Romain Giquaud has given a counterexample to the general form of the question. The bounty is for a solution for locally compact, metrizable rings. (I suspect the answer may be positive with this restriction.)

Let $R$ be a ``topological PID'': a topological ring which is an integral domain in which every principal ideal is closed and every closed ideal is principal.

Is $R$ ``topologically Noetherian'': there is no strictly increasing sequence of closed ideals?

If not, is there a locally compact counterexample?


2 Answers 2


We can study the conjecture for locally compact rings as follows.

In [Kap] a subset $S$ of a topological ring is called algebraically nilpotent if for some $n$, $S^n=0$.

Lemma 1. [Kap, Theorem 2] A locally compact ring with no algebraically nilpotent ideals is the direct sum of a connected ring and a totally disconnected ring. The former is a semi-simple algebra of finite order over the real numbers.

Since Romain Gicquaud's comment rules out the connected ring case, it remains to consider locally compact totally disconnected rings. For this case we have the following structural lemma.

Lemma 2. [Kap, Lemma 4] A locally compact totally disconnected ring has a system of neighborhoods of $0$ which are compact open subrings.

Lemma 3. Let $R$ be a locally compact totally disconnected integral domain. Let $U$ be a compact open subring of $R$ such that $0\in U$ and $1\not\in U$. Then $1+U$ is a multiplicative topological group.

Proof. Since $U$ is a ring, $1+U$ is a multiplicative semigroup. Since $R$ is an integral domain and $0\not\in 1+U$, the multiplicative semigroup $1+U$ is cancellative. Since $1+U$ is compact, it is a multiplicative topological group, see, for instance, [AT, Theorem 2.5.2]. $\square$

Proposition 4. Let $R$ be a locally compact totally disconnected integral domain which is a union of less than $\mathfrak c$ many compact sets. If every closed ideal in $R$ is principal then each proper nonzero ideal $I$ of $R$ is closed, so $R$ is a principal ideal domain.

Proof. Pick an element $x\in R$ such that $(x)=\overline{I}$. By Lemma 2, there exists a compact open subring $U$ of $R$ such that $0\in U$ and $1\not\in U$. There exists a set $A$ of size less than $\mathfrak c$ and the family $\{K_\alpha:\alpha\in A\}$ of compact subsets of $R$ such that $R=\bigcup_{\alpha\in A} K_\alpha$. For each $\alpha\in A$ the family $\{a+U:a\in K_\alpha\}$ is an open cover of a compact set $K_\alpha$. Therefore there exists a finite subset $S_\alpha$ of $K_\alpha$ such that $K_\alpha\subset S_\alpha+U$. Put $S=\bigcup_{\alpha\in A} S_\alpha$. Then $|S|\le |A|\cdot \omega<\mathfrak c$ and $S+U=R$. Let $T$, $T=xR/xU$ be the quotient additive topological group and $q:xR\to T$ be the quotient map. By [AT, Proposition 3.1.23] the topological group $T$ is locally compact. Moreover, $T=xR/xU=q(xR)=q(x(S+U))=q(xS+xU)=q(xS)$, so $|T|<\mathfrak c$. Since a Hausdorff compact space without isolated points has size at least $\mathfrak c$, for instance, by Čech–Pospíšil theorem, see, for instance, [Hod, Theorem 7.19], the group $T$ is discrete, so the set $xU=q^{-1}(0_T)$ is a neighborhood of $0$ in the additive topological group $xR$. Therefore $x+xU$ is a neighborhood of $x$ in $xR$, so there exists a point $y\in I\cap (x+xU)$. That is $y=x+xu$ for some element $u\in U$. By Lemma 3, $1+U$ is a multiplicative topological group, so the element $1+u$ is invertible. Then $x\in I$, so $I=(x)=\overline{I}$. $\square$

Now we can describe the structure of compact topological principal ideal domains in more details applying the following results.

Proposition 5. [War, Theorem 1] Let $R$ be a compact Hausdorff ring, and let $\mathfrak r$ be the radical of $R$. Every ideal of $R$ is closed if and only if $R$ satisfies the ascending chain condition on ideals and every principal ideal of $R$ is closed. Under these circumstances, the topology of $R$ is the $\mathfrak r$-adic topology, either $R=\mathfrak r$ or $R/\mathfrak r$ is isomorphic to the Cartesian product of a finite family of finite simple rings, and consequently there are only finitely many regular maximal, regular maximal left, or regular maximal right ideals in $A$.

