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The definition of polynomial quotient ring requires $\mathbb{Z}_m$ to be a field.

But if $m$ is not prime, then $\mathbb{Z}_m$ is just a ring (not a field). So under what conditions $\mathbb{Z}_m[x]/f(x)$ can still be a ring? ($f(x)$ polynomial over $\mathbb{Z}_m$[x], example $x^n+1$).

Note: I found out this question: Quotient rings over rings that are not fields, and it mentions that any principal ideal ring should suffice, and $\mathbb{Z}_m$ is a commutative principal ideal ring for all $m$, but I'm not sure if that is enough argument to make $\mathbb{Z}_m[x]/f(x)$ a ring.

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  • $\begingroup$ How is $\mathbb{Z}_m[x]/f(x)$ not a ring? $\endgroup$ – lhf Jun 5 '20 at 0:44
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When you write $\Bbb{Z}_m[x]/f(x)$, this is the same thing as writing $\Bbb{Z}_m[x]/(f(x))$. In particular, you are taking the quotient of the ring by the ideal generated by $f(x)$. It is true in general that the quotient of a commutative ring by an ideal is again a ring. E.g. if $A$ is a ring and $\mathfrak{a}$ is an ideal, then $A/\mathfrak{a}$ is a ring. So, what you have written will always be a ring, regardless of the coefficients used. That is, for any ring $A$, $A[x]/(f(x))$ is a ring, for any $f(x)\in A[x]$.

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A quotient of a commutative ring $R$ by an ideal $I \subseteq R$ is always a ring under the following operations and identities: for every $a,b\in R:$

$$(a+I)+(b+I):=(a+b)+I$$ $$(a+I)\cdot(b+I):=ab+I$$ $$0_{R /I}:=0_R+I$$ $$1_{R /I}:=1_R+I.$$

So long as you are modding $\mathbb{Z}_m[x]$ by a principal ideal $(f)$ for some $f\in \mathbb{Z}_m[x],$ the quotient $\mathbb{Z}_m[x]/(f)$ will be a ring.

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