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