Question

If $f\left(x\right)=c_n{x}^n+........c_0$with $c_n\ne 0_F$, what is the gcd of $f\left(x\right)$and $0_F$?

Short answer

The gcd of $\left(f\left(x\right),0\right)$ is c_n^-1f(x)

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 localid="1648077228077" $a\left(x\right),b\left(x\right)$, then $d\left(x\right)$is a monic and $d\left(x\right)$divides localid="1648078903097" $a\left(x\right),b\left(x\right)$. Secondly, if localid="1648078994705" $c\left(x\right)|a\left(x\right)$and localid="1648079005350" $c\left(x\right)|b\left(x\right)$then, $degc\left(x\right)\le degd\left(x\right)$.

Proof part

Let $f\left(x\right)=\sum _{i=0}^n{c}_i{x}^i\in F\left[x\right]$having degree n, where $c_n$is non-zero.

Then, there exists a monic polynomial having the highest degree which divides$f\left(x\right)$ is${c_n}^{-1}f\left(x\right)=\sum _{i=0}^n{c_n}^{-1}{c}_i{x}^i$

It is known that all the polynomials always divide 0, then

$\left(f\left(x\right),0\right)={c_n}^{-1}f\left(x\right)$

Hence, the gcd of $\left(f\left(x\right),0\right)$is${c_n}^{-1}f\left(x\right)$ .