 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 Missildine. In lectures 26 and 25, we introduced the idea of algebraic extensions of fields. Particularly in lecture 26, we developed the theory of algebraic extensions, particularly talk about finite extensions, and how finite field extensions always are algebraic. In lecture 27, we wanted to develop the idea of an algebraically closed field, and thus define the notion of the algebraic closure of a field. Now, we'll do that in the subsequent videos of lecture 27. In this first video, I want to continue to develop some of the properties of algebraic extensions that we're going to need to talk about algebraically closed fields. The first one here, suppose we have a field extension E over F, then the subset E, so I should say the subset of E consisting of all algebraic elements over a field is itself a field. E could have a bunch of algebraic elements over F, but it could also have some transcendental ones. If we only focus on those elements of E, which are algebraic over F, that will give you a subfield of E, which will be a field extension of F, because it will necessarily contain F itself. That's because every element of F is algebraic over F. If we're looking for those elements of E, which are algebraic over F, then all of F will be contained in it. But let's argue why this subset is a subfield. This is then going to be the algebraic subfield of E over F. Imagine we have two elements, alpha and beta, which are algebraic over F contained inside of the field extension E. Then consider the field where you join the elements alpha and beta, and look at that as a field extension over F. Now, what we proved previously in lecture 26 is that the degree of a field extension factors. In particular, we can insert the intermediate field. F sits inside of the field F adjoin alpha which sits inside the field F adjoin alpha beta, like so. Using this intermediate field F adjoin alpha, we can factor the degree of the extension of F adjoin alpha beta over F in the following way. We get the degree F adjoin alpha beta over F adjoin alpha, and then we get F adjoin alpha over F, like so. For which clearly F adjoin alpha over F, because alpha is an algebraic element, this is going to be a finite number. This is gonna be a finite extension. We proved this previously in lecture 26 that a simple algebraic extension is always a finite extension. So this is gonna be a finite number. But what about this? Let's take F adjoin alpha beta over F adjoin alpha. Now, the degree of this extension cannot be worse than the degree of F adjoin beta over F, because if we take the minimal polynomial over here, so the minimal polynomial over F of beta, that has to be divided by, divisible by the F adjoin alpha, the minimal polynomial of beta right there. That is whatever the minimal polynomial of beta is when you have F alpha coefficients, that will divide the minimal polynomial of beta using F coefficients. The more elements you get, the smaller the minimal polynomial it could be. So therefore, the degree of the extension of F alpha beta over F alpha, it can't be worse than the degree of the extension of alpha of F beta over F. It's likely smaller, right? Depending on how alpha beta interact with each other, but it could be that it's just the same thing. And so we get this number right here. But like we said earlier with F alpha over F, F beta over F has to be finite because beta is algebraic over F. So it's minimal polynomial has a finite degree. That's the degree of the extension. So this degree is finite, which implies this one's finite and this one's finite. Well, the product of two finite numbers is gonna be finite. So this degree is finite. And like we showed before, if you have a finite extension, that means it's an algebraic extension, okay? And so this extension, when you take two algebraic elements and join them together over the base field, that it's also going to be an algebraic extension right here. Okay? For which this field F of join alpha beta, it contains alpha plus beta, alpha minus beta, it has alpha times beta, it has alpha divided by beta, because it's a field that has all of those elements. And which case, if you take any of those elements, like take the field F of join alpha times beta, like so, this is a subfield of F of join alpha and beta, but it's an extension field of F, that is, it's this intermediate field that sits in between the two. Now, since this is a finite degree over F, everything inside of it has to also be a finite extension and finite extensions are always algebraic extensions. So this tells us that alpha times beta is an algebraic element. And by similar reasoning here, this is gonna give us that alpha divided by beta is an algebraic element, alpha plus beta, alpha minus beta is an algebraic element. So if you take all of these operations on algebraic elements, they are still gonna be algebraic because F of join that element is a finite extension. And so this tells you that if you take the set of algebraic elements, it's closed under addition, it's closed under subtraction, it's closed under multiplication, division, and therefore we got a subfield, thus proving the theorem that we came out to prove here. And so then I'm gonna present a second theorem here that let K be an algebraic extension over E and let E be an algebraic extension over F. Then the theorem says that K is an algebraic extension over F. This theorem I'm gonna state, but I'm gonna leave this as a proof to the viewer here. And these properties we just proved here will be very useful as we then define the notion of a algebraically close field, which we'll do of course in the next video.