 Welcome back to our lecture series Math 4230, Abstract Outdoor 2 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misseldine. Lecture 16 is going to focus on the idea of factorization, and in particular, we're going to build up to the idea of a unique factorization domain for which we're going to define into the next video, part two of lecture 16 here. Before defining what a unique factorization domain is, we need to actually clarify certain ideas about factorization in commutative rings with unity. Because after all, if we go back to some of our earliest learnings of mathematics beyond basic addition and subtraction and such, some of the first conceptual ideas we learn about the integers, whole numbers as they might have been called, are ideas of factorization to visibility. Where this isn't just like rote calculations where we learn how to add and we can memorize that, or we memorize multiplication tables or things like that. Factorization was one of these first hard problems, and we sometimes draw factorization trees and things like that to find prime factorizations. If we have two different numbers, we can factor them and try to find their greatest common factor or least common multiples, and there's some important applications to understanding these, but factorization and the related concept of visibility are some of the most fundamental yet first conceptual ideas we learn in our mathematics educations. Now, in this lecture 16, and in the next coming lectures as well, we want to start developing the notion of factorization and visibility for the general category of rings, or more specifically, we're gonna focus on commutative rings with unity, and even more specific than that, we're really gonna be talking about integral domains. Because it turns out that outside of an integral domain, factorization can be very strange. It can be a very strange thing. It doesn't exactly behave the way we're used to, at least not from our primary education whatsoever. But even inside an integral domain, things can get a little bit weird as we'll see some examples of that. And so that's what's gonna lead to the idea of a unique factorization domain, for which those are exactly the integral domains where the rules of factorization work the way that we're used to. In other words, they work the way that they do for whole numbers, which is sort of coming up ultimate goal. So I should mention that as we go forward with lecture 16 into the future, when we talk about a ring, that to us is gonna mean a commutative ring with unity. I'll usually try to say it, but it is somewhat of a mouthful. So if I just say ring for short, I want you to assume that in these lectures about factorization, which of course they coincide with chapter 18 in Tom Judson's abstract algebra textbook, this ring means a commutative ring with unity. Because if we don't have commutivity or if we don't have unity, rules of factorization can be very, very alien. And we want to look at those rings that factor the same way that the integers do. That is leading up to this unique factorization domain. So there's some important terms we have to talk about before we can define what unique factorization even means. So imagine we have a ring with unity, call it R, and take two elements A and B that are inside that ring. What does it mean for A to divide B? Well, we say that A divides B if there's some other element C inside the ring such that AC is equal to B. So A divides B if A is a factor or divisor B because there's some product of B involving the number A. So just like we do when we talk about integers, we're gonna use the symbol A dash line or whatever you call that. You're gonna use a line here, A divides B. That's how we read that. And I should mention that this divisibility symbol is gonna form what we call in mathematics a preorder as a relation. A preorder is, it's kind of like the intersection between partial orders and equivalence relations. A preorder is gonna be something that's first of all, reflexive. And so why is this reflexive? Well, it's reflexive because you have unity, right? In particular, A divides A because A can be factored as A times one. So that's one of the reasons why we need to have unity inside of our rings in this situation. But we also have that it's gonna be transitive. Transitive. So if A divides B and B divides C, then we can infer that A divides C. And so why is that? Well, if A divides B, that means there's some AD that's equal to B. If B divides C, that means there's some BE that equals C. And so we can put those together because you can substitute this into there. So we end up with AD times E equals C. But by associativity, you can redo this and you get A times DE is equal to C. So it's gonna be a transitive relationship. Is it a partial order? Is it an anti-symmetric relationship? Well, it kind of depends. I mean, if you look at the de-factorization of positive integers, that does give you a partial order. But if you look at the factorization of integers, then you don't get a partial order because the anti-symmetric property doesn't happen. In the ring of integers, you can have two that and two divides negative two, since negative two is two times negative one. But you also have the negative two divides two because in that situation, two can be factorized negative two times negative one in that situation. And so in general, this divisibility relation on a ring doesn't give you a partial order because the anti-symmetric property doesn't always hold. So like I said earlier, pre-orders, that's actually the intersection between equivalence relations and partial orders. Basically, you're dropping the symmetric or the anti-symmetric. You're just taking reflexive and the transitive relations. That's why it's called a pre-order. It's getting ready to be a partial order, but you don't necessarily have anti-symmetry. Equivalence relationships are also pre-orders because they have symmetry. That's to say, they have the reflexive and transitive properties. And so when it comes to divisibility, if it's not a partial order, it's because of this issue with anti-symmetry. You could have these numbers that divide each other. Now, because of this phenomenon, it actually turns out we should give it a name. If you have two elements in a ring, call it R and S, then we say that R is an associate of S. If there exists a unit U inside the ring such that R equals US, okay? So if R, notice in the situation, we have that S divides R. But since U is a unit, we can actually rewrite this as U inverse R is equal to S. So we see that R divides S. So R divides S and S divides R, they divide each other. That's exactly what an associate is, although this will be our formal definition. Two elements are associates of each other if there is a unit which multiplied by one gives the other. Now, speaking of pre-orders here, the relationship that R is an associate to S is an example of an equivalence relation. An element is an associate to itself because like we said earlier, R equals R times one. It's transitive because being an associate means you divide each other, right? If R is an associate of S, excuse me, if S is the way I phrase it here, if S is an associate of R, that means R divides S, okay? But we also have the other way around, you know? If S divides its associate T, then you're gonna get, in fact, that R divides T. R divides T. So because of the visibility relationship, you're gonna get transitivity for associates as well. But now this is where symmetry comes into play like we said a moment ago, right? Because the unit can come over and become its inverse, so you get U inverse R equals S, you're actually gonna get symmetry as well. So if R is an associate of S, S