 Welcome back to our lecture series Math 4230 abstract algebra two for students at Southern Utah University. As usual be your professor today, Dr. Andrew Misseldine. In this lecture 26, we're going to continue what we started in lecture 25, and specifically we're going to focus on this idea of algebraic extensions. In particular, at the end of lecture 25, we had introduced the idea of the minimal polynomial of an algebraic element from a field extension. You mentioned how the degree of that minimal polynomial gives you the degree of a simple extension. Now in this video, I want to continue developing this idea and particularly start off with two examples of computing minimal polynomials and such. In particular, let's look at this first example. Take the polynomial f of x equals x squared minus 2. Now we're viewing this as a polynomial over the rational field. Now in particular, this polynomial f is an irreducible polynomial. There's two ways you could see it. Since it's a quadratic polynomial, if it's reducible, that happens if and only if it has a linear factor, which happens if and only if it has a root. Now by the rational roots theorem, the only possible rational roots of f of x here would be plus or minus 1 and plus or minus 2. A quick calculation would show that neither of those four numbers work as roots of this polynomial. So that would make it irreducible. But it turns out there's even easier criteria we could use here. We could use Eisenstein's criteria. Because after all, if you look at all of the terms other than the leading and constant terms, they're all coefficient zero. So you could pick any prime you want. In particular, when you look at two there, two divides it before it doesn't. So by Eisenstein's criterion, we get that's an irreducible polynomial. But that's a little bit of overkill in the situation, but still we should be specific here. We have this irreducible polynomial. And we know from past work that if you take the polynomial x squared minus two, its roots are gonna be plus or minus the square root of two. In particular, if we put the square root of two into that, you get the square root of two squared minus two. That'll give you two minus two, which equals zero. And the negative square root will give you something else. I should say it'll give you something similar to what we just did a moment ago. And so we have the two roots which are not rational numbers, right? The square root of two is not rational. This goes back to the days of Euclid who proved the square root of two is an irrational number. And so let's look at the extension field where we adjoin the square root of two to the rational numbers. Well, by Kroniker's theorem, this field q adjoin the square root of two is gonna be isomorphic to the field q adjoin x mod out by the ideal x squared minus two. So we can always build an extension field to contain the root of an irreducible polynomial. And so the field q adjoin the square root of two, you're adjoining the square root of two to q and you're gonna get everything as a consequence of that. Now, if you get the square root of two, you're also gonna get the negative square root of two because that's the additive inverse and this is a field after all. And so in this field, q adjoin the square root of two, we didn't just gain one, but we gained all of the roots of our polynomial x squared minus two. Now also because of previous work, it's important to note that when you look at the field q adjoin the square root of two, that as a rational vector space, it'll be two-dimensional and has a basis one plus the square root of two. That is every element of this field can be written as a plus b times the square root of two where a and b are arbitrary rational numbers like so. And like I also mentioned, the polynomial f of x, it factors into two linear factors, x minus the square root of two and x plus the square root of two. If we view this as a polynomial over q adjoin, let me write that q again, q adjoin the square root of two, adjoin x there. So if you extend the field of coefficients for your polynomials are polynomial f factors by adjoining the square root of two here. Now, before we go any further here, I wanna mention that of course, first of all, I guess I should say that this polynomial x squared minus two is in fact the minimal polynomial of the square root of two, also the negative square root of two, but f of x is the minimum polynomial of the square root of two over the rational numbers. This is the smallest irreducible polynomial with rational coefficients for which the square root of two is a root. And of course, when we talk about the minimal polynomial, we do assume that it's monic in that situation here. And we can see it's pretty easy because if you're a root of an irreducible quadratic, then that has to be your minimal polynomial because the minimal polynomial will divide every polynomial which takes the square root of two as a root. And which case if x squared minus two wasn't the minimal polynomial, that would mean that x minus square root of two would have to be the minimal polynomial because that's the only polynomial of smaller degree. But that would imply the square root of two is a rational number, which it's not. So we've now proven x squared minus two is the minimal polynomial of the square root of two over the rational field. But I wanna make mention that when your root comes from a radical of some kind, in this case, the square root of two, it's very easy to construct the minimal polynomial. We can basically work backwards like we do here. It's like, okay, I wanna come up with a polynomial which has the square root of two as a root. Well, since you have the square root of two here, by construction, the square root of two is a number which squares to be two. So if you take this equation, x equals the square root of two and you square both sides, you're gonna end up with x squared equals two, for which then if you move the two to the other side, you end up with x squared minus two equals zero. And since we chose x to equal that value, we see that x squared minus two is then this minimal