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Let $A$ be a commutative noetherian ring of finite Krull dimension.

Is the set $W = \{p \in Spec(A) \mid A_p \mbox{ is equidimensional}\}$ a dense open subset of $Spec(A)$?

(I guess dense is trivial, because it contains all minimal primes).

And if yes, can I conclude that this $W$ is covered by open sets $D(f)$ such that the ring $A_f$ is equidimensional? (or maybe a smaller dense open subset contained in $W$?)

If it helps, I can also assume that $A$ is catenary, and even universally catenary.

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This is true if $A$ is catenary. We will use that a catenary Noetherian local ring $A$ has a dimension function $d_A \colon \operatorname{Spec} A \to \mathbf Z$ [Stacks, Tags 02I8 and 0ECF]; such a function is well-defined up to adding a constant. Then $A$ is equidimensional if and only if $d_A(\mathfrak p_1) = d_A(\mathfrak p_2)$ for any two minimal prime ideals $\mathfrak p_1,\mathfrak p_2$ of $A$.

However, even for $A$ essentially of finite type over a field, it is not true that $W$ can be covered by standard opens that are spectra of equidimensional rings. I construct one such example below.

It seems likely that there are counterexamples if you drop the catenary hypothesis, but it is a bit hard to make examples. The most useful tool is [Doering–Lequain, Thm. B]; see also [Heinrich] for similar arguments to construct interesting containment graphs of primes in Noetherian rings. (My example below is inspired by one of Heinrich's constructions.)

Definition. For a collection $I$ of minimal primes of $A$, write $Z_I$ for the locus of primes $\mathfrak q$ such that there exist $\mathfrak p_1, \mathfrak p_2 \in I$ with $\mathfrak p_1, \mathfrak p_2 \subseteq \mathfrak q$ and $d_{A_{\mathfrak q}}(\mathfrak p_1) \neq d_{A_{\mathfrak q}}(\mathfrak p_2)$.

If $I$ is the full set of minimal primes of $A$, then simply write $Z = Z_I$. Then this is the complement of the equidimensional locus $W$. Note also that $$Z = \bigcup_{\substack{\mathfrak p_1,\mathfrak p_2\\\text{minimal}}} Z_{\{\mathfrak p_1,\mathfrak p_2\}}.$$ We will also write $Z_{\mathfrak p_1,\mathfrak p_2}$ for $Z_{\{\mathfrak p_1,\mathfrak p_2\}}$.

Lemma. Let $A$ be a catenary Noetherian ring, let $\mathfrak p_1, \mathfrak p_2 \subseteq A$ be minimal prime ideals, and let $I = \mathfrak p_1 + \mathfrak p_2$. Let $\mathfrak q \subseteq \mathfrak r$ be prime ideals in $V(I)$. Then $\mathfrak q \in Z_{\mathfrak p_1,\mathfrak p_2}$ if and only if $\mathfrak r \in Z_{\mathfrak p_1,\mathfrak p_2}$.

Proof. The restriction of the dimension function $d_{A_\mathfrak r}$ to $\operatorname{Spec} A_{\mathfrak q}$ is a dimension function. Thus, we see that $d_{A_{\mathfrak r}}(\mathfrak p_1) \neq d_{A_{\mathfrak r}}(\mathfrak p_2)$ if and only if $d_{A_{\mathfrak q}}(\mathfrak p_1) \neq d_{A_{\mathfrak q}}(\mathfrak p_2)$. $\square$

Corollary. In the situation of the lemma, $Z_{\mathfrak p_1,\mathfrak p_2}$ is an open and closed subset of $V(I)$.

Proof. Note that $Z_{\mathfrak p_1,\mathfrak p_2} \subseteq V(\mathfrak p_1) \cap V(\mathfrak p_2) = V(I)$ by definition of $Z_{\mathfrak p_1,\mathfrak p_2}$. Moreover, the lemma shows that $Z_{\mathfrak p_1,\mathfrak p_2}$ is stable under both generalisation and specialisation of points within $V(I)$. This implies it is open and closed in $V(I)$.

(For example, if $\mathfrak q \in Z_{\mathfrak p_1,\mathfrak p_2}$ and $\mathfrak r \in V(I) \setminus Z_{\mathfrak p_1,\mathfrak p_2}$, then the closed sets $V(\mathfrak q)$ and $V(\mathfrak r)$ have to be disjoint, showing that $D(\mathfrak q)$ is an open neighbourhood of $\mathfrak q$ in $Z_{\mathfrak p_1,\mathfrak p_2}$ since $\mathfrak r \in V(I)\setminus Z_{\mathfrak p_1,\mathfrak p_2}$ was arbitrary. This shows $Z_{\mathfrak p_1,\mathfrak p_2}$ is open in $V(I)$, and similarly the complement $V(I) \setminus Z_{\mathfrak p_1,\mathfrak p_2}$ is open in $V(I)$.) $\square$

Corollay. Let $A$ be a catenary Noetherian ring. Then the equidimensional locus $W \subseteq \operatorname{Spec} A$ is a dense open.

