 Let's do one more example of computing a Galois group and this time we're going to focus on the polynomial 4x to the 5th minus 10x squared plus 5. Compared to some of our previous examples though, this is actually going to be one of the easiest ones to do. Yeah, believe it or not. And we're actually going to show that the Galois group with this polynomial is all of s5. How do we get that? So let's first argue that this is in fact an irreducible polynomial. How do we know that? Well, that follows by Eisenstein's criterion. Notice that 5 divides all of the coefficients besides the leading coefficient. But 5 squared does not divide the constant term. So by Eisenstein's criteria, this thing is an irreducible polynomial. It's going to have, of course, five roots. Let's call one of them alpha. So let's say that f of alpha is equal to zero. This is one of the roots. I should mention, of course, by the intermediate value theorem. The graph of this polynomial must cross the x-axis. And the basic reason for that is the following. Well, if you take the limit as x goes to infinity, then f of x has to go to infinity. So it points up on the right. When you take the limit as x approaches negative infinity, f of x is going to have to go towards negative infinity since the leading term is on. It's an odd monomial. And so by the intermediate value theorem, there's going to be some point at least for which we have a point of intersection. So at least one real root. And so let's suppose that this alpha is, in fact, one of the real roots of this polynomial. So the reason this is relevant, of course, is that if we look at the field, q join alpha colon q, since f is an irreducible polynomial and f is a root of that polynomial, this irreducible polynomial, we get that this degree extension is going to be five. And of course, if we let, for example, e be the splitting field, the splitting field of this polynomial here, f, oh boy, field, then the extension e over q, it factors through q join alpha. So this last one, since it's since it's five, this tells us that the Galois group, so we'll say G is equal to the Galois group of e over f, excuse me, e over q. We see that the Galois group is divisible by five because of this observation right here. So okay, the Galois group is divisible by five. And we also know that G can be visualized as a subgroup of s five. We want to argue it's all of it. So when you look at s five, it's divisible by five, this group, so it has an element of order five. This is follows from Cauchy's theorem. Well, what elements of s five can have order five? It's got to be a five cycle. So you're going to basically get something like one, two, three, four, five is contained in there without the loss of generality, but depending on how we label the five roots, we get that that's in there. So there is at least this five cycle going on there. Now, we want to show that G also contains a two cycle because one can argue that any subgroup of s five that contains a five cycle and a two cycle has to be all of s five. I'm not going to go through the details of that you can actually generalize the argument in fact that if you have any symmetric group whose degree is a prime number P, then you can argue that if it has a P cycle, and it has a two cycle then the group has to be all of five. And again, there's arguments you can go through it basically comes down to without the loss of generality. If it has a two cycle, you can pretend that two cycle is one, two, and since since you have a P cycle, by taking powers, and since it's a prime power, the only power that's going to give of a smaller order is the P power. Without the loss of generality, you can argue that some power of this P cycle will contain something like one, two, and then a bunch of other things like I three, all the way up to IP like so. And so then when you take the product of these two things together, you can argue that two, I three up to IP is inside of the group. Then you can conjugate these things together. If you take one, two, let's call this element right here, sigma. So you can take like sigma to the K times one, two times sigma to the negative K, and assuming I've got everything correct. This basically is going to turn out to be one times I to the K, something like that where I did case Arbitr and you can get everything one, two, one, three, one, four, one, five. And now if you want to construct the two cycle, like say IJ, this is going to be the product of one J times one I times one J. Like so, notice what happens to one right here, one goes to J, J goes to one, so it's left fixed, J goes to one, one goes to I, it's left fixed. So you're going to get this thing right here. And once you have, so this shows you that you have all arbitrary two cycles, and if you have all the two cycles, that's a generating set for SP, we know that. And so that argument applies of course in this S5 case, just want to run that down real quick. So what we have happening in this situation is we have a five cycle, we want to show that there's a two cycle inside of this too. We want to show that there's also a two cycle inside of the Galois group here. Let's do a little bit of calculus. Looking at the derivative of f of x here, you end up with 20x to the fourth minus 20x like so. You can factor out the 20x. That leaves behind xq minus one, for which I can show the factorization of xq minus one, but there's only one real solution to that which is one. So our derivative is going to have two critical numbers at zero and one. I'm going to erase these for a moment. So there's a critical number at zero. There's a critical number at one. If we actually plug in, if we plug in x equals zero into f, you're going to end up with five. So we have the point right here, zero comma five. If you plug in one, you end up with four minus 10, which is negative six plus five, which is negative one. So you end up with this point like this, one comma negative one. If we look at the second derivative now, you're going to end up with 80x cubed minus 20, right? We can factor out the 20. That leaves behind 40x cubed minus one like so. This shows you that there's going to be a point of inflection that sits in between these things somewhere. In particular, if you plug in zero, you end up with negative 20. So this has to be a local maximum. And if you plug in x equals one, you end up with 80 minus 20, which is just 60. So that's going to be, at this point, your concave upwards. So, you know, this is a really, really crude picture. I apologize for that. But the point is, I mean, we could also put this in the graphing calculator and see it very quickly. But given this information we get from the first and second derivative, this graph has to have actually three real roots. And since it has three real roots, that tells us that we have two non real roots. Now those non real roots have to be complex conjugates of each other. And so if you take your roots, right? So we'll say there's like alpha one, alpha two, alpha three, alpha four and alpha five, which these non real roots will say they're alpha four, alpha five. They have to be conjugates of each other. So alpha five is actually equal to alpha four conjugate. And so using complex conjugation, that's going to be an automorphism of the complex numbers within restricts down to it be an automorphism on our splitting field here. Complex conjugation will send a four to a five and vice versa. But as a one, two, three, a one, alpha one, alpha two, alpha three are real numbers. Complex conjugation will do nothing to them. So complex conjugation is going to give us this map alpha four, alpha five inside the Galois group. And like I said, any subgroup of S five that contains a five cycle and two cycle produces all of S five. So this proves in fact that the Galois group for this polynomial is in fact S five. This is one to remember because in the next lecture we're going to talk about the insolvability of the Quintic polynomial. And this will be an example of a polynomial Quintic polynomial that cannot be solved using radicals whatsoever. So take a look in lecture 36 for that one. I do appreciate you watching these videos. 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