Lemma 29.34.20. Let $f : X \to S$ be a morphism of schemes. Let $\sigma : S \to X$ be a section of $f$. Let $s \in S$ be a point such that $f$ is smooth at $x = \sigma (s)$. Then there exist affine open neighbourhoods $\mathop{\mathrm{Spec}}(A) = U \subset S$ of $s$ and $\mathop{\mathrm{Spec}}(B) = V \subset X$ of $x$ such that

1. $f(V) \subset U$ and $\sigma (U) \subset V$,

2. with $I = \mathop{\mathrm{Ker}}(\sigma ^\# : B \to A)$ the module $I/I^2$ is a free $A$-module, and

3. $B^\wedge \cong A[[x_1, \ldots , x_ d]]$ as $A$-algebras where $B^\wedge$ denotes the completion of $B$ with respect to $I$.

Proof. Pick an affine open $U \subset S$ containing $s$ Pick an affine open $V \subset f^{-1}(U)$ containing $x$. Pick an affine open $U' \subset \sigma ^{-1}(V)$ containing $s$. Note that $V' = f^{-1}(U') \cap V$ is affine as it is equal to the fibre product $V' = U' \times _ U V$. Then $U'$ and $V'$ satisfy (1). Write $U' = \mathop{\mathrm{Spec}}(A')$ and $V' = \mathop{\mathrm{Spec}}(B')$. By Algebra, Lemma 10.139.4 the module $I'/(I')^2$ is finite locally free as a $A'$-module. Hence after replacing $U'$ by a smaller affine open $U'' \subset U'$ and $V'$ by $V'' = V' \cap f^{-1}(U'')$ we obtain the situation where $I''/(I'')^2$ is free, i.e., (2) holds. In this case (3) holds also by Algebra, Lemma 10.139.4. $\square$

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