Lemma 16.2.2 (07C6)—The Stacks project

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\} $). $\square$

Comments (6)

Comment #3272 by Dario Weißmann on April 26, 2018 at 14:46

Typo: $a\notin \mathfrak{p}$ should be $a\notin \mathfrak{q}$ .

Comment #3364 by Johan on May 19, 2018 at 17:13

Thanks and fixed here.

Comment #3628 by Kestutis Cesnavicius on October 17, 2018 at 04:37

In the statement of the lemma, one should mention that $\mathfrak{q}$ is a prime ideal of $A$ .

Comment #3728 by Johan on October 23, 2018 at 18:18

Thanks and fixed here.

Comment #6044 by Harry Gindi on April 09, 2021 at 07:36

Should the statement say there exists a in A, a not in q, c≤min(n,m), subsets of cardinality c, etc, such that if ã is a lift of a to the polynomial ring, then ã satisfies the above properties in terms of the jacobian determinant and where ãf_ℓ belongs to the above ideal?

If you don't say something like 'a lift ã of a', you're multiplying an element of the quotient by a polynomial in the polynomial ring, so it doesn't typecheck.

Comment #6183 by Johan on May 01, 2021 at 23:47

I think rather that the equalities are taking place in $A$ -modules and hence they make sense. OK?

There are also:

4 comment(s) on Section 16.2: Singular ideals

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