 In this video, we are going to prove the titular topic for lecture 20 here, which is the so-called lemma of Gauss, Gauss's lemma. Now, before we present Gauss's lemma, there's a definition that's important to bring up in this situation. Suppose D is an integral domain, and we have some polynomial f of x over this integral domain. So f of x looks like a sub n times x to the n plus a n minus one times x to the minus one, all the way down to like some a2x squared plus a1x plus a0. And all of these a's, all these coefficients come from our domain D, like so. So then we define the content of our polynomial f as the GCD of the coefficients a0, a1, a2, a3, all up to a n. And we say that a polynomial is primitive if this GCD is equal to one. Okay? Now, admittedly, we have talked about beforehand how GCDs might not even exist in all integral domains. So this content might be undefined for certain polynomials. But in the setting, of course, of a unique factorization domain, GCDs, arbitrary GCDs do exist. So therefore, every polynomial has a well-defined content. And of course, the notion of a primitive polynomial makes sense always, because if the GCD is equal to one, we call it primitive. So that's why that's the setting for which Gauss's lemma is going to take place in. Suppose that D is a unique factorization domain. So in particular, GCDs always exist, and every polynomial has a well-defined content. Again, it could be one, a primitive polynomial, we like primitive polynomials. Take two polynomials, f and g, that belong to this ring, D adjoined x. Then the content of their product, the content of f times g, is going to equal the content of f times the content of g. So it turns out that over a unique factorization domain, this content function is multiplicative. The product, the content of the product is the product of the contents. We have this multiplicative property, which is really, really nice. So in particular, if f and g are primitive polynomials, this then tells us that the product of two primitive polynomials is primitive. And so that's what often is described as Gauss's lemma, the product of two primitive polynomials is primitive. But it's true for general polynomials that the content of the product is the product of the contents. Okay. And so let's take a look at this. And it turns out that we can actually push Gauss's lemma into the setting of being primitive very, very quickly here. So we have our two polynomials, f of x and g of x, factor out their content, right? If c is the content of f here, that means every coefficient in f is divisible by c. You can factor it out. And then everything that's left behind, all those coefficients where you factor away the t, factor away the c, those numbers have to then be relatively prime because if there was more, if there was another common device or beyond one, we could have factored that out as well. And that would have meant that c was not the greatest common divisor. So what I'm trying to say here is that if you have a polynomial that's imprimitive, you could factor out its content, and then you're going to get content times a primitive polynomial, which we're going to call that f1. And so the same thing with g of x right here, if its content is called d, factor it out, and then g is going to equal its content d times g minus one sub polynomial, and that polynomial will be primitive. And it'll have the same degree as g. Same thing with f1 here. f1 will have the same degree as f because these contents are numbers from the domain d, and therefore there's zero degree polynomials. They're just coefficients in that situation. So f1 and g1 are going to be primitive in this setting. And so when you look at the product f times g, this is going to equal the product cd times f1 times g1. And since f1 and g1 are both primitive, it suffices to prove Gauss's lemma in the setting where the product of primitive polynomials is primitive. Because if, since these polynomials are primitive, if that statement holds, then the content of f1 times g1 is going to be one, that the content there, and then times that by cd, the content of that situation would then be cd, which is then the product of the contents for f and g. So we ought to kick the can down the road, and we can assume that f and g are primitive polynomials going forward here. That makes the argument a lot easier. So again, well, maybe not again, but let's, let's, let's specify the coefficients of f here. So let's say that the coefficients of f will be denoted a sub i, where i then captures the power of x, it's the coefficient of, and the same scheme for g, we will call the coefficients of g, bi, or bj I should call it, where again j is the coefficient, the index of the coefficient, bj is the exponent of x there. Let's suppose that f has degree n and g has degree m like so. Now, let's consider, so we're assuming, we're assuming f and g are primitive polynomials. What if f times g was imprimitive? That is, it's great. The greatest common divisors between the coefficients was something other than one. Well, there's got to be a prime divisor. So let p be a prime