Prove: If $f,g \in \Bbb Z[x]$ then $C(fg) = C(f)C(g)$

Question

If $f,g \in \Bbb Z[x]$ then $C(fg) = C(f)C(g)$. C is the content of a polynomial (greatest common divisor of the coefficients). The proof states that proving: "For any prime, $p$ we have $p|C(fg)$ iff $p|C(f)$ or $p|C(g)$." implies the result. I'm not sure why this is. Can someone explain; is it just because of the prime factorisation? Explanations/examples would be great!

Answer 1

We can complete this as a strong induction proof.

Let $P(n)$ be the statement:

If $n=C(fg),$ then $C(fg)=C(f)C(g).$

To prove $n=1,$ we use the observation that $C(f)C(g)\mid C(fg).$ This is easy to prove by definition. But if $C(fg)=1$ then this means $C(f)C(g)=1.$

Now, given $n>1,$ assume $P(k)$ is true for $k<n,$ and also assume $C(fg)=n.$ Then pick $p\mid n.$ By your note, it must be the case,that $p\mid C(f)$ or $p\mid C(g).$

Assume that $p\mid C(f).$ The same argument works if $p\mid C(g).$

Then $C(f/p \cdot g)=C(fg)/p =n/p<n.$ So, by the induction assumption: $$C(fg)/p=C(f/p \cdot g)=C(f/p)C(g)=C(f)/p\cdot C(g)$$

Multiplying both sides by $p$ gives you $C(fg)=C(f)C(g),$ and we are done.

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This might seem like a lot of work hiding in the author’s note, but it is just a formalization of a standard “descent argument,” following from the property proven. The whole reason the author “waves his hands” at this last part is that it is cumbersome to write out, but a fairly standard argument.