 Welcome back to our lecture series, Math 4230, abstract algebra two for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misseldine. Now, in lecture 23, we are going to continue our conversation about linear algebra, but we're going to take a slight pause for a moment in this video and talk about the notion of Zorn's lima. Now, before I explain what Zorn's lima is, let me remind the viewer or just present to the viewer important definition with regard to partial orders. Imagine we have some partial order on its set, so X is a partially ordered set. Remember that partial order means that we have a reflexive, anti-symmetric, transitive relation. If we take a non-empty subset of our partially ordered set X, call it C, then it's called a chain if it's non-empty and totally ordered, which totally ordered means that if you take any two elements, X and Y inside of the chain C, we get that either X is less than or equal to Y or Y is less than or equal to X. Now, the set X itself, it could itself be totally ordered, in which case every subset is a chain, but in general, that's not what we would expect. That's why we call these things partial orders in the first place. There might be incomparable elements, but a chain is this non-empty, totally ordered subset meaning that every element, which there are some elements in there, every element of the chain is related to each other. All right, so now let's get back to the titular topic here of Zorn's lemma. The next result is titled Zorn's lemma because any good lemma is extremely helpful in proving the results in a variety of locations. I should mention that I can't remember who the quote is belonging to, but there was a mathematician who once said that the greatest honor as a mathematician is to have a lemma named after you because while it's great to have theorems named after you, it means you discovered this great result. The importance of a lemma is that the lemma is used to prove other theorems, other results, and so the lemma in the end is gonna be used more often than theorems themselves per se, so it's this great honor to have a lemma named after you, so way to go Zorn, my hat is tipped to you. But in actual reality, Zorn's lemma isn't really a lemma. That is, it's not provable, at least not in the usual sense, and so let me explain what I mean by that. So it turns out that Zorn's lemma is an axiom of ZFC set theory, which that's a little bit of a lie here, the ZFC set theory here. When it comes to set theory, there are different ways one could axiomatize it. The most standard accepted set theory in mathematics is ZFC, where the C here is short for the idea of choice, the axiom of choice. And there's a couple of different ways one can represent the axiom of choice. Let's see, let me phrase it this way. The axiom of choice is a very important principle in mathematics and I should mention the reason we're included in this video here is that I want, at least I believe that every undergraduate mathematics student should have some exposure to the axiom of choice in some equivalent form or another. So what does the axiom choice mean? I keep on talking about it. Choice essentially states that given any collection of sets, we can choose a representative from each set and such a selection is called a choice function. Now, if you have a finite collection of sets, this is no big deal. Even if you have a countable collection of sets, this is not a big deal because you can use induction to make a selection. But when you start taking uncannable sets, your induction no longer works. Finite methods don't work anymore. You can't really use transfinite efforts to deal with these uncannable sets and therefore you're gonna need something. And if you don't have choice, if you just adopt ZF set theory, which of course is the standard axioms of set theory minus the axiom of choice, if you just adopt ZF set theory as your axioms for a first set theory, I should say, then it's actually possible that choice functions don't exist. You can have this collection of sets for which you cannot make a selection from each and every one of them. And this is why it's called the axiom of choice because you have this infinite collection of sets and you're making a choice from each and every one of those simultaneously. So can you make infinite limited choices simultaneously? Well, without the axiom of choice, that might not be possible. Now, a simpler way of describing the axiom choice is the following. If you take an arbitrary Cartesian product of sets, so you have some sets, say S alpha here, where alpha ranges over some index set here, the axiom of choice tells you that Cartesian products are always non-empty. Assuming that the individual sets S alpha, of course, are not empty themselves. So an arbitrary Cartesian product of non-empty sets is not empty. That gives us, that assumption is the axiom of choice. That's a little bit simpler because any element of this Cartesian product, which we have this alpha tuple, which again, this likely could be an uncannable product, any tuple inside of this set is really a function. It's a choice function. So the fact that this Cartesian product is non-empty would suggest the existence of choice functions. And so when you frame it that way, it's like choice seems somewhat natural. Now I have to confess that, yeah, there are some bizarre consequences of accepting the axiom of choice, like the Bonik Tarski paradox is one of the most famous of all those paradoxes. That is, you can dissect a ball in Euclidean three space into five pieces, then reassemble it. You get two balls of the same volume, right? So you doubled it, and this can be done, of course, in a isometric manner as well. It seems very counterintuitive and we're not gonna delve into the details of that really here, but choice is a very natural thing to accept and set theory because it makes sense that we would want Cartesian products to be non-empty in general. And accepting that, of course, is accepting the