The divisibility relation "$a$ divides $b$", or concisely, $a \vert b$ defined over a commutative integral domain $R$ with identity induces a partial order on the multiplicative semigroup $R/R^{\times}$ where $R^{\times}$ is the unit group of $R$. The inclusion relation $I \subset J$ is a partial order on the set of principal ideals of $R$ and the two corresponding partially ordered sets are anti-isomorphic in a natural sense.

If $R$ is a GCD domain, then $D(R) \Doteq (R/R^{\times}, \vert)$ is a bounded distributive lattice, the joint $a \vee b$ being the lowest common multiple of $a$ and $b$ and the meet $a \wedge b$ being the greatest common divisor of $a$ and $b$. We call $D(R)$ the divisibility lattice of $R$.

If $R$ is any ring (commutative or not, with identity or not and possibly with zero divisors), the set $\mathcal{I}(R)$ of ideals of $R$ is partially ordered by the inclusion and the operations $I \vee J \Doteq I + J$ and $I \wedge J \Doteq I \cap J$ turns $L(R) \Doteq (\mathcal{I}(R), \subset)$ into a complete and modular lattice. We call $L(R)$ the ideal lattice of $R$.

If $R$ is a principal ideal domain, then $D(R)$ and $L(R)$ are naturally anti-isomorphic. More generally, if $R$ is a Bézout domain, then $D(R)$ is anti-isomorphic to the lattice $L_{\text{principal}}(R)$ of its principal ideals.

Question 1. Are there remarkable results that bind the structure of $R$ to lattice properties of $D(R)$ and $L(R)$? In particular, what can be said about $R$ if one of the two lattices is
$(i)$ complete or, $(ii)$ distributive or, $(ii)$ complemented or, $(iii)$ boolean?

Here are the few things I know. The lattice $D(R)$ is complete if and only if $R$ satisfies the ascending chain condition on principal ideals (ACCP). Moreover, there exist non-atomic Bézout rings, e.g., the ring of entire functions (think about $\sin(\frac{z}{2^k})$ for $k \ge 0$). Hence there exists a ring $R$ which is a GCD domain and such that $D(R)$ is not complete. If the lattice $L(R)$ is distributive for a commutative ring $R$, then $R$ is called arithmetical and this is equivalent to say that the localization $R_{\mathfrak{m}}$ of $R$ at $\mathfrak{m}$ is a uniserial ring for every maximal ideal $\mathfrak{m}$ of $R$. (An arithmetical domain is the same as Prüfer domain). The Jaffard-Ohm-Kaplansky Theorem states that the positive cone of every lattice-ordered Abelian group arises as the semigroup of divisibility of a Bézout domain.

The answers of Sridhar Ramesh and Luc Guyot show that for a unique factorization domain $R$, the lattice $D(R)$ is complemented if and only if $R$ is a field, and $D(R)$ is Boolean in this case. In addition, $D(R)$ is totally ordered if and only if $R$ is a discrete valuation ring or a field.

My second question is

Question 2. Have $D(R)$ and $L(R)$ been studied or classified for specific classes of rings such as principal ideal domains (PIDs), unique factorizations domains (UFDs), Dedekind domains and rings of integers in particular?

The answers of Sridhar Ramesh and Luc Guyot show that the situation is extremely tamed for UFDs, since the cardinality of the set of primes of $R$ (up to multiplication by units of $R$) is a complete invariant of isomorphism for $D(R)$. Indeed, $D(R)$ identifies with the lattice of $\mathbb{N}$-valued functions with finite support on the set of primes (modulo units), plus an additional top element corresponding to $0$. Any cardinal can be achieved as the cardinality of the set of primes of some UFD (think about $k[X_i; i \in I]$ for $k$ a field and any set $I$) and actually it is even true for a PID (think of $k[X]$ for a $k$ an arbitrary algebraically closed field).

Any references, specifically comprehensive surveys or textbooks, will be appreciated!


