 Welcome back to our lecture series math 4230 abstract algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. So as you recall, in lecture 16, we talked about the idea of a unique factorization domain. In lecture 17, we're going to continue this discussion talking about principal ideal domains and the two concepts are related to each other. We'll define what a principal ideal domain is just one second. It is in fact going to be a domain, so an integral domain meaning that we have a commutative ring with unity such that the cancellation axiom is satisfied, there's no proper divisors of zero. You recall that a unique factorization domain said that it's an integral domain for which all non-zero, non-unit elements of the domain have a unique prime factorization. And this is deeply connected to the idea of a principal ideal domain, which we will prove that at the end of lecture 17, not in this video though. So let's get to the titular topic here. We say that an integral domain, D, is a principal ideal domain, often called a PID for short. If all of the all of the ideals in the domain are principal, remember a principal ideal meaning that particularly in a commutative ring will be ideals of the form. So we denote it as this, we'll use in the same notation we did for cyclic groups, right? It's the ideal generated by a single element in a commutative ring with unity. The principal ideal will look like everything of the form a times r were r is an arbitrary element of the ring. Some people will of course denote principal ideals using parentheses, but we made the choice in this lecture series to stick with these bracket, these angle brackets for principal ideals. Now have we seen principal ideal domains? Yes, yes we have. In fact, the poster child of a principal ideal domain is going to be the integers, which using the division algorithm we've argued that every ideal in the integers has to be a principal ideal. A similar argument can also be used to show that the Gaussian integers z join the imaginary unit i also forms a principal ideal domain. Every ideal in the ring of Gaussian integers can be shown to be a principal ideal. And again, the main reason for this is because alternative version of the division algorithm is applicable in the ring of Gaussian integers. And we'll see in fact in the next lecture, lecture 18, that any ring that has a division algorithm is necessarily going to be a principal ideal domain. But that's something we'll talk about of course in the next lecture. If we look at our examples, z join the square root of negative three. This was an important example for us because this was an example of an integral domain that was not a unique factorization domain. And we showed that it wasn't a UFD because it didn't have unique GCDs or LCMs and things like that. Properties that a unique factorization domain had to have. It is also true that this domain, this integral domain z join the square root of negative three is not a principal ideal domain. And the issue comes down to sort of the same thing about GCDs that if you take the ideal generated by four and two plus two times the square root of negative three. If this ideal, which is clearly generated by two elements, if it was generated by a single element, then that single generator would have to be the GCD of four and two plus two square root of negative three, which we showed in a previous video that, in fact, those two elements do not have a GCD. So that's why it wasn't a unique factorization domain. And this is actually sort of a foreshadowing to what we're going to see at the end of this lecture here that every principal ideal domain is actually a unique factorization domain. Because, among other things, the principal ideal property in a situation like this would enforce the existence of a GCD. So if you don't have GCDs, then you can't be a principal ideal domain, okay? So let's mention a proposition here about principal ideal domain. So let D be a PID, and suppose we have two elements, A and B, that belong to the domain here, then the greatest common divisor, and likewise the least common multiple, of A and B exist. And in particular, the greatest common divisor D can be written as a linear combination of A and B. So this is a property that we used all the time back in Math 42.20, Aptac to Aldera 1, for which, if you took two integers A and B, that we would commonly use the fact that the two numbers could have its GCD expressed as a linear combination of the two numbers A and B. This is a property that's demonstrated by all principal ideal domains. The proof of this proposition, I'm going to leave as an exercise to the viewer here. Another important property about principal ideal domains is the following. Let D, again, be a principal ideal domain, and consider an element P that generates a non-zero principal ideal in the domain here. We claim that this principal ideal generated by P is a maximal ideal if and only if P is an irreducible element. So remember what the vocabulary there means. So what does it mean to be a maximal ideal? A maximal ideal means that if you take the whole ring and you take this maximal ideal, if there's any ideal that sits in between the maximal ideal and the ring itself, then it must be that the ideal is equal to the maximal one, or the ideal is equal to the whole ring itself. So that's what a maximal ideal is. What does it mean to be an irreducible element? It's called an irreducible element has to do factorizations. If P is equal to some product A times B, then what has to happen is that one of the factors had to be a unit, and the other ones it's associate, or either A or B was a unit in this factorization. The other one that wasn't a unit would have to be an associate of the number P in that situation. So it turns out that the concept of divisibility is closely related to the concept of ideals inside of integral domains. And so we can often make connections between ideals and factorization, such as in the principal ideal domain, we get that maximal ideals, because every ideal is principal in this domain. Maximum ideals coincide to its generator being irreducible. This is an if and only if statement would go in