If \(p\) is an irreducible polynomial and \(fg \equiv 0 \bmod p\), prove that either \(f \equiv 0 \bmod p\) or \(g \equiv 0 \bmod p\). Give an example which shows that this is false if \(p\) is not irreducible.
Solution
The congruence \[ fg\equiv0\pmod p \] means \(p\mid fg\). Since \(p\) is irreducible in the PID \(F[x]\), it is prime. Hence \[ p\mid f\quad\text{or}\quad p\mid g. \] Equivalently, \[ f\equiv0\pmod p \quad\text{or}\quad g\equiv0\pmod p. \]
Irreducibility is necessary. For example, take \[ p=x^2,\qquad f=g=x. \] Then \(fg=x^2\equiv0\pmod{x^2}\), but neither \(f\) nor \(g\) is congruent to \(0\) modulo \(x^2\).