Prove that if px is reducible then there exists a polynomial

Prove that if p(x) is reducible, then there exists a polynomial [f(x)] in F[x]/(p(x)), such that [f(x)] is a zero divisor

Solution

If p | f, then f 0 (mod p(x)) as required, so assume that p - f. This implies that (p, f) = 1 (because p is irreducible, the only divisors of p are units and associates of p; we know however, that p - f and hence no associate of p divides f either). Because fg 0 (mod p(x)) we know that p | fg, and hence p | g. Thus g 0 (mod p(x)) as required.