polynomial. It's the smallest polynomial we created here. And therefore the minimal polynomial of the square root of two over the rational field is x squared minus two. And I should emphasize that this is the minimal polynomial over the rational field because if you want the minimal polynomial of the square root of two over the queue of join, the square root of two, then you just get x minus the square root of two. So it matters which field you're referring to. So if you're looking in the field, queue join the square root of two, then the minimal polynomial is x minus the square root of two. If you're talking about the rational field, then you're gonna get this x squared minus two in that situation. So the base field matters. And that's why the minimal polynomial does depend on the fields you're looking at. And this was a pretty simple example. Let's turn up the heat a little bit. Now let's look at the polynomial x squared minus 16x, excuse me, x to the fourth minus 16x squared plus four. This is a rational polynomial. Clearly it is like lies irreducible. Now, of course, Eisenstein's criteria doesn't apply here because when you look at the constant term there, you have a four. There's no other prime we could use other than two, but that doesn't work. You can still use, so while you can't use Eisenstein's criteria, you can still use the rational roots test for which because of the rational roots test, the only possible roots are gonna be plus or minus one, plus or minus two, plus or minus four. And I'll leave it to the viewer. Just pause it right now if you want to. Just to check, do any of those, do any of those six numbers work as roots? None of them do. Now this is a degree four polynomial. There is the possibility that maybe it factors into two irreducible quadratics, right? I mean, that is a possibility of course, but with this one, one could check that that is not the case whatsoever. This polynomial is in fact irreducible. Like I said, I'll leave that up to an exercise to the viewer here to verify such a thing. I claim that the roots of this polynomial are gonna be plus or minus the square root of eight plus or minus two root 15, like so. So what here of course we mean is that there's a choice. You can choose plus or minus inside the radical. You can also choose plus or minus outside the radical. So there are four possible roots that I'm describing, but that makes sense because we have a degree four polynomial we would anticipate we have these four roots here. Now in this situation, let's take one of these roots. So I'm gonna choose positive and positive in both situations. So take the number of the square root of eight plus two times the square root of 15. Take Q adjoin that irrational number right there. Well by Kroniger's theorem, this field where you take a root of this irreducible polynomial and join it to the rationals, this will be the same thing as taking the polynomial ring Q adjoin X and modding out by this irreducible polynomial X to the fourth minus 16 X squared plus four. And so this gives us a field for which this polynomial G of X now has a root in it. And it doesn't just have one root, it actually does have two roots because if our field includes the square root of eight plus two square root of 15, it also includes its additive inverse which was one of the other roots, negative square root of eight plus two root 15 there. Both of those belong there. So this polynomial when you go to the extension field then gains two of its roots, but does it have the other two? I claim that that's a no, that if you take the square root of eight minus two square root of 15 or its negation those do not belong to this field. And so to try to see why that is, think of the following situation. What would be a basis for this field, all right? We've seen previously that we can, when it comes to a simple extension like this, if we take something like f adjoin alpha you can always take as a basis the following set one alpha squared, alpha cubed, alpha to the fourth and then you keep on going. That by itself will give you a spanning set. And of course, by the pruning theorem every spanning set can be pruned down into a basis. So at some point, well, if you take with a simple extension of course I should say that this is an algebraic element. If you have a transcendental element that statement might have some issues but we're focusing on algebraic elements here. If you take a simple algebraic extension we always have a basis by using the powers of the element. And this will eventually lead to a dependence relation. That's where this minimal polynomial comes from. So as the square root of eight plus two root square root of 15 is a root of that irreducible degree four polynomial I do know that if I look at the powers of our root here so you have one, I'm gonna just call this element alpha for the sake of simplicity in our language here. So if you can take as your basis one alpha this here would be alpha squared this would be alpha cubed. Now I can stop at alpha cubed because if I add alpha to the fourth then I now have a dependency relation because of the minimal polynomial we had before. Because if you take G of alpha we know this is zero it's a root of the polynomial. So this tells us that alpha to the fourth minus 16 alpha squared plus four is equal to zero which then tells us that alpha to the fourth is equal to 16 alpha squared minus four. So any power of alpha four or larger can be rewritten as a linear combination of the smaller power. So this does in fact give us a basis. Now I do wanna simplify things a little bit here because when you look at these alphas right when you square this term right here that is when you square alpha and this is alpha squared right here you're gonna get eight plus two square root of 15 for which eight's a rational number you could subtract from that two's a rational number you could divide by that. And so taking alpha squared is essentially the same thing as taking the square root of 15. So this field will contain a square root of 15 like so. And then when you come to alpha cubed right here again you can rewrite this thing