Proof for an adapted version of Gauss' Lemma

Question

I am self learning abstract algebra. Today I watched a youtube video explaining the proof for an adapted version of Gauss' Lemma
In his proof, he claims that for any polynomial $f = gh \in \mathbb{Z}[x]$ and for any prime $p$, where $g,h \in \mathbb{Z}[x].$ If $p$ divides all the coefficients of $f$, then $p$ must divide all the coefficients of $g$ or $h$ .
His argument is simply as following :
After taking mod $p$, $f=gh$ will turn into $0=g'h'$(mod$p$). Then $p$ must divide all the coefficients of $g$ or $h$.
I am very doubtful about his argument. This works well if $f$ is a integer, but seem not to work if $f$ is a polynomial. Since both $g$ and $h$ are in $\mathbb{Z}[x]$, we can expand $g$ and $h$as $\sum_{i=0}^{n}a_{i}x^i$ and $\sum_{j=0}^{m}b_{j}x^j$ respectively. Then $f = \sum_{i=0}^n \sum_{j=0}^m a_ib_jx^{i+j}$.
Then we can see If $p$ divides all the coefficients of $f$, then $p$ must divides $\sum_{i+j=q}a_ib_j$ for any $q\in[0,m+n]\cap\mathbb{Z}$
But this will not give us the desired result, which states $p$ has to divide either {${a_i}$}$_{i=0}^n$ or {${b_j}$}$_{j=0}^m$
My question is how could one derive his claim or is his claim correct at all?

Answer 1

If you accept that $(\Bbb Z/p\Bbb Z)[X]$ is an integral domain, then $f'g'=0$ in $(\Bbb Z/p\Bbb Z)[X]$ implies either $f'=0$ in $(\Bbb Z/p\Bbb Z)[X]$ or $g'=0$ in $(\Bbb Z/p\Bbb Z)[X]$. (Here, $f'$ and $g'$ denote the images of $f$ and $g$ in $(\Bbb Z/p\Bbb Z)[X]$.)

For a more naive approach, if $f$ and $g$ are both non-zero modulo $p$, each has a "first" coefficient that is nonzero modulo p. So $p\nmid a_r$ and $p\mid a_i$ for $i<r$, and $p\nmid b_s$ and $p\mid b_j$ for $j<s$. Then the coefficient of $x^{r+s}$ in $fg$ equals $a_rb_s$ plus some multiples of $p$ and so is not a multiple of $p$.