## 16.2 Singular ideals

Let $R \to A$ be a ring map. The singular ideal of $A$ over $R$ is the radical ideal in $A$ cutting out the singular locus of the morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$. Here is a formal definition.

Definition 16.2.1. Let $R \to A$ be a ring map. The singular ideal of $A$ over $R$, denoted $H_{A/R}$ is the unique radical ideal $H_{A/R} \subset A$ with

$V(H_{A/R}) = \{ \mathfrak q \in \mathop{\mathrm{Spec}}(A) \mid R \to A \text{ not smooth at }\mathfrak q\}$

This makes sense because the set of primes where $R \to A$ is smooth is open, see Algebra, Definition 10.137.11. In order to find an explicit set of generators for the singular ideal we first prove the following lemma.

Lemma 16.2.2. Let $R$ be a ring. Let $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\mathfrak q \subset A$ be a prime ideal. Assume $R \to A$ is smooth at $\mathfrak q$. Then there exists an $a \in A$, $a \not\in \mathfrak q$, an integer $c$, $0 \leq c \leq \min (n, m)$, subsets $U \subset \{ 1, \ldots , n\}$, $V \subset \{ 1, \ldots , m\}$ of cardinality $c$ such that

$a = a' \det (\partial f_ j/\partial x_ i)_{j \in V, i \in U}$

for some $a' \in A$ and

$a f_\ell \in (f_ j, j \in V) + (f_1, \ldots , f_ m)^2$

for all $\ell \in \{ 1, \ldots , m\}$.

Proof. Set $I = (f_1, \ldots , f_ m)$ so that the naive cotangent complex of $A$ over $R$ is homotopy equivalent to $I/I^2 \to \bigoplus A\text{d}x_ i$, see Algebra, Lemma 10.134.2. We will use the formation of the naive cotangent complex commutes with localization, see Algebra, Section 10.134, especially Algebra, Lemma 10.134.13. By Algebra, Definitions 10.137.1 and 10.137.11 we see that $(I/I^2)_ a \to \bigoplus A_ a\text{d}x_ i$ is a split injection for some $a \in A$, $a \not\in \mathfrak q$. After renumbering $x_1, \ldots , x_ n$ and $f_1, \ldots , f_ m$ we may assume that $f_1, \ldots , f_ c$ form a basis for the vector space $I/I^2 \otimes _ A \kappa (\mathfrak q)$ and that $\text{d}x_{c + 1}, \ldots , \text{d}x_ n$ map to a basis of $\Omega _{A/R} \otimes _ A \kappa (\mathfrak q)$. Hence after replacing $a$ by $aa'$ for some $a' \in A$, $a' \not\in \mathfrak q$ we may assume $f_1, \ldots , f_ c$ form a basis for $(I/I^2)_ a$ and that $\text{d}x_{c + 1}, \ldots , \text{d}x_ n$ map to a basis of $(\Omega _{A/R})_ a$. In this situation $a^ N$ for some large integer $N$ satisfies the conditions of the lemma (with $U = V = \{ 1, \ldots , c\}$). $\square$

We will use the notion of a strictly standard element in $A$ over $R$. Our notion is slightly weaker than the one in Swan's paper [swan]. We also define an elementary standard element to be one of the type we found in the lemma above. We compare the different types of elements in Lemma 16.3.7.

Definition 16.2.3. Let $R \to A$ be a ring map of finite presentation. We say an element $a \in A$ is elementary standard in $A$ over $R$ if there exists a presentation $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and $0 \leq c \leq \min (n, m)$ such that

16.2.3.1
\begin{equation} \label{smoothing-equation-elementary-standard-one} a = a' \det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c} \end{equation}

for some $a' \in A$ and

16.2.3.2
\begin{equation} \label{smoothing-equation-elementary-standard-two} a f_{c + j} \in (f_1, \ldots , f_ c) + (f_1, \ldots , f_ m)^2 \end{equation}

for $j = 1, \ldots , m - c$. We say $a \in A$ is strictly standard in $A$ over $R$ if there exists a presentation $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and $0 \leq c \leq \min (n, m)$ such that

16.2.3.3
\begin{equation} \label{smoothing-equation-strictly-standard-one} a = \sum \nolimits _{I \subset \{ 1, \ldots , n\} ,\ |I| = c} a_ I \det (\partial f_ j/\partial x_ i)_{j = 1, \ldots , c,\ i \in I} \end{equation}

for some $a_ I \in A$ and

16.2.3.4
\begin{equation} \label{smoothing-equation-strictly-standard-two} a f_{c + j} \in (f_1, \ldots , f_ c) + (f_1, \ldots , f_ m)^2 \end{equation}

for $j = 1, \ldots , m - c$.

The following lemma is useful to find implications of (16.2.3.3).

