 In this video, we're going to prove the main result of lecture 17, that every principle ideal domain is in fact a unique factorization domain, or more particular, if D is a principle ideal domain, then D is a unique factorization domain. I guess that doesn't specify much more, but we're going to be talking about this domain D. It's a PID by assumption. We have to prove it's a unique factorization domain. Now, of course, to be unique, to be a unique factorization domain, there's sort of two things have to happen. First of all, of course, it does have to be an integral domain in the first place, but that's a non-issue for a principle ideal domain because both of those definitions assume the ring in question is an integral domain. So then the idea comes with unique factorization, which really unique factorization breaks up into two parts as well. So to have unique factorization, first of all, you have to have factorization. Every non-zero non-unit element has a factorization of irreducibles. That's the first part. So there's an existed statement of factorization, but then the unique part says that two different factorizations of the same element are really equivalent to each other with regard to equivalence of factorization. So they're not really different factorizations, you know, up to relabel or up to order of factors, associates of irreducibles and things like that. So that's how we're going to proceed forward. We're going to first prove that if you take a non-zero non-unit element of a ring, we're going to call in our PID, of course. We're going to call that element A, it belongs to D. If we take a non-zero non-unit element A, we have to first prove it has a factorization consisting of irreducible elements. Then we're going to prove that that same element A has a unique factorization, that two different factorizations are actually the same thing. So let's take this element A. So the first thing that we're going to claim here is that A has an irreducible factor. Clearly, to have an irreducible factorization, you need to have all these different irreducible factors, but does it even have one? Okay, that's a very important concept. And so that's the first claim we're going to prove here. So if A was an irreducible, it would be a factor of itself, because you can just factor A as one times A, so we'd be done. So we can assume that A is, in fact, it's a reducible element. Therefore, it has a factorization factor A into A1 times B1, where neither A1 nor B1 was a unit. And so what that then means because of this factorization, the principal ideal generated by A is contained inside the principal ideal generated by A1. And this containment must be proper. Since B1 is not an associate, well, let's say this way, since A1 is not a unit, that means B1 can't be an associate of A, but also goes the other way around. Since B1 is not a unit, A1 is not an associate. So we know that this ideal is not equal to this ideal, because principal ideals are equal to each other if and only if the two generated elements are associated to each other. So we have a proper containment. We also know that this is not the whole domain, because A1 itself is not a unit. Now, if A1 was an irreducible element, we would be done. So let's suppose that A1 is not irreducible. So it also has a factorization since it's a reducible element into a product of A2 and B2, where again we can assume that A2 and B2 are not units. Let me bring the words up here. And if by similar reasoning, this tells us that the principal ideal generated by 1 is properly contained in the principal ideal generated by A2. Because again, A2 is not an associate of A1. And again, this is not the whole domain itself, because A2 is not a unit. So if A2 was an irreducible element, then we could stop because that's the irreducible factor of A by transitivity of factorization here. So if A2 is not irreducible, that makes it reducible, and then we can continue to do this process. So we continue to factor, factor, factor. So we take a factor of A, call it A1, then we take a factor of A2, excuse me, a factor of A1, which we call A2, then we can take a factor of A2, call it A3. And if these elements are continuously not being irreducible as the reducible elements, then we have an ascending chain of ideals. At each step, this containment grows properly. And so we have an infinite ascending chain that never stabilizes. That's not possible in a principal ideal domain, because principal ideal domains are noetherian. And noetherian rings satisfy the ascending chain condition by definition. Therefore, that tells us that at some point this process has to stabilize. There actually is one of these factors, some An, that's in fact irreducible, excuse me. It's going to be irreducible. And we're commutative and everything. So what we're going to have here is that A factors as An. And then we have, of course, have all these B's floating around here, B1, B2, all the way up to BN, like so. And put those together, we do have a factorization