1
$\begingroup$

A domain is a valuation ring if and only if the partially ordered set of ideals in this ring is totally ordered.

Does this equivalence hold when we restrict to prime ideals only? I.e. is a domain $R$ a valuation ring, if the partially ordered set $\operatorname{Spec}(R)$ is totally ordered?

$\endgroup$
4
$\begingroup$

No. Here is a counter-example deduced from Bourbaki, Commutative Algebra, Ch. V, Valuation Rings, § 4, exercise 7 d):

Let $K$ be an algebraically closed field with characteristic $0$, $B=K[X,Y]$. The polynomial $$P(X,Y)=X(X^2+Y^2)+X^2-Y^2$$ is irreducible, hence generates a prime ideal in $B$. $B/PB$ is a noetherian domain with Krull dimension $1$, so its localisation $A=\bigl(B/PB)_{(X,Y)}\;$ at the maximal ideal generated by the canonical images of $X$ and $Y$ is a local noetherian domain of dimension $1$ and henceforth $\operatorname{Spec}A$ is totally ordered by inclusion.

However this ring is not integrally closed, so it can't be a valuation ring.

$\endgroup$
  • $\begingroup$ For short, any noetherian local domain of dimension one and which is not a DVR is a counterexample. $\endgroup$ – user26857 Jun 9 '18 at 20:13
  • 1
    $\begingroup$ Yes. The single problem is to have one at hand. $\endgroup$ – Bernard Jun 9 '18 at 20:19
  • 1
    $\begingroup$ Dear Bernard, to find a local noetherian domain of dimension one which is not a DVR is the easiest task in the world: just take the local ring of a singularity of a curve. The ring $A$ of this answer is an example of this procedure. Of course Bourbaki cannot say so because He hasn't published His Magnum Opus on Algebraic Geometry... yet :-). Anyway, +1. $\endgroup$ – Georges Elencwajg Oct 19 '18 at 21:03
  • 1
    $\begingroup$ Nitpick: all fields are integrally closed, so this hypothesis is superfluous! You probably meant to write that $K$ is algebraically closed, but that is irrelevant for the answer. We don't need $\operatorname {char}(K)= 0$ either: the only needed condition on the field $K$ is $\operatorname {char}(K)\neq 2$, which ensures the irreducibility of $P$. $\endgroup$ – Georges Elencwajg Oct 19 '18 at 21:17
1
$\begingroup$

Another example of a $1$-dimensional Noetherian local domain which isn’t a valuation ring is $k[[x^2,x^3]]$, whose integral closure is $k[[x]]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.