# Can a prime in a Dedekind domain be contained in the union of the other prime ideals?

Suppose $R$ is a Dedekind domain with a infinite number of prime ideals. Let $P$ be one of the nonzero prime ideals, and let $U$ be the union of all the other prime ideals except $P$. Is it possible for $P\subset U$?

As a remark, if there were only finitely many prime ideals in $R$, the above situation would not be possible by the "Prime Avoidance Lemma", since $P$ would have to then be contained in one of the other prime ideals, leading to a contradiction.

The discussion at the top of pg. 70 in Neukirch's "Algebraic Number Theory" motivates this question.

Many thanks,

John

Yes, it is possible.
According to Claborn's theorem1 any abelian group is the class group of some Dedekind ring.
Take a Dedekind ring $R$ whose class group is isomorphic to $\mathbb Z$ and freely generated by the ideal $I$. Since $I=\mathfrak m_1 \mathfrak m_2\ldots \mathfrak m_N$ with all $\mathfrak m_i$'s maximal, one of those maximal ideals, call it $\mathfrak m$, must be without torsion. I claim that $\mathfrak m$ is contained in the union of the other maximal ideals of $R$.
Indeed, take an arbitrary nonzero $f\in \mathfrak m$ and decompose $(f)$ into a product of primes :
$$(f)=\mathfrak m^r.\prod \mathfrak n_i^{r_i}$$ ( $\mathfrak n_i\neq \mathfrak m, \quad$almost all $r_i=0$)
You can't have all the $r_i=0$, else $(f)=\mathfrak m^r$ would imply that $\mathfrak m$ is torsion in the class group.
Since $f$ is in all the maximal ideals $\mathfrak n_i$ with $r_i\neq0$ , the claim is proved : $\mathfrak m$ is contained in the union of the other maximal ideals of the Dedekind ring $R$.

An easy warm-up John (rightfully) evokes the prime avoidance theorem. It is easy to see that this theorem doesn't hold for infinitely many primes. For example consider the product ring $R=\mathbb Q^{\mathbb N}$ and the maximal ideals $\mathfrak m_n=\{(q_i)\in R | q_n=0\}\subset R$ . Then for the ideal $I=\mathbb Q^{(\mathbb N)}$ of almost zero sequences we have $I \subset \bigcup \mathfrak m_n$ although $I\nsubseteq \mathfrak m_n$ for each $n$.
This easy counterexample doesn't answer John's actual (more precise and more demanding) question .

Thanks to Jyrki who accurately pointed out (in a now tactfully deleted comment!) that my previous version incorrectly assumed that in Claborn's theorem I could take primes as free generators of the class group .

A mistake in a book (added later) In the book Algebraic Number Theory mentioned by John in his question the author describes (on page 66) a generalized localization. He starts with a completely general commutative ring $A$ and a completely arbitrary set $X\subset\text{Spec}(A)$ of prime ideals of $A$. He remarks that the complement $S=\text{Spec}(A)\setminus \bigcup \{\mathfrak p|\mathfrak p\in X\}$ is a multiplicative set and considers the ring of fractions $A(X)=S^{-1}A$. He writes that the only primes $\mathfrak q\subset A$ that survive in $A(X)$ are those which are subsets $\mathfrak q\subset \bigcup \{\mathfrak p|\mathfrak p\in X\}$, and this is absolutely correct. However he adds that in the case of a Dedekind ring $A$ the surviving ideals are those $\mathfrak p \in X$ . This claim (repeated page 70) is not true, as shown by taking for $A$ our $R$ above and for $X$ the set of all maximal ideals in $R$ different from $\mathfrak m$: that ideal $\mathfrak m$ survives in $R(X)$ although it is not an element of $X$ : $\mathfrak m \notin X$ by the very choice of $X$.
Congratulations to John for catching this very subtle little mistake made by a great arithmetician in a great book.

1 C. R. Leedham-Green: The class group of Dedekind domains, Trans. Amer. Math. Soc. 163 (1972), 493-500 ; doi: 10.1090/S0002-9947-1972-0292806-4, jstor.

