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

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?

• $$\Bbb Z_n:=\Bbb Z/n\Bbb Z$$ , for any non-prime $\;n\;$ .\ – DonAntonio Apr 2 '14 at 18:19
• Polynomial rings over nondomains? – André Nicolas Apr 2 '14 at 18:20
• I meant nondomains besides $\mathbb Z/n\mathbb Z$. – Nishant Apr 2 '14 at 18:20
• @Nishant - Look, several respondents have given you nice examples of rings with 1 that are not integral domains, ywt none of them seems to satisfy you. So, I would ask what is the statement about this type of ring that you are trying to disprove, and why, exactly, are the examples provided not good enough. – Chris Leary Apr 2 '14 at 21:26
• All right. The problem is as follows: if two elements of $R$ generate the same ideal, then one is a unit multiple of the other. I don't immediately see why this would be true for non-domains, but I haven't been able to find counterexamples in any of the non-domains given. – Nishant Apr 2 '14 at 23:08

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.

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.

• Ooh, that's a good one! Unfortunately, I don't think it helps me with my problem... – Nishant Apr 2 '14 at 20:37
• What is your problem? Knowing that would make it easier to find a good ring. – Thomas Credeur Apr 3 '14 at 2:07
• Is the following statement true for a nondomain: if $r$ and $s$ generate the same principal ideal, then $r=us$, for some unit $u\in R$. – Nishant Apr 3 '14 at 3:49
• Maybe try $\mathbb{Z}_4 \times \mathbb{Z}_6$? I haven't worked through the details but I am pretty sure $((2,2)) = ((2,4))$, even though one isn't a unit multiple of the other. – Thomas Credeur Apr 3 '14 at 4:09
• Nope, they differ by $(1, -1)$, which is a unit. – Nishant Apr 3 '14 at 4:14

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.