to make it a little bit cleaner. If you have alpha cubed that's essentially the same thing as having the square root of 120 plus 30 times the square root of 15 which is gonna be linearly independent from this radical we had before like so. And so you'll notice that when it came back to this element here the square root of eight minus two root 15 that when you look at these radicals there's no combination that's gonna change the radicand the term inside the radical here. The only thing that can change the possible radicands is when we're taking powers of alpha here which like we said we'll start off with the square root of eight plus two root 15 we can produce a square root of 15 and we can produce the square root of 120 plus 30 square root 15 but besides those we can only take linear combinations of these radicals for which you're not gonna produce that one right there. All right so when we had joined alpha one of the roots of our polynomials we gained two of the roots in its additive inverse but we are still missing the other two roots of that polynomial. So if you were to factor g of x over the field q of join this q of join alpha here you don't get a linear factorization. In particular you're gonna get the following g of x which remember was x to the fourth minus 16x squared plus four over q of join alpha this would factor as x squared minus eight plus two square root of 15 and x squared minus eight minus two times the square root of 15. So you get that factorization. Notice that this field q of join alpha does contain the field q of join the square root of 15 right because the square root of 15 will be contained in there. So this is a valid factorization over q of join the square root of 15. Okay so when you get a larger field the fact that the polynomials can potentially factor more well into smaller pieces. But it turns out that this factor can also still factor further because our field contains a square root of eight plus two root 15. And so treating this as a difference of squares we can factor it like this and we can factor it like this. Although this term right here one can argue is still gonna be irreducible in the q of join alpha right here. And this came to what we described earlier. This is quadratic polynomial. If it factored it has a root but as we already observed there is no root of the polynomial for this polynomial right here. So this then gives you the complete factorization of g in that situation. We didn't quite get linear factors. And so the minimal polynomial of the minimum polynomial of our element alpha depends on the field right? If we're looking at the field q our minimal polynomial turned out to be x to the fourth minus 16x squared plus four. If we look at some intermediate field q would join the square root of 15 then in that situation your polynomial mu will factor at your minimal polynomial will be x squared minus eight plus two times the square root of 15 because you can factor that way. But of course if you just join all of alpha then your minimal polynomial alpha will just be x minus alpha. The minimal polynomial depends on which field we're looking at here. And be aware that joining one root of a polynomial doesn't necessarily give you all of the roots of the polynomial. If you call, if we call alpha this element we keep on going with alpha equals the square root of eight plus two root 15. Let's introduce another element called beta which beta is the square root of eight minus two root 15. In that situation you have these two distinct roots. And if you want your polynomial g of x to factor into linear factors then your field would have to be alpha, a join, q would join alpha and beta. If you get those two roots you get the other two but one root isn't enough to get all of them. But like I said earlier, if we have a number which is formed using radicals we can actually utilize those radicals to form the minimal polynomial over the resultant field here. So take the number of the square root of eight plus two times the square root of 15. Since this is a square you can square both sides and which case on the left hand side you get x squared on the right hand side you'll get eight plus two square root of 15. Now if you move this entire number to the right hand side you end up with x squared minus eight minus two square root of 15. And that then gave you the minimal polynomial over q would join the square root of 15. Which in the middle of the very first line if you took x minus alpha equals zero that gives you the middle polynomial over q would join alpha. All right, so we have to keep on going. It's like okay if the square root of 15 is not in my field it's not in the rational field we keep on going. So I'm gonna move just the eight to the other side. So we get x squared minus eight equals two times the square root of 15. We're now gonna square both sides again. On the left hand side we will have the foil. So you're gonna get x squared times x squared which is x to the fourth. You're gonna get negative eight times negative eight which is positive 64. Then we're gonna get x squared times negative eight and negative eight times x squared that gives you a negative 16x squared. And then on the right hand side when we square things two squared is four. Square root of 15 squared is 15. Four times 15 is 60. If you subtract 60 from both sides you end up with g of x equals zero. And so then we were able to construct the minimal polynomial mu of our number alpha as then we found this minimal polynomial over the rational field that we see of course here. And so this example then provides to us basic examples of how we can actually construct the minimal polynomial of a number constructed using radicals very much in this way. But of course are all roots of polynomials constructed from radicals? Well, that's actually a fundamental question that drove a lot of the Galois theory that we're gonna be seeing in the future. And we're gonna discover in the future that this process is not necessarily reversible that there do exist polynomials whose solutions cannot be constructed from radicals which makes studying them a lot harder.