Question

Let $\mathbf{p}\left(x\right)$be irreducible in $\mathbf{F}\left[x\right]$. If$\left[f\left(x\right)\right]\mathbf{\ne }\left[0_F\right]$ in$\raisebox{1ex}{$\mathbf{F}\left(x\right)$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$ and $\mathbf{h}\left(\mathbf{x}\right)\in \mathbf{F}\left[x\right]$, prove that there exists$\mathbf{g}\left(\mathbf{x}\right)\in \mathbf{F}\left[\mathbf{x}\right]$ such that$\left[f\left(x\right)\right]\left[g\left(x\right)\right]\mathbf{=}\left[h\left(x\right)\right]$ in $\raisebox{1ex}{$\mathbf{F}\left(x\right)$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$. [Hint: Theorem 5.10 and Exercise 12(b) in Section 3.2.]

Short answer

It is proved there exists$g\left(x\right)\in F\left[x\right]$ such that$\left[f\left(x\right)\right]\left[g\left(x\right)\right]=\left[h\left(x\right)\right]$ in $\raisebox{1ex}{$F\left(x\right)$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$.

Step-by-step solution

Statement of Theorem 5.10

Theorem 5.10states consider that$F$ as a field and a nonconstant polynomial in $F\left[x\right]$. Then the statements are equivalent as follows:

1. $p\left(x\right)$is irreducible in $F\left[x\right]$.
2. $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$willbe a field.
3. $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$will be an integral domain.

Show that there exists gx∈Fx such that [fx][gx]=[hx] in F(x)(px)

According to theorem 5.10, $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$will be a field, and as a result, $\left[f\left(x\right)\right]\ne 0$indicates that there is any $\overline{f}\left(x\right)\in F\left[x\right]$in which ${\left[f\left(x\right)\right]}^{-1}=\left[\overline{f}\left(x\right)\right]$. Then with every $h\left(x\right)\in F\left[x\right]$by letting $g\left(x\right):=\overline{f}\left(x\right)h\left(x\right)$. In $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$, we have the following:

$$\begin{gathered} \left[f\left(x\right)\right]\left[g\left(x\right)\right]=\left[f\left(x\right)g\left(x\right)\right] \\ =\left[f\left(x\right)\overline{f}\left(x\right)h\left(x\right)\right] \\ =\left(\left[f\left(x\right)\right]\left[\overline{f}\left(x\right)\right]\right)\left[h\left(x\right)\right] \\ =\left[h\left(x\right)\right] \end{gathered}$$

Hence, it is proved there exists$g\left(x\right)\in F\left[x\right]$ such that$\left[f\left(x\right)\right]\left[g\left(x\right)\right]=\left[h\left(x\right)\right]$ in $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$.