The Stacks project

92.16 Deformations of ring maps and the cotangent complex

This section is the continuation of Deformation Theory, Section 91.2 which we urge the reader to read first. We start with a surjective ring map $A' \to A$ whose kernel is an ideal $I$ of square zero. Moreover we assume given a ring map $A \to B$, a $B$-module $N$, and an $A$-module map $c : I \to N$. In this section we ask ourselves whether we can find the question mark fitting into the following diagram
\begin{equation} \label{cotangent-equation-to-solve} \vcenter { \xymatrix{ 0 \ar[r] & N \ar[r] & {?} \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[u]^ c \ar[r] & A' \ar[u] \ar[r] & A \ar[u] \ar[r] & 0 } } \end{equation}

and moreover how unique the solution is (if it exists). More precisely, we look for a surjection of $A'$-algebras $B' \to B$ whose kernel is an ideal of square zero and is identified with $N$ such that $A' \to B'$ induces the given map $c$. We will say $B'$ is a solution to (

Lemma 92.16.1. In the situation above we have

  1. There is a canonical element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^2_ B(L_{B/A}, N)$ whose vanishing is a sufficient and necessary condition for the existence of a solution to (

  2. If there exists a solution, then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{B/A}, N)$.

  3. Given a solution $B'$, the set of automorphisms of $B'$ fitting into ( is canonically isomorphic to $\mathop{\mathrm{Ext}}\nolimits ^0_ B(L_{B/A}, N)$.

Proof. Via the identifications $\mathop{N\! L}\nolimits _{B/A} = \tau _{\geq -1}L_{B/A}$ (Lemma 92.11.3) and $H^0(L_{B/A}) = \Omega _{B/A}$ (Lemma 92.4.5) we have seen parts (2) and (3) in Deformation Theory, Lemmas 91.2.1 and 91.2.2.

Proof of (1). Roughly speaking, this follows from the discussion in Deformation Theory, Remark 91.2.8 by replacing the naive cotangent complex by the full cotangent complex. Here is a more detailed explanation. By Deformation Theory, Lemma 91.2.7 and Remark 91.2.8 there exists an element

\[ \xi ' \in \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/A'}, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{A/A'} \otimes _ A^\mathbf {L} B, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \]

(for the equalities see Deformation Theory, Remark 91.2.8 and use that $\mathop{N\! L}\nolimits _{A'/A} = \tau _{\geq -1} L_{A'/A}$) such that a solution exists if and only if this element is in the image of the map

\[ \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A'}, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{B/A'}, N) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \]

The distinguished triangle ( for $A' \to A \to B$ gives rise to a long exact sequence

\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{B/A'}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \to \mathop{\mathrm{Ext}}\nolimits ^2_ B(L_{B/A}, N) \to \ldots \]

Hence taking $\xi $ the image of $\xi '$ works. $\square$

Comments (2)

Comment #7112 by Amnon Yekutieli on

A minor comment: I believe the ideal of the ring should be square zero.

The same condition is on the ideals and in (this diagram), but there the discussion is of first order thickenings, which is another way of saying that these ideals have square zero.

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