axiom of choice. Now, what does this have to do with Zorn's lemma, okay? Well, in the framework of ZF set theory, the axiom of choice as we framed it over here is logically equivalent to Zorn's lemma. So if one was learning set theory, one probably would accept some version of choice that resembles the existence of choice functions, but then you could prove Zorn's lemma then as a theorem of ZFC set theory. But they are logically equivalent. You could go the other way around. You could adopt Zorn's lemma as your version of axiom of choice. And then using Zorn's lemma, you could prove the existence of choice functions, all right? That's what it means for the two topics to be logically equivalent. And so that's why I meant that Zorn's lemma is really not provable. You accept choice, you're accepting Zorn's lemma. But if you accept Zorn's lemma, you have to accept choices too. As well, you can't separate the two. Another principle worth noting here is the well-ordering principle. We've talked about the well-ordering principle for like the integers or the natural numbers, I should say, which really the well-ordering principle is just saying that the natural numbers with its usual ordering is a well-ordered set. That's not the well-ordering principle I'm referring to in this situation. With regard to this conversation of Zorn's lemma and choice, the well-ordering principle I mean is that every set has a well-ordering for which, yeah, sure, for finite sets, that's easy. For countable sets, induction. But for uncountable sets, how do you put a well-ordering on the real numbers? It can't be the standard ordering you have on the real numbers. Is there an ordering you can put on the real numbers such that every set has a least element regardless of the subset, right? That's not so easy. And again, the well-ordering principle is equivalent to the axiom choice because placing a well-ordering principle, a well-ordering on the real numbers really comes down to using choice functions and such. And so I'm gonna return to our algebraic narrative that when it comes to abstract algebra in the algebraic setting, the version of choice that we love the most is actually Zorn's lemma. Algebraics really, really, really like Zorn's lemma. So as an algebraist, I really have no objections to the axiom choice. Yeah, sure, there are some weird paradoxes that come about because of the axiom choice. But if I forfeit the axiom choice, I have to give up Zorn's lemma as well. And Zorn's lemma has some very important applications to algebra, which we're gonna talk about in this video and the next. And so as such, we're going to proceed in this lecture series, accepting Zorn's lemma as our version of axiom choice. And therefore we have to accept all of the conditions that come with it, kind of like if you're a genie in the Disney Aladdin's franchise, right? You can't have all of that great omnipotent power without the lamp itself. We have to accept the rules of that power. And so now let's then say, what is Zorn's lemma? I keep on talking about what is it? Perhaps you've already read it on the screen while I've been going on and on about the axiom choice, but let's talk about it together now. Zorn's lemma says the following, if X is a non-empty partially ordered set in which every chain, chains were defined previously, right? Non-empty totally ordered subsets. If X is a non-empty partially ordered set for which every chain has an upper bound, then X itself has a maximal element. So let me remind you some other terminology that has to do with order theory that you may or may not remember here. So what is an upper bound? So he said that if every chain has an upper bound, then well, what's an upper bound? Well, an upper bound when you have a partial order is the following. Recall that an upper bound U on some subset C of a partial ordered X means that X is less than or equal to U for all X and U. So you have an upper bound, all right? I'm not claiming that this is a least upper bound or anything like that, but Zorn's lemma says that if every chain has an upper bound, then the partially ordered set has a maximal element. What's a maximal element again? Well, a maximal element, we'll call it say Mu inside of the X means that for all X in the partially ordered set such that Mu is less than or equal to X actually implies that Mu is equal to X. So a maximal element means that there is no element in the partially ordered set that is larger than Mu. Now, there could be multiple maximal elements. This is the difference between a maximum. Maximum element says that everything is smaller than it. A maximal element says that there's nothing bigger than it, which is a slight distinction. Clearly a maximum element is maximal because it's the unique maximal element, but maximal, there could be multiple maximal elements in a partially ordered set. So Zorn's lemma tells us that if we, let me say one more time, if you have a non-empty partially ordered set such that every chain has an upper bound, then the partial ordered set has a maximal element. And that can be a very, very useful condition. And so in this video, I wanna provide one example to show how Zorn's lemma is a very useful tool when you are looking at infinite dimensional algebra here. That is algebra that's not necessarily bounded by finite conditions. And that could mean things like no-theorem rings. No-theorem rings do satisfy finite condition. You have the ascending chain condition that's equivalent to all ideals being finitely generated. Artinian rings are a similar situation. Yeah, when you have some type of finite control on your algebra, you might not need Zorn's lemma, but there are situations where really you can't bind down the infinite within a finite condition. Therefore, Zorn's lemma is sort of like our last tool in that situation. So let me prove a very famous, very important theorem. Let R be a ring. And just as a reminder in our lecture series when we talk about a ring, we mean a commutative