Question

If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{F}$and$\mathit{a}\mathbf{\ne }\mathit{b}$ , show that $\mathit{x}\mathbf{+}\mathit{a}$and $\mathit{x}\mathbf{+}\mathit{b}$are relatively prime in role="math" localid="1648078640717" $\mathit{F}\left[x\right]$.

Short answer

It is proved that $x+a$and $x+b$are relatively prime in $F\left[x\right]$.

Step-by-step solution

Definition

Let $a\left(x\right),b\left(x\right)\in F\left[x\right]$with $a\left(x\right),b\left(x\right)\ne 0$, then $d\left(x\right)$ is the gcd of $a\left(x\right),b\left(x\right)$, then,$d\left(x\right)$ is a monic and$d\left(x\right)$ divides $a\left(x\right),b\left(x\right)$. Secondly, if role="math" localid="1648078962209" $c\left(x\right)|a\left(x\right)$and $c\left(x\right)|b\left(x\right)$then $degc\left(x\right)\le degd\left(x\right)$.

Proof part

It is known that the only polynomial of degree 0 is 1. So, the greatest common divisor of $x+a$and $x+b$should be a factor of common monic divisor of degree 1.

It is known that,

$deg\left(f\left(x\right)g\left(x\right)\right)=deg\left(f\left(x\right)\right)+deg\left(g\left(x\right)\right)\in F\left(x\right)$

Then $\frac{x+a}{x+c}$has a degree 0, so the quotient can be of the form $q\in F$, then we have:

$$\begin{gathered} x+a=q\left(x+c\right) \\ =qx+qc \end{gathered}$$

By using uniqueness of representations of polynomial, it can be concluded that $q=1$, so we get $c=a$.

Similarly, we have $c=b,a=b$which is a contradiction.

Hence, the gcd of $\left(x+a,x+b\right)=1$.

Which implies that $x+a$and $x+b$are relatively prime in $F\left(x\right)$.