Proof.
We define $\text{Inf}_ x$ as the functor
\[ \text{Mod}_ A \longrightarrow \textit{Sets},\quad M \longrightarrow \text{Inf}(x'_ M/x) = \text{Lift}(\text{id}_ x, A[M]) \]
mapping $M$ to the group of infinitesimal automorphisms of the trivial deformation $x'_ M$ of $x$ to $\mathop{\mathrm{Spec}}(A[M])$ or equivalently the group of lifts of $\text{id}_ x$ in $\mathit{Aut}_\mathcal {X}(x'_ M)$. We define $T_ x$ as the functor
\[ \text{Mod}_ A \longrightarrow \textit{Sets},\quad M \longrightarrow \text{Lift}(x, A[M]) \]
of isomorphism classes of infinitesimal deformations of $x$ to $\mathop{\mathrm{Spec}}(A[M])$. We apply Formal Deformation Theory, Lemma 90.11.4 to $\text{Inf}_ x$ and $T_ x$. This lemma is applicable, since (RS*) tells us that
\[ \textit{Lift}(x, A[M \times N]) = \textit{Lift}(x, A[M]) \times \textit{Lift}(x, A[N]) \]
as categories (and trivial deformations match up too).
Let $(x, A' \to A)$ be a deformation situation. Consider the ring map $g : A' \times _ A A' \to A[I]$ defined by the rule $g(a_1, a_2) = \overline{a_1} \oplus a_2 - a_1$. There is an isomorphism
\[ A' \times _ A A' \longrightarrow A' \times _ A A[I] \]
given by $(a_1, a_2) \mapsto (a_1, g(a_1, a_2))$. This isomorphism commutes with the projections to $A'$ on the first factor, and hence with the projections to $A$. Thus applying (RS*) twice we find equivalences of categories
\begin{align*} \textit{Lift}(x, A') \times \textit{Lift}(x, A') & = \textit{Lift}(x, A' \times _ A A') \\ & = \textit{Lift}(x, A' \times _ A A[I]) \\ & = \textit{Lift}(x, A') \times \textit{Lift}(x, A[I]) \end{align*}
Using these maps and projection onto the last factor of the last product we see that we obtain “difference maps”
\[ \text{Inf}(x'/x) \times \text{Inf}(x'/x) \longrightarrow \text{Inf}_ x(I) \quad \text{and}\quad \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow T_ x(I) \]
These difference maps satisfy the transitivity rule “$(x'_1 - x'_2) + (x'_2 - x'_3) = x'_1 - x'_3$” because
\[ \xymatrix{ A' \times _ A A' \times _ A A' \ar[rrrrr]_-{(a_1, a_2, a_3) \mapsto (g(a_1, a_2), g(a_2, a_3))} \ar[rrrrrd]_{(a_1, a_2, a_3) \mapsto g(a_1, a_3)} & & & & & A[I] \times _ A A[I] = A[I \times I] \ar[d]^{+} \\ & & & & & A[I] } \]
is commutative. Inverting the string of equivalences above we obtain an action which is free and transitive provided $\text{Inf}(x'/x)$, resp. $\text{Lift}(x, A')$ is nonempty. Note that $\text{Inf}(x'/x)$ is always nonempty as it is a group.
$\square$
Comments (0)