Let $A$ be an integral domain with quotient field $K$. We say that $A$ is a valuation ring if for any $0 \neq x \in K$, $x$ or $\frac{1}{x}$ lies in $A$. Then $A$ is necessarily a local ring. If $A$ is a valuation ring, then any ring $B$ between $A$ and $K$ is also valuation ring. In fact, $B$ is necessarily the localization of $A$ at one of its prime ideals (if $\mathfrak m$ is the maximal ideal of $B$, then $B$ is the localization of $A$ at $A \cap \mathfrak m$). This also implies something cool: since the ideals of a valuation ring are lineary ordered (easy to prove), so are all the intermediate rings between $A$ and $K$.

This recent question (Can a valuation ring properly contains another valuation ring with the same field of fractions?) got me thinking about something. The question I linked to is whether such a $B$ as I've indicated above, may be anything other than $A$ and $K$.

The answer to that question is yes: let $A$ be a valuation ring with $1 < \textrm{Dim } A < \infty$ (they exist), and let $\mathfrak p$ be a prime ideal of $A$ which is not maximal and not zero. Then $A_{\mathfrak p}$ cannot be equal to $A$ or $K$; if it were equal to one of those, then in particular they would have to be isomorphic, and they would then have the same Krull dimension. However, here we have $\textrm{Dim } K = 0 < \textrm{Dim } A_{\mathfrak p} < \textrm{Dim } A$.

This argument relies on the impossibility of these rings being isomorphic, because of their distinct dimensions. So my question is: what about the infinite dimensional case?

Does there exist an infinite dimensional valuation ring $A$, and a nonzero nonmaximal prime ideal $\mathfrak p$ of $A$, such that $A$ and $A_{\mathfrak p}$ are abstractly isomorphic?

First of all, for a commutative ring $A$ with infinite Krull dimension, it can certainly happen that $A_{\mathfrak p} \cong A_{\mathfrak q}$ for prime ideals $\mathfrak p \subsetneq \mathfrak q$. For example, $k$ a field, and $$A = k[X_1, X_2, ...], \mathfrak q = (X_1,X_3,X_5, ...), \mathfrak p = (X_3,X_5, ...)$$ I only know one example of an infinite dimensional valuation ring (https://math.berkeley.edu/~ogus/Math%20_256A--08/bigval.pdf), and tomorrow I'm gonna try to play around with the prime ideals of this ring. Right now, I'm too tired but I wanted to write my thoughts down in case I forgot, and also in the hope that someone who knows more about valuation rings can point me in the right direction.

  • 1
    $\begingroup$ "Abstractly" I mean some isomorphism, rather than a canonical isomorphism. $\endgroup$
    – D_S
    Mar 21, 2016 at 13:09

1 Answer 1


Yes, this can definitely happen. Let $G\mapsto A(G)$ be any operation which takes a totally ordered abelian group $G$ and gives you a valuation ring $A(G)$ with value group $G$ which is functorial with respect to isomorphisms (that is, an (order-preserving) isomorphism $G\to H$ induces an isomorphism $A(G)\to A(H)$ which turns valuations of elements $A(G)$ into valuations of elements of $A(H)$ via $\alpha$, and this respects composition; this will be satisfied by any operation that can reasonably be called a "construction on totally ordered abelian groups"). There are many such operations; for instance, you can use Hahn series with coefficients in any field.

Now suppose $G$ is a totally ordered abelian group, $S\subset G$ is an upward-closed set of positive elements such that $a+b\in S$ implies $a\in S$ or $b\in S$, and $\alpha:G\to G$ is an automorphism such that $\alpha(S)$ is a proper subset of $S$. It is easy to construct such triples $(G,S,\alpha)$. For instance, you can take $G$ to be freely generated as an abelian group by elements $g_n$ for $n\in\mathbb{Z}$, ordered such that each $g_n$ is positive and larger than every integer multiple of $g_{n-1}$. You can then let $S$ be the set of positive elements which have a nonzero coefficient of $g_n$ for some $n\geq 0$, and let $\alpha:G\to G$ be given by $\alpha(g_n)=g_{n+1}$.

Given such a triple $(G,S,\alpha)$, note that the set of elements of $A(G)$ whose valuation is in $S$ form a prime ideal $\mathfrak{p}\subset A(G)$. Furthermore, $\alpha$ induces an automorphism $\alpha_*:A(G)\to A(G)$, and $\mathfrak{q}=\alpha_*(\mathfrak{p})$ is just the set of elements of $A(G)$ whose valuation is contained in $\alpha(S)$. In particular, this means $\mathfrak{q}\subset\mathfrak{p}$. Then $\alpha_*$ induces an isomorphism between the localizations $A(G)_\mathfrak{p}$ and $A(G)_\mathfrak{q}$. But since $\mathfrak{q}\subset\mathfrak{p}$, $A(G)_\mathfrak{p}$ can itself be considered as a (nontrivial) localization of $A(G)_\mathfrak{q}$. So $A(G)_{\mathfrak{q}}$ is a valuation ring which is abstractly isomorphic to a localization of itself at a nonzero prime ideal, namely $\mathfrak{p}A(G)_\mathfrak{q}$.


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