## 15.46 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 10.109.7 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 15.46.1. 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. On the other hand J-1 implies J-0 for domains and even reduced rings as such a ring is regular at the minimal primes. Here is a characterization of the J-1 property.

Lemma 15.46.2. 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 immediately from Topology, Lemma 5.16.5. $\square$

Lemma 15.46.3. 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 15.46.2 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 10.105.3. Thus by Algebra, Lemma 10.67.6 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 10.105.7. $\square$

Lemma 15.46.4. 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 10.117.1. Then $R$ is regular by Algebra, Lemma 10.162.4. $\square$

Lemma 15.46.5. 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 10.139.9 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 10.161.10. $\square$

Lemma 15.46.6. 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 $\kappa (\mathfrak p) \subset L$ 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 15.46.3 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 10.41.4. 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 15.46.5 and our choice of $R'$. Then we apply Lemma 15.46.4 to see that $S$ is J-0 as desired. $\square$

Comment #2266 by Mattias on

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

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