# Equidimensional locus of a noetherian ring

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.

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

• 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! – the L Dec 15 '19 at 11:24
• 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 $V(\text{only } \mathfrak p)$). – R. van Dobben de Bruyn Dec 15 '19 at 21:13