Hypothesis :
Let $R$ be a finite commutative ring with unity.
Two polynomials $P,Q \in R[X]$ are said to be equivalent over $R$ if : $$\forall r \in R, P(r)=Q(r).$$ If so, we note $P \sim Q$.
My question : is it true that every element from $R[X]/\sim$ can be represented by a polynomial with degree less than $|R|$ ? Is $R[X]/\sim$ still a ring ?
My thoughts : the answer seems to be positive if $R$ is a field and $|R|=p$, $p$ being prime, thanks to Frobenius' endomorphism. Indeed, for such a ring, $ \forall r \in R, r^{|R|}=r$. Then it is easy to reduce every polynomial to a representative with degree less than $|R|$.
If $R$ is an integral domain, then $R$ is a commutative field. Thus $|R|=p^n$ with $p$ prime and $n \in \mathbb{N}$.
The case $n=1$ has already been done. But i'm already blocked when $n>1$ and have absolutely no idea how to tackle the question entirely. May be it has something to do with Euclidean domains ?
Thank you.
Let $|R|=n$. The polynomial $$\prod_{a\in R}(X-a)$$ evaluates to zero. So $X^n\sim$ a polynomial of degree $<n$. So $X^{n+k}\sim$ a polynomial of degree $<n+k$ and so inductively $\sim$ a polynomial of degree $<n$.