root 15 root five plus c root three minus d root 15, like so. Then what does what does sigma do in this situation? Sigma conjugates the square root of three for which, in that situation, then you're going to get a minus b root five. Sigma does nothing to that. You're going to get a negative c root three. But then the square root of 15 will be conjugated again. Because again, the square root of 15 is just the square root of five times the square root of three. Tau negated the square root of five. Sigma is going to negate the square root of three. And so when you do them together, you're going to actually do nothing to the square root of 15, right? So notice that the elements that got their coefficients changed. The b is now a negative b and the c is now a negative c on the square root of five and the square root of three. The square root of 15 actually was left alone. It has the original coefficient here. And so if this is supposed to be some fixed element, so this is what sigma does. Sigma will conjugate the square root of three and square root of five but not conjugate the square root of 15. Tau, remember, conjugates the square root of five and square root of 15. And sigma conjugates the square root of three and the square root of 15. Alright, so that's what's happening here. If this were a fixed element, this is supposed to equal a plus b root five plus c root three plus d root 15. Like before, we'd see that you'd have to have b and c both equal zero so they're going to vanish. And so we see that the fixed field, the fixed field e sub, the subgroup generated by sigma Tau here, this is going to fix the field q adjoin the square root of 15. The square root of 15 is the thing fixed by this element. And so we're going to call this field F3 for the, for the sake of discussion here. And so I want us then to see what are the, the lattices that we have been constructed with regard to the Galois correspondence. Let me draw the fields over here. So our top field was q adjoin the square root of three and the square root of five. There's the base field of q. And then we've discovered there were three intermediate fields. We had q adjoin the square root of three, which we called F1. We have q adjoin the square root of five, which we called F2. And we just recently discovered there actually is a third field, which in hindsight makes total sense. We have q adjoin the square root of 15. And so then when we draw this picture, this picture looks exactly like the lattice for the Klein 4 group, for which, just so we're aware, if we look at the Galois group of e over q, that gives us the full Galois, the full Galois group, which is v4 in that situation. If we look at the Galois group of e over e, that'll just be the identity. And so you get one right there. And then we have some intermediate fields based upon fixed fields. There's the subgroup sigma here. There's a subgroup generated by sigma. There's the subgroup generated by tau. Remember that tau, the subgroup generated by tau actually fixes the square root of three. And the subgroup generated by sigma, it fixes the square root of five. And then the subgroup generated by sigma tau, it fixes the square root of 15. And so we see this picture here as well. Right? If I mentioned the degrees of these extensions, these are all degree two extensions. And we see the same correspondence happening over here. We have two, two, two, two, two and two. And so we see this correspondence between the lattices. Now this example can be a little bit deceptive, because the lattice for the Klein 4 group is the same top down versus bottom up. The lattice upside down, you get the exact same picture. So it can give you this misconception that the correspondence looks like this. This field corresponds with this subgroup. And this field corresponds with this subgroup. The bottom coincides with the top and the top coincides with the bottom. So you do flip this thing upside down. And so therefore, this degree, this extension coincides with this one down here. You flip it upside down. And this extension coincides with this extension right here. And lastly, this extension coincides with this one. But there's so much stinking symmetry happening right now, it can be a little bit deceptive. Now this example, we worked with the square roots of three and the square roots of five, but there's nothing particularly special about those square roots. In fact, the example could have been done where you take the square root of some number p and the square root of some other number q, which basically we just require that p and q are not themselves perfect squares. We require that the GCD between p and q is one. So we want them to be a coprime, for which, you know, this would work if p and q are both prime numbers like three and five. And this actually gives us what we refer to as a biquadratic extension of the rational numbers. And it turns out that with the right assumptions that the Galois group of a biquadratic extension of the rational numbers will always be the client for group. So if you kind of botched some of those assumptions I just mentioned, then you'll get a subgroup of the client for group, or I guess maybe it's a quotient group, but whatever one it is, you might get like a Z2 as your Galois group. And these are biquadratic extensions of the rational numbers. If you do other fields, you can get things similar to this. And so this is a very well understood and one of the simplest types of Galois groups to study that are beyond just a simple extension. So this is a good example to get started with, despite the fact that one might be confused with the Galois correspondence because the lattices have symmetry from top to bottom there.