If each subring of the fraction field of an integral domain $\mathcal{R}$ (and properly containing $\mathcal{R}$) is a PID, is $\mathcal{R}$ a PID?

Question

Let $\mathcal{R}$ be an $\textrm{integral domain}$ such that for all $\mathcal{S}\underset{\text{containing $1$}}{\underset{\text{subring}}\subseteq\textrm{Frac}(\mathcal{R}})$

with $$ \mathcal{R}\subsetneq{\mathcal{S}}\subset \textrm{Frac}{(\mathcal{R})}$$ is a PID $(\textrm{principal ideal domain})$.

Conjecture: $\mathcal{R}$ is a PID.

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We know the converse is true i.e for a PID $\mathcal{R}$ and any subring $S$ of $\textrm{Frac}(\mathcal{R}) $ containing $1$ and contained in $\mathcal{R}$ is a PID.

A particular case: Any subring $\mathcal{S}$ of $\mathbb{Q}$ containing $1$ is a PID as $\mathbb{Z}\subset S\subset \textrm{Frac}(\Bbb{Z}) =\Bbb{Q}$

Answer 1

This is incorrect. See the comments.

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your question is whether every proper localisation of $R$ being a PID means $R$ is a PID.

The localisation of a localisation is a localisation, so we can consider the partial order of all localisations of $R$, with natural homomorphisms between them. The localisation of a PID is a PID, so we know that the subset of this order which are not PIDs is downwards-closed. Your question is then whether this partially ordered subset can have a maximal element (an element that cannot be properly localised into another non-PID, but which is not a PID itself).

Assuming the $R$ we start with is not a PID, this partial order of non-PIDs will be non-empty. And if we have a chain in this partial order, the direct limit that the homomorphisms give us will be another localisation of $R$. So if the localisations of $R$ in this chain being non-PIDs means their direct limit is not a PID, that would give an upper bound on every chain in the partial order of non-PID localisations. So by Zorn's lemma there would be maximal element in this partial order (and you get one for any non-PID integral domain $R$).

Answer 2

The conjecture does not hold true. There are both Noetherian and non-Noetherian counter-examples.

Note 1. Let $R$ be an integral domain. From now on, we call a ring $S$ an overring of $R$, if $S$ contains $R$ and $S$ is in turn contained in the field of fractions of $R$.

Note 2. We shall take for granted the following facts. Every valuation domain is Bézout. A valuation domain is therefore Noetherian if and only if it is a discrete valuation domain (DVR). An overring of a valuation domain is in turn a valuation domain. If $R$ is a valuation domain of Krull dimension $0 < d < \infty$, then every proper overring of $R$ has dimension at most $d - 1$ [1, Theorem 10.1, (i) and (ii)].

As already noted by Wojowu in a comment (at the time the question was still on Mathoverflow), any non-discrete $1$-dimensional valuation domain $R$ satisfies OP's condition on overrings without being a PID. Indeed, we have:

Claim 1 [1, Exercise 10.5]. Let $R$ be an integral domain and let $K$ be its field of fractions. Then the following are equivalent:

• The ring $R$ is a valuation domain of Krull dimension $1$.
• The ring $R$ is a maximal proper subring of $K$.

If $R$ is a non-discrete $1$-dimensional valuation domain then $R$ is not a PID since it is not a DVR. Furthermore, it trivially satisfies OP's condition on overrings because of the above claim.

Example 1. [1, Exercise 10.11] Let $K$ be a field and let $X$ and $Y$ be indeterminates. Let $\alpha$ be an irrational positive real number. Let $\nu: K[X, Y] \rightarrow \mathbb{R} \cup \{\infty\}$ be the map defined by $\nu(\sum c_{m, n}X^m Y^n) = \min \{m + \alpha n \, \vert\, c_{m, n} \neq 0\}$. Then $\nu$ defines a valuation of $K(X, Y)$ with value group $\mathbb{Z} + \mathbb{Z}\alpha$. Let $R = \{ f \in K(X, Y) \, \vert \, \nu(f) \ge 0 \}$ be the valuation domain defined by $\nu$. Then $R$ has Krull dimension $1$ but its value group has rational rank $2$, hence $R$ isn't a DVR as DVRs have their value group isomorphic to $\mathbb{Z}$.

Example 2. (inspired by user Wojowu and this MSE answer) Let $K$ be a field and let $R = K[[\mathbb{Q_{\ge 0}}]]$ the ring of formal power-series of the form $\sum_{q \in \mathbb{Q}_{\ge 0}} a_q X^q \, (a_q \in K)$. Then $R$ is a $1$-dimensional valuation domain with value group $\mathbb{Q}$. In particular $R$ is not discrete. See also this MSE answer for other examples of this kind.

If $R$ is a local domain satisfying OP's condition on overrings, then localizing $R$ at prime ideals shows that the Krull dimension of $R$ is at most $2$.

Until now, the given examples are all local (valuation domains are local Bézout domains) and non-Noetherian. Here is an example that is a $1$-dimensional local and Noetherian domain, but which is not integrally closed.

Example 3 (inspired by this MSE answer). Let $K$ be field and let $R = K[[X^2, X^3]]$ be the ring of formal power-series over $K$ of the form $a + X^2f$ with $a \in K, f \in K[[X]]$. It is not difficult to check that every proper overring of $R$ contains $X$ or $X^{-1}$. This shows that $K[[X]]$ is the only proper overring of $R$ properly contained in the fraction field $K((X))$ of $R$.

