Question

Let $\mathbf{p}\left(\mathbf{x}\right)$be irreducible is F$\left[x\right]$. If $\left[\mathbf{f}\left(\mathbf{x}\right)\right]\ne \left[0_F\right]$in F$\left(x\right)$$/$$\left(\mathbf{p}\left(\mathbf{x}\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 $\left[\mathbf{f}\left(\mathbf{x}\right)\right]\left[\mathbf{g}\left(\mathbf{x}\right)\right]=\left[\mathbf{h}\left(\mathbf{x}\right)\right]$in F$\left(x\right)$$/$$\left(\mathbf{p}\left(\mathbf{x}\right)\right)$ . [Hint: Theorem 5.10 and Exercise 12(b) in Section 3.2.]

Short answer

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

Step-by-step solution

Statement of Theorem 5.10

Theorem 5.10states consider that $F$as a field and $\mathbf{p}\left(\mathbf{x}\right)$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) $F$$\left(x\right)$localid="1648810427406" $/\left(\mathbf{p}\left(\mathbf{x}\right)\right)$will be a field.

(3) $F$$\left(x\right)$$/$$\left(\mathbf{p}\left(\mathbf{x}\right)\right)$will be an integral domain.

Show that there exists gx∈Fx such that fxgx=hx in   Fx/px.

According to theorem 5.10, $F$$\left(x\right)$$/$$\left(p\left(x\right)\right)$will be a field, and as a result $\left[f\left(x\right)\right]\ne 0$, indicates that there is any $\overline{f}\left(x\right)\in F\left[x\right]$in which $\overline{f}\left(x\right)\in F\left[x\right]$. Then with every $h\left(x\right)\in F\left[x\right]$by letting $g\left(x\right):=\overline{f}\left(x\right)h\left(x\right)$. In $F$$\left(x\right)$$/$$\left(p\left(x\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 $F$role="math" localid="1648809848814" $\left(x\right)/\left(p\left(x\right)\right)$ .