 Welcome back to our lecture series math 42-30 abstract algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. Now in lecture 34, we introduced the fundamental theorem of Gaoua theory and established this very important Gaoua correspondence between subfields of a field and the subgroups of the Gaoua group. And we did some examples and we computed two Gaoua groups of order 4. In lecture 35, I want to do some more computations of Gaoua groups. These ones will be a little bit more involved, so do stick with me through these ones. And so we want to establish what the Gaoua groups are for these polynomials and then just demonstrate what the correspondence between the fields and the subgroups is going to look like. So in this video, we're going to tackle the polynomial x equals x cubed plus 5x plus 5. And so when we look at this polynomial, is it an irreducible polynomial? The answer is, in fact, yes, this is an irreducible polynomial. How do I know that? Well, one option of showing it, since it's a degree 3 polynomial, I could just check all of the possible rational roots. We're viewing this as a polynomial over the rational numbers. So this belongs to q adjoint x. And so we can look at all the possible rational roots. By the rational roots theorem, this would be plus or minus 1, plus or minus 5. And so it would be too exhausting to check all of those, especially if you synthetic division. But a much cleaner argument actually would just be Eisenstein's criterion. Notice that all of the coefficients besides the leading coefficients are divisible by the prime number five. But five squared does not divide the constant term. So this is, in fact, irreducible using Eisenstein's criteria and the prime five. So we do get this as irreducible. And so not knowing exactly what it is yet, we're going to let e be the splitting field. We're going to let e be the splitting field of our polynomial f right here. Okay. And so because of that, since f is an irreducible polynomial, the three roots of f are all conjugates of each other inside of the algebraic closure. And so every field automorphism is going to permute conjugates to conjugates. And so there is this action going on here. So we can, we can take say g to equal the Galois group, the Galois group of e over f. So it's the Galois group of the polynomial. We can actually view this as a subgroup of s3, which of course, s3 has ordered six. And now the first thing we can do is let's take a single root of the polynomial. Okay. Let's say that alpha is a root of the polynomial. So alpha belongs to e such that f of alpha, which is going to equal alpha cube plus five alpha plus five is equal to zero. Okay. So in particular, this relationship does tell us that alpha cube is equal to negative five times alpha plus one. This is the relationship we can infer from this polynomial here. So that's what we know about alpha. Now if we wanted to use something like the cubic formula, we could come up with the exact expression of what alpha would look like plus it's two other conjugates using square roots and cube roots and things like that. That gets very, very complicated, but one could do that. I'm trying to compute what the Galois group is from a more theoretical point of view that I don't necessarily need to know what alpha is to know what the Galois group of this polynomial is going to be. So with our field e, there is one very important subfield we now can establish. So there's going to be this subfield, the subfield q adjoint alpha. All right. If you look at this subfield q adjoint alpha over q, this is going to be a degree three extension. Okay. How do we know that it's degree three extension because it's a root of a irreducible degree three polynomial. All right. So noticing now that because of this, because we know this, we can also do one better. Notice that e over q, these degrees can factor as e over q adjoint alpha times q adjoint alpha over q for which because we know the second one, we're going to get three times e and q adjoint alpha like so. And that's, so we know this degree is divisible by three, but because this is, because e is a splitting field over the rational numbers, this is going to be a Galois extension over the rational numbers, normal extensions and Galois extensions are one of the same thing because characteristic zero fields are separable. This is the order of our Galois group G. So this tells us that three divides the order of the Galois group and we're contained inside of S three. So how many subgroups of S three are divisible by three. So this is, this is now where the situation we're in right now. We get that G is going to equal either all of S three, or it's going to equal the alternating group a three, which of course is just in this case just a cyclic group of order three. And so with this, with this in mind here, we now can stay a state a very general principle. If you have an irreducible cubic polynomial over the rational numbers, then your Galois group is either the cyclic group of order three or the symmetric group S three. Those are your two options. And then comes down to deciding which of those two options are you in, right? Are there three automorphisms or are there six? It depends. Does this thing, does this element here alpha, does it generate the other conjugates or does it not, right? If you have all, if alpha can generate all the other conjugates, then in fact you get that your Galois group is going to be Z three in that situation. If not, if there are, if there's a conjugate that doesn't belong here, that means we have to extend the field one more time because the question comes down to, is this extension one or two? If it's, if this extension has degree one, then our Galois group is going to be Z three. If our, if it's two, then our Galois group is going to be S three. And again, that just comes down to, can the other conjugates be generated by this element or not? And it depends on the polynomial, right? And another strategy that some people try to do here is we can develop this idea of the discriminant, right? With the quadratic, with the quadratic polynomials, you know, you have X equals negative B plus or minus the square root of B squared minus four AC, all over two A, right? This discriminant tells you a lot