 So we're called the definition of an algebraically closed field. We say a field is algebraically closed if every non-trivial polynomial has a root. For which, if every non-trivial polynomial has a root, that means every non-trivial polynomial factors into linear factors are, we say, every polynomial splits. The fundamental theorem of algebra states that the complex number field is an algebraically closed field and for most of us, that's probably the one and only algebraically closed field that we're aware of at this moment. I want to construct another one because if you look at the field Q versus the field C, yeah, Q does live inside the complex numbers, but the complex numbers in addition to having algebraic elements over Q, like the complex numbers include i, the square root of 3. These are algebraic numbers over Q. There's other things like pi, e, which are these transcendental numbers, which are not algebraic over Q since they're transcendental. Let's take a, Blackboard a in fact, to denote the set of all algebraic elements in the complex number field over Q itself. We proved previously that if you take the collection of algebraic elements in a field extension, that itself forms a field extension, a subfield that lives in the middle of these two, which by construction necessarily is an algebraic extension. So take a to be the set of all algebraic elements of Q, I should say all the algebraic elements of C over Q. This field we can show is in fact algebraically closed as well. And in fact, this field a, it's commonly referred to as the field of algebraic numbers. This is in fact the smallest field which contains Q and is algebraically closed. What we want to do in this video is talk about this idea here. This idea of being the smallest algebraically closed field, this leads to the notion of what we call the algebraic closure of Q, which is what we mean by a here. And for an arbitrary field F, we will define F bar to be this algebraic closure. That is, so that is F bar is an algebraic extension of F, which is an algebraically closed field. What do we mean by here by an algebraic extension? Is there more than one that could do this? We're going to see that uptie some morphism that that's not the case, that an algebraic closure by definition is an algebraically closed algebraic extension of the field F, but uptie some morphism, only one such field can do that. And so we're claiming from before that this field a is the algebraic closure of the rational numbers. It is the set of all algebraic numbers, which are algebraic over the rational numbers there. So before we prove, because one thing we have to argue is that do algebraic closures even exist, there is an issue going on there. We'll prove that in just a second. And that proof will utilize Zorn's lemma. But before we do that, I want to prove the following lemma. So let F be a field, then E is an algebraic closure. And I'm using the word and here as opposed to the, because I'm not claiming they're unique yet. They will be, but we'll prove that in a second. So E is an algebraic closure of F, if and only if E is a maximal algebraic extension of F. Okay, so close algebraic closures are maximal algebraic extensions. What does maximal algebraic extension mean here? That first of all, E is an algebraic extension, but any other extension of E would contain something that's not algebraic over F. That is, it's transcendental. So let's look at the details of that argument right here. So let E be an algebraic closure now by a previous proposition 21-1-21 in our lecture series, which we proved that in a previous video. Let me just remind you what that says here. Proposition 21-1-21 says that any algebraically closed field has no proper algebraic extension. Okay, so if E, the statement right here is algebraically closed fields have no proper algebraic extensions, which then makes E over F a maximal extension. Okay, so by definition, an algebraically closed, an algebraic closure is an algebraically closed field, right? So since a closure is algebraically closed, this proposition implies there's no proper algebraic extensions. That means it's a maximal algebraic extension. So that direction is already taken care of. The other direction we need to go about is that if you have a maximal algebraic extension, that then makes it the algebraic closure, all right? So suppose that E is a maximal algebraic extension of F. Consider the polynomial P of X, which belongs to the ring E of joint X. And suppose this is a non-constant polynomial. And also suppose that this doesn't have a root in E, okay? Because if we have an algebraically closed field, then that means this polynomial would have a root. Because again, we're trying to prove here that since E is a maximal algebraic extension, that it's an algebraic closure. An algebraic closure is an algebraic extension. We already got that part that is algebraically closed. So every polynomial has a root, okay? Well, if this polynomial P doesn't have a root, by Kroniker's theorem, there does exist an algebraic field