Question

Show that a nonzero polynomial in${\mathrm{\mathbb{Z}}}_p\left[x\right]$ has exactly p-1associates.

Short answer

It is proved that h(x) has exactly p-1 associates.

Step-by-step solution

Number of elements in cyclic group

It is known that the set${\mathrm{\mathbb{Z}}}_p$ has p-1 non-zero elements.

Apply this definition in the polynomial ${\mathrm{\mathbb{Z}}}_p\left[x\right]$ then it has p-1 associates.

Steps for non-zero associates in the given polynomial 

Let $h\left(x\right)=\sum _{n=0}^m{a}_n{x}^n$ where $a_m\ne 0$.

Let two non-zero distinct elements $r,s\in {\mathrm{\mathbb{Z}}}_p$then, it gives rh(x)$\ne$sh(x) .

Consider that the mth term is written as $r{a}_m$and $s{a}_m$.

If we write $r{a}_m=s{a}_m$then, it gives $a_m=r^{-1}s{a}_m$ and after comparing both sides, it gives $r^{-1}s=1$.

Therefore, it implies that r=s which is a contradiction.

Hence, h(x) have exactly p-1 associates.