Question

Let $f\left(x\right),g\left(x\right)\in F\left[x\right]$, both non-zero, and let $d\left(x\right)$be their gcd. If $h\left(x\right)$is a common divisor of $f\left(x\right)$and $g\left(x\right)$of the highest possible degree, then prove that $h\left(x\right)=cd\left(x\right)$for some non-zero $c\in F$.

Short answer

It is proved that$h\left(x\right)=cd\left(x\right)$ for some non-zero $c\in F$.

Step-by-step solution

Statement of corollary 4.9

Corollary 4.9 considers F as a field and $a\left(x\right),b\left(x\right)\in F\left|x\right|$, both non-zero. The greatest common divisorof a(x) and b(x) such that if d(x) satisfies the following conditions.

1. d(x)|a(x) and d(x)|b(x)
2. When c(x)|a(x) and c(x)|b(x) , then c(x)|d(x) .

Show that h(x)=cd(x) for some non-zero c∈F

According to corollary 4.9, h(x)=d(x) implies that there is some $a\left(x\right)\in F\left[x\right]$in which d(x). This condition is satisfied by d(x). From the definition, it may deduce that deg(h(x)) = deg(h(x)) implying that deg(a(x)) and therefore, a(x) = $a_0$because h(x) is considered to have the maximum possible degree. d(x) would be a non-zero polynomial and therefore, $a_0\ne 0$because f(x) and g(x) are both non-zero.

As a result, h(x) = $h\left(x\right)={a_0}^{-1}d\left(x\right)$.

Hence, it is proved thath(x)=cd(x) for some non-zero $c\in F$.