extension over E, call it K, such that P of X has a root. The construction basically comes down to you take E of joint X and mod out the ideal generated by this irreducible polynomial P of X. That field K will be a field extension of E, but it contains a root of P, okay? Now, we know that by assumption E is algebraic over F. By construction, K will be algebraic over E. Because K right here is just E adjoin the root, we'll call it alpha, of this polynomial P of X here. So it's an algebraic extension. Now, also by a previous result, we did this in the very first video of lecture 27. That was known as theorem 21, 118. That theorem told us that if K is algebraic over E and E is algebraic over F, then K has to be algebraic over F as well, okay? That would contradict, so K being algebraic over F contradicts the maximality of E being an algebraic extension of F. So what must have happened is that K and E would actually have to be the same thing, really. But that means E had a root, we get a contradiction. So the maximality of the algebraic extension here tells us that we actually were algebraically closed. So a field E is an algebraic closure of F if and only if it's a maximal algebraic extension. This is very important observation because this is going to be the main argument we use in the forthcoming Zorn's Lemma argument. So remember, by Zorn's Lemma, the basic template we argue is the following. We come up with a partially ordered set, argue that partially ordered set is non-empty. Then we take an arbitrary chain inside of that partially ordered set and argue that it has an upper bound inside of the set X. Then we invoke Zorn's Lemma that tells us that since every chain has an upper bound, there exists a maximal element inside the set. And then that maximal element we're going to argue is an algebraic closure. And so here we go. The main theorem of this lecture is that every field has an algebraic closure. So let F be some arbitrary field. Let X be the set of all algebraic extensions of F. Now there is a slight little caveat I have to make mentioned here. That technically speaking, this set X is not a set. It's actually a proper class, which I don't want to delve into the deep, deep set theory going on here. But a system of symbols can be introduced in correspondence with F of joint X to avoid the set theoretic problem that's going on here. That is, I can make X into a set and not a proper class. But I'm going to hide the details that most of us won't be too concerned about. The issue is that we can do it and I'm just going to make it for simplicity's sake, suppress those details because it goes beyond the scope of this abstract algebra lecture series. So we have this set X of field extensions for which these are all fields that contain F and we can order those fields by set containment. This set is non-empty because F belongs to the set. F is an algebraic extension of F itself. It's a trivial one, but it is still an algebraic extension, so X is non-trivial. It's non-empty. That has to be checked. Now let's see be an arbitrary chain of fields inside of this set X. So if we take the union of all of those fields, so if F prime is an arbitrary element of the chain, take the union of all of these different F primes, call that set E. Now I claim that E, the union of these ascending fields, is itself a field. All right? We've made that argument before from time to time, so I'm not going to say too much about the details, but this set E is in fact a field. It contains zero because each of these elements contain zero. If you take two elements inside of E here like alpha and beta, something like that, well, that means there was some field alpha, some field, excuse me, F prime that contains alpha. There was some field F double prime that contains beta, but this is a chain, so these things, this is totally ordered. So without the loss of generality, we can assume that F prime is contained inside of F double prime. So alpha and beta both belong to F double prime. As F double prime is a field, it'll contain F plus beta, excuse me, alpha plus beta, alpha minus beta, alpha times beta, alpha divided by beta. It's got all the field stuff, so yet the union of ascending fields is going to be itself a field. Okay, now being a field does not necessarily mean you belong to X. You first of all have to be a field that is an algebraic extension of F. Now be aware that each of these fields F prime contains F. They're all algebraic extensions over F. So each of these fields contains F, so E is going to contain F as well. So clearly E is an extension of F. We want to then prove it's an algebraic extension. So let's look at the details there. Like I said, clearly F is contained inside of E since every field in C contains it. C is non-empty, of course. It's a non-empty chain. So let's take an arbitrary element of E. Now, if there's an element alpha inside of E, that means there was some field F prime in the chain C that also contained alpha, okay? But F prime over F is an algebraic extension, which means that alpha is an