both directions here. So let's first suppose that the principal ideal in question is a maximal ideal. We have to then show it's irreducible. So consider a factorization of P into two elements, A and B. Well, since P can be factored as A times B, this product, A times B, belongs to the principal ideal generated by A, of course. And therefore that shows us that the ideal generated by P is contained inside the ideal generated by A. But okay, P was a maximal ideal. So if you have an ideal that sits in between a maximal ideal and the whole ring, then it has to be that the ideal equals P or the ideal equals the whole ring. Now in the first case, two principal ideals are equal to each other if and only if they're associates to one another. So if P and A were associates, that would imply that B was a unit. Thus showing, okay, so that's the first case. The second case is that if a principal ideal is the whole ring, in particular since this is a ring with unity, the number one is in there, that could only happen if A was a unit. So in either case, either A is a unit because that principal ideal is the whole domain or B is a unit because A and P are associates of each other. And as this was an arbitrary factorization of the element P, this then indicates that P was an irreducible element. So maximal ideals generated by a maximal principal ideals always coincide with irreducible elements. So let me kind of highlight something before we go down to the next part. In the first direction of this proof, we did not use that the domain was a principal ideal. We just started a principal ideal domain. We just started with a maximal principal ideal. And so I want to sort of summarize that this important proposition, which could be useful in other contexts. We have that a maximal principal ideal P implies that P is in fact irreducible. If we have any commutative ring with unity, we can make this argument here. A maximal principal ideal always coincides with an irreducible element. The importance, of course, of a principal ideal domain is that the reverse direction also happens, that all ideals are principal. And let's see that direction just right now. So this time, suppose that our element P is an irreducible element. And then we want to show that the principal ideal generated by P is maximal. So consider an ideal that sits between this principal ideal and the whole domain itself. Well, now since the domain is a principal ideal domain, this ideal that sits in between the two itself must also be principal. So it's generated by a single element called that generator A. Well, since the principal ideal generated by P is contained inside the ideal I, that means the element P belongs to the principal ideal generated by A. For which if P belongs to this principal ideal, there has to be some element of the ring called B such that P is equal to A times B. So notice we did have a factorization of P now. Well, P is an irreducible element. And so the only factorization that could happen is that either A is a unit or B is a unit. If A was a unit, like we argued similarly, then this ideal would be all of the domain D. The other possibility is that B is a unit, which then would imply that A is an associate. And if two elements are associates, then their principal ideals are the same. Okay? And so because A is an associate or a unit, being associates means the ideals are the same. If A was a unit, then it's the whole ring. And so that's the only possible ideal that can sit in between the two. And therefore we see that this is in fact a maximal ideal. Okay? A very important corollary of this result is that if you have a principal ideal domain D, every irreducible element inside of the domain is also a prime element. And so I want to remind us what we've seen before. The notion of a prime element and an irreducible element because of this corollary, which again we'll prove this in just one second, is actually in fact equivalent inside of a principal ideal domain. Because we already know that in an integral domain, primes imply irreducible. So prime elements imply that they're irreducible elements. This of course is in any integral domain. That's always the direction that we have. We've proven of course in a unique factorization domain that every irreducible element is prime. So that's why we're justified in calling it a prime factorization as opposed to irreducible factorization. This same property is also true for principal ideal domains that every irreducible element is in fact a prime element. And so the proof of it is very short compared to what we just did, which is why we call it a corollary. So let's take an element P, which is inside the principal ideal domain that is irreducible. By the previous result, we then see that the principal ideal generated by P is a maximal ideal. And by a previous result, not in this video, but in a previous lecture in this lecture series, we show that every maximal ideal is a prime ideal, which then means that the principal ideal generated by P is a prime ideal. But principal ideals that are prime are necessarily prime elements, and therefore proving the conclusion there. Every irreducible element is a prime element in a PID. So in PIDs and in UFDs, primes and irreducibles are equivalent to each other. But in general, we have seen examples of, we've have seen examples of, I should say irreducible elements that were not prime. Of course, our favorite example is this integral domain, Z adjoin the square root of negative 3. We did produce irreducible elements in that ring, which were not prime. As such, one of those elements we produced was the element 2. The number 2 is an irreducible element in this ring, but it wasn't prime. So we take this element here, the principal ideal generated by 2, it's in fact not a maximal ideal, because of course you can extend it into 2 adjoin 1 plus the square root of negative 3. This isn't the whole ring, but it's also a larger one. It's not a maximal ideal, but it is generated by an irreducible element. So in general, if you have an integral domain, irreducible elements don't necessarily coincide with maximal ideals, but they do in fact do that in a principal ideal domain.