Lemma 16.5.2. Let $R \to \Lambda$ be a ring map. Let $I \subset R$ be an ideal. Assume that

1. $I^2 = 0$,

2. $\Lambda /I\Lambda$ is a filtered colimit of smooth $R/I$-algebras, and

3. $R \to \Lambda$ is flat.

Let $\varphi : B \to \Lambda$ be an $R$-algebra map with $B$ smooth over $R$. Let $J \subset IB$ be a finitely generated ideal such that $\varphi (J) = 0$. Then there exists $R$-algebra maps

$B \xrightarrow {\alpha } B' \xrightarrow {\beta } \Lambda$

such that $B'$ is smooth over $R$, such that $\alpha (J) = 0$ and such that $\beta \circ \alpha = \varphi \bmod I\Lambda$.

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 10.38.11) 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

$C = B[x_1, \ldots , x_ m]/(b_ i - \sum a_{ij} x_ j)$

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

$C \to B'/J' \to \Lambda$

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

$\psi (b_ i) = \sum a_{ij} \xi _ j + \theta _ i$

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

$\psi (h) = \psi (\sum \epsilon _ i b_ i) = \sum \epsilon _ i a_{ij} \xi _ j + \sum \epsilon _ i \theta _ i = 0$

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

Comment #4044 by Kęstutis Česnavičius on

The assumption $\varphi(J) = 0$ seems to be missing from the statement. It is used in the proof.

Comment #4134 by on

Thank you very much for pointing this out. The fix is here.

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