algebraic element over F. And so since alpha was chosen arbitrarily, every element of E is algebraic over F. So that means E over F is an algebraic extension. Therefore, E belongs to the set X. So we've now shown that, well, clearly by construction, right, E is the union of all of this F prime. So if you take an individual field F prime inside the chain, E contains it. And so therefore, C, excuse me, E is an upper bound for the chain C. And since C was an arbitrary chain, that shows that every chain inside of X has an upper bound. By Zorn's lemma, that then tells us that the set X has a maximal element, that maximum element, we're gonna call it F bar. Now, what does F bar mean here? It's a maximum element of X. X is the set of all algebraic extensions of F. Therefore, F bar is a maximal algebraic extension of F. As we just saw with the previous lemma, maximal algebraic extensions are algebraic closures. Therefore, F bar is an algebraic closure of F. And we have then that algebraic closures then, in fact, exist. That's the main result I want us to get here. The next result, and this is where we're gonna end our lecture with, lecture 27, is that algebraic closures of a field are isomorphic. To prove this corollary, we're gonna first prove the following theorem here, that let K be an algebraic extension of the field F, then there exists an injective homomorphism, we'll call it phi from K into F bar, where F bar here is an algebraic closure of F. They do exist now, such that the function phi when restricted to the base field F is the identity map. So given any algebraic extension of F, there exists an embedding of it inside of an algebraic closure, for which at least fixed the base field like so. And so every algebraic extension can be placed inside of an algebraic closure. Now, if you have two different algebraic closures, well, an algebraic closure, we'll call it, we have F bar versus F double bar, right? Well, these are both algebraic extensions of F, both maximum algebraic extensions of F. So if the previous theorem is true, that means that F double bar can be embedded inside of F bar, which F bar can also be embedded inside of F double bar like so, and therefore we can infer from these that these injective field homomorphisms do have to in fact be isomorphisms. Again, algebraic extensions are maximal for algebraically closed fields. So the corollary follows from this theorem right here, very nicely, and the proof of this theorem is I'm actually gonna leave it up as an exercise to the viewer right here to prove this, to prove this isomorphism theorem for these algebraic extensions here. And now I'm gonna give you a nice proof, a nice hint on this proof here because the proof depends again on Zorn's lemma. So to do Zorn's lemma there's the ingredients we always have to go through, we need a partially ordered set. So the above theorem is proven using Zorn's lemma and the set X, which X is gonna be the set of injective homomorphisms of the form. F goes from E to F bar, where E is then a subfield of K because we're trying to build one for K here. And when you restrict your function F to F, you get just the identity in that situation. Now it's like, how in the world are you gonna order this set? So it's a set of functions. All the previous Zorn's lemma examples we've done in this lecture series, we did sets of sets, but now we have a set of functions. Now be aware that functions themselves are sets, right? Because a function first and foremost is a relation. A function F is a relation on the set. So if your function F is from E to F bar, that means it's a relation, it's a special type of relation, but it's a relation on the set E cross F bar. And so a relation in particular is a subset of the Cartesian product. For which E cross F bar, since E is a subfield of K, each of these Cartesian products is a subset of K cross F bar. So each and every one of these functions is a subset of K cross F bar. And therefore that tells us that the set X, which is the set of all these functions here, is a subset of the power set K cross F bar for which the power set has a natural ordering to it. It's a partial order set. You just, you order it by set containment. As X is a subset of that, you're gonna borrow the ordering from there. So to prove this one, your best bet is to actually think of the functions as sets of ordered pairs for which you have the restriction that if the first coordinate is from the field F, the second coordinate is gonna be from field F. And if you do it that way, you can improve this using Zorn's limit like we've done all the other ones. But like I said, I'm gonna leave that as a, as an exercise to the viewer here, which then brings us to the end of lecture 27. Thanks for watching. If you learned anything about algebraically closed field or algebraic closures, please like these videos, subscribe to the channel to see more videos like this in the future. And if you have any questions, please post them in the comments below and I will answer them as soon as I can. Thank you everyone. Bye.