In An elementary characterisation of Krull dimension and A short proof for the Krull dimension of a polynomial ring, Coquand, Lombardi, and Roy give an elementary characterization of Krull dimension, which inductively makes use of one of two notions of the "boundary" of a subvariety, given as follows:

Let $R$ be a commutative ring, and $x\in R$. \begin{align*} & \operatorname{upper boundary} R^{\{x\}} \mathrel{:=} R/I^{\{x\}}, && I^{\{x\}} \mathrel{:=} xR + (\sqrt{0}:x) \\ & \operatorname{lower boundary} R_{\{x\}} \mathrel{:=} S_{\{x\}}^{-1}R, && S_{\{x\}} \mathrel{:=} x^{\mathbb{N}}(1+xR) \end{align*} where $(\sqrt{0}:x)$ is the ideal quotient of the nilradical, and $x^{\mathbb{N}}(1+xR) = \{x^n(1+rx) \mathrel\vert \text{$n\in\mathbb{N}$, $r\in R$}\}$.

Upon inspection, $\mathrm{Spec}(R^{\{x\}})$ is $V(x) \cap \overline{\mathrm{Spec}R\setminus V(x)}$, and $\mathrm{Spec}(R_{\{x\}})$ is a localization (not quite open) that is disjoint from the locus $V(x)$. Also, both are trivial exactly when $x\in R^\times \cup \sqrt{0}$.

However, I do not have good intuition for these subschemes.

  1. How to think about these boundary schemes? Do they represent anything in particular?

  2. Do these constructions appear anywhere else in the literature? I have not been able to find anything.

  3. Are they commutative, in that $R^{\{x\}\{y\}} = R^{\{y\}\{x\}}$ and $R_{\{x\}\{y\}} = R_{\{y\}\{x\}}$?

I suspect they are commutative, but am unable to prove it, and I have reservations stemming from the fact that permutations of a regular sequence are not necessarily regular.

  1. Are these very natural constructions? I.e. would it be worth studying them in more detail, in specific cases, or are they primarily instrumental in the characterization of Krull dimension?

I am willing to restrict to cases where $R$ is integral and Noetherian or has a finitely generated function field. It seems best to consider first the $R^{\{x_0\}...\{x_k\}}$, where $x_0, ..., x_k$ form a regular sequence, but I was not able to get much further with this assumption.

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    $\begingroup$ The sense in which these are 'boundaries' is purely order-theoretic (with respect to the specialization ordering on the spectrum), if that helps. I suggest working this out in the case where R is a valuation ring. $\endgroup$ Jun 15, 2020 at 20:25
  • $\begingroup$ @HarryGindi I got a sense that something like this was the case, from the first reference, but I'm not used to thinking about lattices and partial orders, so I didn't gain much from this. I know a la here this captures most of Zariski topology, but somehow that doesn't console me. $\endgroup$ Jun 15, 2020 at 20:31

2 Answers 2


This isn’t really an answer, just some musings.

First, the lower boundary. $S_{\{x\}}$ is the multiplicative set generated by the multiplicative sets $T_{\{x\}} := \{x^n : n\in \mathbb{N}\}$ and $U_{\{x\}} := \{1+rx : r\in R\}$. Localizing at a multiplicative set has the effect of “turn these elements into units”, or “cut out any prime ideals that contain these elements”. So localizing at $T_{\{x\}}$ has the effect of turning $x$ into a unit — i.e. replace $R$ with $R[x^{-1}]$. On the other hand, it is well-known that an element $x\in R$ is in the Jacobson radical of a commutative ring if and only if every element of $U_{\{x\}}$ is a unit. So localizing at $U_{\{x\}}$ has the effect of putting $x$ into the Jacobson radical of the ring — i.e. all the maximal ideals of the resulting ring will now contain $x$. This is puzzling, since it means that localizing at $S_{\{x\}}$ should simultaneously turn $x$ into a unit (i.e. throw out any prime ideals that contain it) and an element of the Jacobson radical of the ring (i.e. put it into all the maximal ideals). These would seem contradictory.

Next, the upper boundary. As far as I can tell, the effect of modding out $I^{\{x\}}$ should be to make $x$ nilpotent, since it will identify $rx^2$ with a nilpotent element for all $r\in R$. So that is like putting $x$ into all the prime ideals of $R$. I guess what it does to the prime spectrum is to cut out any prime ideals that don’t contain $x$.

Maybe it’s instructive to see what these operations do to to a prime number $p$ of $\mathbb Z$. We have $I^{\{p\}} = p{\mathbb Z}$, so ${\mathbb Z}^{\{p\}} = {\mathbb F}_p,$ the field of $p$ elements. On the other hand, $S_{\{p\}}$ inverts $p$ and also every number that is congruent to $-1$ mod $p$. By Fermat’s Little Theorem, for any prime number $q$ other than $p$, we have that $-q^{p-1}$ is such a number, whence $q$ is a unit in the resulting ring. Hence, ${\mathbb Z}_{\{p\}} = \mathbb Q$.

Next, let’s do to the element $x$ of the ring $R=\mathbb{Z}[x]$. Similar to the above, we have $I^{\{x\}}$ = xR, so $R^{\{x\}} = \mathbb Z$. On the other hand, inverting $S_{\{x\}}$ has the effect of both of inverting $x$ and inverting every polynomial in $x$ that has a constant term of $1$. Kind of a weird ring. I think it has a lot more prime ideals than $\mathbb Z$ does (like the principal ideal generated by $2x^2 + x + 2$?), but I can prove that it has Krull dimension 1 anyway.

I’m pretty sure they are commutative.

I haven’t quite seen these constructions before. They remind me a bit of computations regarding multiplicity and analytic spread.

  • $\begingroup$ Thanks. Still need to think a bit more about your remarks, but it's helpful. It doesn't change your conclusion, but wouldn't $S_{\{p\}}$ invert $p$ and the numbers congruent to $1$ mod $p$? I also suspected they were commutative, but couldn't prove, and had reservations relating to regular sequences (I'm editing to add). $\endgroup$ Jun 15, 2020 at 19:54

This is also not really an answer, but it explains an alternative version of the characterisation that I find easier to work with. Let $P_d(R)$ be the polynomial ring over $R$ in variables $x_0,\dotsc,x_d$, and order the monomials lexicographically. Say that $f\in P_d(R)$ is comonic if the lowest monomial has coefficient one. Say that an $R$-algebra homomorphism $\phi\colon P_d(R)\to R$ is thin if the kernel contains a comonic polynomial. After a little translation, the results of Coquand and Lombardi say that $R$ has dimension $\leq d$ iff every such homomorphism is thin.


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