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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?

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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.

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

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  • $\begingroup$ 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 ? $\endgroup$ – Matthieu Romagny Sep 2 '19 at 20:33

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