about the nature of the solution because this discriminant B squared minus four AC can tell, you know, if that's a perfect square, then you'll have rational solutions. I guess if it's zero, you'll just have one. It's still going to be rational solutions to that situation though. If it's negative, you're going to have some imaginary solutions. If it's, if it's positive, but not a perfect square, you'll get irrational solutions that are real. So the discriminant tells you about the nature of the solutions, which in the context of Galois theory, it tells you about field extensions. Does the root of your polynomial leave the rationals? And if it does, does it go to the reals? Does it go to the complex? This tells you about the degree of the extension, right? If the discriminant is a perfect square, you're going to get that the quadratic extension is only degree one. But if the discriminant is not a perfect square inside the rational field, of course, then your quadratic extension will be degree two, right? So just like quadratic polynomials, this notion of a discriminant can help us out here a lot because the discriminant does make sense in this context. If we generalize the theory for which then if that discriminant turns out to be a perfect square, we're going to get that the Galois group is z three. And if it's not a square, then we're going to see the Galois group is s three. And so that's how you can handle in a purely algebraic manner. But again, I don't want to, I don't want to develop the theory of generalized discriminants in this lecture series right now. So I'm going to utilize a little bit of topology to help us out here. So thinking of this polynomial as a real polynomial, notice that as we take the limit as x approaches infinity, you're going to get that f of x approaches infinity, right? And then if you take the limit as x approaches negative infinity, we're going to see that f of x approaches negative infinity that situation. So we think of it as a graph, right? Our function, because the leading term is a cubic here, it's going to point up on the right hand side point down on the left hand side. So by the intermediate value theorem, if we think of this as a real value polynomial, it's going to give us that there's some point of intersection somewhere with the x axis. So there is some x intercept, that would be a root of the polynomial. So this polynomial, because it's a cubic polynomial does have a real has a has one real root at least, okay? There could be more real roots, there could, but there's at least one. So in fact, we know that there's going to be either one real root, or no, there's one real root or two real roots. So thinking of our polynomial here for a moment. So let me just sketch it down again. So our polynomial was x cubed plus five x plus five. How was it five x squared? I already forgot. Nope, x cubed plus five x plus five. Okay, there we go. So we know it has at least one real root. If we take its derivative, right, this becomes three x squared plus five. In the real number system, x squared is always non-negative. If times three, it's still non-negative. If you add five to it, this thing is always positive. Okay? And so in particular, f prime here is going to be always positive. And since it's always positive, that translates to saying that the function f is going to be always increasing. These of course are consequences of the mean value theorem. And so because the function's always increasing, this tells us that has one and only one real root. Now for different cubic polynomials, you could have three real roots. That's possibility, but you have one real root. Okay? And so these again, these are consequences of the intermediate value theorem, which gave us the first real root. The mean value theorem is basically telling us there's not a second real root. So we have just the one real root. But there are three roots to the polynomial. By the fundamental theorem of algebra, this polynomial, if you view it as a complex polynomial, it has three roots. And as a consequence of the fundamental theorem of algebra, the complex roots are going to come in conjugate pairs. So let's say that alpha here was the single real root. All right? It's the only real root of this polynomial. And so then we're going to take these two other roots, right? So we have two complex roots. I should say two non-real roots. Every real number is a complex number, of course. There's two non-real roots. We will call one of them beta. And then the other one I'm just going to call it beta conjugate. Okay? Because they're going to come in conjugate pairs like so. And so then the splitting field, the splitting field E, then is going to be formed by taking Q adjoin alpha and beta and beta bar like so. Okay? Now be aware that whenever you adjoin two roots of a polynomial, you automatically, excuse me, if you include all roots except for one, you automatically get the last one in there as well. So this field, we actually can't identify with Q adjoin alpha and Q adjoin beta like so. And so I want us to think about the possible automorphisms. One automorphism you have is that this is an automorphism from E to E, is that you can conjugate these things, right? So you're going to have that sigma since alpha to beta. And there has to be a three cycle from what we discussed before. So if alpha goes to beta, then that tells us that beta would have to go to beta bar. And then that would have to go back to alpha when we're done. So I actually can write these things as permutations. So we can write this as the permutation alpha, beta, beta gamma. So that's got to be an element of our group G. You also would have its inverse, sigma inverse, which is going to be alpha, beta bar, beta that belongs to G. I also can then take, we're going to take tau here to be a map from E to E for which this map is going to be complex conjugation. All right. So this is the map that's going to send beta to beta bar. And then alpha actually is left fixed because alpha is a real number in that situation. So we're going to end up with this. Notice that tau is a map of order two, excuse me, it's order two, sigma is a map of order three. And so clearly because our subgroup has, our group has elements of order three and two by Lagrange's