§ Gauss Lemma for polynomials
- Let such that where . Then we claim that there exists such that .
- For example, suppose and , such that and these fractions are in lowest form. So, and .
- Take common demoniator, so we can then find things the denominator divides to write as a product in . For example, we know that . This can be obtained by rearranging the product as . We wish to perform a similar rearrangement, by first writing as , and then pairing up and to get the final integer . After pairing up, each of the pairs and are clearly integers.
- Take common demoniator in and write it as a fraction: , and similarly .
- We claim that the denominator of , does not divide the numerator of , . This can be seen term-by-term. does not divide since which was assumed to be in lowest form, and a real fraction. Similarly for all terms in the numerator.
- Since the product which we write as fractions as is integral, we must have that divides the numerator. Since does not divide the first factor , it must divide the second factor . Thus, the polynomial is therefore integral [ie, ].
- By the exact same reasoning, we must have divides the product . Since does not divide , it must divide (a 0 b 1 + (a 1 b 0)x) and therefore is integral.
- Thus, we can write where .
- This generalizes, since we never used anything about being linear, we simply reasoned term by term.
§ Alternate way to show that the factorization is correct.
- Start at .
- Rewrite as
- Suppose is a prime factor of . Then reduce the above equation mod . We get . Since is an integral domain, we have that one of or vanishes, and thus divides one of the two.
- This works for all prime divisors of the denominators, thus we can "distribute" the prime divisors of the denominators across the two polynomials.
- Proof that is an integral domain: note that is a field, thus is a Euclidean domain (run Euclid algorithm). This implies it is integral.