divisor dividing all of the coefficients of f times g. Now, because f and g, well, let's just look at f for a moment. Since f is a primitive polynomial, p cannot divide all of the coefficients of f. In particular, there's got to be a smallest index where a smallest integer, again with regard to the subscripts here, there's some smallest number r such that p doesn't divide ar. Because again, if there wasn't, that is, if p divided everything, then f would be imprimitive, which by assumption it's not. Similarly for g, since g is primitive, there is at least one coefficient of g that's not divisible by this prime number p. And so choose s to be the smallest index where that happens. So we're assuming that p does not divide ar and p does not divide b sub s there. And this follows from the primitivity of f and g. So now we're going to look at the coefficient of the monomial x to the r plus s in the product f of x times g of x. Where does that come from? Well, the coefficient of x to the r plus s power, let's call it c sub r plus s, it's going to be computed by the convolution product here. We're going to take the sums of all of the possible products a i times b j, where the indices i and j when they add together is equal to the quantity r plus s. So then if you look at all the possibilities, you could have a zero times b times r plus s a one times b r plus s minus one. Going down here, you have a times r plus s minus one times b one, you'll have ar plus s times b zero. And I confess that because of the degree of the polynomial, it might be that like r plus s is larger than one of the degrees, it could be larger than n or m or something like that. That's okay. Some of these coefficients could in fact be zero in the product. It's not a problem by including them here. In particular, the product ar times b s is going to be in that combination because clearly r plus s adds up to be r plus s. It's in the name there. And so notice the assumptions we have here. By assumption, we have p divides a zero, a one, all the way up to ar minus one. Because remember, ar was the first coefficient of f that wasn't divisible by p. Likewise, by assumption, p divides b zero, b one, b two, all the way to b s minus one. Because again, b s is the smallest coefficient of g s. I should say the first coefficient of g s that's not divisible by p. And so if you take all of these things here, so when you look at this one, oh, a zero is divisible by p. So is a one all the way up to all of the a's up to ar are going to be divisible by p. But from the other direction, b zero is divisible by p, b one is divisible by p, and so all the way up all of those coefficients b's except for b s there are going to be divisible by p. So everything here, this is sum of products which each of those a's is divisible by p. And over here is a sum of products where each of the corresponding b's are divisible by p. So if you put all of those things together, that sum is divisible by p. Okay. And then by assumption, we also have that cr plus s is divisible by p, because p is a prime divisor of the content of f times g. So it divides all of the coefficients. So if we move this sum to the other side of the equation, clean this up a little bit, we're going to have cr plus s minus each and every one of these. Well, actually let me make a little bit simple. We're going to move this one to the other side like so. Because each and every one of these things is divisible by p, you're going to have that cr plus s minus ar b s is divisible by p. But then there again, cr plus s is divisible by p by assumption. So then this eventually gets us to the conclusion that p divides ar times b s. But wait a second. p is a prime number. So Euclid's lemma applies. And why by prime number, I don't necessarily mean it's like an integer. It's a prime in the field. And I say by Euclid's lemma, but honestly speaking here, Euclid's lemma is taken as the definition of a prime number. I guess I say a field a moment ago, d is a unique factorization domain here. We do have these prime elements. Prime and irreducible elements are exactly the same thing. But prime elements are defined by this Euclid's lemma property here. So the only way that a prime element could divide a product is that p divides ar or p divides b s. That's the only thing that could happen. But by construction, I should say by selection of ar and b s, that doesn't happen. We get a contradiction. And what was the contradiction? The contradiction was that the polynomial f of x times g of x, its content had a prime divisor. There is a prime that divides all of the coefficients of f times g. We don't have that. That's a contradiction. So there is no prime divisor of the content of f times g. Now the coefficients of f times g, like all the polynomials here, they come from a unique factorization domain, d. This is a unique factorization domain. Therefore, each of those coefficients has prime divisors. We have unique factorization. So what that means is our content has to be a unit. It's a unit for this polynomial, which then tells us that the GCD is equal to 1. And therefore, f