Let $f(x)\in \mathbb{Z}[x]$. The set content of $f(x)=a_{n}x^{n}+ \cdots +a_{0}$ is defined as the greatest common divisor of $a_{0}, ..., a_{n}$ and is denoted by cont$(f(x))$.
I want to show that cont$(f(x)g(x))=$cont$(f(x))$cont$(g(x))$. I saw a proof of it, but I have some doubts. This is:
We know that $$\text{mcd}\left( \frac{a}{\text{mcd}(a, b)}, \frac{b}{\text{mcd}(a, b)}\right)=1$$ without loss of generality let us cont$(f(x))=1$ and cont$(g(x))=1$ where:
\begin{align*} f(x) = & a_{n}x^{n}+ \cdots +a_{0} \\ g(x) = & b_{n}x^{n}+ \cdots +b_{0} \end{align*}
Supose that $f(x)g(x)\neq 1$ then there is no unity $p \in \mathbb{Z}$ that divides all the coefficients of $f(x)g(x)$. Well, it is not very clear to me why such a unit satisfies that.
Let $i$ the smallest integer such that $p \nmid a_{i}$ and $j$ the largest integer such that $p \nmid b_{j}$ and take the coeficient $x^{i+j}$ in $f(x)g(x)$, from where
\begin{equation*} f(x)g(x)=a_{0}b_{i+j}+a_{1}b_{i+j-1}+\cdots +a_{i-1}b_{j-1}+a_{i}b_{j-1}+\cdots +a_{i+j-1}b_{1}+a_{i+j}b_{0} \end{equation*}
In the summands to the left of $a_{i}b_{j}$ the factor $a_{k}$, with $k<i$, is divisible by $p$, and to the right of $a_{i}b_{j}$ the factor $b_{k}$ with $k<j$ is divisible by $p$. So $p \mid a_{i}b{j}$ i.e. $p \mid a_{i}$ or $p \mid b{j}$ but this contradicts our previous assumption. Therefore, it cannot be the case that $f(x)g(x)\neq 1$, then $f(x)g(x)= 1$ now cont$(f(x)g(x))=$cont$(f(x))$cont$(g(x))$.
Firstly, the proof is correct? and secondly, I am left with the doubt why the factors $a_{k}$ and $b_{k}$ are divisible by $p$, in addition to the doubt I mentioned before.
Any help is welcome!
If there is some prime $p$ that divides each of the coefficients of $f(x)g(x)$, then we have $f(x)g(x) \equiv 0 \pmod p$. But $(\Bbb Z / p \Bbb Z)[x]$ is an integral domain (this can easily be seen by looking at the product of the leading coefficients of $f$ and $g$), so that means either $f(x) \equiv 0 \pmod p$ or $g(x) \equiv 0 \pmod p$, contradicting your hypothesis that $\operatorname{cont}(f) = \operatorname{cont}(g)=1$.