where An is irreducible. Going forward, we're going to call this element P1. And this is because we want to think of it as a prime element in a principal ideal domain, primes and irreducibles are actually one of the same things. And so this proves that A does in fact have an irreducible factor. Now, I wanted to make a comment before we go on here. We didn't really use the fact that the ring D was a principal ideal domain. We just argued that it was a noetherian domain. So in this situation, a noetherian domain implies you have irreducible factors. Not necessarily an irreducible factorization, but every non-zero non-unit element had an irreducible factor because of the sinian chain condition, which a PID does satisfy. But more broadly, a noetherian ring, a noetherian domain would have the property that every non-zero non-unit element has a irreducible factor. So we're then able to factor A using P1, which is irreducible, and C1. Now, if C1 was a unit, we would be done because then this would be an irreducible factorization, because that actually would imply that A itself was irreducible. If you're an associate of an irreducible element, you yourself are irreducible. So let's assume that C is not a unit. If C was itself an irreducible element, then A would be a product of two irreducibles, so we would be done. So we then might want to consider what happens if C is a irreducible element. But if C1 is reducible, even if it's irreducible, that's fine. If it's reducible, you at least have an irreducible factor by the previous claim. So because C1 is not a unit, it's also not zero because that would imply A was zero, C1 has an irreducible factor, call it P2. And so we don't even have to consider whether C1 is itself irreducible because it could be that C1 and P2 are actually the same element and C2 is one. That's a possibility. But that then changes the factorization of A to be A equals P1 times P2 times C2, for which P1 and P2 are known to the irreducibles. Well, is C2 a unit? If C2 were a unit, then this would be an irreducible factorization of A because P2 times C2 would be, well, which is of course equal to C1 would have been irreducible in that situation. So if C2 is not a unit, then we could continue this process. But does this process ever stop? The answer has got to be yes. Because as we look at this chain that we've created, the ideal generator by A is contained inside the ideal generator by C1, which is then contained in the ideal generator by C2, which is then contained in the ideal generator by C3. And so every time that C is not a unit, this ideal grows. And we get bigger one and a bigger one and a bigger one. And so if C never becomes a unit, then we have an infinite ascending chain that never stabilizes, which contradicts the ascending chain condition. Which again, principal ideal domains satisfy the ascending chain condition because they're noetherian. So what this means is eventually this ideal is going to stabilize. And at that moment of stabilization, the number C had to in fact be a unit, in which case then a unit times an irreducible is still an irreducible. And so we then get this factorization into irreducible elements. A equals P1 times P2, all the way up to PR, where R was the step where this process terminated with this ascending chain. All right. So this then shows that A has a prime factorization. A does have a factorization inside of A, has a factorization into irreducible elements. So again, highlighting what we saw here, we haven't really used the fact that our ring is a principal ideal domain except that a principal ideal domain is noetherian. So if you have an oetherian domain, I want you to be aware that an oetherian domain implies the existence of prime factorizations. Although in a general domain, we don't necessarily know that irreducibles are always prime, so we should call it an irreducible factorization in that situation. So you can see of course the use we have of the ascending chain condition that principal ideals have. The ascending chain condition gives us that we have these prime factorizations for every nonzero nonunit element. So what about the uniqueness of factorization? So we know that PIDs like every noetherian domain have irreducible factorizations. Why are they unique? So the uniqueness of factorizations where we're going to see where the principal ideal component comes out here. Suppose we have A still, which is a nonzero nonunit element. And let's say it has two distinct factorizations into irreducible elements. One of them is P1 times P2 all the way to PR. Another one is Q1 times Q2 all the way to QS. And each of the PIs and the QJs are irreducible elements inside of the domain D. And for the loss of generality, or without the loss of generality, let's assume that R is less than or equal to S. We're going to prove that R equals S and that there's the one-to-one correspondence between the irreducibles that these are actually associates of each other. All right? So how are we going to do that? Well, if