- Let $z(x) \in Z[X]$ such that $z(x) = p(x) q(x)$ where $p(x), q(x) \in Q[X]$. Then we claim that there exists $p'(x), q'(x) \in Z[x]$ such that $z(x) = p'(x) q'(x)$.
- For example, suppose $p(x) = a_0 / b_0 + a_1 x / b_1$ and $q(x) = c_0 / d_0 + c_1 x / d_1$, such that $p(x)q(x) \in \mathbb Z[x]$ and these fractions are in lowest form. So, $b_i \not | a_i$ and $d_i \not | c_i$.
- Take common demoniator, so we can then find things the denominator divides to write as a product in $\mathbb Z$. For example, we know that $9/10 \cdot 20 / 3 = 6$. This can be obtained by rearranging the product as $(9/3) \cdot (20/10) = 3 \cdot 2 = 6$. We wish to perform a similar rearrangement, by first writing $9/10 \cdot 20 / 3$ as $(9 \cdot 20)/(10 \cdot 3)$, and then pairing up $10 \leftrightarrow 20$ and $3 \leftrightarrow 9$ to get the final integer $(9/3) (20/10) = 6$. After pairing up, each of the pairs $(9/3)$ and $(20/10)$ are clearly integers.
- Take common demoniator in $p(x)$ and write it as a fraction: $p(x) = (a_0 b_1 + (a_1 b_0)x) / b_0 b_1$, and similarly $q(x) = (c_0 d_1 + (c_1 d_0)x)/d_0 d_1$.
- We claim that the denominator of $p(x)$, $b_0 b_1$
the numerator of $p(x)$, $(a_0 b_1 + (a_1 b_0)x)$. This can be seen term-by-term. $b_0 b_1$ does not divide $a_0 b_1$ since $a_0 b_1 / b_0 b_1 = a_0 / b_0$ which was assumed to be in lowest form, and a real fraction. Similarly for all terms in the numerator.*does not divide* - Since the product $p(x)q(x)$ which we write as fractions as $(a_0 b_1 + (a_1 b_0)x) (c_0 d_1 + (c_1 d_0)x) / (b_0 b_1)(d_0 d_1)$ is integral, we must have that $b_0 b_1$ divides the numerator. Since $b_0 b_1$
the first factor $(a_0 b_1 + (a_1 b_0)x)$, it*does not divide*the second factor $(c_0 d_1 + (c_1 d_0)x)$. Thus, the polynomial $q'(x) \equiv (c_0 d_1 + (c_1 d_0)x)/b_0 b_1$ is therefore integral [ie, $q'(x) \in Z[x]$].*must divide* - By the exact same reasoning, we must have $d_0 d_1$ divides the product $p(x)q(x)$. Since $d_0 d_1$ does not divide $(c_0 d_1 + (c_1 d_0)x)$, it must divide (a
*0 b*1 + (a*1 b*0)x) and therefore $p'(x) \equiv (a_0 b_1 + (a_1 b_0)x)/(d_0 d_1)$is integral. - Thus, we can write $z(x) = p'(x) q'(x)$ where $p'(x), q'(x) \in \mathbb Z[x]$.
- This generalizes, since we never used anything about being linear, we simply reasoned term by term.

- Start at $p(x)q(x) = (a_0 b_1 + (a_1 b_0)x) (c_0 d_1 + (c_1 d_0)x) / (b_0 b_1)(d_0 d_1)$.
- Rewrite as $p(x)q(x) \cdot (b_0 b_1)(d_0 d_1) = (a_0 b_1 + (a_1 b_0)x) (c_0 d_1 + (c_1 d_0)x)$
- Suppose $\alpha$ is a prime factor of $b_0$. Then reduce the above equation mod $\alpha$. We get $0 \equiv_\alpha (a_0 b_1 + (a_1 b_0)x) (c_0 d_1 + (c_1 d_0)x)$. Since $\mathbb Z/\alpha \mathbb Z[x]$ is an integral domain, we have that one of $(a_0 b_1 + (a_1 b_0)x)$ or $(c_0 d_1 + (c_1 d_0)x)$ vanishes, and thus $p$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 $Z/\alpha Z[x]$ is an integral domain: note that $Z/\alpha Z$ is a field, thus $Z/ \alpha Z[x]$ is a Euclidean domain (run Euclid algorithm). This implies it is integral.