Lemma 16.5.1. Let $R \to \Lambda$ be a ring map. Let $I \subset R$ be an ideal. Assume that

1. $I^2 = 0$, and

2. $\Lambda /I\Lambda$ 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

$A \to B/J \to \Lambda$

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 10.136.13) 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 10.19.1). Thus $A' = B/J$ with $J \subset IB$ finitely generated (see Algebra, Lemma 10.6.3). $\square$

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