The Stacks project

Proof. We prove this by verifying condition (4) of Lemma 98.20.3. Let $(\xi _ n)$ and $(R_ n)$ be as in Remark 98.20.2 such that $\mathop{\mathrm{Ker}}(R_ m \to R_ n)$ is an ideal of square zero for all $m \geq n$. Set $A = R_1$ and $x = \xi _1$. Denote $M_ n = \mathop{\mathrm{Ker}}(R_ n \to R_1)$. Then $M_ n$ is an $A$-module. Set $R = \mathop{\mathrm{lim}}\nolimits R_ n$. Let

Then $\tilde R \to A$ is surjective with kernel $M = \prod M_ n$. There is a map $R \to \tilde R$ and a map $\tilde R \to A[M]$, $(r_1, r_2, r_3, \ldots ) \mapsto (r_1, r_2 - r_1, r_3 - r_2, \ldots )$. Together these give a short exact sequence

of deformation situations, see Remark 98.21.5. The associated sequence of kernels $0 \to \mathop{\mathrm{lim}}\nolimits M_ n \to M \to M \to 0$ is the canonical sequence computing the limit of the system of modules $(M_ n)$.

Let $o_ x(\tilde R) \in \mathcal{O}_ x(M)$ be the obstruction element. Since we have the lifts $\xi _ n$ we see that $o_ x(\tilde R)$ maps to zero in $\mathcal{O}_ x(M_ n)$. By assumption (5)(b) we see that $o_ x(\tilde R) = 0$. Choose a lift $\tilde\xi $ of $x$ to $\mathop{\mathrm{Spec}}(\tilde R)$. Let $\tilde\xi _ n$ be the restriction of $\tilde\xi $ to $\mathop{\mathrm{Spec}}(R_ n)$. There exists elements $t_ n \in T_ x(M_ n)$ such that $t_ n \cdot \tilde\xi _ n = \xi _ n$ by Lemma 98.21.2 part (2)(b). By assumption (5)(a) we can find $t \in T_ x(M)$ mapping to $t_ n$ in $T_ x(M_ n)$. After replacing $\tilde\xi $ by $t \cdot \tilde\xi $ we find that $\tilde\xi $ restricts to $\xi _ n$ over $\mathop{\mathrm{Spec}}(R_ n)$ for all $n$. In particular, since $\xi _{n + 1}$ restricts to $\xi _ n$ over $\mathop{\mathrm{Spec}}(R_ n)$, the restriction $\overline{\xi }$ of $\tilde\xi $ to $\mathop{\mathrm{Spec}}(A[M])$ has the property that it restricts to the trivial deformation over $\mathop{\mathrm{Spec}}(A[M_ n])$ for all $n$. Hence by assumption (5)(a) we find that $\overline{\xi }$ is the trivial deformation of $x$. By axiom (RS*) applied to $R = \tilde R \times _{A[M]} A$ this implies that $\tilde\xi $ is the pullback of a deformation $\xi $ of $x$ over $R$. This finishes the proof. $\square$