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
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.
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
for some a' \in A and
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
for some a' \in A and
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
for some a_ I \in A and
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
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
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