Question

Prove Theorem 4.10.

Short answer

Theorem 4.10 is proved.

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 theorem 4.10

Theorem 4.10states$F$ as a field and$a\left(x\right),b\left(x\right),c\left(x\right)\in F\left[x\right]$ . When $a\left(x\right)\overline{)b\left(x\right)c\left(x\right),a\left(x\right)}$, and$b\left(x\right)$ are relatively prime, then$a\left(x\right)\overline{)c\left(x\right)}$ .

Consider that$a\left(x\right),b\left(x\right),c\left(x\right)\in F\left[x\right]$ in which$a\left(x\right)\overline{)b\left(x\right)c\left(x\right)}$ and$\left(a\left(x\right),b\left(x\right)\right)=1$ . According to theorem 4.8, there are$u\left(x\right),v\left(x\right)\in F\left[x\right]$ in which$1=a\left(x\right)u\left(x\right)+b\left(x\right)v\left(x\right)$ .

Multiply by$c\left(x\right)$ on both sides of the above equation as follows:

$c\left(x\right)=a\left(x\right)c\left(x\right)u\left(x\right)+b\left(x\right)c\left(x\right)v\left(x\right)$

There exists any$r\left(x\right)\in F\left[x\right]$ in which$b\left(x\right)c\left(x\right)=a\left(x\right)r\left(x\right)$ because$a\left(x\right)\overline{)b\left(x\right)c\left(x\right)}$ . Therefore,

$\begin{array}{rcl}c\left(x\right)& =& a\left(x\right)c\left(x\right)u\left(x\right)+b\left(x\right)c\left(x\right)v\left(x\right)\\ & =& a\left(x\right)c\left(x\right)u\left(x\right)+a\left(x\right)r\left(x\right)v\left(x\right)\\ & =& a\left(x\right)\left[c\left(x\right)u\left(x\right)+r\left(x\right)v\left(x\right)\right]\end{array}$

As a result,$a\left(x\right)\overline{)c\left(x\right)}$ .

Hence, theorem 4.10 is proved.