 Consider the following example of a polynomial ring. Let's take a ring whose coefficients are the ring of integer's mod 6. And consider the following two polynomials, f of x equals 2x squared plus 4 and g of x equals 3x. If I consider the product of these two polynomials, f of x times g of x, well, this means 2x squared plus 4 times 3x. By the usual rules of polynomial multiplication we can distribute. You're going to end up with a 6x cubed, 3 times 2. And then you're going to get 12x, 4 times 3. But as we're working mod 6, these coefficients are both zero, and thus we actually produce the zero polynomial. So the product f of x times g of x is equal to zero. So we can see here that r of x, the polynomial ring, might not be an integral domain. In particular, if the coefficient ring is not a domain, then we can mimic this type of situation right here and show that r of x is likewise not going to be a domain. Generalizing this example, we can actually see that if r is not a domain, then r of x is not going to be a domain. So it's not just a z6 situation. The next theorem, which we're going to see in just a second, reverses this process. In particular, we're going to make the argument that a polynomial ring is a domain if and only if its coefficient ring is a domain. And the way we do this is we're going to use the degree function of a polynomial. Remember, the degree of a polynomial is the largest power of the indeterminate x that has a non-zero coefficient. With regard to polynomial rings, this forms a norm, an additive norm on that ring. So suppose that r is a domain, then the polynomials f and g, f of x and g of x are polynomials in the ring r of x, then we can see that the degree of f times g is going to equal the degree of f plus the degree of g. So therefore, on the polynomial ring r of x, when your coefficients come from a domain, this will form an additive norm. And using that additive norm, we can then show that the coefficient ring is a domain if and only if the polynomial ring is a domain. And in this situation, I'm not assuming that the ring r is commutative, so I'm using domain in the general sense. It doesn't necessarily have to be a commutative domain, a.k.a. an integral domain. Commutivity is not going to be used here. So note that if either f or g are the zero polynomial, then we define previously the degree of the zero polynomial to be negative infinity. So notice if you take negative infinity plus something. If you take negative infinity plus negative infinity, that's negative infinity. If you take negative infinity plus a natural number, that's negative infinity. But then over here, if you have zero times something, you're going to get zero, and that's going to be negative infinity. So okay, if one of the polynomials was zero, this formula is trivially true because of how negative infinity interacts with things. So we can assume that f and g are non-zero polynomials called the coefficients of f a, say its degree is n, and then say the coefficients of g are bi, and its degree is m. So then by the formal product of polynomials, if we take the product f times g, this is going to be the sum where i ranges from zero to n plus m here, where of course these numbers are combined together by the usual convolution that comes about by the distributive laws and things like that. In particular, I care about the leading term. The leading term in this product is going to be the product of the leading term from f times the leading term of g. The biggest power is going to come from the sum from their biggest powers, f and g. And its coefficient is then just going to be the product of the leading coefficient of f times the leading coefficient of g. Now since the coefficients come from a domain, the product of a n times b n will not be zero, and that's how I know this is in fact going to be the leading term. If you did have some proper divisors of zero inside of your ring, then it could be the possibility that you take a product, the leading coefficient could be zero, so the things that would have been the leading coefficient could be then zero, it disappears and so you get a smaller term. That's something we saw on the previous example, right, the product of two non-zero polynomials was zero, in fact it could get smaller. But for a domain, your leading term is going to be then the product of the leading terms from f and g, and that product can't be zero because we're in a domain. And so what this tells us is that the leading term will be non-zero because these two polynomials are non-zero as well. And so then looking at the degree, you're going to have n plus m, which n was the degree of f, m is the degree of g, and so then the degree of the new polynomial of product is going to be n plus m. This proves the first assertion that when your coefficient ring is a domain, then the degree function is an additive norm in that situation. So consequently, the product f times g is not the zero polynomial as its leading coefficients, its leading coefficients non-zero. The other coefficients, yeah, they could be zero, right, but we know that the leading coefficients non-zero, so this is not the zero polynomial, so the product of two polynomials f times g is non-zero if they're both non-zero polynomials. So if r is a domain, then