Question

If $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is relatively prime to $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, prove that there is a polynomial$\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ such that $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\equiv }{\mathbf{1}}_{\mathbf{F}}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{d}\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right)}$

Short answer

It is proved that there is a polynomial$g\left(x\right)\in F\left[x\right]$ such that $f\left(x\right)g\left(x\right)\equiv 1_F\left(\mathrm{mod}p\left(x\right)\right)$.

Step-by-step solution

Use Theorem 4.8

By theorem 4.8 which is says that Let $\mathit{F}$be a field and $\mathit{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathit{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$, not both zero. Then there is a unique greatest common divisor $\mathit{d}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of $\mathit{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and $\mathit{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$. Furthermore, there are (not necessarily unique) polynomials $\mathit{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and role="math" localid="1654236554645" $\mathit{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$such thatrole="math" localid="1654236551134" $\mathit{d}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Here, role="math" localid="1654236419930" $f\left(x\right)$and $p\left(x\right)$are relatively prime.

Therefore, by above theorem, there are$g\left(x\right),h\left(x\right)\in F\left[x\right]$such that $f\left(x\right)g\left(x\right)+p\left(x\right)h\left(x\right)=1$

Therefore, we can write$f\left(x\right)g\left(x\right)=1-p\left(x\right)h\left(x\right)$.

Use Corollary 5.5

By corollary 5.5 which is says that Let $\mathit{F}$be a field and $\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$a polynomial of degree n in $\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$, and consider congruence modulo $\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

If $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$and $\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is the reminder when $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is divide by $\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, then $\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{r}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}$

Using above result, we have that $f\left(x\right)g\left(x\right)\equiv 1\left(\mathrm{mod}p\left(x\right)\right)$

Hence proved.