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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" (see for example http://commalg.wiki-site.com/index.php/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

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+1, interesting question! However, we should probably add the requirement that $P\neq0$, to avoid the answer being trivially yes. –  Zev Chonoles Jul 14 '11 at 5:19
    
@Zev Chonoles - thanks I've edited that. –  John M Jul 14 '11 at 5:22

3 Answers 3

up vote 17 down vote accepted

Yes, it is possible.
According to Claborn's theorem 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.

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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
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@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 :-(

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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 number rings and 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.

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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
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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

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