## 16.5 The lifting problem

The goal in this section is to prove (Proposition 16.5.3) that the collection of algebras which are filtered colimits of smooth algebras is closed under infinitesimal flat deformations. The proof is elementary and only uses the results on presentations of smooth algebras from Section 16.3.

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

\[ A \to B/J \to \Lambda \]

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).
$\square$

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

$I^2 = 0$,

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

$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 $I^2 = 0$.
$\square$

slogan
Proposition 16.5.3. Let $R \to \Lambda $ be a ring map. Let $I \subset R$ be an ideal. Assume that

$I$ is nilpotent,

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

$R \to \Lambda $ is flat.

Then $\Lambda $ is a filtered colimit of smooth $R$-algebras.

**Proof.**
Since $I^ n = 0$ for some $n$, it follows by induction on $n$ that it suffices to consider the case where $I^2 = 0$. Let $\varphi : A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. We have to find a factorization $A \to B \to \Lambda $ with $B$ smooth over $R$, see Algebra, Lemma 10.126.4. By Lemma 16.5.1 we may assume that $A = B/J$ with $B$ smooth over $R$ and $J \subset IB$ a finitely generated ideal. By Lemma 16.5.2 we can find a (possibly noncommutative) diagram

\[ \xymatrix{ B \ar[rr]_\alpha \ar[rd]_\varphi & & B' \ar[ld]^\beta \\ & \Lambda } \]

of $R$-algebras which commutes modulo $I$ and such that $\alpha (J) = 0$. The map

\[ D : B \longrightarrow I\Lambda ,\quad b \longmapsto \varphi (b) - \beta (\alpha (b)) \]

is a derivation over $R$ hence we can write it as $D = \xi \circ \text{d}_{B/R}$ for some $B$-linear map $\xi : \Omega _{B/R} \to I\Lambda $. Since $\Omega _{B/R}$ is a finite projective $B$-module we can write $\xi = \sum _{i = 1, \ldots , n} \epsilon _ i \Xi _ i$ for some $\epsilon _ i \in I$ and $B$-linear maps $\Xi _ i : \Omega _{B/R} \to \Lambda $. (Details omitted. Hint: write $\Omega _{B/R}$ as a direct sum of a finite free module to reduce to the finite free case.) We define

\[ B'' = \text{Sym}^*_{B'}\left(\bigoplus \nolimits _{i = 1, \ldots , n} \Omega _{B/R} \otimes _{B, \alpha } B'\right) \]

and we define $\beta ' : B'' \to \Lambda $ by $\beta $ on $B'$ and by

\[ \beta '|_{i\text{th summand }\Omega _{B/R} \otimes _{B, \alpha } B'} = \Xi _ i \otimes \beta \]

and $\alpha ' : B \to B''$ by

\[ \alpha '(b) = \alpha (b) \oplus \sum \epsilon _ i \text{d}_{B/R}(b) \otimes 1 \oplus 0 \oplus \ldots \]

At this point the diagram

\[ \xymatrix{ B \ar[rr]_{\alpha '} \ar[rd]_\varphi & & B'' \ar[ld]^{\beta '} \\ & \Lambda } \]

does commute. Moreover, it is direct from the definitions that $\alpha '(J) = 0$ as $I^2 = 0$. Hence the desired factorization.
$\square$

## Comments (0)