Exercise from Atiyah-Macdonald, Chapter 1, 2.iv)

Assume that $fg$ is not primitive. Then the ideal of coefficients of $fg$ is contained in a maximal one, say $\mathfrak m$. In $(A/\mathfrak m)[x]$ we have $f\ne 0$ and $g\ne 0$, so $fg\ne 0$, a contradiction.