is an associate of R. So being an associate forms an equivalence relation like we see over here, you have these association classes inside of a ring with respect to these factorizations. And the association classes are the obstruction from divisibility being a partial or if you're not, you're gonna be anti-symmetric only if no elements have different associates, if you're only associates to each other. Like if you look at the natural numbers, they don't have any associates because their only unit is one. If your only unit is one, then of course you have no other associates. But like in the integers, you do get the unit negative one, which is really typical. And so in general, we're not gonna get a partial over there, but this studying of associates is really important. When it comes to factorization, there's really no reason to distinguish between associates. So when we talked about two before, right? You know, two can factor as negative two times negative one. This factorization is kind of trivial. It's like, oh, you just replaced two with an associate. If one of your factors is a unit, it kind of doesn't matter in terms of the factorization. And so that's then gonna lead to the idea of an irreducible element inside of a ring. So if M is an element of commutative ring with unity, we say that M is reducible if every factorization of M involves a unit, which is equivalent to saying that every factorization of M involves an associate of N. So in other words, if M equals AB, so if you have a factorization of M, irreducible means that either A was a unit or B was a unit. Remember the set R star here represents the units of the ring. Related to the idea of irreducible elements is the idea of a reducible element. A reducible element means it has a factorization where the two elements involved are not units. And so in some regard, a factorization of reducible element then gets you something smaller, right? We were factoring 12 before. 12 is a reducible element inside of the integers because we can factor this four times three. Four is likewise reducible because we can factor this two times two. Three, on the other hand, is an irreducible element because every factorization of three involves either one or negative one. In other words, it involves the three itself or it's associate negative three. And the same thing can be said for two. Two and three are gonna be these irreducible elements because there's no factorization of them that gets you something simpler than what you already had. Every factorization involves a unit or an associate. Now, don't we usually call two and three prime numbers? Why are we calling them irreducibles? Why don't we just call this a prime element? Well, it turns out in ring theory, we have a notion of a prime element and it's actually defined differently than irreducibility. In primary school, when we learn about a prime, what we actually defined is an irreducible number, not a prime number, which I confess irreducible numbers a little bit more of a mouthful for little children. But nonetheless, when we define prime numbers in primary school, we're like, oh, three is a prime because every factorization of three is one times three. We're ignoring negative numbers because students often haven't learned about negative numbers. But because the negatives are just associates to positive, that's okay, we can ignore them because the inclusion of negative numbers, the inclusion of the negative unit, negative one, doesn't really change the factorization, it's irrelevant. And so we define prime numbers actually to be irreducible numbers, okay? So what's a prime number then? What's the real definition of a prime? So we say that P is a prime element of a commutative ring with unity R if every product, if any product A, B and R has the property that P divides it, then either P divided A or P divided B. So say that again, we say that a number, or what, a number, what's a number, right? A number is just an element of a ring, I guess, right? You know, if we have some element of a ring that divides a product, that element is prime if for whenever it divides the product, it divides either the first factor or the second factor. Now, this statement right here is a resemblance of what's commonly referred to in number theory as Euclid's lemma. Euclid's lemma says that if a prime number, speaking of integers right now, if a prime number divides a product, then it divides one of the factors in the product. So we're actually taking Euclid's lemma to be the definition of a prime in a general ring, right? If an element's prime is whenever it divides a product, it must divide one of the factors. So if we were to use the language of ring theory, which we now have on the screen, we would actually say that Euclid's lemma is the following, irreducible numbers or prime numbers, at least in the ring of integers. We're gonna see that in general, that's not always the case. Now, in the case of an integral domain, it is true that primes are always irreducible, but in a later video, I'll actually provide you an example of an irreducible element in an integral domain that's not a prime element. That's not the case for the ring of integers, which is why in primary school, we can abuse the term irreducible versus prime, but in general domains, we have to be careful we distinguish between these objects, but in integral domain, primes are always irreducibles. Prime being a prime implies you're irreducible for the following reason. Imagine we have a factorization of a prime element, we're assuming P is prime here. So to be irreducible, we have to prove that either A is a unit or B is a unit, okay? Well, it's always true that P divides itself, right? Because you can always factor an element as itself times one. So P does divide P times one, okay? We also know that P divides AB, right? Because I mean, it's equal to AB, that's what we're trying to say right here. Since P divides itself and itself is equal to AB, we know that P divides AB. Why is that important? Well, that's part of the definition of being prime. Since P can be factored as A times B and P divides itself, P divides AB. Because it's prime, then by the Euclid property, the definition of a prime, it must divide A or it must divide B. We are in a commutative ring, so it really doesn't make any difference which factor we use, because the first one could be the second one doesn't matter. Without the Lawson General, we may assume that P divides the factor A. Now, if P divides A, that means there exists some element C inside of the integral domain D, such that PC is equal to A. Well, again, that's what divisibility means. That's the definition. And therefore, when we come back to the original factorization, P is equal to AB. Since P divided A, that means that we can replace A with PC times B. We're in an associative ring here, so we actually end up with this thing right here. This is P times CB, like so. And so notice the factorization. We have P is equal to P times CB. Wait a second, we're in an integral domain, right? We haven't used that fact yet. Well, because we have a P times one is equal to a P times CB, in integral domains, we can cancel, we left cancel, and so we end up with CB is equal to one. That implies that B was a unit. Remember, we said P divided A. Well, the one that P didn't divide actually was a unit. And so since the factorization of P, this was an arbitrary factorization, always involved a unit, that implies that P is an irreducible element. So in integral domains, primes are always irreducible, but be careful. In the future, we will see an example of an element, but at the end of this lecture, in fact, we'll see an example of an element which is irreducible, but not prime.