Proof. By definition, the complement $Z = \operatorname{Spec} A \setminus W$ is the finite union of $Z_{\mathfrak p_1,\mathfrak p_2}$. By the corollary above, each $Z_{\mathfrak p_1,\mathfrak p_2}$ is closed in $\operatorname{Spec} A$, hence so is their union. Then $W$ is open, and it is dense because it contains the open loci $$V\Big(\text{only } \mathfrak p\Big) := D\left(\bigcap_{\substack{\mathfrak q \subseteq A \text{ minimal}\\\mathfrak q \neq \mathfrak p}} \mathfrak q\right)$$ of primes $\mathfrak r$ such that $\mathfrak p$ is the only minimal prime contained in $\mathfrak r$, for any minimal prime $\mathfrak p \subseteq A$. $\square$


The open set $\bigcup_{\mathfrak p} V(\text{only } \mathfrak p) \subseteq W$ is clearly covered by spectra of equidimensional rings, since it a disjoint union of irreducible components (we exactly removed all intersections between irreducible components of $\operatorname{Spec} A$). However, it is not always possible to cover $W$ by spectra of equidimensional rings:

Example. Let $C = k[u,v,w,x,y]/(uy)$ be the union of $\mathbf A^4_{u,v,w,x} = V(y)$ and $\mathbf A^4_{v,w,x,y} = V(u)$ along $\mathbf A^3_{v,w,x} = V(u,y)$. Let $B = S^{-1}C$, where $S$ is the complement of $(u,y) \cup (u-1,v,w,x) \cup (w,x,y-1)$. Then $B$ is a semilocal ring with three maximal ideals $(u,y)$, $(u-1,v,w,x)$, and $(w,x,y-1)$ of height $1$, $4$ and $3$ respectively. The irreducible components $V(y)$ and $V(u)$ of $\operatorname{Spec} A$ have dimension $4$ and $3$ respectively, and are glued along the closed point $V(u,y)$ that has codimension $1$ in both components.

Finally, let $A = B_b$ for some element $b \in B$ vanishing at $(u-1,v,w,x)$ and $(w,x,y-1)$ but not $(u,y)$; for example $b = (u-1)(y-1)$. Then $\operatorname{Spec} A$ has irreducible components $V(y)$ and $V(u)$ of dimension $3$ and $2$ respectively, and they meet in $V(u,y)$ which has codimension $1$ in each component.

Since $A$ is a localisation of the equidimensional ring $C$, all localisations $A_{\mathfrak p}$ are equidimensional. In particular, $W = \operatorname{Spec} A$. However, any open neighbourhood of $V(u,y)$ meets infinitely many curves through $(u-1,v,w,x)$ in the first copy of $\mathbf A^4$ and infinitely many surfaces containing $(w,x,y-1)$ in the second copy of $\mathbf A^4$. In other words, $U$ contains infinitely many closed points of maximal height in each component of $\operatorname{Spec} A$. Therefore, $U$ has one component $U \cap V(y)$ of dimension $3$ and one component $U \cap V(u)$ of dimension $2$. $\square$

What's going on is the following: we can locally extend the dimension function to a function $d \colon \operatorname{Spec} A \to \mathbf Z$ [Stacks, Tag 02IC]; in this example we can do it globally because $A$ is essentially of finite type over $k$. The arguments above then show that on each connected component of $W$ we must have $d(\mathfrak p_1) = d(\mathfrak p_2)$ for any minimal primes $\mathfrak p_1, \mathfrak p_2$. The dimension is recovered as $\max(d(\mathfrak p)-d(\mathfrak q)\ |\ \mathfrak p \subseteq \mathfrak q)$. However, we do not know what the values of the dimension function at closed points is.

In the example above we constructed two components that meet in a point having the same codimension in each (forcing $d(\mathfrak p_1) = d(\mathfrak p_2)$), but the two components have closed points of different heights.


References.

[Doering–Lequain] A. M. d. S. Doering; Y. Lequain, The gluing of maximal ideals. Spectrum of a Noetherian ring. Going up and going down in polynomial rings, Trans. Am. Math. Soc. 260, p. 583-593 (1980). Available online through the AMS.

[Heinrich] K. Heinrich, Some remarks on biequidimensionality of topological spaces and Noetherian schemes, J. Commut. Algebra 9.1, p. 49-63 (2017). arXiv:1403.5814

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  • $\begingroup$ Thanks! This is great. Just to be clear, so for a catenary ring, W always contains a smaller dense open set which is covered by spectra of equidimensional rings? I will study you answer and then accept it. This looks great! $\endgroup$ – the L Dec 15 '19 at 11:24
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    $\begingroup$ Ah, actually for that statement you don't need catenary. Any Noetherian scheme has a dense open subset that is a disjoint union of irreducible components. Just remove all the intersections of pairs of components (this is my set $V(\text{only } \mathfrak p)$). $\endgroup$ – R. van Dobben de Bruyn Dec 15 '19 at 21:13

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