 Welcome back to our lecture series, Math 4230, Abstract Outdoor 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. So in lecture 21, we're ready to get a pretty important theorem that we've been building up towards for quite a while now. In fact, what we're going to prove in this video is that if D is a unique factorization domain, which remember a unique factorization domain means that it's first and foremost an integral domain for which every non-zero non-unit elements in the domain has a unique factorization into prime elements. Of course, in a UFD, prime elements and irreducible elements are one of the same thing. So if you have a unique factorization domain D, then the polynomial ring D adjoin X is likewise a unique factorization domain. We've mentioned previously that a lot of properties of the ring D can be carried over to the polynomial ring as well. So if D has unity, then DX has unity. If D is commutative, then DX is commutative. We've seen previously if D was a domain, then DX is a domain. And we kind of got stuck there for a while. We were able to show that if F was a field, then F adjoin X is a Euclidean domain, which then implies it's a principal ideal domain and a unique factorization domain. But as of course, we've been interested in factorizations of polynomials in, well, we've been interested in factorizations in integral domains to begin with. So UFDs are the places to be. But how about factorizations of polynomials? We learned from Gauss's lemma that factorizations of polynomials over UFD coincide with factorizations of polynomials over the field of fractions. And since the the field of fractions is necessarily a field, F adjoin X in that situation is going to be a Euclidean domain. In particular, it's a unique factorization domain. And so we can push that back on to D. So really, this is going to be a consequence of Gauss's lemma here. So to be a unique factorization domain, we have to prove that every nonzero, non-unit element of DX has a unique factorization. So we have to prove the factorization exists and that it's in fact unique. So we don't have to worry about zero because you don't get unique factorization for zero. We have to also worry about units, right? Now, we know from previous work that if there's any units in DX, they actually belong to D themselves. That is the only units in a polynomial ring, they would have to have been coefficients, constant polynomials to begin with. So any unit in DX actually already belongs to D. And so in particular, if F is a constant polynomial, then it's really just an element of D in which case it has a unique factorization, unless it's a unit, but whatever, we don't have to worry about that. So the constant polynomials were good. So we have to consider a polynomial F whose degree is strictly greater than zero. That is, it's positive. And we know it can't be units. So with such a polynomial, does it have a unique factorization? Well, like I mentioned earlier, the field of fractions is going to be relevant in this conversation because the field of fractions, which itself is a unique factorization domain because it's a principal ideal domain because it's a Euclidean domain, F of X there. We know that since F of X is not a unit, it's clearly not zero, it does have a prime factorization inside of F a joint X. So let's say this is our factorization. F can be factored as P1X times P2X times P3X all the way up to PNX, for which each of these polynomials are going to be irreducible elements of F a joint X. Now this is where Gauss's limit comes into play here. If we have a factorization of F over the fraction field, then we can, without the loss of generality, suppose that each and every one of these polynomials Pi are actually polynomials inside of DX. But they might not necessarily be irreducible anymore over DX as they were over FX. So that's the thing we have to be concerned about, which we could factor out their content. It could be, because after all, the content of a polynomial is going to be an element of D, which every element of D, except for zero, is a unit inside of F. So with regard to the factorization in F a joint X, it doesn't care about content, because content is always a unit. But in D, that content might not be a unit. And so factor the content out of F, let's say the content is A. We may, without the loss of generality, assume that each and every one of these polynomials P1, P2 up to Pn are primitive polynomials. That is, their content is trivial, which remember the content is the GCD of all of the coefficients. So we have some factor A in front. So we have these, we have these polynomials P1 through Pn, which are primitive polynomials inside of DX. And we know they're irreducible over FX, and we factor out the content. So that's where we are right now. Now, as PIX is primitive in DX, we have seen that PIX is necessarily going to be irreducible in DX, if and only if it's irreducible. Did I say that one right? These polynomials are going to irreducible in DX, if and only if they're irreducible in FX, because of the primitivity statement there. Because after all, if PI had some factorization in DX, say we factored into some type of, you know, some other polynomial, some Q1 times Q2 or something like this, if this was a factorization in DX, whoops, into DX, then this would be a factorization inside of FX, which PIX was irreducible there. So one of these would have to be a unit. And therefore, one of these elements would have to have been, would have to have belonged to the domain D, for which if we could factor out some non-unit element away from PI right here, that means it wouldn't have been primitive. So the primitivity of a polynomial in DX means it's irreducible in DX, if and only if it's irreducible in FX. So okay, so we can then say that each and every one of these polynomials PIX is irreducible. We know it's irreducible in FX, so therefore it's going to be irreducible in DX, because these are in fact primitive polynomials. Now, as D is a unique factorization domain, the content of the polynomial has a unique factorization. So A can factor into A1, A2, A3, up to AK, right? And so these are going to be irreducible factors inside of D. They're