Question

(a) Let $\mathit{f}\left(x\right)\mathbf{.}\mathit{g}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$. If role="math" localid="1648080019147" $\mathit{f}\left(x\right)\mathbf{|}\mathit{g}\left(x\right)$and$\mathit{g}\left(x\right)\mathbf{|}\mathit{f}\left(x\right)$ , show that $\mathit{f}\left(x\right)\mathbf{=}\mathit{c}\mathit{g}\left(x\right)$for some non-zero$\mathit{c}\mathbf{\in }\mathit{F}$ .

(b) If $\mathit{f}\left(x\right)$and$\mathbf{g}\left(x\right)$ in part (a) are monic, show that $\mathit{f}\left(x\right)\mathbf{=}\mathit{g}\left(x\right)$.

Short answer

(a) It is proved that $f\left(x\right)=cg\left(x\right)$.

(b) It is proved that$f\left(x\right)=g\left(x\right)$ .

Step-by-step solution

Solution to Part (a)

It is given that $f\left(x\right).g\left(x\right)\in F\left[x\right]$and , $f\left(x\right)|g\left(x\right)$,$g\left(x\right)|f\left(x\right)$then there exists , $a\left(x\right),b\left(x\right)\in F\left[x\right]$such that $g\left(x\right)=a\left(x\right)f\left(x\right)$and , $f\left(x\right)=b\left(x\right)g\left(x\right)$combine $g\left(x\right)$and $f\left(x\right)$.

$f\left(x\right)=a\left(x\right)b\left(x\right)f\left(x\right)$

This implies that,

$deg\left(f\left(x\right)\right)=deg\left(a\left(x\right)\right)+deg\left(b\left(x\right)\right)+deg\left(f\left(x\right)\right)$

The degree of $a\left(x\right),b\left(x\right)$is zero because both are constant polynomials.

Particularly, $bx=c$for any non-zero c. Then, we get:

$f\left(x\right)=cg\left(x\right)$

Solution to Part (b)

For the polynomials $f\left(x\right)$and $g\left(x\right)$, assume that$f_n$and $g_n$are the leading coefficients of $f\left(x\right)$and $g\left(x\right)$respectively.

Assume that $f\left(x\right)$and $g\left(x\right)$are monic polynomials, then the leading coefficients will be 1, that is $f_n=g_n=1$, from part (a), the leading coefficient of $cg\left(x\right)$is c, then we get its value will be 1, that is $c=1$. Then, we have $f\left(x\right)=g\left(x\right)$.