The Stacks project

15.47 The singular locus

Let $R$ be a Noetherian ring. The regular locus $\text{Reg}(X)$ of $X = \mathop{\mathrm{Spec}}(R)$ is the set of primes $\mathfrak p$ such that $R_\mathfrak p$ is a regular local ring. The singular locus $\text{Sing}(X)$ of $X = \mathop{\mathrm{Spec}}(R)$ is the complement $X \setminus \text{Reg}(X)$, i.e., the set of primes $\mathfrak p$ such that $R_\mathfrak p$ is not a regular local ring. By the discussion preceding Algebra, Definition 10.110.7 we see that $\text{Reg}(X)$ is stable under generalization. In this section we study conditions that guarantee that $\text{Reg}(X)$ is open.


Definition 15.47.1. Let $R$ be a Noetherian ring. Let $X = \mathop{\mathrm{Spec}}(R)$.

  1. We say $R$ is J-0 if $\text{Reg}(X)$ contains a nonempty open.

  2. We say $R$ is J-1 if $\text{Reg}(X)$ is open.

  3. We say $R$ is J-2 if any finite type $R$-algebra is J-1.

The ring $\mathbf{Q}[x]/(x^2)$ does not satisfy J-0, but it does satisfy J-1. On the other hand, J-1 implies J-0 for Noetherian domains and more generally nonzero reduced Noetherian rings as such a ring is regular at the minimal primes. Here is a characterization of the J-1 property.

Lemma 15.47.2. Let $R$ be a Noetherian ring. Let $X = \mathop{\mathrm{Spec}}(R)$. The ring $R$ is J-1 if and only if $V(\mathfrak p) \cap \text{Reg}(X)$ contains a nonempty open subset of $V(\mathfrak p)$ for all $\mathfrak p \in \text{Reg}(X)$.

Proof. This follows from Topology, Lemma 5.16.5 and the fact that $\text{Reg}(X)$ is stable under generalization by Algebra, Lemma 10.110.6. $\square$

Lemma 15.47.3. Let $R$ be a Noetherian ring. Let $X = \mathop{\mathrm{Spec}}(R)$. Assume that for all primes $\mathfrak p \subset R$ the ring $R/\mathfrak p$ is J-0. Then $R$ is J-1.

Proof. We will show that the criterion of Lemma 15.47.2 applies. Let $\mathfrak p \in \text{Reg}(X)$ be a prime of height $r$. Pick $f_1, \ldots , f_ r \in \mathfrak p$ which map to generators of $\mathfrak pR_\mathfrak p$. Since $\mathfrak p \in \text{Reg}(X)$ we see that $f_1, \ldots , f_ r$ maps to a regular sequence in $R_\mathfrak p$, see Algebra, Lemma 10.106.3. Thus by Algebra, Lemma 10.68.6 we see that after replacing $R$ by $R_ g$ for some $g \in R$, $g \not\in \mathfrak p$ the sequence $f_1, \ldots , f_ r$ is a regular sequence in $R$. After another replacement we may also assume $f_1, \ldots , f_ r$ generate $\mathfrak p$. Next, let $\mathfrak p \subset \mathfrak q$ be a prime ideal such that $(R/\mathfrak p)_\mathfrak q$ is a regular local ring. By the assumption of the lemma there exists a non-empty open subset of $V(\mathfrak p)$ consisting of such primes, hence it suffices to prove $R_\mathfrak q$ is regular. Note that $f_1, \ldots , f_ r$ is a regular sequence in $R_\mathfrak q$ such that $R_\mathfrak q/(f_1, \ldots , f_ r)R_\mathfrak q$ is regular. Hence $R_\mathfrak q$ is regular by Algebra, Lemma 10.106.7. $\square$

Lemma 15.47.4. Let $R \to S$ be a ring map. Assume that

  1. $R$ is a Noetherian domain,

  2. $R \to S$ is injective and of finite type, and

  3. $S$ is a domain and J-0.

Then $R$ is J-0.