ring with unity. So there is a number one here. In this proof, you don't actually need commutivity so I could drop that condition whatsoever. But this proof does use the existence of unity. If you don't have unity, this theorem can be false. There exist counter examples there. But be aware that when we talk about rings in this lecture series, a ring is a commutative ring with unity, all right? So if R is a ring and I is an ideal inside of R, then there exists a maximal ideal of, a maximal ideal called M, such that I is contained inside of the maximal ideal. Remember a maximal ideal is a proper ideal of the ring because the ring itself is an ideal. Maximum ideal is a proper ideal of the ring that is not contained in any other proper ideal. Sounds like a maximal element in a Zorn's lemma sense. It is. So given any ideal, there exists a maximal ideal that contains it. In particular, if you take the ideal, just the zero ideal, because zero ideal is an ideal for every ring, that's a proper ideal. It is then contained in, it is contained in a maximal ideal. So this actually shows the existence of maximal ideals. And I should mention here, of course, that given any ideal, I'm assuming that, I'm assuming that I is a proper ideal. Clearly R itself is not contained in a maximal ideal. Proper ideals are contained in maximal ideals. And since the zero ideal can be extended to a maximal ideal, every ring with unity has a maximal ideal. All right, so how are you gonna prove this? Well, the short answer is we're gonna do an argument with Zorn's lemma. Let X be the set of all proper ideals of R that contain I. I claim that this set is non-empty because I itself is contained inside of that. I is a proper ideal that contains itself. So X is a non-empty set. We can order the elements of X by set containment. That is, we just have the usual set containment symbol, that is a partial order on sets. And this set X is ideals, you have ideals inside of ideals, and they all contain I. So that is a partial order on the set. It's non-empty because it contains I itself. So let's then take an arbitrary chain inside of X. So that is, it's just some non-empty, totally ordered subset. Let's say that the elements of it are J prime or something like that, that's what we're gonna describe down here. So what we're then gonna do is we're gonna take the union of the chain. So if you allow J prime to range over all of the arbitrary elements inside of our chain, take the union of each and every one of these things, okay? Call that element J. We claim that J is an ideal of R. Now, this argument we've actually done previously in our lecture series when we talked about no-theorem rings and we took an ascending chain of ideals, the union of an ascending chain of ideals is an ideal. In this situation though, it's slightly different, right? Cause when we've had the no-theorem situation, we had an ideal sitting inside of an ideal, sitting inside of an ideal, all right? There's sort of like this discreet progression, kind of like the natural numbers. We identified each ideal with a natural number. That's not what we're doing here. This chain, this chain C in fact could have cardinality equal to that of the continuum. It could be even larger than that, right? This could have any potential cardinality whatsoever, not necessarily accountable. So we are gonna reproduce the argument, but really it is just the exact same argument that we had before. The fact that we had countable chains before versus now potentially uncannable chains, arbitrary chains doesn't change any bit of the argument, but nonetheless we're gonna include it again. Why is the union of a chain of ideals equal to an ideal? For the following reason. Note first of all, that there does exist something inside this chain. When we talk that this chain is not empty, that's part of the definition. So there's some ideal inside of it. Now that ideal, J prime, which is an ideal, it contains the zero element. Therefore, since one of the ideals that forms J contains zero, and we're taking the union of those ideals, J itself contains zero. All right, so it contains zero. We have to also show that J is closed under addition. So take two arbitrary elements inside of this set J, call them A and B. Then how did A and B possibly get in there? Cause after all, J is a union of ideals. So there has to be some ideal, J prime in the chain that contains the element A. And likewise, there has to be some other ideal. Maybe it's the same ideal, maybe it's not, whatever. There has to be some ideal J double prime that's inside the chain that contains B, like so. Now remember, C is a chain. So it's a totally ordered subset. So every element in C is comparable. Which means that either J prime is less than or equal to J double prime. That's a typo right there. Or J prime contains J double prime. One of those two possibilities has to happen. And so without the loss of generality, we're gonna assume that J double prime is contained inside of J prime. Which that means since B was inside of J double prime and A is inside of J prime, that both A and B are contained inside of J prime. But J prime is itself an ideal. So it's closed under addition. So A plus B is contained inside of J prime. Now as J is the union of all of these ideals, that means if A plus B is inside J prime, it's inside of J. So J is closed under addition. What about ideal closure? Well, if you take an arbitrary element of the ring R, sticking with an arbitrary element A, right? It's contained in some ideal. We'll still call that J prime. Well, since A is in J prime and A J prime is an ideal, that means A times R is inside of J prime because it's an ideal. But J prime is a subset of J. So if A R is inside J prime, that means A R is inside of J. In which case then this shows that J is an ideal. Now I do wanna pause here and make a quick comment here that this proof that J was an ideal did assume that we were in a commutative ring. Proving that an ideal, proving a set is