For an ideal $\mathfrak a\subset A,$ define $S(\mathfrak a)=\{f\in A\mid f\not\in x, \forall x\in D(\mathfrak a)\};$ namely, $S(\mathfrak a)$ is the set of elements that do not belong to any prime ideal which does not contain $\mathfrak a.$
Then the question is:

For which ring $A$ is the following true?
If $\mathfrak a$ is finitely generated and if $\mathfrak p$ is a prime ideal containing $\mathfrak a,$ then $S(\mathfrak a)\cap\mathfrak p\not=\emptyset.$

My attempt:
When $\mathfrak a$ is generated by a single element $f,$ then clearly $f\in S(\mathfrak a)\cap\mathfrak p.$
By induction on the number of generators of $\mathfrak a=\left(f_1,\cdots,f_n\right),$ I found elements $x_1,\cdots,x_n$ such that $x_i\in S(\mathfrak a_i)\cap\mathfrak p, \forall i,$ where $\mathfrak a_i=\left(f_1,\cdots,\hat f_i,\cdots,f_n\right).$ And, if some $x_i\in S(f_i),$ then this $x_i$ belongs to the intersection in question, so we assume that $x_i\not\in S(f_i).$
Then we have $\alpha_i:=x_1\cdots\hat x_i\cdots x_n\not\in S(f_j), \forall j\not=i,$ and $\alpha_i\in S(f_i).$ Thus I tried to somehow glue together these $\alpha_i$ to produce one element in the intersection in question, to no avail.
However, if there are only finitely prime ideals of $A,$ then I managed to produce a weighted sum of these $\alpha_i$ which does belong to $S(\mathfrak a)\cap\mathfrak p.$

P.S. My attemps above ignore completely the prime-ness of $\mathfrak p,$ and is valid for every ideal containing $a,$ which led me to believe that I must have missed something.

Any help will be greatly appreciated; thanks in advance.

Thanks to Jake Levinson, I now know that the proposition is not true when $A$ is Noetherian of dimension $\ge2,$ and I have shown its validity when $A$ is a PID or when $A$ has finitely many prime ideals. So the remaining case is when $\operatorname{dim}(A)=1,$ and $A$ is not a PID, and has infinitely many prime ideals.

  • $\begingroup$ I know I can accept answers, I am just considering the remaining cases that are not yet solved. Maybe others will give better answers, who knows? In any case, thanks for the reminder. $\endgroup$
    – awllower
    Nov 16, 2014 at 3:45
  • $\begingroup$ Well, the answer plus the comments explored the cases: $A$ noetherian of dimension $\ge2,$ and Dedekind domains. But what about Noetherian rings of dimension $1,$ which are not integrally closed? In any case, if Jake Levinson edited and integrated the answer, I will definitely accept that. Or maybe I will edit that answer. :p And thanks for reminding me of the acception of this great answer, by the way. $\endgroup$
    – awllower
    Nov 16, 2014 at 8:01

1 Answer 1


This is not true. For example, let $A$ be Noetherian and let $\mathfrak a$ be an ideal of height $2$ or more. So in this case the set $D(\mathfrak a)$ of "prime ideals which do not contain $\mathfrak a$" includes every prime ideal of height $1$ (or $0$).

In particular if $f \in S(\mathfrak a)$, then $f$ is a unit. So $S(\mathfrak a) \cap \mathfrak{p} = \emptyset$ for any proper ideal $\mathfrak{p}$.

As an example, let $\mathfrak a = (x,y)$ in $k[x,y]$.

Note: It's worth pointing out that the cases you found proofs for are somewhat exceptional, -- principal ideals, and rings with only finitely-many prime ideals. If $A$ is assumed Noetherian, then principal ideals have height at most $1$, and the "finitely-many prime ideals" condition implies $\dim A \leq 1$.

As per your request, I will consider the case where $\mathfrak{a}$ is itself a prime ideal, of height one (and $A$ is Noetherian). In this case it follows that $S(\mathfrak{a}) \cap \mathfrak{p} = S(\mathfrak{a}) \cap \mathfrak{a}$, for any choice of $\mathfrak{p} \supseteq \mathfrak{a}$. (Note that every non-unit is contained in some height-one prime ideal). In particular $f \in S(\mathfrak{a}) \cap \mathfrak{a}$ if and only if $\mathfrak{a}$ is the unique height-one prime ideal containing $f$, that is, $\mathrm{div}(f) = n \cdot \mathfrak{a}$ for some $n\geq 1$. Equivalently, $\mathfrak{a}$ is a torsion element in the class group of $A$. (Such elements are rare in general.)

Sometimes these elements $f$ will exist for all ideals $\mathfrak{a}$ of $A$: for example, if $A$ is the ring of integers of a number field, then its class group is finite, so some power of $\mathfrak{a}$ is a principal ideal.

On the other hand, there are counterexamples even if $A$ is still taken to be a Dedekind domain -- just take one whose class group has elements of infinite order, For instance, let $A = \mathbb{C}[x,y]/(y^2 - x^3 - 1)$. There are ideals $\mathfrak{a}$ in $A$ no power of which is principal (in fact, such ideals are dense in $\mathrm{Spec}(A)$.) But $f \in S(\mathfrak{a}) \cap \mathfrak{a}$ implies that $\sqrt{(f)} = \mathfrak{a}$, hence by ideal factorization $(f) = \mathfrak{a}^n$ for some $n$.

  • $\begingroup$ What about the case $A$ is of dimension $1,$ but has infinitely many prime ideals, and is not a PID either? For example, $A=\mathbb Z(\sqrt{-5})$ has class number $2,$ so is not a P.I.D. but it is a Dedekind domain, hence of dimension $1.$ (It evidently has infinitely many prime ideals.) And thanks for this helpful answer! $\endgroup$
    – awllower
    Nov 11, 2014 at 3:15
  • 2
    $\begingroup$ @awllower: In the case $A = \mathbb{Z}[\sqrt{-5}]$, or any other number field's ring of integers, $S(a) \cap \mathfrak{p}$ will be nonempty. The reason is that, by the finiteness of the class group, every ideal becomes principal when raised to some power. In particular, for any finite collection of prime ideals, there are elements $f \in A$ that are contained in those primes and no others. I suspect there may be counterexamples in other rings, though -- for example $\mathbb{C}[x,y] / (y^2 - x^3 - 1)$, which has a class group with elements of infinite order. $\endgroup$ Nov 11, 2014 at 14:45
  • $\begingroup$ @awllower: Also you're welcome! $\endgroup$ Nov 11, 2014 at 14:46
  • 2
    $\begingroup$ @awllower: Okay, yes, I think there is a counterexample -- let $a$ be a prime ideal in $\mathbb{C}[x,y]/(y^2 - x^3 - 1)$ that is of infinite order in the class group. If there exists $f \in S(a) \cap a$, then rad$(f) = a$ since no other prime ideal contains $f$; and by ideal factorization in Dedekind domains, this means $(f) = a^n$ for some $n$, a contradiction. $\endgroup$ Nov 11, 2014 at 15:11
  • 1
    $\begingroup$ Thanks for the good answer. If you integrated all the results into the answer body, then I will accept it. Thanks again. :) $\endgroup$
    – awllower
    Nov 16, 2014 at 8:04

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