 In lecture 25 of this series, we prove the following statement. That if you have a simple field extension, so F adjoined alpha over the base field F, where alpha is an algebraic element, that is, alpha is algebraic over F, then the degree of the extension, that is, what's the dimension of F adjoined alpha as an F vector space, that's going to equal the degree of the minimal polynomial of alpha over F. And we did that by showing that the set 1 alpha, alpha squared, alpha cubed, alpha fourth all the way up to alpha N minus 1 is a basis for the field extension F adjoined alpha over F. Because F adjoined alpha, we can view as an F vector space. And so we claim this is the basis. The idea, of course, was that if you take all of these powers, 1 alpha, alpha squared, alpha cubed, continue on for forever. This gives you a spanning set. Of course, it can be pruned down. In particular, this comes from the minimal polynomial. Because the minimal polynomial of alpha here is going to look something like, oh, well, you have some coefficient Cn alpha, excuse me. No, we'll take alpha to the N right here, plus Cn minus 1, alpha to the N minus 1, all the way down to some constant term C. And this equals 0. Well, since all these things equal 0, you can take these things, move them to the right-hand side, and divide by Cn, of course, on both sides. So we end up with alpha to the N equals 1 over Cn. And then a bunch of other things. I guess we have a minus. I'll stick the minus in front, negative 1 over Cn, Cn minus 1, alpha to the N minus 1, all the way down to C0. So we can then write alpha to the N like this. And by induction, every larger power works as well. So we get this nice basis when it comes to a simple algebraic extension. So in particular, the degree of alpha join alpha over F is finite if alpha is an algebraic element. The reverse direction is also true. That, of course, I have two meanings by this. First, if you take a simple transcendental extension, that will always be infinite. But in particular, what we're going to talk about right now is that if you have a finite extension, a finite simple extension, that's got to be algebraic. So suppose that the degree of F join alpha over F is a finite number. Let's call it N in that situation. So then consider the set 1 alpha, alpha squared, all the way up to alpha to the N. What we have to argue is we have to argue that the element alpha is algebraic. So there is some polynomial which is satisfied by alpha. That is, alpha is a root of that polynomial whose coefficients come from F. So consider the set 1 alpha, alpha squared, all the way up to alpha to the N. This set, which contains N plus 1 many elements, necessarily must be linearly dependent. After all, if the degree is N, this is the dimension of a vector space. If you take N plus 1 vectors, that has to be a dependent set. So there's got to be some dependence relationship on this set. So in particular, there's going to exist some relationship, some dependence relationship, C0 plus C1 alpha plus C2 alpha squared, all the way up to CN alpha to the N, that equals zero. And then we do the same basic thing we did up here. We can solve for alpha to the N in this situation. Now this, of course, will work if CN is a non-zero number. If CN is zero, then this term vanishes, but then you have the next term, CN minus 1, alpha to the N minus 1, and then you repeat the process there, and you can keep on going down, down until you eventually find this algebraic relationship, in which case alpha is going to be an algebraic element. Now, of course, I know in this situation that CN is not going to be zero, because if it were, and you have some smaller power, that's going to give you the minimal polynomial of alpha is actually smaller than N. And as we proved earlier, this degree is the same as the degree of the minimal polynomial. But nonetheless, this equation right here, this dependence relation turns into an algebraic relation. Alpha is an algebraic element in that situation, okay? Let's look at an important proposition here. Basically, just summarizing what we just said here, if a field extension E over F is finite, meaning that it has a finite degree. So E as an extension field over F has finite dimension, then E over F is an algebraic extension, all right? So we mentioned above this for simple extensions. So if you take a simple extension, F would join alpha over F. If this is finite, then that implied that the element alpha was algebraic. But what if we don't necessarily have a simple extension? What if there's multiple elements we've adjoined to F to create E? Is it still gonna be algebraic? And the idea is still true, right? Because if you take any element alpha inside of the extension field E, consider the field F adjoined alpha. It'll be a subfield of E right there. Well, if you look at F adjoined alpha, this is going to be a F subspace of E. Therefore, its dimension as an F subspace must be smaller than the dimension of E. So the degree of F adjoined alpha over F is less than or equal to the degree of E over F. Which by assumption, the degree of E over F is finite. So that implies this one is finite as well. And thus by the previous argument, this shows that alpha is algebraic over F. Now, as alpha was an arbitrary element of the field extension E, since every element of E is algebraic, we then say that the extension E over F is algebraic. So finite extensions always imply algebraic extensions. From this, we can then infer that a transcendental extension must be infinite because finite extension is always algebraic. But I don't want you to confuse this to think that algebraic extensions cannot be infinite. You can. What you could do is, for example, you could throw in an infinite amount of linearly independent algebraic elements onto a field and that would still be algebraic even if the extension is infinite dimensional there. So for example, take S to be the set of all square roots of prime numbers over the rational numbers, of course. So we could take like the square root of two, square root of three, square root of five, square root of seven, you get the idea there. Take that set to be S. Then take the field Q adjoined S over the rational numbers. This is going to be a infinite dimensional vector space because your basis in this situation is in fact going to be S if you include one in there. So this field Q adjoined S is gonna be the span. I should write that out. We're gonna get the span of one, the square root of two, the square root of three, the square root of five, the square root of seven, et cetera. There's an infinite set here. But if you drop any one of these square roots, you're gonna lose something. Because the fact that they're all prime, there's not gonna be any way I can produce like the square root of 11 using the square roots of other prime numbers. So every single one of these is necessary. We've now constructed a field extension of Q, which actually has a countable basis. But each element is going to in fact be algebraic. As each element we adjoined is itself algebraic, anything we produce with them, any linear combinations of algebraic elements will also be algebraic. And so Q adjoined S over Q is an algebraic even though it's an infinite extension. But be aware, so when it comes to infinite extensions, infinite field extensions, they could be algebraic, they could be transcendental, but what we can say is that finite extensions are always algebraic. And this is a very useful property when we study field extensions.