Lemma 16.5.1. Let R \to \Lambda be a ring map. Let I \subset R be an ideal. Assume that
I^2 = 0, and
\Lambda /I\Lambda is a filtered colimit of smooth R/I-algebras.
Let \varphi : A \to \Lambda be an R-algebra map with A of finite presentation over R. Then there exists a factorization
where B is a smooth R-algebra and J \subset IB is a finitely generated ideal.
Proof.
Choose a factorization
A/IA \to \bar B \to \Lambda /I\Lambda
with \bar B standard smooth over R/I; this is possible by assumption and Lemma 16.3.5. Write
\bar B = A/IA[t_1, \ldots , t_ r]/(\bar g_1, \ldots , \bar g_ s)
and say \bar B \to \Lambda /I\Lambda maps t_ i to the class of \lambda _ i modulo I\Lambda . Choose g_1, \ldots , g_ s \in A[t_1, \ldots , t_ r] lifting \bar g_1, \ldots , \bar g_ s. Write \varphi (g_ i)(\lambda _1, \ldots , \lambda _ r) = \sum \epsilon _{ij} \mu _{ij} for some \epsilon _{ij} \in I and \mu _{ij} \in \Lambda . Define
A' = A[t_1, \ldots , t_ r, \delta _{i, j}]/ (g_ i - \sum \epsilon _{ij} \delta _{ij})
and consider the map
A' \longrightarrow \Lambda ,\quad a \longmapsto \varphi (a),\quad t_ i \longmapsto \lambda _ i,\quad \delta _{ij} \longmapsto \mu _{ij}
We have
A'/IA' = A/IA[t_1, \ldots , t_ r]/(\bar g_1, \ldots , \bar g_ s)[\delta _{ij}] \cong \bar B[\delta _{ij}]
This is a standard smooth algebra over R/I as \bar B is standard smooth. Choose a presentation A'/IA' = R/I[x_1, \ldots , x_ n]/(\bar f_1, \ldots , \bar f_ c) with \det (\partial \bar f_ j/\partial x_ i)_{i, j = 1, \ldots , c} invertible in A'/IA'. Choose lifts f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n] of \bar f_1, \ldots , \bar f_ c. Then
B = R[x_1, \ldots , x_ n, x_{n + 1}]/ (f_1, \ldots , f_ c, x_{n + 1}\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c} - 1)
is smooth over R. Since smooth ring maps are formally smooth (Algebra, Proposition 10.138.13) there exists an R-algebra map B \to A' which is an isomorphism modulo I. Then B \to A' is surjective by Nakayama's lemma (Algebra, Lemma 10.20.1). Thus A' = B/J with J \subset IB finitely generated (see Algebra, Lemma 10.6.3).
\square
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