r of x is likewise a domain. Now if we combined the comments we've made so far with the last example we just talked about, then it's very clear that the other direction is also true. Because if r of x is itself a domain but the coefficient ring wasn't a domain, then we could create a product of elements whose product is zero but they were non-zero elements. Essentially, because we can view r of x as a subring of, excuse me, r can be viewed as a subring of r of x, if the larger ring has no devices of zero, the smaller ring can't have devices of zero as well, so we do get this if and only if statement. r is a ring, if and only if, excuse me, r is a domain, if and only if rx is a domain as well. Now in this video we proved that if the coefficient ring is a domain, then the polynomial ring is domain if and only if. We also said things about the polynomial ring is commutative if only if the coefficient ring is commutative, it has unity if and only if the coefficient ring has unity. So it can be very, very tempting to assume that any ring theoretic property on the coefficient ring is then inherited on the polynomial ring if and only if. But that's actually not the case. Before we get too ambitious here, we might be tempted to say that r of x is in fact a field if and only if r is a field. Or if we don't worry about commutivity, we might wonder, oh, r of x is going to be a skew field, a division ring if and only if r is a skew field, a.k.a. division ring. And that's not the case. This is grossly incorrect and I want to show you why that is right now. Suppose that r of x is in fact a skew field so every non-zero element has a multiplicative inverse and consider the polynomial f of x. Just some polynomial, right? If in fact r of x was also a skew field, that means we're taking some non-zero polynomial here. That means f would have to be a unit so it has some multiplicative inverse f inverse so that f times f inverse is equal to 1. Well, if we look at the degree of this thing, f times f inverse is supposed to equal 1. The degree of 1 as it's a constant polynomial would then be 0. But conversely, because this is an additive norm, the degree of f times f inverse should be the degree of f added to the degree of f inverse. Now these are both natural numbers here. In particular, they're integers but they cannot be negative. So we have the sum of two natural numbers that adds up to be 0. It has to imply that both of the numbers were already 0 so in particular the degree of f was 0 and f was a constant polynomial. So if you have a polynomial ring whose coefficients come from a skew field, the only way a polynomial could be a unit is if it's a constant polynomial. That is it's just a number from r already. So since we can naturally view r as a subring of rx here, this tells us that the units of rx are just the units of r itself. So if r is a skew field and so therefore the units is everything except for 0, that's the only units you get in r of x as well. So in particular, r of x is never a division ring even when r is itself a division ring. Now one has to be very, very careful with this situation because if you take coefficients from a domain, which of course a skew field is a domain, if you take your coefficients from a domain then the degree function is additive, it's a norm and that's how we argued that there's no new units in the polynomial ring, only the units that were already in the coefficient ring. But like the example we saw earlier with z6, if your coefficients rings are not a domain, then it's possible that when you take the product of two polynomials, the degree didn't necessarily have to be additive. We had an example where the product of two polynomials was 0, could you have it so that the product of two polynomials was a constant but not 0 and therefore potentially could be a unit? Sure you can, let me give you an example here. So consider the polynomial 2x plus 1 inside the polynomial ring whose coefficients are integers mod 4. If I take this polynomial and I square it by the usual multiplicative rules, 2x plus 1 quantity squared is 4x squared plus 4x plus 1 and as you reduce that mod 4, you just get 1. So 2x plus 1 is a unit and it's actually its own multiplicative inverse. So this actually is very different than what we saw before, right? And why are these things consistent with each other? Well, we mentioned before that the degree of a polynomial function is going to be an additive norm if the coefficient ring is a domain, which for the skew field, that would be a domain. If your ring of coefficients is not a domain, then a lot of weird stuff can happen like this and therefore a polynomial could be a unit. And so this is to suggest that if your coefficients are not a domain, then polynomial rings can get kind of weird and they don't necessarily factor the way they're used to, that you might be used to. And so in this lecture series, we're going to restrict our attention only to those polynomial rings whose coefficients come from a domain and therefore we have the, we do in fact have this additive norm, the degree. And so we want to study factorization of polynomials in that setting. It's not that this setting is never relevant, it absolutely is, but that's going to go beyond the scope of this lecture series.