The Stacks project

Lemma 87.4.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of ( Write $B = A[x_1, \ldots , x_ r]^\wedge /J$ (Lemma 87.2.2) and let $\mathop{N\! L}\nolimits _{B/A}^\wedge = (J/J^2 \to \bigoplus B\text{d}x_ i)$ be its naive cotangent complex ( The following are equivalent

  1. $B$ is rig-smooth over $(A, I)$,

  2. the object $\mathop{N\! L}\nolimits _{B/A}^\wedge $ of $D(B)$ satisfies the equivalent conditions (1) – (4) of More on Algebra, Lemma 15.84.10 with respect to the ideal $IB$,

  3. there exists a $c \geq 0$ such that for all $a \in I^ c$ there is a map $h : \bigoplus B\text{d}x_ i \to J/J^2$ such that $a : J/J^2 \to J/J^2$ is equal to $h \circ \text{d}$,

  4. there exist $b_1, \ldots , b_ s \in B$ such that $V(b_1, \ldots , b_ s) \subset V(IB)$ and such that for every $l = 1, \ldots , s$ there exist $m \geq 0$, $f_1, \ldots , f_ m \in J$, and subset $T \subset \{ 1, \ldots , n\} $ with $|T| = m$ such that

    1. $\det _{i \in T, j \leq m}(\partial f_ j/ \partial x_ i)$ divides $b_ l$ in $B$, and

    2. $b_ l J \subset (f_1, \ldots , f_ m) + J^2$.

Proof. The equivalence of (1), (2), and (3) is immediate from More on Algebra, Lemma 15.84.10.

Assume $b_1, \ldots , b_ s$ are as in (4). Since $B$ is Noetherian the inclusion $V(b_1, \ldots , b_ s) \subset V(IB)$ implies $I^ cB \subset (b_1, \ldots , b_ s)$ for some $c \geq 0$ (for example by Algebra, Lemma 10.62.4). Pick $1 \leq l \leq s$ and $m \geq 0$ and $f_1, \ldots , f_ m \in J$ and $T \subset \{ 1, \ldots , n\} $ with $|T| = m$ satisfying (4)(a) and (b). Then if we invert $b_ l$ we see that

\[ \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B_{b_ l} = \left( \bigoplus \nolimits _{j \leq m} B_{b_ l} f_ j \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} B_{b_ l} \text{d}x_ i \right) \]

and moreover the arrow is isomorphic to the inclusion of the direct summand $\bigoplus _{i \in T} B_{b_ l} \text{d}x_ i$. We conclude that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ is $b_ l$-power torsion and that $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ becomes finite free after inverting $b_ l$. Combined with the inclusion $I^ cB \subset (b_1, \ldots , b_ s)$ we see that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ is $IB$-power torsion. Hence we see that condition (4) of More on Algebra, Lemma 15.84.10 holds. In this way we see that (4) implies (2).

Assume the equivalent conditions (1), (2), and (3) hold. We will prove that (4) holds, but we strongly urge the reader to convince themselves of this. The complex $\mathop{N\! L}\nolimits _{B/A}^\wedge $ determines an object of $D^ b_{\textit{Coh}}(\mathop{\mathrm{Spec}}(B))$ whose restriction to the Zariski open $U = \mathop{\mathrm{Spec}}(B) \setminus V(IB)$ is a finite locally free module $\mathcal{E}$ placed in degree $0$ (this follows for example from the the fourth equivalent condition in More on Algebra, Lemma 15.84.10). Choose generators $f_1, \ldots , f_ M$ for $J$. This determines an exact sequence

\[ \bigoplus \nolimits _{j = 1, \ldots , M} \mathcal{O}_ U \cdot f_ j \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ U \cdot \text{d}x_ i \to \mathcal{E} \to 0 \]

Let $U = \bigcup _{l = 1, \ldots , s} U_ l$ be a finite affine open covering such that $\mathcal{E}|_{U_ l}$ is free of rank $r_ l = n - m_ l$ for some integer $n \geq m_ l \geq 0$. After replacing each $U_ l$ by an affine open covering we may assume there exists a subset $T_ l \subset \{ 1, \ldots , n\} $ such that the elements $\text{d}x_ i$, $i \in \{ 1, \ldots , n\} \setminus T_ l$ map to a basis for $\mathcal{E}|_{U_ l}$. Repeating the argument, we may assume there exists a subset $T'_ l \subset \{ 1, \ldots , M\} $ of cardinality $m_ l$ such that $f_ j$, $j \in T'_ l$ map to a basis of the kernel of $\mathcal{O}_{U_ l} \cdot \text{d}x_ i \to \mathcal{E}|_{U_ l}$. Finally, since the open covering $U = \bigcup U_ l$ may be refined by a open covering by standard opens (Algebra, Lemma 10.17.2) we may assume $U_ l = D(g_ l)$ for some $g_ l \in B$. In particular we have $V(g_1, \ldots , g_ s) = V(IB)$. A linear algebra argument using our choices above shows that $\det _{i \in T_ l, j \in T'_ l}(\partial f_ j/ \partial x_ i)$ maps to an invertible element of $B_{b_ l}$. Similarly, the vanishing of cohomology of $\mathop{N\! L}\nolimits _{B/A}^\wedge $ in degree $-1$ over $U_ l$ shows that $J/J^2 + (f_ j; j \in T')$ is annihilated by a power of $b_ l$. After replacing each $g_ l$ by a suitable power we obtain conditions (4)(a) and (4)(b) of the lemma. Some details omitted. $\square$

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