Lemma 16.2.8. Let R \to A \to \Lambda be ring maps with A of finite presentation over R. Assume that H_{A/R} \Lambda = \Lambda . Then there exists a factorization A \to B \to \Lambda with B smooth over R.
Proof. Choose f_1, \ldots , f_ r \in H_{A/R} and \lambda _1, \ldots , \lambda _ r \in \Lambda such that \sum f_ i\lambda _ i = 1 in \Lambda . Set B = A[x_1, \ldots , x_ r]/(f_1x_1 + \ldots + f_ rx_ r - 1) and define B \to \Lambda by mapping x_ i to \lambda _ i. To check that B is smooth over R use that A_{f_ i} is smooth over R by definition of H_{A/R} and that B_{f_ i} is smooth over A_{f_ i}. Details omitted. \square
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Comment #6045 by Harry Gindi on
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