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Everyone is familiar with schemes whose topological space consists of two points $x$, $y$ such that $x$ specializes to $y$ (e.g. any discrete valuation ring). I've been trying to come up with an example of a scheme whose topological space would consist of $3$ points such that $x\leadsto y\leadsto z$, yet couldn't. Is anyone familiar with an example of such a scheme? And if not, can it be proven that such a scheme cannot exist?

Serpentes
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1 Answers1

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You're looking for a valuation ring of rank $2$. If $A$ is a valuation ring, then the ideals of $A$ are totally ordered by inclusion, and in particular the prime ideals are totally ordered by inclusion. The total number of prime ideals of $A$ is then equal to $\mathrm{dim}(A)$ (Krull dimension), and this is exactly equal to the rank of the value group $\Gamma_A=K^\times/A^\times$ as a totally ordered group where $K$ is the fraction field of $A$. If $A$ is a DVR then the value group is isomorphic to $\mathbb{Z}$, which is rank $1$ which corresponds to the fact that $\mathrm{Spec}(A)$ has exactly two prime ideals.

So if you take a valuation ring of rank $2$, then the Krull dimension of $A$ is $2$ and the only prime ideals of $A$ form a chain $0\subset\mathfrak p\subset\mathfrak m$ with $\mathfrak m$ the maximal ideal of $A$. I won't construct a rank $2$ valuation ring myself here, this is well documented elsewhere e.g. here, so you can check out their example to get the example you're looking for.

TY Mathers
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  • Couldn't we have a bunch primes $\mathfrak p$ sitting between $0$ and $\mathfrak m$? – Chris Jun 13 '25 at 22:12
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    Nope! For a valuation ring the ideals are totally ordered, i.e. for any $I,J$ you either have $I\subseteq J$ or $J\subseteq I$, so this can't happen – TY Mathers Jun 13 '25 at 22:13