• Georges - Many thanks for your help. The discussion in Neukirch (pg 70) seems to tacitly assume that above is not possible, so I really appreciate your clarification! – John M Jul 14 '11 at 13:23
• This is very nice. – Akhil Mathew Jul 14 '11 at 13:59
• Unfortunately, this small problem also invalidates his Proposition (11.6) (pg 70) without some further hypotheses to rule out cases like the above. The sequence would not then be exact, or even a chain complex. – John M Jul 14 '11 at 17:34
• You are right, John: the mentioned mistaken result in the book unfortunately seems to be actually used in the Proposition you quote . – Georges Elencwajg Jul 14 '11 at 21:40
• @Georges: I find it interesting that you link to a paper of Leedham-Green on my website which gives the second published proof of Claborn's Theorem. You could also have linked to Claborn's original proof: math.uga.edu/~pete/claborn66.pdf. Alternately, I myself have a soft spot for the third published proof: math.uga.edu/~pete/ellipticded.pdf. – Pete L. Clark Jul 17 '11 at 13:46

If $R$ is the ring of integers $O_K$ of a finite extension $K$ of $\mathbf{Q}$, then I don't think this can happen. The class of the prime ideal $P$ is of finite order in the class group, say $n$. This means that the ideal $P^n$ is principal. Let $\alpha$ be a generator of $P^n$. Then $\alpha$ doesn't belong to any prime ideal other than $P$, because at the level of ideals inclusion implies (reverse) divisibility, and the factorization of ideals is unique.

This argument works for all the rings, where we have a finite class group, but I'm too ignorant to comment, how much ground this covers :-(

• I'm upvoting this answer: Jyrki would obviously have completely solved the question had he known (remembered?) Claborn's theorem. – Georges Elencwajg Jul 14 '11 at 11:38
• @Georges: Thanks. You're too kind. My exposure to Dedekind domains is limited to number rings and the simplest coordinate rings. I had never heard of Claborn's result. Thanks for the link! – Jyrki Lahtonen Jul 14 '11 at 12:48
• Jyrki - thanks for your help! – John M Jul 14 '11 at 13:19

This answer refers to the contributions of Jyrki and Georges: assume that a maximal ideal $P$ of a Dedekind domain $R$ is NOT contained in the union of all other maximal ideals. Then there exists an element $f\in P$ such that $v_P(f)=n>0$ for the discrete valuation attached to $P$ and $v_Q(f)=0$ for all $Q\neq P$. Now $P^n$ consists of those elements $r\in R$ such that $v_P(r) \geq n$. Thus for every $r\in P$ we get $r=fs$ with $s\in R$. Hence $P^n =fR$. Jyrki has already shown that if $R$ has torsion class group, then no maximal ideal contained in the union of all others can exist.

So: a maximal ideal of $R$ contained in the union of all other maximal ideals exists if and only if the class group of $R$ is not torsion.

• Thanks! That's interesting. – John M Jul 14 '11 at 13:24
• I think this is the best answer. Note also that the condition that the class group be torsion is also equivalent to every ring intermediate between $R$ and its field of fractions is a localization: see e.g. $\S 22.2$ of math.uga.edu/~pete/integral.pdf. (When I get the chance, I may add the result of this answer to this section: it is interesting!) – Pete L. Clark Jul 17 '11 at 13:36
• @Georges: there is indeed a little typo in Hagen's answer: it should say $P^n = \{r \in R \ | \ v_P(r) \geq n\}$ [edit: I took the liberty of fixing this]. But then if you have an element $f \in P^n$ with valuation exactly $n$ at $P$ and zero at every other prime, it must indeed generate $P^n$. To restate what Hagen and Jyrki have shown: a maximal ideal $P$ in a Dedekind domain is not contained in the union of all other maximal ideals iff the class of $P$ has finite order in $\operatorname{Pic} R$. – Pete L. Clark Jul 17 '11 at 13:39
• Thanks to Pete for his answers to my ( now deleted) questions. I'm upvoting Hagen this minute. – Georges Elencwajg Jul 17 '11 at 14:37
• Dear Pete, I think your notes are great because they are not contained in the union of all the other books on commutative algebra :-) – Georges Elencwajg Jul 17 '11 at 17:55