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I have two questions in the following explanation:

"Consider the quotient ring $S/3=Z_3[x]/(3,x^{n-1}+x^{n-2}+...+1)$. The canonical S/3-representative of $a\in Z[x]$ is the unique polynomial $b\in Z[x]$ of degree at most $n-2$ with coefficients in $\{-1,0,1\}$ such that;

$a\equiv b $ $(mod~(3, x^{n-1}+x^{n-2}+...+1))$. We write $\underline S3(a)$ for the canonical S/3 -representative of $a $. We write $S3(a)$ when the choice of representative is not normative".

Question 1: How do we take modulo of : (something) $(mod~(3, x^{n-1}+x^{n-2}+...+1))$ ? Do we first take (something) mod 3, then take $(mod~( x^{n-1}+x^{n-2}+...+1))$ of it afterwards?

Question 2: What does normative mean in the context above?

Esra
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  • The order in which you mod out by something doesn't matter, which can be proven with the Correspondence Theorem. –  Dec 06 '20 at 21:27
  • First time I’ve seen the word “normative” ! And I’m a geezer. My guess is that it’s saying that other people may choose different representatives, without risking disapproval from the author. In other words, I don’t think it’s a mathematical word at all. – Lubin Dec 06 '20 at 23:20
  • @JetChung yes you are right, the order of mod does not change anything but my main question here is how do we take the mod with two parameters? I never saw something like this. I mean I can take just mod 3 or just mod p(x) or mod 3 then mod p(x) but never seen mod (3, p(x)) where p(x) is a polynomial? – Esra Dec 07 '20 at 10:47
  • @Esra Yes, if you consider the ring $\mathbb{Z}[x]$ of integer polynomials, then the quotient ring $\mathbb{Z}[x]/(3, x^{n-1} + x^{n-2} + \ldots + 1)$ is formed by first setting $3 = 0$, then $x^{n-1} + x^{n-2} + \ldots + 1 = 0$ (or the other way around if you'd like). –  Dec 07 '20 at 18:08
  • Also, I believe the ring $S/3$ should be $\mathbb{Z}[x]/(3, \ldots)$ and not $\mathbb{Z}_3[x]/(3, \ldots)$ as this is redundant. –  Dec 07 '20 at 18:10
  • With $,g = (x^n-1)/(x-1),,$ a normal form of $f$ is $,f\bmod g,,$ calculated $\bmod 3$, or more simply $,(f\bmod x^n-1)\bmod g$ where we first reduce using the rewrite rule $,x^n\equiv 1,,$ then eliminate $,x^{n-1},$ by $,x^{n-1}\equiv x^{n-1}-g = -x^{n-2}-\cdots -x-1,,$ see the method of simpler multiples. – Bill Dubuque Dec 07 '20 at 20:00

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