Examples of Commutative Rings with $1$ that are not integral domains besides $\mathbb Z/n\mathbb Z$?

Question

I'm doing a problem where I'm trying to find a counterexample to some statement about commutative rings with $1$ when the ring is not a domain. I've tried looking at $\mathbb Z/n\mathbb Z$, for composite $n$, but I haven't been able to find a counterexample yet (maybe the statement is also true for nondomains, but I don't see any reason why that would be the case), so I was wondering: what are some other examples of unital commutative rings that are not domains?

Answer 1

For a very generic example: ${\mathbb Z}[X,Y]/(XY)$. This is in a certain sense the simplest non-domain in that you can build an obvious map from this ring to every (unital commutative) ring that is not a domain.

Edit. This may go way too far, but let me formalize the informal claim above that this is the simplest non-domain.

Consider the category ${\cal C}$ of (unital commutative) rings with two chosen elements, i.e., objects are triples $(R,r,s)$ with $R$ a ring and $r, s \in R$; morphisms are ring homomorphism preserving the two elements. The object $({\mathbb Z}[X,Y],X,Y)$ is the initial object of ${\cal C}$.

Now look at the full subcategory ${\cal D}$ of ${\cal C}$ on the objects $(R,r,s)$ where $R$ is a non-domain and $r, s \in R$ are non-zero elements such that $rs = 0$. The object $({\mathbb Z}[X,Y]/(XY),X,Y)$ is the initial object of ${\cal D}$. So in this sense $({\mathbb Z}[X,Y]/(XY)$ is the smallest non-domain.

Edit. And an example adressing the question by the OP in the comments ("In a non-domain $R$, if $(r) = (s)$ is then $r = us$ for some unit $u$?") constructed in a similar fassion.

Look at the ideal $I = (X - UY, Y - VX)$ of ${\mathbb Z}[X,Y,U,V]$ and consider the ring $R = {\mathbb Z}[X,Y,U,V]/I$. In this ring $(X) = (Y)$, since $Y = VX \in (X)$ and $X = UY \in (Y)$. And $U$ is not a unit, since $R/(U) \cong {\mathbb Z}[X,Y,U,V]/(X - UY, Y - VX, U) \cong {\mathbb Z}[V]$ which is not the trivial ring.

Answer 2

Given a set $S$, the power set of $S$ with addition given by symmetric difference and multiplication given by intersection, is a commutative ring with unity, but as long as $S$ has $2$ or more elements, $ \mathcal P \left({S}\right)$ is not an integral domain.

Answer 3

Whenever $R$ and $S$ are two nontrivial commutative rings with $1$, the product ring $R\times S$ has an identity, namely $(1,1)$, and is commutative; on the other hand, all elements of the form $(r,0)$ and $(0,s)$ are zero-divisors, so that $R\times S$ is not a domain. (Do you know what I mean by $R\times S$?) This is related to the $\mathbb Z/n\mathbb Z$ example; by the Chinese remainder theorem for rings, every ring of the form $\mathbb Z/pq\mathbb Z$ with $p$ and $q$ distinct primes, for example, is isomorphic to $\mathbb Z/p\mathbb Z\times\mathbb Z/q\mathbb Z$ and is therefore not a domain. Similarly, the product of any number of commutative rings with $1$ is not a domain.

Here's a less algebraic example: the ring of continuous functions from $\mathbb R$ to $\mathbb R$ is commutative and has an identity -- the constant function $1$ -- but has many zero divisors.