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. Prove that the gcd ofrole="math" localid="1648552537733" $f\left(x\right)h\left(x\right)$ and$g\left(x\right)$ is the same as the gcd of $h\left(x\right)$and$g\left(x\right)$ .

Short answer

It is proved that the gcd of $f\left(x\right)h\left(x\right)$and $g\left(x\right)$is the same as the gcd of $h\left(x\right)$ and g(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 the gcd of fxhx and gx is the same as the gcd of hx and gx 

Consider that$\left(h\left(x\right),g\left(x\right)\right)\overline{)\left(f\left(x\right)h\left(x\right),g\left(x\right)\right)}$because$\left(h\left(x\right),g\left(x\right)\right)\overline{)g\left(x\right)\left(h\left(x\right),g\left(x\right)\right)\overline{)h\left(x\right)}}$

and $h\left(x\right)\overline{)f\left(x\right)h\left(x\right)}$.

According to theorem 4.8, there exists $u\left(x\right),v\left(x\right),r\left(x\right),s\left(x\right)$in which $\left(h\left(x\right),g\left(x\right)\right)=u\left(x\right)h\left(x\right)+v\left(x\right)g\left(x\right)$and $1=r\left(x\right)f\left(x\right)+s\left(x\right)g\left(x\right)$.

Compute that $\left(h\left(x\right),g\left(x\right)\right)$as follows:

$\begin{array}{rcl}\left(h\left(x\right),g\left(x\right)\right)& =& \left(h\left(x\right),g\left(x\right)\right)1\\ & =& \left[u\left(x\right)h\left(x\right)+v\left(x\right)g\left(x\right)\right]\left[r\left(x\right)f\left(x\right)+s\left(x\right)g\left(x\right)\right]\\ & =& u\left(x\right)r\left(x\right)h\left(x\right)f\left(x\right)+u\left(x\right)s\left(x\right)h\left(x\right)g\left(x\right)+v\left(x\right)r\left(x\right)g\left(x\right)f\left(x\right)+v\left(x\right)s\left(x\right)g{\left(x\right)}^2\\ & =& \left[u\left(x\right)r\left(x\right)\right]\left[h\left(x\right)f\left(x\right)\right]\left[u\left(x\right)s\left(x\right)h\left(x\right)+v\left(x\right)r\left(x\right)f\left(x\right)+v\left(x\right)s\left(x\right)g\left(x\right)\right]g\left(x\right)\\ & =& \left[u\left(x\right)r\left(x\right)\right]\left[h\left(x\right)f\left(x\right)\right]+\left[u\left(x\right)s\left(x\right)h\left(x\right)+v\left(x\right)\right]g\left(x\right)\end{array}$

It follows that $\left(h\left(x\right)f\left(x\right),g\left(x\right)\right)\overline{)h\left(x\right),g\left(x\right)}$because$\left(h\left(x\right)f\left(x\right),g\left(x\right)\right)\overline{)\left(h\left(x\right)f\left(x\right)\right)}$ and $\left(h\left(x\right)f\left(x\right),g\left(x\right)\right)\overline{)g\left(x\right)}$.

It concludes that$\left(h\left(x\right)f\left(x\right),g\left(x\right)\right)=\left(h\left(x\right),g\left(x\right)\right)$ .

Hence, it is proved that the gcd of$f\left(x\right)h\left(x\right)$ and$g\left(x\right)$ is the same as the gcd of$h\left(x\right)$ and$g\left(x\right)$ .