constant terms, so we can't factor them anymore in DX either. Therefore, the polynomial FX does have a factorization into primes, into irreducible elements. You factor the content, you factor the polynomial in FX, and so this gives us a prime factorization. That's the first step to be a unique factorization domain. Every non-zero non-unit element needs a factorization. Why is it unique? Well, that's what we're going to do on the next slide right here. Why are these factorizations unique? So suppose that F, the same polynomial here, has two factorizations inside of DX right here, right? So one of the factorizations was the one we saw a moment ago. So you factor the content A1 through AK, you factor the polynomial P1 through PNK, for which we can assume that each and every one of these polynomials is irreducible in DX. Therefore, it's primitive in DX, and then the content is also factored into its prime divisors inside of D. We're going to do the same thing with B right here, right? You took out all of its content. Therefore, the Qs are each primitive polynomials, and the content can also be factored uniquely like so. Each of these polynomials, P1, P2 through PN, and Q1, Q2 through QM, we can assume that these are proper polynomials. That is, they're non-constant polynomials. We know that none of their degrees are zero. Again, they're non-constant. So these polynomials, Pi and Qj, they are proper polynomials. They're degrees greater than zero. And the numbers A through the As and the Bs, these are elements of the domain D. So they're constant polynomials, their degrees equal to zero. So that's the factorization we have there. And again, we also know that the P's and the Q's are primitive polynomials over DX, because they're irreducible. Now, again, that's what I said just right there. Since they're irreducible, they were primitive. Now, as such, Pi X and Qi, or Qj X have to be irreducible over FX as well, because again, this is Gauss's lemma. The factorization over DX is the same as the factorization over FX, essentially. I mean, there are certainly other factorizations you can do in FX, but it can always be pushed back onto DX. And so by Gauss's lemma, we know these polynomials, they're primitive and they're irreducible in the Euclidean domain FX. Okay, so let's set A just to be the content here and B to be the content of the second factorization. Then in particular, these are units inside of FX because they're neither one of them is zero, because that would imply FX was zero. They're constant polynomials, those are all units inside of FX join X. So then the above factorizations of F have to be unique, because FX join X, we've proven, is a Euclidean domain, so therefore it is a unique factorization domain. So when we look at this up here, if we just call this thing right here A and this thing right here B, these two factorizations, because A and B are just units here, this factorization of F inside of FX join X, we have unique factorization there. So there's a one-to-one correspondence between the factors, the irreducible factors of the P's and the irreducible factors involved with the Q's. So we're going to get things like N equals M and we can then say without the loss of generality that P1 is an associate to Q1, P2 is an associate to Q2 all the way down, so that PN is an associate of QN like so. Because again, we have unique factorization in that situation. All right, so where are we now here in our writing? Now as each of these polynomials PI and QI were primitive in DX, it actually holds that they must have been equal to each other. Because if there are associates, if P1 was an associate to Q1 X here, that means we can time something, we have a unit so that P1 X times a unit times Q1 X, we have equality in that situation, where of course a unit would necessarily have to be a constant polynomial. So as this element U is an element of D, we can basically kick this thing out here and so we can just assume that the Q's and the J's are actually equal to each other, canceling these things out because DX is in fact a domain, it then boils down to the A's versus the B's, right? So we then have A equals B, for which yeah, we push this U over here into this part of the factorization. So these are elements, these are these elements A equals B is inside of a unique factorization domain. So this is two different factorizations of the same element, A and B are the same element. So by uniqueness, we're going to have to have that these factorizations are the same, they have the same number of factors, K equals L, and there's this one-to-one correspondence between association in that situation, and this improves that we have a unique factorization of FX here. So therefore, the polynomial ring D adjoin X is in fact a unique factorization domain. Now it's worthy to note that if we take the previous theorem and apply induction to it, we get a stronger corollary. For example, if D is a unique factorization domain, then D adjoin any number of indeterminants will likewise be a unique factorization domain. And why does this follow by induction? Well, this ring can be written as DX adjoin 1, X2, all the way up to X in minus 1, and you can think of that as the coefficient ring when you adjoin on a new variable. So if you have it in the first step, then you have it in the second step, the third step, the fourth step, it follows by induction. So if with any finite number of indeterminants adjoin to a unique factorization domain, you still have a UFD. So this will include Z adjoin X, a field adjoin X, Y, a field adjoin three variables, or field adjoin four variables, you get the idea. These are all going to be unique factorization domains, but they're not going to be principal ideal domains, right? Z adjoin X has an ideal to join X, that is, you have two an ideal generated by them. That's not an ideal. Excuse me, it is an ideal, it's not a principal ideal, and therefore ZX is not a principal ideal domain. F adjoin two variables X and Y, you can take the ideal generated by the two variables X and Y, that's not going to be a principal ideal. And so these are good ways to construct unique factorization domains that are not principal ideal domains.