In a comment, @user26857 gives a remarkable example satisfying OP's condition on overrings. It is a valuation domain of Krull dimension $2$ (and value group isomorphic to $\mathbb{Z}^2$ endowed with lexicographical order). It is therefore not a PID, although it is a Bézout domain with a principal maximal ideal:

Example 4 (@user26857). Let $p$ be a rational prime and let $\mathbb{Z}_{(p)} = \{\frac{r}{s} \, \vert \, r \in \mathbb{Z}, s \in \mathbb{Z} \setminus p\mathbb{Z} \}$. Then ring $R = \mathbb{Z}_{(p)} + X \mathbb{Q}[[X]]$ is a $2$-dimensional valuation domain with rank $2$ whose only proper overring strictly contained in the field of fractions $Q((X))$ of $R$ is $\mathbb{Q}[[X]]$, that is, a DVR. The first part of the assertion follows from [1, Theorem 10.1.(iv) and the Remark before Theorem 11.2], i.e., the construction of the composite of $\mathbb{Q}[[X]]$ and $\mathbb{Z}_{(p)}$; the second part can be checked directly.

It follows from Claim 1 that every $2$-dimensional valuation domain with value group $\mathbb{Z}^2$ (with lexicographical order) satisfies OP's condition on overrings.

Lemma 1. If $R$ is non-local and satisfies OP's condition on overrings, then $R$ is a $1$-dimensional Prüfer domain.

Proof. We claim that an overring of $R$ is integrally closed. It is obvious for proper overrings of $R$. The domain $R$ is the intersection of those overrings by [1, Theorem 4.4.7]. Localize at prime ideals to prove the assertion about the dimension.

If $R$ is a Noetherian Prüfer domain, that is a Dedekind domain, and if $R$ contains at least one non-zero prime element, then $R$ satisfies OP's condition on overrings if and only if $R$ is a PID. Indeed, the ideal class group of $S^{-1}R$ is identifies with the ideal class group of $R$ for every monoid $S$ generated by some non-zero prime elements of $R$.

Here is a Noetherian counter-example directly inspired by the Zoe Allen's answer.

Claim 2. There is a Dedekind domain $R$ satisfying OP's condition but which is not a PID.

Proof. Let $D$ be a Dedekind domain with a non-trivial finite cyclic ideal class group $C$, say $D = \mathbb{Z}[\sqrt{-5}]$ with $C \simeq \mathbb{Z}/2\mathbb{Z}$. Let $S$ be a multiplicative subset of $D$ whose intersection with every maximal ideal of $D$ is empty and which is maximal with respect to this property. (Such a set $S$ exists by Zorn's lemma; the monoid generated the units and by the non-zero prime elements of $D$ satisfies the desired property, but isn't necessarily maximal). Then $R := S^{-1}D$ is a Dedekind domain with ideal class group $C \neq \{1\}$, hence it is not a PID. By construction, $T^{-1}R$ is a PID for every multiplicative subset $T$ of $R$ which contains at least one non-zero non-unit element. As every overring of $R$ that properly contains $R$ is of this form [2, Proposition 1; the result is shown to hold more generally for Krull domains], the proof is complete.

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About the converse.

It is an easy exercise to show that:

• every overring of a Bézout domain $R$ is a localization $S^{1}R$ of $R$ with respect to some multiplicative subset of $R$.
• the localization of a Bézout domain (resp. a PID) with respect to any multiplicative subset is again a Bézout domain (resp. a PID).

The following is now immediate:

Claim 3. Let $R$ be a PID. Then every overring of $R$ is a PID.

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Towards a characterization.

Claim 4. Let $R$ be an integral domain which satisfies OP's condition on overrings. Then the Krull dimension of $R$ is at most $2$. If $R$ is local, then either

• $R$ is integrally closed, in which case $R$ is a valuation domain, or
• the integral closure of $R$ in its field of fraction is a PID, in which case $R$ is $1$-dimensional.

If $R$ is not local, then $R$ is a Prüfer domain of Krull dimension $1$. If $R$ is Noetherian and not local, then $R$ is a Dedekind domain.

Proof. If $R$ is not local, then the corresponding statement has already been proved with Lemma 1. In particular, the Krull dimension of $R$ is $1$ in this case. Assume now that $R$ is local. If $R$ is not integrally closed, then its integral closure within its field of fractions is a PID, by assumption. As PIDs are $1$-dimensional, the Krull dimension of $R$ is $1$ [1, Exercise 9.2, the dimension of an integral domain and its integral closure coincide]. If $R$ is local and integrally closed, it is a Prüfer domain since all its proper overrings are integrally closed (see the Integral closure section). The ring $R$ is therefore a valuation domain since it coincides with its localization $R_{\mathfrak{m}}$, with $\mathfrak{m}$ the maximal ideal of $R$. (The latter ring is a valuation domain, see (see the Localization section)). Localizing at non-maximal prime ideals of $R$, if any, we immediately infer that the Krull dimension of $R$ is at most $2$.

Claim 5. Every $1$-dimensional valuation domain satisfies OP's condition on overrings. A $2$-dimensional valuation domain satisfies OP's condition if and only if its value group is order isomorphic to $\mathbb{Z} \times G$, endowed with the lexicographical order, for some non-trivial additive subgroup $G$ of $\mathbb{R}$.

Claim 6. Let $R$ be a Noetherian $1$-dimensional domain which is not integrally closed. Let $\tilde{R}$ be the normalization of $R$, that is, the integral closure of $R$ with in its field of fractions. If $R$ satisfies OP's condition on overrings then $\tilde{R}$ is semiloca and the $R$-module $\tilde{R}/R$ is simple. In particular, the ring $R$ is a Bass domain (see [3, Theorem 2.1]).

To be continued.

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[1] H. Matsumura, "Commutative Ring Theory", 1986.
[2] L. Claborn, "Every abelian group is a class group", 1966.
[3] L. Levy, R. Wiegand, "Dedekind-like behavior of rings with $2$-generated ideals", 1985.