Question

Question: Prove or disprove: If $p\left(x\right)$ is irreducible in $F\left[x\right]$ and $f\left(x\right)g\left(x\right)\equiv 0_F\left(\text{mod}p\left(x\right)\right)$, then $f\left(x\right)\equiv 0_F\left(\text{mod}p\left(x\right)\right)$ or $f\left(x\right)\equiv 0_F\left(\text{mod}p\left(x\right)\right)$.

Short answer

It is proved $f\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$ or $g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$.

Step-by-step solution

Given part

It is given $p\left(x\right)$ is irreducible in $F\left(x\right)$ and $f\left(x\right)g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$ .

Proof

It is given $f\left(x\right)\equiv g\left(x\right)\left(\text{mod}p\left(x\right)\right)$.

Then, by the definition $p\left(x\right)|\left(f\left(x\right)-g\left(x\right)\right)$, which implies

$p\left(x\right)|f\left(x\right)$ or $p\left(x\right)|g\left(x\right)$.

Again, applying the definition, we have

$f\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$

or

.$g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$

Hence, it is proved $f\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$or $g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$.