Question

If $p\left(x\right)$ is reducible in $F\left[x\right]$, prove that there exist $f\left(x\right)g\left(x\right)\in F\left[x\right]$ such that $f\left(x\right)\cancel{\equiv }{0}_F\left(\text{mod}p\left(x\right)\right)$ and $g\left(x\right)\cancel{\equiv }{0}_F\left(\text{mod}p\left(x\right)\right)$ but $f\left(x\right)g\left(x\right)\equiv 0_F\left(\text{mod}p\left(x\right)\right)$.

Short answer

It is proved there exists $f\left(x\right)g\left(x\right)\in F\left[x\right]$, such that $f\left(x\right)\cancel{\equiv }{\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$and $g\left(x\right)\cancel{\equiv }{\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$, but $f\left(x\right)g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$.

Step-by-step solution

Corollary statement

Assume that $p\left(x\right)$ is a non-zero polynomial of degree n in $F\left[x\right]$ and assume that congruence modulo $p\left(x\right)$.

1. When $f\left(x\right)$ is divided by $p\left(x\right)$, we have $\left[f\left[x\right]\right]=\left[r\left(x\right)\right]$, where $r\left(x\right)$ is the remainder and $f\left(x\right)\in F\left[x\right]$.

1. Let S is the set of all polynomials whose degrees be less than the degree of $p\left(x\right)$, then each congruence class modulo $p\left(x\right)$ is also the class of some polynomial in S, where each class is distinct.

Proof

It is given that $p\left(x\right)\in F\left[x\right]$ is reducible;then,there are two functions,$f\left(x\right)$and $g\left(x\right)$, such that $0<\text{deg}f\left(x\right),\text{deg}g\left(x\right)<\mathrm{deg}p\left(x\right)$ and $p\left(x\right)=f\left(x\right)g\left(x\right)$.

Then, by the corollary stated in step 1, we have

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

And

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

Then, using the definition, we have $f\left(x\right)g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$.

Hence, it is proved there exists $f\left(x\right)g\left(x\right)\in F\left[x\right]$, such that $f\left(x\right)\cancel{\equiv }{\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$and $g\left(x\right)\cancel{\equiv }{\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$, but $f\left(x\right)g\left(x\right)\equiv {\mathbf{0}}_F\left(\text{mod}p\left(x\right)\right)$.