Question

Let $f\left(x\right),g\left(x\right),h\left(x\right)\in F\left[x\right]$, with$f\left(x\right)$ and$g\left(x\right)$ relatively prime. If $f\left(x\right)\overline{)h\left(x\right)}$and$g\left(x\right)\overline{)h\left(x\right)}$ , prove that $f\left(x\right)g\left(x\right)\overline{)h\left(x\right)}$.

Short answer

It is proved that f(x) g(x)|h(x) .

Step-by-step solution

Statement of theorem 4.8

Theorem 4.8states that$F$ is a field and$a\left(x\right),b\left(x\right)\in F\left[x\right]$ , both non-zero. Then, there exists a uniquegreatest common divisor$d\left(x\right)$of$a\left(x\right)$ and $b\left(x\right)$. Also, there exist polynomials (not unique)$u\left(x\right)$ and$v\left(x\right)$ in which$d\left(x\right)=a\left(x\right)u\left(x\right)+b\left(x\right)v\left(x\right)$. .

Show that fxgxhx

Theorem 4.10considers $F$as a field and $a\left(x\right),b\left(x\right),c\left(x\right)\in F\left[x\right]$.

When $a\left(x\right)\overline{)b\left(x\right)c\left(x\right),a\left(x\right)andb\left(x\right)}$, are relatively prime, then $a\left(x\right)\overline{)c\left(x\right)}$.

There exists$u\left(x\right),v\left(x\right)\in F\left[x\right]$ in which$1=f\left(x\right)u\left(x\right)+g\left(x\right)v\left(x\right)$ because $\left(f\left(x\right),g\left(x\right)\right)=1$according to theorem 4.8.

There are$r\left(x\right),s\left(x\right)\in F\left[x\right]$ in which$h\left(x\right)=r\left(x\right)f\left(x\right)$ and$h\left(x\right)=s\left(x\right)g\left(x\right)$ because$f\left(x\right)\overline{)h\left(x\right)}$ and $g\left(x\right)\overline{)h\left(x\right)}$.

Calculate that $h\left(x\right)$as follows:

$\begin{array}{rcl}h\left(x\right)& =& h\left(x\right)1\\ & =& h\left(x\right)\left[f\left(x\right)u\left(x\right)+g\left(x\right)v\left(x\right)\right]\\ & =& h\left(x\right)f\left(x\right)u\left(x\right)+h\left(x\right)g\left(x\right)v\left(x\right)\\ & =& s\left(x\right)g\left(x\right)f\left(x\right)u\left(x\right)+r\left(x\right)f\left(x\right)g\left(x\right)v\left(x\right)\\ & =& \left[f\left(x\right)g\left(x\right)\right]\left[s\left(x\right)u\left(x\right)+r\left(x\right)v\left(x\right)\right]\end{array}$

As a result,$f\left(x\right)g\left(x\right)\overline{)h\left(x\right)}$ .

Hence, it is proved that$f\left(x\right)g\left(x\right)\overline{)h\left(x\right)}$ .