Lemma 16.2.4. Let $R$ be a ring. Let $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and write $I = (f_1, \ldots , f_ m)$. Let $a \in A$. Then (16.2.3.3) implies there exists an $A$-linear map $\psi : \bigoplus \nolimits _{i = 1, \ldots , n} A \text{d}x_ i \to A^{\oplus c}$ such that the composition

$A^{\oplus c} \xrightarrow {(f_1, \ldots , f_ c)} I/I^2 \xrightarrow {f \mapsto \text{d}f} \bigoplus \nolimits _{i = 1, \ldots , n} A \text{d}x_ i \xrightarrow {\psi } A^{\oplus c}$

is multiplication by $a$. Conversely, if such a $\psi$ exists, then $a^ c$ satisfies (16.2.3.3).

Proof. This is a special case of Algebra, Lemma 10.15.5. $\square$

Lemma 16.2.5 (Elkik). Let $R \to A$ be a ring map of finite presentation. The singular ideal $H_{A/R}$ is the radical of the ideal generated by strictly standard elements in $A$ over $R$ and also the radical of the ideal generated by elementary standard elements in $A$ over $R$.

Proof. Assume $a$ is strictly standard in $A$ over $R$. We claim that $A_ a$ is smooth over $R$, which proves that $a \in H_{A/R}$. Namely, let $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, $c$, and $a' \in A$ be as in Definition 16.2.3. Write $I = (f_1, \ldots , f_ m)$ so that the naive cotangent complex of $A$ over $R$ is given by $I/I^2 \to \bigoplus A\text{d}x_ i$. Assumption (16.2.3.4) implies that $(I/I^2)_ a$ is generated by the classes of $f_1, \ldots , f_ c$. Assumption (16.2.3.3) implies that the differential $(I/I^2)_ a \to \bigoplus A_ a\text{d}x_ i$ has a left inverse, see Lemma 16.2.4. Hence $R \to A_ a$ is smooth by definition and Algebra, Lemma 10.134.13.

Let $H_ e, H_ s \subset A$ be the radical of the ideal generated by elementary, resp. strictly standard elements of $A$ over $R$. By definition and what we just proved we have $H_ e \subset H_ s \subset H_{A/R}$. The inclusion $H_{A/R} \subset H_ e$ follows from Lemma 16.2.2. $\square$

Example 16.2.6. The set of points where a finitely presented ring map is smooth needn't be a quasi-compact open. For example, let $R = k[x, y_1, y_2, y_3, \ldots ]/(xy_ i)$ and $A = R/(x)$. Then the smooth locus of $R \to A$ is $\bigcup D(y_ i)$ which is not quasi-compact.

Lemma 16.2.7. Let $R \to A$ be a ring map of finite presentation. Let $R \to R'$ be a ring map. If $a \in A$ is elementary, resp. strictly standard in $A$ over $R$, then $a \otimes 1$ is elementary, resp. strictly standard in $A \otimes _ R R'$ over $R'$.

Proof. If $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ is a presentation of $A$ over $R$, then $A \otimes _ R R' = R'[x_1, \ldots , x_ n]/(f'_1, \ldots , f'_ m)$ is a presentation of $A \otimes _ R R'$ over $R'$. Here $f'_ j$ is the image of $f_ j$ in $R'[x_1, \ldots , x_ n]$. Hence the result follows from the definitions. $\square$

Lemma 16.2.8. Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation over $R$. Assume that $H_{A/R} \Lambda = \Lambda$. Then there exists a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$.

Proof. Choose $f_1, \ldots , f_ r \in H_{A/R}$ and $\lambda _1, \ldots , \lambda _ r \in \Lambda$ such that $\sum f_ i\lambda _ i = 1$ in $\Lambda$. Set $B = A[x_1, \ldots , x_ r]/(f_1x_1 + \ldots + f_ rx_ r - 1)$ and define $B \to \Lambda$ by mapping $x_ i$ to $\lambda _ i$. To check that $B$ is smooth over $R$ use that $A_{f_ i}$ is smooth over $R$ by definition of $H_{A/R}$ and that $B_{f_ i}$ is smooth over $A_{f_ i}$. Details omitted. $\square$

Comment #3552 by Dario Weißmann on

Typo just after the proof of lemma 07C6: ... in a $A$ over $R$.

Comment #6712 by Nana7mi on

I'm sorry but why is 'A_{f_i} is smooth over R by definition of H_{A/R}' in the proof of lemma 07EU? I can only see that if A_{f_i} is smooth over R, then f_i \in H_{A/R}.

Comment #6911 by on

Definition 16.2.1 says that $H_{A/R}$ cuts out the non-smooth locus of $A$ over $R$ in the spectrum of $A$. In order to parse the definition please read a little bit in Algebra, Section 10.137.

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).