Question

Prove that every non-zero$f\left(x\right)\in F\left[x\right]$ has a unique monic associate in$F\left[x\right]$ .

Short answer

It is proved that the monic associate of $f\left(x\right)=\sum _{n=0}^m{a}_n{x}^{n}ish\left(x\right)=\sum _{n=0}^m{a}_n^{-1}{a}_n{x}^n$ .

Step-by-step solution

Monic associate

It is given that the polynomial is$f\left(x\right)\in F\left[x\right]$ . Let $f\left(x\right)=\sum _{n=0}^m{a}_n{x}^n$ where the value of $a_n\ne 0$.

Steps for Monic associate 

Let $h\left(x\right)=\underset{n=0}{\overset{m}{\sum a_n^{-1}{a}_n{x}^n}}$is a monic associate of the given$f\left(x\right)$ where $f\left(x\right)=a_{m}g\left(x\right)$.

Consider that $g\left(x\right)$is a monic associate of the function$f\left(x\right)$ .

Therefore, some$c\in F\left[x\right]$ such that $f\left(x\right)=cg\left(x\right)$.

Now, apply uniqueness of representation $g\left(x\right)$ can be written as $g\left(x\right)=\underset{n=0}{\overset{m}{\sum b_n{x}^n}}$ where $b_m=1$ such that, $a_n=c{b}_n$ for all$0\le n\le m$ .

When we put $n=m$ then, it gives$c=a_m$ and for all $0\le n<m$ it gives $a_n=a_m{b}_n$ with $b_n=a_m^{-1}{a}_i$ then it gives$g\left(x\right)=h\left(x\right)$ .

Hence, the monic associate of $f\left(x\right)=\sum _{n=0}^m{a}_n{x}^n$ is $h\left(x\right)=\sum _{n=0}^m{a}_n^{-1}{a}_n{x}^n$.