Product of two primitive polynomials

Solution 1:

To summarize the WP reference I gave in a comment: supposing $fg$ is not primitive, form the quotient of $R$, and consequently $R[x]$, by any maximal ideal (any prime ideal will do too) of$~R$ containing all coefficients of$~fg$. Then $fg$ is killed but neither $f$ nor $g$ is; however this is impossible in $K[x]$ where $K$ is the quotient field (or quotient integral domain) of $R$ by the mentioned ideal, since $K[X]$ is an integral domain when $K$ is.

Solution 2:

The $\Leftarrow$ part of this falls under "Gauss's lemma", for which Wikipedia has a nice proof here:

http://bit.ly/18TjLgb

This proof should, however, be extended to an arbitrary ring.

This works for that situation:

http://bit.ly/19PhGDs

In order to prove the other direction by contrapositive, suppose that $f=\sum_{i=0}^n r_i x^i$ is not a primitive polynomial. Then each coefficient of its product with $g=\sum_{i=1}^n s_ix^i$ is a member of the ideal generated by the coefficients of $f$. That is, we may state that if $f\, g=\sum_{i=1}^{n+m} t_ix^i$, then $\langle t_1,t_2,...t_{m+n} \rangle \subseteq \langle r_1,r_2,...r_{n} \rangle$. It follows that $f\,g$ is not a primitive polynomial.

EDIT: never mind, guess I'm late for the party