Proof. After replacing $S$ by $S_ g$ for some nonzero $g \in S$ we may assume that $S$ is a regular ring. By generic flatness we may assume that also $R \to S$ is faithfully flat, see Algebra, Lemma 10.118.1. Then $R$ is regular by Algebra, Lemma 10.164.4. $\square$

Lemma 15.47.5. Let $R \to S$ be a ring map. Assume that

  1. $R$ is a Noetherian domain and J-0,

  2. $R \to S$ is injective and of finite type, and

  3. $S$ is a domain, and

  4. the induced extension of fraction fields is separable.

Then $S$ is J-0.

Proof. We may replace $R$ by a principal localization and assume $R$ is a regular ring. By Algebra, Lemma 10.140.9 the ring map $R \to S$ is smooth at $(0)$. Hence after replacing $S$ by a principal localization we may assume that $S$ is smooth over $R$. Then $S$ is regular too, see Algebra, Lemma 10.163.10. $\square$

Lemma 15.47.6. Let $R$ be a Noetherian ring. The following are equivalent

  1. $R$ is J-2,

  2. every finite type $R$-algebra which is a domain is J-0,

  3. every finite $R$-algebra is J-1,

  4. for every prime $\mathfrak p$ and every finite purely inseparable extension $L/\kappa (\mathfrak p)$ there exists a finite $R$-algebra $R'$ which is a domain, which is J-0, and whose field of fractions is $L$.

Proof. It is clear that we have the implications (1) $\Rightarrow $ (2) and (2) $\Rightarrow $ (4). Recall that a domain which is J-1 is J-0. Hence we also have the implications (1) $\Rightarrow $ (3) and (3) $\Rightarrow $ (4).

Let $R \to S$ be a finite type ring map and let's try to show $S$ is J-1. By Lemma 15.47.3 it suffices to prove that $S/\mathfrak q$ is J-0 for every prime $\mathfrak q$ of $S$. In this way we see (2) $\Rightarrow $ (1).

Assume (4). We will show that (2) holds which will finish the proof. Let $R \to S$ be a finite type ring map with $S$ a domain. Let $\mathfrak p = \mathop{\mathrm{Ker}}(R \to S)$. Let $K$ be the fraction field of $S$. There exists a diagram of fields

\[ \xymatrix{ K \ar[r] & K' \\ \kappa (\mathfrak p) \ar[u] \ar[r] & L \ar[u] } \]

where the horizontal arrows are finite purely inseparable field extensions and where $K'/L$ is separable, see Algebra, Lemma 10.42.4. Choose $R' \subset L$ as in (4) and let $S'$ be the image of the map $S \otimes _ R R' \to K'$. Then $S'$ is a domain whose fraction field is $K'$, hence $S'$ is J-0 by Lemma 15.47.5 and our choice of $R'$. Then we apply Lemma 15.47.4 to see that $S$ is J-0 as desired. $\square$

Comments (5)

Comment #2266 by Mattias on

Missing punctuation towards the end of the first paragraph: "under generalization In the section"

Comment #6612 by on

Why is it that rings with empty Reg(X) are ruled out from being J-0? I've tried to glean from the theory here why it might be, but I think I am underinformed.

Also, doesn't the wording of Lemma 07P8 imply that J-1 rings never have an empty regular locus? (The wording: The ring R is J-1 if and only if V(𝔭)∩Reg(X) contains a nonempty open subset[...] ) I thought part of the point was that J-1 rings are allowed to have empty regular loci.

Comment #6613 by Laurent Moret-Bailly on

About the comment following Definition 07P7: J-1 implies J-0 for nonzero reduced rings. In fact, in the rest of the section at least, the J-0 condition is considered mostly for domains. This suggests that a better definition for J-0 might be that contains a dense open.

Comment #6854 by on

In response to comments #6612 and #6613: I have fixed what Laurent said. I did not make the change in definition as this would be too confusing since many people have it this way and it is how Matsumura defined these conditions -- I have added a reference to his book. As Laurent points out, there is no good reason why J-0 is the way it is; rather it is a historical artifact. I do not think that Lemma 15.47.2 is stated wrongly as it clear assumes the prime is in . So if , then there is nothing to check.

My edits are here.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07P6. Beware of the difference between the letter 'O' and the digit '0'.