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Let $X$ be a Noetherian scheme and let $U\subseteq X$ be a dense open subset. In general, we may have

$$\text{dim}(U)<\text{dim}(X).$$

Indeed, a discrete valuation ring $R$ with maximal ideal $\mathfrak{m}=(\pi)$ has dimension $1$, but the dense open subset $D(\pi)\subseteq\text{Spec}(R)$ has dimension $0$.

Is it true, that the inequality

$$\text{dim}(U)\geq\text{dim}(X)-1$$

always holds? I suspect this is the case, but I'm not able to prove it.

donhansis
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  • I think this can fail for basic open sets in the affine setting. If we have $R$ a Noetherian ring and $\dim R > 1$, it can be the case that the non-vanishing locus $D(f)$ for some $f \in R$ can be small. For example, see this MathOverflow answer https://mathoverflow.net/a/68388/159030. I am not sure if $D(x)$ is dense in this example, however... – walkar May 07 '25 at 13:39
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    It is not dense, since $x\in R$ is idempotent and hence $D(x)=V(1-x)$ closed. – donhansis May 07 '25 at 14:14
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    If there is a counter-example, I suspect that it would have to be some non-catenary noetherian ring. Perhaps something like this can be used. – Daniel May 07 '25 at 15:02
  • I think it’s true. It’s enough to show that for any local Noetherian domain $R$ and any $f \in R$, then $R_f$ has dimension at least $\dim(R)-1$, which I think holds (although I don’t recall the proof right now…). – Aphelli May 07 '25 at 15:23
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    At any rate, Daniel’s comment is correct: by https://math.stackexchange.com/questions/798347/noetherian-ring-with-finitely-many-height-n-primes , if a counter-example exists, it is non-catenary. – Aphelli May 07 '25 at 15:58

1 Answers1

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Edit: as pointed out in the comments, this answer assumes, as the OP does, that $X$ has finite Krull dimension (otherwise the question is not well-posed). This assumption does not necessarily hold for affine integral Noetherian schemes.

However, it holds when $X$ is the spectrum of a Noetherian local ring (to which we always reduce the problem), and a byproduct of our proof is that the bound also holds when $X$ has infinite Krull dimension – ie in this case, any dense open subscheme has infinite Krull dimension as well.

Yes, the claim is true. Because $X$ is Noetherian, we may assume that it is affine, then that it is integral (by considering an irreducible component, and the dimension doesn’t see nilpotents), and that $U$ is a principal open subset.

So we are reduced to the following situation: given a Noetherian domain $R$ of dimension $d$ and a nonzero $f \in R$, does one have $\dim{R_f} \geq d-1$? We prove that this is the case by induction over $d$.

If $d \leq 1$, this is clear, so we may assume $d \geq 2$. If there is a maximal ideal $\mathfrak{m}$ of $R$ with height $d$ and not containing $f$, then $\dim{R_f} \geq \dim{R_{\mathfrak{m}}}=d$ so we are done.

Since $R$ has a maximal ideal of height $d$, we may therefore assume that $R$ is local of dimension $d$. Fix a maximal chain $\{0\}=\mathfrak{p}_0 \subset \mathfrak{p}_1 \subset \mathfrak{p}_2 \subset \ldots \subset \mathfrak{p}_d=\mathfrak{m}$ of prime ideals in $R$.

By the MSE question I linked (because $d \geq 2$), $R$ contains infinitely many primes between $\mathfrak{p}_0$ and $\mathfrak{p}_2$, and those therefore have height exactly one. Since height one primes containing $f$ are in bijection with minimal prime ideals of $R/fR$, of which there are finitely many, we can find a prime ideal $\mathfrak{q} \subset \mathfrak{p}_2$ of height one not containing $f$, but which is a piece of a chain of prime ideals in $R$ of length $d$.

Then $\dim{R/\mathfrak{q}}=d-1$, so $\dim{(R/\mathfrak{q})_f} \geq d-2$ by the induction hypothesis, ie $\dim{R_f/\mathfrak{q}R_f} \geq d-2$. Since $\mathfrak{q}R_f$ is a nonzero prime ideal in the domain $R_f$, one has $\dim{R_f} \geq 1+ \dim{R_f/\mathfrak{q}R_f} \geq 1+d-2=d-1$, QED.

Aphelli
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  • Very nice argument! Minor remark: Noetherian schemes (even irreducible affine ones) can be infinite-dimensional, so you have to phrase the claim in the second paragraph more carefully. In any case, once you reduce to the local case, this subtlety disappears. – Leobeth May 07 '25 at 23:33
  • @Leobeth: thank you! Your remark is also worth pointing out. I ignored the issue because the OP did so as well, but I shouldn’t have! – Aphelli May 08 '25 at 11:13