theorem, it's got to have order six. So this in fact tells us that the Gaoua group in this situation is in fact going to be all of S3. Since we know the Gaoua group is now S3, let's think of the Haase diagram associated to S3. So as a group, we have S3. There's the alternating group that lives inside of it. So just the three cycles and identity. We're going to have our transpositions. So, you know, you have the subgroups here by one, two, y, one, three, and two, three, like so. And these all sit atop of the identity, like so. And so if we label the degrees of these extensions, this is three, two, two, two, and this one is two, three, three, three. So as we start gathering the subfields of our field here, then it's got to be true that the lattice for the subfields is going to look like S3 right here, although it's going to be flipped upside down. Okay, so that is the top, the top group coincides with the bottom field and vice versa. So our diagram is going to look something like this, where E is going to be on the top, Q is going to, I'm going to, I'm going to move it down just a teeny bit like so. So we're going to have E over here as our top field, and then Q is going to be at the bottom. And then we're going to flip all of these things. There's going to be fields that go like this and down like this for which we can then describe their degrees. These will be each degree three extensions. These will be degree two extensions. And then we also have this other field over here for which its extensions will be two and three. So those degrees are all flipped upside down there. So we want to figure out who these other fields are. So we have to look for a field. Some of the subfields of E are going to be degree three extensions of Q that we can see right here. Now we already have a candidate for one of them. One of them is going to be Q a joint alpha. Okay. So Q joint alpha is degree three extension of Q. And so that's got to be one of these three. And that also gives us natural candidates of who the other two are going to be. Because we could take Q a joint beta, or we could take Q a joint beta conjugate. Because those are all each roots of the same minimal polynomial. And thus attaching any one of them to Q will give a degree three extension of Q. Now if you had a field that contains two of these, like you've got alpha and beta, you'll automatically get the third one. So if you have two roots of this polynomial, you necessarily have to have the third three roots. And the reason for that is because if you have two of the roots, then the minimal polynomial has to factor as x minus alpha, x minus beta. And then the last factor has to be linear. So it would have to be x minus beta bar. So you have all three roots there. So you can't stick any of the roots together. So any Q a joint alpha beta, that's going to be E, right? This is going to be Q a joint alpha beta like so. So we have to have these are all distinct conjugate fields that are going to be isomorphic to each other. Excellent. Excellent. So what is then the field that's going to coincide with this subgroup A3 over here? It has to be a quadratic extension of the rational numbers here. But it's going to be a degree three extension up here. Okay. So really, we know essentially what this is. This is going to be the whole field fixed by A3. What is A3 going to be there? What is A3 in that situation? A3, of course, as a subgroup, it's going to be 1, 1, 2, 3, and then 1, 3, 2. But when we write these 1, 2, 3s, we really should think of it as, oh, okay, I'm going to take my roots alpha beta bar there, and then you have alpha beta bar beta. That's really what the transmissions are, the permutations, excuse me, in this situation are. And so honestly, these other fields, I can improve upon them. So how we should be thinking of as the following way, right? If alpha is fixed, it's going to be fixed by the map that sends beta to beta bar. This was actually the complex conjugation that we had from before. What's going to fix beta? This is going to be a map that sends alpha to alpha bar. And then what's going to fix alpha bar? This is going to be a map that sends alpha to beta. And so as permutations, that's what this thing is going to look like. That's good. Everything will be induced from them. So we have to look for things that are going to be fixed by alpha beta beta bar, okay? So can we say better than just it's the fixed field associated to A3? Absolutely we can. And this is where that theory of discriminance really comes to play here. So while I'm not going to develop the full theory, I'm going to say the following. If you have a cubic polynomial of the form x cubed plus ax plus b, then its discriminant will have the form negative 4a cubed minus 27b squared. And in particular, if this number is a perfect square, then it shows that the Galois group will be contained inside of A3. If this number is not a perfect square inside of the rationals, then your Galois group will not be contained inside of A3. So it's going to be a subgroup of S3 that's not contained inside of A3. We already know, we've already established that the group here is going to be S3. And so this right here, this quadratic extension, because after all, this is going to be a quadratic extension, this is going to be the field q adjoin the square root of our discriminant. So we just have to compute what that is. Now in this example, our a and b value are both 5. So we end up with negative 4 times 5 cubed minus 27 times 5 squared. Factoring this, this becomes 25 times negative 47, in which case 25 is a perfect square. So really, we're adjoining the square root of negative 47. That's what this other field is going to be. That seems kind of mysterious as it's sort of like weighed by magic wand, discriminants. But one can further develop this concept of discriminants for Galois groups and help you fill in those pieces. But even if, even if we struggle to find a good primitive element to represent this field, I did know it is all of those elements of the field that are fixed by this permutation. And for which case I can investigate that, I could find it without even the discriminants. But for the sake of time, I just wanted to mention this discriminant approach to finish our puzzle.