Let $F$ be a field and let $p(x) , a_1(x) ,a_2(x), . . . . ,a_k(x) \in F[x]$ where $p(x)$ is irreducible over $F$ . If $p(x) | a_1(x) a_2(x) . . . . a_k(x)$ , then $p(x)$ divides $a_i$ for some $i$.
I tried the following :
Since, $p(x) | a_1(x) a_2(x) . . . . a_k(x)$ , there exists a non-zero polynomial $g(x)$ , such that $a_1(x) a_2(x) . . . . a_k(x) = g(x)p(x)$ where $g(x) \in F[x]$,
$ \Rightarrow g(x) = \dfrac{a_1(x) a_2(x) . . . . a_k(x)}{p(x)}$ ,
Now for $g(x)$ to belong in $F[x]$ , $p(x)$ must divide some $a_i(x)$.( That's what I think)
Is the above statement correct ?
It's just an intuition as if $p(x)$ doesn't divide one of the $a_i $'s , then $g(x)$ cannot belong to $F[x]$,
Please correct me where I am wrong.
Since $F$ is a field, then $F[x]$ is a Euclidean Domain, in particular, a PID. This means that irreducible is equivalent to prime.
The condition that $p(x) \mid a_1 (x) \cdots a_k (x)$ then is equivalent to $a_1 (x) \cdots a_k (x) \in (p(x))$. The latter is a prime ideal so one of the $a_i(x) \in (p(x))$ which again, is equivalent to $p(x) \mid a_i (x)$.
Well, yes, you are correct, but for what is probably the wrong reason. You haven't used the irreducibility of $p$ at all. Note that if you just take $a_i (x) = x$ for each $i$, then certainly, $p(x) = x^2$ will divide the product, provided $k \ge 2$.