The Stacks project

Proof. If we can prove the lemma in case $J = (h)$, then we can prove the lemma by induction on the number of generators of $J$. Namely, suppose that $J$ can be generated by $n$ elements $h_1, \ldots , h_ n$ and the lemma holds for all cases where $J$ is generated by $n - 1$ elements. Then we apply the case $n = 1$ to produce $B \to B' \to \Lambda $ where the first map kills of $h_ n$. Then we let $J'$ be the ideal of $B'$ generated by the images of $h_1, \ldots , h_{n - 1}$ and we apply the case for $n - 1$ to produce $B' \to B'' \to \Lambda $. It is easy to verify that $B \to B'' \to \Lambda $ does the job.

Assume $J = (h)$ and write $h = \sum \epsilon _ i b_ i$ for some $\epsilon _ i \in I$ and $b_ i \in B$. Note that $0 = \varphi (h) = \sum \epsilon _ i \varphi (b_ i)$. As $\Lambda $ is flat over $R$, the equational criterion for flatness (Algebra, Lemma 10.39.11) implies that we can find $\lambda _ j \in \Lambda $, $j = 1, \ldots , m$ and $a_{ij} \in R$ such that $\varphi (b_ i) = \sum _ j a_{ij} \lambda _ j$ and $\sum _ i \epsilon _ i a_{ij} = 0$. Set

with $C \to \Lambda $ given by $\varphi $ and $x_ j \mapsto \lambda _ j$. Choose a factorization

as in Lemma 16.5.1. Since $B$ is smooth over $R$ we can lift the map $B \to C \to B'/J'$ to a map $\psi : B \to B'$. We claim that $\psi (h) = 0$. Namely, the fact that $\psi $ agrees with $B \to C \to B'/J'$ mod $I$ implies that

for some $\xi _ i \in B'$ and $\theta _ i \in IB'$. Hence we see that

because of the relations above and the fact that $I^2 = 0$. $\square$