# Polynomial quotient ring $\mathbb{Z}_m[x]/f(x)$ when $\mathbb{Z}_m$ is only a ring (not a field)?

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.

• How is $\mathbb{Z}_m[x]/f(x)$ not a ring? – lhf Jun 5 '20 at 0:44

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 $$f(x)\in A[x]$$.
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.