# When the set of prime ideals is linearly ordered

Imposing structure on the poset of prime ideals of a ring $R$ is one way to gain a hold onto its structure. The poset of prime ideals of $R$ is simply a fancy term for the set of prime ideals of $R$, partially ordered by inclusion. Usually this set is not totally ordered: in the ring of integers $\Z$ for instance, the prime ideals $(2)$ and $(3)$ cannot be compared by inclusion. It seems to me that requiring the poset of primes to be totally ordered is a strict condition indeed.

Here is one type of domain in which the prime ideals are totally ordered: the valuation domain.

## What is a valuation domain?

A valuation domain is an integral domain $R$ such that for every $a,b\in R$, either $a\mid b$ or $a\mid a$. Put another way, a valuation domain is such that the poset of principal ideals is totally ordered. The name 'valuation domain' arises because they can actually also be defined via valuations on fields.

So why are the prime ideals of a valuation domain linearly ordered? Suppose $P$ and $Q$ are two prime ideals in a valuation domain. We suppose that $P\not\subseteq Q$, and we would like to prove that $Q\subseteq P$. So, take some $p\in P$ such that $p\not\in Q$. If $q\in Q$, then either $p\mid q$ or $q\mid p$ because we are working in a valuation domain. However, because $p\not\in Q$, it is impossible that $q\mid p$. Therefore $p\mid q$ which implies that $q\in P$.

Since $q$ was an arbitrary element of $Q$, we see that $Q\subseteq P$. The astute reader probably noticed something: we never really used the fact that $P$ and $Q$ were primes. So, although it is true that in valuation domains, the prime ideals are linearly ordered by inclusion, the assumption of valuation domain is actually too strong.

## The weaker condition

Instead of requiring $R$ to be a valuation domain, one could require that for every $a,b\in R$, there exists a positive integer $n$ such that $a\mid b^n$ or $b\mid a^n$. Evidently, in such a ring the prime ideals must still be linearly ordered.

Ayman Badawi in his paper 'On domains which have prime ideals that are linearly ordered' noticed that these two conditions are equivalent, and proved the following theorem:

Theorem. [Badawi] For a commutative ring $R$ the following are equivalent:

1. The prime ideals of $R$ are linearly ordered
2. The radical ideals of $R$ are linearly ordered
3. Each proper radical ideal of $R$ is prime
4. The radical ideals of principal ideals of $R$ are linearly ordered
5. For every $a,b\in R$ there exists a positive integer $n$ such that $a\mid b^n$ or $b\mid a^n$.

The proof is elementary and we already gleaned the essence of the argument of proving (1) from (5).

## Back to valuation domains

As we have seen, in a valuation domain, the prime ideals are linearly ordered. What about the converse? If we restrict ourselves to greatest common divisor domains (GCD domains), then the converse is true. Stated precisely:

Theorem. Let $R$ be a GCD domain. If the set of prime ideals of $R$ is totally ordered, then $R$ is a valuation domain.
Proof. We use the equivalent conditions of Badawi's theorem. To show that $R$ is a valuation domain, it suffices to prove that for every $a,b\in R$ that are nonzero nonunits, either $a\mid b$ or $b\mid a$. To do so, we let $d = \gcd(a,b)$. If $d$ is associated to either $a$ or $b$, then the conclusion is implied. Otherwise, we suppose that $d$ is not associated to $a$ or $b$.

Under this assumption, $a' = a/d$ and $b' = b/d$ are both not units. However, by assumption that the prime ideals of $R$ are totally ordered, without loss of generality there exists a positive integer $m$ such that $a' | b'^m$. But $\gcd(a',b') = 1$ and hence $\gcd(a', b'^m)=1$. Therefore, $a'$ is a unit after all, which is a contradiction.

The proof that we have given is Badawi's although the theorem was known before his paper.

Of course, this raises the question: what if we drop the assumption that $R$ is a GCD domain? Can you think of an integral domain $R$ whose primes are totally ordered by inclusion, and yet $R$ is not a valuation domain? At the moment, I cannot think of one but I would be very interested to see one.