an ideal in a commutative ring is a little bit simpler than a non-commutative ring. But if you drop commutivity, you can modify this argument to still show that J is a two-sided ideal. But again, as we're only interested in commutative rings in this lecture series, I'm gonna take the simpler argument while assuming that R was commutative. And therefore, I don't have to worry about being closed on the left and on the right, okay? But that argument can be modified for that setting. All right, so now that we have that the union of all of the ideals in the chain is itself an ideal, I wanna argue that this is a proper ideal, okay? And this is where the unity comes into play. J is a proper ideal because it doesn't contain one. If one was inside of the ideal, in fact, if any unit is inside of an ideal, it has to be the whole ring. So if J doesn't contain one, then it has to be proper. Because if it doesn't contain one, that's an element of the ring, so that means it's proper. But if one was in there, it would then be improper. So J is a proper ideal if and only if it doesn't contain one. Now let's think to the contrary, what if one was in J? Well, since J is the union of all of those ideals, J prime from the chain, if one was in J, that means one of the elements in the chain had to contain one. But if J prime contained one, that means it actually would be equal to R. Because if J prime contained one, it would have to contain the principal ideal gender by one, which is itself R. And clearly J prime is a subset of R. So we get equality in that situation. J prime equals R. And that would make it an improper ideal, which means it wouldn't be a member of X. Therefore, none of the J primes has a one. Therefore, when we take the union, you can't magically create a new element. It had to be in one of them along the way. So since none of the ideals contain one, that means their union doesn't contain one, and that makes J a proper ideal. J is then an element of the set X. Well, if J is itself a proper ideal, that means the chain C has an upper bound. Because J is larger than every element inside of this chain, right? Because by construction, J prime will be less than or equal to the union of the J primes, right? Which is by definition J. So J is an upper bound for the chain. Therefore, we now get to Zorn's lemma. Because this was an arbitrary chain, this means that since every chain, since every chain has a upper bound, by Zorn's lemma, X has a maximal element. Well, this is a maximal element with regard to set containment, and it has to be a proper ideal. That's exactly what a maximal ideal is. So therefore, by Zorn's lemma, every ring with unity has a maximal ideal. In particular, this maximal ideal will contain the ideal I. So before we close this video on Zorn's lemma, I wanna summarize the previous proof and basically provide a template that in an abstract algebra setting, when you're proving something by Zorn's lemma, Zorn's lemma arguments basically always have the following template, the following six steps. And I want you to go back to the proof that we went through to prove the existence of maximal ideals containing I and show that each and every one of these steps are there. So the first step is we have to declare the appropriate partially ordered set for the proof. So we have to declare some set X, which is gonna be our partially ordered set. And so we have to tell the reader what that is, what is the partial order. In the previous proof, we did that, we said that X is the set of all proper ideals of our containing I, and it's ordered by set containment. So you declare what is the set that you're considering. Then you have to argue that the partially ordered set is non-empty. This was a required statement for Zorn's lemma, the partially ordered set has to be non-empty. In the previous proof, we argued that the ideal I itself belongs to X. Then, number three, we consider it arbitrary chain C and then nominate an upper bound for that chain. We'll call that nominee J. Now typically, like in the previous proof, typically the nominee for the upper bound is going to be the union of all of the elements in there. Because after all, this is the case when your set X is ordered by set containment, then you can take unions. But oftentimes, if you're in some type of like lattice situation, you would take the join, the compositum of each of those elements and that's your upper bound. That might not always be appropriate, of course, but oftentimes we're making arguments where this set X is ordered, partially ordered by set containment, therefore the union will pretty much always do it for you. You're then gonna nominate this upper bound. So that's what we did there. We said in the proof, we claim, well, we define J to be the union of all of the ideals inside of C and then we claimed that it was an ideal itself. So then step four, you have to demonstrate that J belongs to the set X and therefore is an upper bound for C. So that's what we did there. We showed that the union of a chain of ideals, it's itself an ideal. We argued because of unity that it was a proper ideal, therefore it belongs to the set X and then because it was the union of a chain, it was necessarily an upper bound. Step five, you then invoke Zorn's lima to then produce a maximum element of the set X, okay? And then step six, so we did that since every chain had an upper bound by Zorn's lima, X has a maximal element, which has to be a maximal ideal containing I. And then step six, we didn't really see this in the previous proof there. You had to argue that the existence of a maximal element that proves the theorem that's in the questions that you're asking in question, and which case for us in the previous proof, our question was, do maximal ideals exist or not? So step six was sort of trivialized in that situation, but in the next video, we'll provide an example where the maximum element doesn't end the proof, there's still some more that has to be considered there.