you take the first irreducible factor P1 in the first factorization of A, well, because A can be factored using these P's, that means P1 divides A. But A is also equal to the factorization Q1 times Q2 all the way up to QS. Now, we've previously shown that in a principal ideal domain, irreducible elements are prime elements. And therefore, a prime element, by induction, if it divides a product, then it has to divide one of the factors. And so let's say that P1 divides the irreducible factor QJ. One of those, it has to divide one of those by basically just prime elements by definition are those elements that satisfy Euclid's lemma. It has to divide one of the factors. So up to, you know, we can relabel things, right? So up to a rearrangement, we can move QJ up to the front and we'll then call that one Q1. And then the previous Q1, we can call that now QJ, whatever, just relabel things. We can assume that P1 actually divides Q1. And that therefore will make them associates because Q1 itself is an irreducible element. And so if we have a factorization here, Q1 equals P1 times something B, I will call B1. These are have to be associates because P1 and Q1 are both irreducibles of each other. So that's the first one. So up to a unit, let's actually, I guess I noticed down below I called that unit U1 for unit there. So if you have that, you can then replace the Q1 on the Q factorization with P1 times U1. You can cancel out the P's since we are in a domain. We have the cancellation principle. So then our factorization reduces down to this. These two things equal each other. We're going U1 is some unit here. But then the same principle happens as well. Q2 divides this factorization. Therefore it has to divide this factorization. So because irreducibles are prime elements, then that means P2 divides one of the elements QJ. Without the loss of general order, we can assume that P2 divides Q2 so that Q2 factors as P2 times some unit U2. And so then we can cancel P2 from both sides. We end up with P3 times a bunch of irreducibles up to PR. We then have U1, U2, which as U1 and U2 are both units, they're product as a unit. We then have Q3 all the way up to QS. And then we can keep on repeating this process, constantly canceling out all of the irreducibles in this factorization. We cancel in now with some QJ until we eventually get the following here that one on the left-hand side, because we cancel out all of the irreducible elements PI, is then equal to the following, which of course U1 all the way up to UR. That product is a unit because each of those U's is a unit. But then we also have these elements right here. But wait a second. If you take the product of QR plus one up to QS. If you multiply that by U1 up to UR, that equals one. This tells us that the product QR plus one up to QS, that product is a unit. And each of the individual factors in that would have to be units themselves, which means they weren't irreducibles. That would be a contradiction. That contradiction only happens if S is strictly larger than R. So we would have to conclude that S is in fact equal to R. We have the exact same number of irreducibles in the two factorizations. And this cancellation process we did earlier then determines the one-to-one correspondence between the irreducible factors PI with the irreducible factors QJ. And this one-to-one correspondence was always an association correspondence. And so up to reordering and up to association, it turns out these two prime factorizations were in fact unique. This then proves that every principal ideal domain is a unique factorization domain. And if we summarize, we did prove that every noetherian domain will have irreducible factorizations. But without the principal ideal assumption, we can't actually prove that a noetherian domain is necessarily going to be a unique factorization domain. In fact, there exist counter examples to the contrary, which we can have a noetherian domain that is not a unique factorization domain. Of course, our favorite example is going to be Z adjoined the square root of negative 3. We know this thing is not a unique factorization domain, but as we mentioned earlier, this is in fact a noetherian domain. So in a noetherian domain, you have irreducible factorizations, but they're not necessarily unique. In a principal ideal domain, those factorizations are in fact unique, which is what makes it a UFD. So every PID is always going to be a UFD. So the class of PIDs sits inside of the class of UFDs, but this containment is in fact proper. There do exist unique factorization domains that are not principal ideal domains. And an example of this will be provided some other time, not in Lecture 17, which we actually bring to a close right now. Thanks for watching. Like these videos if you've learned anything about principal ideal domains. Subscribe to the channel to see more videos like this in the future. And post any questions you have in the comments below, and I'll be glad to answer them when I can. Bye.