Simple example of non-arithmetic ring

Question

Can anyone provide a simple concrete example of a non-arithmetic commutative and unitary ring (i.e., a commutative and unitary ring in which the lattice of ideals is non-distributive)?

Answer 1

HINT $\ \: $ Distributivity easily yields that a finitely generated ideal is $\:1\:$ if it contains a cancellable element $\rm\:u\:$ that is $\rm\:lcm$-coprime to the generators. For example, for a $2$-generated ideal $\rm\:(x,y)$

LEMMA $\:\ $ If $\rm\ x,\:y\:$ and cancellable $\rm\:u\:$ are elements of an arithmetical ring then $$\rm\ \begin{array}{} (u)\cap(x)\ =\ (u\:x)\\ \rm (u)\cap(y)\ =\ (u\:y)\end{array}\ \ \ and\ \ \ (u) \subseteq (x,y)\ \ \Rightarrow\ \ (x,y) = 1$$

Proof $\rm\ \ (u) = (u)\cap(x,y) = (u)\cap(x) + (u)\cap(y) = u\ (x,y)\:$ so $\rm\:(x,y)=1\:$ by cancelling $\rm\:u\:.$

REMARK $\ $ Thus to prove that a domain is not arithmetical it suffices to exhibit elements that violate the Lemma. That is easy, e.g. put $\rm\ u = x+y\ $ for $\rm\ x,y \in \mathbb Q[x,y]\:,\: $ or $\rm\ x,\:y=2\in \mathbb Z[x]\:.$

Arithmetical domains are much better known as Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfying: $\rm CRT$ (Chinese Remainder Theorem) for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\ $ or the $\rm\: GCD\cdot LCM\:$ law: $\rm\: (A + B)\ (A \cap B) = A\ B\:,\ $ or $\:$ "contains $\rm\Rightarrow$ divides" $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for finitely generated $\rm\:A\:$ etc. It's been estimated that there are over $100$ known characterizations, e.g. see my prior answer for close to $30$ interesting such.

Answer 2

In the ring $K[X,Y]$, where $K$ is a field and $X$ and $Y$ are indeterminates, we have

$$(X+Y)\cap\Big((X)+(Y)\Big)\not\subset\Big((X+Y)\cap (X)\Big)+\Big((X+Y)\cap (Y)\Big).$$

[Thank you to Bill Dubuque for having pointed out a catastrophic typo!]

Answer 3

A commutative integral domain is "arithmetic" in the sense that you specify iff it is a Prüfer domain, i.e., iff every nonzero finitely generated ideal is invertible. This class of domains is famously robust: there is an incredibly long list of equivalent characterizations: see e.g. the beginning of this paper for some characterizations. For a proof that a domain is arithmetic iff its finitely generated ideals are invertible, see e.g. Theorem 6.6 of Larsen and McCarthy's text Multiplicative Theory of Ideals.

Note in particular that a Noetherian domain is Prüfer iff it is Dedekind, i.e., iff it is integrally closed and of Krull dimension at most one. Therefore examples of rings with non-distributive lattice of ideals abound, e.g.:

For any field $k$, $k[t_1,\ldots,t_n]$, $n \geq 2$. (The dimension is greater than one.)
For any nonfield Noetherian domain $k$, $k[t_1,\ldots,t_n]$, $n \geq 1$. (The dimension is greater than one.)
$\mathbb{Z}[\sqrt{-3}]$, $k[t^2,t^3]$ for any field $k$. (The rings are not integrally closed.)

And so forth...