# Simple example of non-arithmetic ring

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)?

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+1. Out of curiosity, how did this question arise? How did you encounter arithmetical rings? –  Bill Dubuque Aug 19 '11 at 2:02

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.

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Hi Bill! Great answer! +1 –  Pierre-Yves Gaillard Aug 18 '11 at 18:02
@Bill I have an irrelevant question which is nevertheless relevant to my interests. Why do you always use the \rm version of your math variables? Is it just a personal, stylistic choice? –  barf Aug 18 '11 at 19:46
Yes, I find MathJax far too ugly any other way @barf –  Bill Dubuque Aug 18 '11 at 20:11
Great answer, indeed. Also, thank you for the link to the other answer. –  Gonzalo Medina Aug 19 '11 at 0:58

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...

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+1. Much better answer! (I should add: much better than mine, in case there will be other answers...) –  Pierre-Yves Gaillard Aug 18 '11 at 12:00
@Bill: The paper linked to in my answer cites references where the equivalences are proved. If you (or the OP, or anyone else) do not have access to any of these references but are still interested in the proof, probably someone here could help you out. (Sorry for not wanting to argue about whether something is "simple": surely it's best for you to leave your own answer, as you have.) –  Pete L. Clark Aug 18 '11 at 19:06
@Bill: Actually I said that I do not want to argue about whether my answer is "simple" (or whether the word "simple" in the OP's answer is meant to apply to the ring itself or to the proof that it is not arithmetic, etc.). Again, I apologize. If none of the answers given so far are satisfactory to the OP, he will surely tell us so. –  Pete L. Clark Aug 18 '11 at 19:22
What's the beef here? Surely there's a better way to resolve this... –  The Chaz 2.0 Aug 18 '11 at 19:45
@Gonzalo If you continue your study of these rings I highly recommend that you do study the many diverse characterizations. They are both important and fascinating (that's why I typed $27$ of them into my answer last Decemeber!) But they are a bit overkill for the simple question that you posed. Note: not all of the $27$ I mentioned are in the Bazzoni and Glaz paper that I cite in the comments there. There are also many other interesting characterizations in the literature. Perhaps more than for any other algebraic structure. –  Bill Dubuque Aug 19 '11 at 1:20

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!]

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Thank you for the example (simple and concrete). –  Gonzalo Medina Aug 19 '11 at 0:58