# Looking for a simple one-dimensional noetherian domain whose regular locus is not open

In the wikipedia webpage for "excellent ring", one finds the following.

If R is the subring of the polynomial ring k[x1,x2,...] in infinitely many generators generated by the squares and cubes of all generators, and S is obtained from R by adjoining inverses to all elements not in any of the ideals generated by some xn, then S is a 1-dimensional Noetherian domain that is not a J-1 ring as S has a cusp singularity at every closed point, so the set of singular points is not closed, though it is a G-ring. This ring is also universally catenary, as its localization at every prime ideal is a quotient of a regular ring.

I am not sure what the "elements not in any of the ideals generated by some xn" are, because no xn lies in R. Also I am not able to prove noetherianity. In fact, I am not sure that the example has all the claimed properties.

In Exposé XIX of the volume "Travaux de Gabber" in Astérisque 363-364, there is an example of a one-dimensional noetherian domain whose regular locus is not open, with ordinary double points as singularities at closed points.

I understand this latter example, but it is much more complicated than the former and I'd really like to find an example simple enough to be presented in a colloquium-style talk.

Can anyone help me understand the wikipedia example, or find an example in the same vein?

This example appears as Example 1 in the following reference:

Melvin Hochster, Non-openness of loci in Noetherian rings, Duke Math. J. 40 (1973), 215–219. MR311653. ZBL0257.13015. DOI: 10.1215/S0012-7094-73-04020-9.

In fact, the example in question is a special case of the general result quoted below.

[…] $$K$$ denotes a field and all otherwise unspecified tensor products are taken over $$K$$. If $$R$$ is a $$K$$-algebra, we say that $$R$$ is absolutely Noetherian over $$K$$ if for every overfield $$L \supset K$$, $$L \otimes R$$ is Noetherian. If $$R$$ is a localization of a finitely generated $$K$$-algebra, then $$R$$ is absolutely Noetherian over $$K$$. $$R$$ is absolutely a domain over $$K$$ if, likewise, each $$L \otimes R$$ is a domain, and $$P$$ is absolutely prime if, equivalently, either $$R/P$$ is absolutely a domain or each $$L \otimes P$$ is prime (in $$L \otimes R$$).

[…]

Proposition 1. Let $$\{R_i\}_{i \in I}$$ be a family of absolutely Noetherian $$K$$-algebras indexed by an infinite set $$I$$, and for each $$i \in I$$, let $$P_i$$ be a nonzero absolutely prime ideal of $$R_i$$. Assume that each $$R_i$$ is absolutely a domain.

Let $$R' = \bigotimes_{i \in I} R_i$$. Then $$R'$$ is a domain and for each $$i$$, $$P_iR'$$ is prime. Moreover, if $$S = R' - (\bigcup_i P_iR')$$ and $$R = S^{-1}R'$$, then $$R$$ is a Noetherian domain whose maximal ideals are in one-to-one correspondence with $$I$$ via the map $$i \mapsto P_iR$$. In addition, each nonzero element of $$R$$ belongs to only finitely many maximal ideals, and for any maximal ideal $$P_iR$$ of $$R$$ $$R_{P_iR} \cong (L_i \otimes R_i)_{P_i^e},$$ where $$L_i$$ is a certain extension field of $$K$$ and $$P_i^e$$ is $$P_i(L_i \otimes R_i)$$.

In particular, if for each $$i \in I$$, $$R_i$$ is a subring of a polynomial ring over $$K$$ and is generated by a finite nonempty set of forms of positive degree and $$P_i$$ is the ideal generated by these forms, then the hypotheses of the first paragraph are satisfied, and the conclusions of the second paragraph hold. Moreover, the local rings of $$R$$ are algebro-geometric in this case.

In Example 1, Hochster applies the Proposition to the situation where $$I$$ is the set of positive integers and $$K$$ is an arbitrary field, in which case he sets $$R_i = K[x_i^2,x_i^3]$$ and $$P_i = (x_i^2,x_i^3)R_i$$ for each $$i$$. Then, $$R_{P_iR} \cong L_i[x_i^2,x_i^3]_{P_i^e}$$ for every $$i$$, which is a non-normal local domain of dimension one. By the Proposition these are the localizations of $$R$$ at nonzero prime ideals, and the only regular point in $$\operatorname{Spec}R$$ is the generic point corresponding to the zero ideal.

It is useful to note the following consequence of Proposition 1, also in Hochster's paper:

Proposition 2. Let $$\mathscr{P}$$ be a property of local rings and suppose there is an algebro-geometric local ring $$(R_1,P_1)$$ over a field $$K$$ such that

1. $$R_1$$ is absolutely a domain,
2. $$P_1$$ is absolutely prime, and
3. for every overfield $$L \supset K$$ $$(L \otimes R_1)_{P_1^e}$$ fails to have property $$\mathscr{P}$$.

Suppose that each field $$L \supset K$$ has $$\mathscr{P}$$. Then there is a locally algebro-geometric Noetherian domain $$R$$ over $$K$$ in which the $$\mathscr{P}$$ locus is not open.

This gives a nice way to construct rings that are locally excellent but not excellent, for instance.

• Thank you! This perfectly answers my question. – Matthieu Romagny Sep 3 '19 at 13:02

The standard method they must be referring to is this: suppose given a field $$k$$ and geometrically integral finite type $$k$$-algebras $$R_n$$ and maximal ideals $$\mathfrak m_n \subset R_n$$. Then one constructs $$R$$ as the localization of $$S = \text{colim} (R_1 \otimes_k R_2 \otimes_k R_3 \otimes_k \ldots \otimes_k R_n)$$ at the multiplicative set of elements $$f$$ which are not contained in $$\mathfrak q_i = \text{colim}_{n > i} (R_1 \otimes_k \ldots \otimes_k \mathfrak m_i \otimes_k \ldots \otimes_k R_n)$$ for any $$i$$. Note that $$\mathfrak q_i$$ is a prime ideal and that the singularity of $$S$$ at $$\mathfrak q_i$$ is (more or less) the same as the singularity of $$R_i$$ at $$\mathfrak p_i$$. Namely, $$S_{\mathfrak q_i}$$ is a localization of $$L_i \otimes_k R_{\mathfrak m_i}$$ for some geometrically integral field extension $$L_i/k$$. Every prime of $$R$$ corresponds to a prime of $$S$$ contained in $$\mathfrak q_i$$ for some $$i$$. Then it follows that $$R$$ is Noetherian because all prime ideals are finitely generated (for example).

This was explained to me by János Kollár on a hike in Utah some day (as something that commutative algebraists do). I hope others will provide references to the literature or give better answers.

• Thank you! So the wikipedia page actually means to say, "adjoining inverses to all elements not in any of the ideals $m_n=(x_n^2,x_n^3)$" right ? – Matthieu Romagny Sep 2 '19 at 20:33