Differences between polynomial quotient rings $\mathbb{Z}_m[x]/(x^n+1)$ and $\mathbb{Z}_m[x]/(x^n-1)$

Question

As based on the definition of the polynomial quotient ring

$\mathbb{Z}_m[x]/(x^n+1) = \left\{a_{n-1}x^{n-1}+\cdots+a_1x+a_0:a_i\in\mathbb{Z}_m\right\}$,

does that imply that $\mathbb{Z}_m[x]/(x^n+1) = \mathbb{Z}_m[x]/(x^n-1)$, as any negative coefficient in the elements of $\mathbb{Z}_m[x]/(x^n-1)$ will be shifted to be between 0 and $m-1$?

Or is there any fundamental difference between $\mathbb{Z}_m[x]/(x^n+1)$ and $\mathbb{Z}_m[x]/(x^n-1)$ under the same $\mathbb{Z}_m$?

Thank you.

Answer 1

I re-read your question a couple of times and your question seems to stem from your belief that $Z_m[x]/(x^n-1)$ should be equal (isomorphic) to $Z_m[x]/(x^n+1)$, which is not necessarily the case (although I can't come up with an instance where it is in deed the case). Anyways, take a look at this example, maybe it will shed some light on the topic. Take $Z_3[x]/(x^2+1)$ and $Z_3[x]/(x^2-1)$. We have that $(x^2+1)$ is irreducible since $x^2 \equiv 0$ or $1$ and so $x^2+1 \equiv 1$ or $2$, but $(x^2-1)=(x-1)(x+1)$. Hence, $Z_3[x]/(x^2+1)$ is a field but $Z_3[x]/(x^2-1)$ is not.

Answer 2

The misunderstanding may be because of a misuse of the symbol $=$. It's technically incorrect to say that $$\mathbb{Z}_m[x]/(x^n-1)=\left\{a_{n-1}x^{n-1}+\cdots+a_{1}x+a_0 \;|\; a_i\in \mathbb{Z}_m\right\}$$ because the left hand side is a set of cosets of $(x^n-1)$ in $\mathbb{Z}_m[x]$ and the right hand side is a finite set of polynomials with coefficients in $\mathbb{Z}_m$. It may be more excusable to write use the symbol $\cong$: $$\mathbb{Z}_m[x]/(x^n-1)\cong\left\{a_{n-1}x^{n-1}+\cdots+a_{1}x+a_0 \;|\; a_i\in \mathbb{Z}_m\right\},$$ however this is still not really right becuase the right hand side lacks a ring structure.

I suppose you mean this: let $R$ be the ring whose underlying set is $\left\{a_{n-1}x^{n-1}+\cdots+a_{1}x+a_0 \;|\; a_i\in \mathbb{Z}_m\right\}$ with the usual addition and multiplication of polynomials, but such that any instances of $x^{k+n}$ ($k\ge 0$) are replaced with $x^k$. This is a perfectly fine ring and we do indeed have $\mathbb{Z}_m[x]/(x^n-1)\cong R$ using the map given by $x+(x^n-1)\mapsto x$. We can do the same thing for $\mathbb{Z}_m[x]/(x^n+1)$, defining the ring $Q$ in a similar way to $R$ and getting $\mathbb{Z}_m[x]/(x^n+1)\cong Q$.

Note here that $R$ and $Q$ have the same underlying set but they are not in general isomorphic as rings, and so it is incorrect to write $R\cong Q$ or even $R=Q$. crystal_math's answer probably has the easiest counterexample to this claim: when $m=3$ one of these is a field while the other is not.