Intereting Posts

Cancellation problem: $R\not\cong S$ but $R\cong S$ (Danielewski surfaces)
Examples of $\mathcal{O}_X$-modules that are not quasi-coherent sheaves
Using the same limit for a second derivative
Discriminant of a binary quadratic form and Jacobi symbol
Moment generating function of a stochastic integral
Intuitive understanding of why the sum of nth roots of unity is $0$
Optimization of relative entropy
Branches of analytic functions
Prove or Disprove: If every nontrivial subgroup of a group $G$ is cyclic, then $G$ is cyclic.
Proving the Kronecker Weber Theorem for Quadratic Extensions
Probability/Combinatorics Problem. A closet containing n pairs of shoes.
Radius of convergence of entire function
In GCD domain every invertible ideal is principal
Proof that every metric space is homeomorphic to a bounded metric space
Computing $n$-th external power of standard simplectic form

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.

- normalization of multiplicative subset of domain
- Fermat's Last Theorem and Kummer's Objection
- Prime divisors of $k^2+(k+1)^2$
- Is such a field element an element of a subring?
- The ring of integers of the composite of two fields
- How to prove that there exist infinitely many integer solutions to the equation $x^2-ny^2=1$ without Algebraic Number Theory

Many thanks,

John

- Primary decomposition of a monomial ideal
- How to arrive at Ramanujan nested radical identity
- When is a product of two ideals strictly included in their intersection?
- Poles of a sum of functions
- An ideal that is maximal among non-finitely generated ideals is prime.
- Does this “extension property” for polynomial rings satisfy a universal property?
- Nilradical of polynomial ring
- Tensor product, Artin-Rees lemma and Krull intersection theorem
- A question about a proof of a weak form of Hilbert's Nullstellensatz
- Group theory proof of existence of a solution to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$

Yes, it is possible.

According to Claborn’s theorem^{1} 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.

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 🙁

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.

- $x_n$ is the $n$'th positive solution to $x=\tan(x)$. Find $\lim_{n\to\infty}\left(x_n-x_{n-1}\right)$
- Given an axiom in a formal system, can we always find another formal system in which this axiom is a theorem?
- Is it possible to alternate the law of mathematics?
- Permutations of a string with duplicate characters
- Fibonacci Generating Function of a Complex Variable
- Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$
- $\tau$ and grouping of prime numbers
- Part (b) of Exercise 13 of first chapter of Rudin's book “Functional Analysis”
- Teenager solves Newton dynamics problem – where is the paper?
- Vanishing of the first Chern class of a complex vector bundle
- Find the limit $\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}$?
- Example of a non-splitting exact sequence $0 → M → M\oplus N → N → 0$
- Biggest powers NOT containing all digits.
- Prove $2^{1092}\equiv 1 \pmod {1093^2}$, and $3^{1092} \not \equiv 1 \pmod {1093^2}$
- Compositeness of $n^4+4^n$