Given a ring $R$, let us denote by $L(R)$ the lattice of two-sided ideals of $R$ for which the infimum and supremum are given by $\inf(I, J) = I \cap J$ and $\sup(I, J) = I + J$.

Such lattices are complete and modular. If $R$ is a principal ideal domain (see original MSE question/answer) or if $R$ is the ring of integers of a number field [3, page 135], then $L(R)$ is distributive.

The following claims classify the lattices of principal ideal domains (PIDs).

Claim 1. Two PIDs have isomorphic ideal lattices if and only if they have equinumerous sets of prime ideals.

Claim 1 is, I believe, immediate. (In [2], it is proven that $L(\mathbb{Z}[\sqrt{-5}]) \simeq L(\mathbb{Z})$; this should give an idea as how to proceed.)

So are the following claims.

Claim 2. Let $\kappa$ be a cardinal number. Then there is a PID with exactly $\kappa$ prime ideals.

Proof of Claim 2. For $\kappa = 1$, take $R = \mathbb{Q}$. For $2 \le \kappa < \omega$, consider $\kappa - 1$ distinct prime numbers $p_1, \dots, p_{\kappa - 1}$ in $\mathbb{Z}$ and set $S \Doteq p_1\mathbb{Z} \cup \cdots \cup p_{\kappa - 1}\mathbb{Z}$. Then the localization $\mathbb{Z}_S$ of $\mathbb{Z}$ is a PID with exactly $\kappa$ maximal ideals. For any $\kappa \ge \omega$, set $R = k[X]$ for $k$ an algebraically closed field of cardinality $\kappa$. (For $\kappa = \omega$, setting $R = \mathbb{Z}$ is a simpler and natural choice.)

Claim 3. The ideal lattice $L(R)$ of a principal ideal domain $R$ is totally ordered if and only if $R$ is a field or discrete valuation ring. It is Boolean if and only if $R$ is a field.

The question as to whether $L(R)$ is complemented was settled by Robert Blair in [1, Theorems 1 and 2] in a broader setting:

Blair's Theorem. Let $R$ be a (not necessarily commutative, possibly without identity) ring. Then the following are equivalent:

$(i)$ The lattice $L(R)$ of two-sided ideals partially ordered by inclusion is complemented.

$(ii)$ The ring $R$ is a direct sum of minimal two-sided ideals.

$(iii)$ The ring $R$ is isomorphic with a direct sum of simple rings.

The question as to whether $L(R)$ is distributive for more general rings $R$ was also tackled by Robert Blair in [1, Section 5]. Here is one of Blair's results.

Blair's Lemma. (Lemma 11) If $R$ is a primitive ring whose lattice $L_r(R)$ of right ideals is distributive, then $R$ is a division ring.

[1] R. Blair, "Ideal lattices and the structure of rings", 1952.
[2] H. Subramanian, "Principal ideal in the ideal lattice", 1972.
[3] G. Birkhoff, "Lattice theory", 1948.


A PID is a unique factorization domain, so the divisibility lattice (identifying associate elements, of course) has the same ordering as the set of $\mathbb{N}$-valued functions with finite support on some (possibly infinite) set of primes, plus an additional top element corresponding to 0. Actually, I suppose it is the reverse of this in your convention, where $a \leq b$ corresponds to "$b$ divides $a$".

It is thus distributive, but not Boolean (except in the trivial case where the set of primes is empty; i.e., when the PID is a field); the only complemented elements are the top and bottom element, and indeed, if $a \wedge b = 0$ (in the sense that $lcm(a, b) = 0$), then at least one of $a$ and $b$ is itself $0$.

  • $\begingroup$ Nice description! Note that the divisibility lattice of a field, which is a very special PID, is Boolean. $\endgroup$ – Luc Guyot Oct 31 '17 at 10:25
  • $\begingroup$ Good catch. That is of course the only exception; the case where the set of primes is empty. $\endgroup$ – Sridhar Ramesh Oct 31 '17 at 19:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.