Affine square-zero lifting obstruction, torsor, and automorphisms

Present the surjection $A^{\mathrm{\prime }}\to A$ by the polynomial $A^{\mathrm{\prime }}$-algebra with no variables, namely $A^{\mathrm{\prime }}$ itself. Its conormal complex is $[I\mathrm{/}{I}^2\to 0]=I[1]$. Thus $L_{A\mathrm{/}{A}^{\mathrm{\prime }}}$ has $H^0=0$, $H^{-1}=I$ and lies in degrees at most $-1$. Since $B$ is flat over $A$, $C$ also lies in those degrees and $H^{-1}(C)=B{\otimes }_{A}I$. Derived Hom into $N[0]$ consequently gives ${\mathrm{Ext}}_B^0(C,N)=0$ and ${\mathrm{Ext}}_B^1(C,N)={\mathrm{Hom}}_B(B{\otimes }_{A}I,N)$. Indeed its part in degrees at most $-2$ has Hom into an injective resolution of $N$ zero in degrees below two; truncation at $-1$ therefore gives precisely these assertions. Let $\gamma$ be the latter Hom element $b\otimes i\mapsto bc(i)$. This map is well defined by the assumed $A$-linearity.

By [F1], classes in ${\mathrm{Ext}}_B^1(K,N)$ are exactly square-zero $A^{\mathrm{\prime }}$-algebra extensions of $B$ by the specified $N$. The map to ${\mathrm{Ext}}_B^1(C,N)$ induced by transitivity records their induced map on $I$. To check this explicitly, pull an extension back along $A\to B$. Present $A$ over $A^{\mathrm{\prime }}$ using the no-variable presentation from step 1.1; evaluation of $i\in I$ is its image in the extension kernel. The conormal description in [F1] therefore gives exactly $i\mapsto c_E(i)$, or $b\otimes i\mapsto b{c}_E(i)$ after extending coefficients to $B$. The naturality calculation in [F1] shows this is the actual canonical cotangent map, independent of the presentation. Hence the required marked classes are exactly the fiber over $\gamma$, with no additional map choices or compatibility conditions left over.

The canonical triangle is $C\to K\to L\to C[1]$. Apply derived Hom into $N$. Its long exact sequence contains ${\mathrm{Ext}}_B^1(K,N)\to {\mathrm{Ext}}_B^1(C,N)\stackrel{\mathrm{\partial }}{\to }{\mathrm{Ext}}_B^2(L,N)$. Define $o=\mathrm{\partial }\gamma$. Exactness says $o=0$ if and only if the fiber in step 2.1 is nonempty, proving both directions of the existence assertion. This constructs the obstruction using the full cotangent complex; the generally false substitution of the two-term naive complex into an untruncated transitivity triangle has not been made. Canonical transitivity and the comparison in [F2], together with the explicit restriction computation of step 2.1, prove presentation independence of $o$.

By [F1] an automorphism is $e\mapsto e+D(\bar{e})$, with $D:B\to N$ an $A^{\mathrm{\prime }}$-derivation. Such derivations are exactly $A$-derivations because the structural map $A^{\mathrm{\prime }}\to B$ factors through the surjection $A^{\mathrm{\prime }}\to A$. Composition adds $D$ and the inverse uses $-D$. The universal property of differentials gives ${\mathrm{Der}}_A(B,N)={\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}A},N)={\mathrm{Ext}}_B^0(L,N)$ by [F3]. This proves the asserted automorphism group with its group law, and it fixes the prescribed kernel pointwise.

The preceding part of that sequence is ${\mathrm{Ext}}_B^0(C,N)\to {\mathrm{Ext}}_B^1(L,N)\to {\mathrm{Ext}}_B^1(K,N)\to {\mathrm{Ext}}_B^1(C,N)$. Its first term vanishes by step 1.1, so ${\mathrm{Ext}}_B^1(L,N)$ injects and its image is exactly the kernel of restriction. Translation in this abelian group therefore acts freely and transitively on each nonempty fiber. Under [F1] it is the action obtained by adding square-zero extension classes through the fiber-product and kernel-sum construction. Thus the marked isomorphism classes form precisely the stated torsor, rather than a quotient of that group.

In the flat-deformation specialization the required kernel is $N=B{\otimes }_{A}I$ and $\gamma$ is the identity of this module. For any flat $A^{\mathrm{\prime }}$-algebra lift $E$ with $E\mathrm{/}IE=B$, tensor $0\to I\to A^{\mathrm{\prime }}\to A\to 0$ over $A^{\mathrm{\prime }}$ with $E$: exactness gives $0\to I{\otimes }_{A^{\mathrm{\prime }}}E\to E\to B\to 0$. Here $I{\otimes }_{A^{\mathrm{\prime }}}E=I{\otimes }_{A}B$ because $I^2=0$, and the structure map sends $i$ to $1\otimes i$. This proves the precise kernel specialization. It does not presume a lift exists: existence is decided by step 3.1. In particular no expression $B{\otimes }_A{A}^{\mathrm{\prime }}$ is used, since the given surjection need not have an $A$-algebra section. The zero-kernel and zero-module cases are included by the same exact sequence and zero fiber. All assertions follow.