of x is a primitive polynomial. So that gives us then the proof of Gauss's lemma. Now there's a very important corollary to Gauss's lemma. And honestly, this is how we are going to think of Gauss's lemma in this lecture series, this corollary right here. So let d be a unique factorization domain. Therefore, the Gauss's lemma we saw in the previous slide will apply. And let f be the field of fractions over that domain, d. Let f be a polynomial in d adjoin x. So this is a polynomial whose coefficients come from the UFD. If there exist polynomials, which call them alpha and beta in f adjoin x. So alpha and beta have coefficients coming from the field of fractions. If there's polynomials f and beta, alpha beta, so it's at f of x, factors as alpha x times beta x, then there necessarily exist polynomials a and b over d adjoin x, such that the degree of alpha is equal to the degree of a, and the degree of beta is equal to the degree of b. And we have that f of x equals a adjoin x and b adjoin x. All right. So before proving this statement, I want to kind of explain what's going on here. So Gauss's lemma and its corollary that we're going to see here on the screen, they're going to tell us, and you might have to use a little bit of an induction argument here as well. But Gauss's lemma plus its corollary says that if a polynomial factors over a field, then we have essentially the same factorization over the corresponding integral domain, or I should say over the corresponding unique factorization domain. That is both factorizations when considered factorizations in f adjoin x here, because after all alpha and beta are in f adjoin x, but a and b are in d adjoin x, but d adjoin x is a sub is a subring of f adjoin x. So these are two factorizations in f adjoin x, which basically f adjoin x here is a unique factorization domain. Therefore, there's only one factorization up to up to uniqueness, of course. And so with regard to f adjoin x, these are essentially the same factorization, they're equivalent factorizations. But in particular, this factorization is in fact a de factorization, as opposed to the previous one, which is a f factorization. So over f adjoin x, it doesn't care, these are the same factorizations, we have a unique factorization domain. But every f factorization can be turned into a de factorization. So factoring polynomials over a unique factorization domain is equivalent to factoring it over its field of fractions. And so essentially, it doesn't matter which setting you're in when it comes to factoring, whether you're over the field of fractions or unique factorization domain. And so what this empowers us to do is things like the remainder theorem, the factor theorem, the number of roots theorem, those are theorems which we proved over a field, a field f. Well, if we take some subring of f, it's necessarily going to be, of course, a domain. If we take some subfield, excuse me, some subdomain of f such that f is still the field of fractions here, because if you take it too small, right, like the integers inside of the complex numbers here, well, this is a field, this is a domain, but the field of fractions is not c, it would be the rational numbers in that situation. If you have a domain whose field of fractions is given as f, the factorizations are the same. And therefore, things like the factor theorem can be pushed from the rationals into the integers. The remainder theorem, the number of roots theorems, and some other things that we'll talk about in the future, it really doesn't matter whether we're talking about the field or the unique factorization domain because of this consequence right here, Gauss's lemma. The factorization over the unique factorization domain is the same as the factorization of the field with regard to the polynomials. If you take, for example, the factor theorem that says the roots coincide with each other, right? That is roots of a polynomial correspond with linear factors. We prove that for fields, but if we have a factorization over the field with a linear factor, we get a root there, then we can push that into a factorization over the domain. And so therefore, the factor theorem can be carried over in that setting as well. For example, take the polynomial f of x equals, we're going to take a doozy here, x to the fifth plus 5x to the fourth minus 8x cubed minus 14x squared plus 6x plus nine. Take this polynomial as an example. I claim that one of the roots of this polynomial is going to be three halves. So it has a root. Now this is it, you can think of this as an integer polynomial. Its roots might be in a larger ring, like three halves, but we still have a factorization of the polynomial because, because by the factor theorem, if we look at the field of fractions of the integers, which is the rational numbers, yeah, you're going to have, you have a root, a rational root, three halves, that'll then give us a, that'll give us a linear factor over the rational numbers. What would that look like? Well, I can use synthetic division to ascertain that. So by synthetic division, let's just do this one real quick. You're