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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