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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.

nicomezi
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  • The remark I want to make is that one can often do a bit better. The case of $R=\Bbb{Z}_n$ is a well studied one. Chinese remainder theorem reduces it to the prime power case in general, and falling factorials give you the result. It's a bit too long to write out fully, but the following example gives you a taste of it. Consider the case $n=2^3$. The values of the polynomial $f_4(x)=x(x-1)(x-2)(x-3)$ when $x\in\Bbb{Z}$ are all divisible by $4!=2^3\cdot3$. Therefore any polynomial in $\Bbb{Z}_8[x]$ is $\sim$ to a polynomial of degree $<4$. – Jyrki Lahtonen Aug 26 '17 at 05:31
  • For more details about the case of $\Bbb{Z}_n$, see here. – Jyrki Lahtonen Aug 26 '17 at 05:38
  • @Jyrki Lahtonen, thank you for those interesting precisions. I will look into it. – nicomezi Aug 26 '17 at 05:45

1 Answers1

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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 $<n$.

Angina Seng
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  • Great ! I have not thought about such a construction. I was overthinking this problem apparently. Thank you ! – nicomezi Aug 25 '17 at 19:14