Proposition 6. [War, Theorem 2] Let $R$ be a topological ring with identity, and let $\mathfrak r$ be the radical of $R$. The following conditions are equivalent.

1)) $R$ is compact Hausdorff, and every left ideal of $R$ is closed.

2)) $R$ is compact Hausdorff and Noetherian.

3)) $R$ is compact Hausdorff and satisfies the ascending chain condition on closed left ideals.

4)) $R$ is Noetherian, the topology of $R$ is the $\mathfrak r$-adic topology, $R$ is complete for that topology, $\bigcap_{n=1}^\infty \mathfrak r^n=\{0\}$, and $R/\mathfrak r$ is a finite ring.


[AT] Alexander V. Arhangel'skii, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[Hod] R. Hodel, Cardinal Functions I, in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.

[Kap] Irving Kaplansky, Locally compact rings, Am. J. Math. 70:2 (Apr 1948) 447-459.

[War] Seth Warner, Compact rings, Mathematische Annalen 145 (Feb 1962) 52-63.

  • $\begingroup$ Okay, this looks promising. Let me think it through. $\endgroup$
    – Nik Weaver
    Sep 22 at 22:18
  • $\begingroup$ First question, how does Theorem 1 apply to our situation? Does topological PID imply the ascending chain condition on ideals? $\endgroup$
    – Nik Weaver
    Sep 23 at 0:32
  • 1
    $\begingroup$ Well, good luck, udacha! $\endgroup$
    – Nik Weaver
    Sep 24 at 13:54
  • 1
    $\begingroup$ The case where $|R| \leq 2^{\aleph_0}$ would satisfy me. $\endgroup$
    – Nik Weaver
    Sep 25 at 23:03
  • 1
    $\begingroup$ This is fantastic, thank you! $\endgroup$
    – Nik Weaver
    Sep 26 at 23:26

I think the answer to your question is negative.

Let $\Omega$ be a non-empty open subset of $\mathbb{C}$. Choose $(z_n)_n$ a sequence of points in $\Omega$ such that, for all $n$ $$z_n \not\in \overline{\cup_{k \neq n} z_k}$$ (for example a sequence of points tending to a boundary point of $\Omega$).

Define $I_n \subset \mathrm{Hol}(\Omega)$ to be the set of holomorphic functions vanishing at $z_k$ for all $k \geq n$.

As is "well-known" there exists an holomorphic function $f_n$ whose zeros are simple and exactly the $z_k$ for $k \geq n$. Then $I_k = (f_k)$. and the $I_k$ form a strictly increasing sequence of closed ideals of $\mathrm{Hol}(\Omega)$.

Note that $\mathrm{Hol}(\Omega)$ is a PID according to your definition and has a topology defined by uniform convergence over compact subsets (i.e. it is a Fréchet space).

  • 2
    $\begingroup$ Your assumption on $(z_n)$ is not exactly the right one. Write $Z=\{z_n:n\ge 0\}$ (and assume $n\mapsto z_n$ is injective). Your assumption, as now formulated, is that $Z$ is discrete. But the assumption you need is that $Z$ is closed discrete, and this is indeed the case when you write "for example a sequence of points tending to a boundary point". For instance if $z_n$ is a sequence tending to a point of $\Omega$ (not among the $z_n$), then the assertion "there exists a holomorphic..." is not correct. $\endgroup$
    – YCor
    Sep 19 at 15:43
  • 4
    $\begingroup$ Nice! I think this is correct, modulo YCor's adjustment. What if we restrict to locally compact rings? $\endgroup$
    – Nik Weaver
    Sep 19 at 15:53
  • $\begingroup$ @NikWeaver: You can already rule out the possibility that $R$ is a $\mathbb{R}$-algebra as local compactness will imply that $R$ is a finite dimensional vector space. But for more general rings, I have no clue. Maybe you should precise the kind of applications you have in mind. $\endgroup$ Sep 20 at 12:03
  • 2
    $\begingroup$ @RomainGicquaud: Right. What I need is for the implication to be true for all locally compact, metrizable rings. (But if you figure this out, maybe wait a day before posting, to give me a chance to assign a bounty.) $\endgroup$
    – Nik Weaver
    Sep 20 at 12:23

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