Lemma 10.137.20. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a smooth ring map. Then there exists elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i} \cong S_ i/IS_ i$ for some (standard) smooth ring $S_ i$ over $R$.
Proof. By Lemma 10.137.10 we find a collection of elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i}$ is standard smooth over $R/I$. Hence we may assume that $\overline{S}$ is standard smooth over $R/I$. Write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ as in Definition 10.137.6. Choose $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$. Set $S = R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, x_{n + 1}\Delta - 1)$ where $\Delta = \det (\frac{\partial f_ j}{\partial x_ i})_{i, j = 1, \ldots , c}$ as in Example 10.137.8. This proves the lemma. $\square$
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Comment #773 by Keenan Kidwell on