going to have two, five, negative eight, negative 14, six and nine. We are going to divide by the number three halves, like so we're testing for a root, bring down the two, you get three halves times two, which is three plus five, which is eight times three halves is going to give you 12 minus eight, which is equal to four times three halves, which is equal to six, negative 14 plus six is then going to give us a negative eight, three halves times negative eight is going to give us a negative 12, like so negative 12 plus six is then going to give us a negative six, three halves times negative six is going to give us a negative nine, which adds up to be zero. So we didn't find, we didn't fact find a root. And so this root then gives us a factorization. So our polynomial f of x factors as x minus three halves times, we're going to get two x to the fourth plus eight x cubed plus four x squared minus eight x minus six, like so this is a factorization over the rational numbers. But I want you to notice here that this polynomial, its content is two, if you view it as an integer polynomial, of course, you have two, eight, four, negative eight, negative six, these are all even numbers, we can factor out a two. And so when you factor out that two, you have this factor of two, that's now you could put in front of the polynomial, which then you could redistribute here to clear the denominators, you also get the factorization of two x minus three times x to the fourth plus four x cubed plus two x squared minus four x minus three, sorry, kind of crowded there. And so we ought to turn a factorization of the rational numbers into a factorization of the integers. And the beauty of Gauss's lemma is this always happens if you are factoring polynomials over a unique factorization domain. So let's then see the details of this very argument here. Okay, so let me scooch back up just a little bit so we can see our two statements here. So we have a factorization over F, F equals alpha times beta, and we have a factorization over D where F equals A times B, like so. So as alpha and beta are polynomials over the field of fractions, if we look at basically the least common multiple between all of the fractions in alpha, let's call that least common multiple, you know, least common denominator C, you could times alpha by C, which C would be a number in D, which since it's unique factorization domain, least common multiples do exist in that situation. Then that gives us that C times alpha will be a polynomial in the ring DX. If we do the same thing for beta here, we could look at the least common denominator D, and we times everything in that by D. This will give you D beta is a polynomial in DX, like so. And so because of that, we're then going to, well, consider the following, take C alpha X to equal C one AX, and then you have D beta X is equal to D one BX. Okay. So just so you're aware in this situation, we are given alpha beta, we don't have A B, we have to construct with the polynomials A, B, or R, and that's what we're trying to do right here. So you have the polynomial C alpha X, you have the polynomial D beta X, these belong to D join X. And so then factor out their content. So let's say the content of C alpha is C one, and the content of D beta is D one, that then gives us primitive polynomials A and B, which will also belong to the ring D join X. In particular, the primitive polynomials A and B will have the same degrees as alpha and beta, respectively. So if I take the polynomial C D times F of X, this would then be the same thing as C alpha times D beta, which then is C one A times D one B, for which then by Gauss's lemma, the content of this product is C one D one. And therefore C one D one is the content of CD, which admittedly, since you took F of X and times it by CD, the content of CDF is going to be CD times whatever the content of F was. In particular, CD is going to divide C one D one. So there's some number E inside of our domain D such that CDE is equal to C one D one like so. All right, for which then if you cancel CD from from both sides of this equation right here, because we can write this as C one, excuse me. No, that was the right direction. We want, we can rewrite C one D one as CDE A of X times B of X like so DX is an integral domain. So the CDs cancel and we end up with F of X equals E times A X times B X here. So this then gives us a factorization since AX belongs to DX E times AX is going to belong to DX. And you have B right here. So we might have this factor of E in play here, but in particular, the factorization over the field of fractions turned into a factorization over the unique factorization domain D. And so this is a very powerful result known as Gauss's lemma. And this is where of course we're going to end our lecture. Thanks for watching. If you learned anything about factorization of polynomials over arbitrary unique factorization domains and fields, like these videos, please subscribe to the channel to see more videos like this in the future and post any questions in the comments below. If you have any, I'll be glad to answer them.