The Stacks project

16.6 The lifting lemma

Here is a fiendishly clever lemma.

Lemma 16.6.1. Let $R$ be a Noetherian ring. Let $\Lambda $ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$ and $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. Suppose we have $R$-algebra maps $R/\pi ^2R \to \bar C \to \Lambda /\pi ^2\Lambda $ with $\bar C$ of finite presentation. Then there exists an $R$-algebra homomorphism $D \to \Lambda $ and a commutative diagram

\[ \xymatrix{ R/\pi ^2R \ar[r] \ar[d] & \bar C \ar[r] \ar[d] & \Lambda /\pi ^2\Lambda \ar[d] \\ R/\pi R \ar[r] & D/\pi D \ar[r] & \Lambda /\pi \Lambda } \]

with the following properties

  1. $D$ is of finite presentation,

  2. $R \to D$ is smooth at any prime $\mathfrak q$ with $\pi \not\in \mathfrak q$,

  3. $R \to D$ is smooth at any prime $\mathfrak q$ with $\pi \in \mathfrak q$ lying over a prime of $\bar C$ where $R/\pi ^2 R \to \bar C$ is smooth, and

  4. $\bar C/\pi \bar C \to D/\pi D$ is smooth at any prime lying over a prime of $\bar C$ where $R/\pi ^2R \to \bar C$ is smooth.

Proof. We choose a presentation

\[ \bar C = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m) \]

We also denote $I = (f_1, \ldots , f_ m)$ and $\bar I$ the image of $I$ in $R/\pi ^2R[x_1, \ldots , x_ n]$. Since $R$ is Noetherian, so is $\bar C$. Hence the smooth locus of $R/\pi ^2 R \to \bar C$ is quasi-compact, see Topology, Lemma 5.9.2. Applying Lemma 16.2.2 we may choose a finite list of elements $a_1, \ldots , a_ r \in R[x_1, \ldots , x_ n]$ such that

  1. the union of the open subspaces $\mathop{\mathrm{Spec}}(\bar C_{a_ k}) \subset \mathop{\mathrm{Spec}}(\bar C)$ cover the smooth locus of $R/\pi ^2 R \to \bar C$, and

  2. for each $k = 1, \ldots , r$ there exists a finite subset $E_ k \subset \{ 1, \ldots , m\} $ such that $(\bar I/\bar I^2)_{a_ k}$ is freely generated by the classes of $f_ j$, $j \in E_ k$.

Set $I_ k = (f_ j, j \in E_ k) \subset I$ and denote $\bar I_ k$ the image of $I_ k$ in $R/\pi ^2R[x_1, \ldots , x_ n]$. By (2) and Nakayama's lemma we see that $(\bar I/\bar I_ k)_{a_ k}$ is annihilated by $1 + b'_ k$ for some $b'_ k \in \bar I_{a_ k}$. Suppose $b'_ k$ is the image of $b_ k/(a_ k)^ N$ for some $b_ k \in I$ and some integer $N$. After replacing $a_ k$ by $a_ kb_ k$ we get

  1. $(\bar I_ k)_{a_ k} = (\bar I)_{a_ k}$.

Thus, after possibly replacing $a_ k$ by a high power, we may write

  1. $a_ k f_\ell = \sum \nolimits _{j \in E_ k} h_{k, \ell }^ jf_ j + \pi ^2 g_{k, \ell }$

for any $\ell \in \{ 1, \ldots , m\} $ and some $h_{i, \ell }^ j, g_{i, \ell } \in R[x_1, \ldots , x_ n]$. If $\ell \in E_ k$ we choose $h_{k, \ell }^ j = a_ k\delta _{\ell , j}$ (Kronecker delta) and $g_{k, \ell } = 0$. Set

\[ D = R[x_1, \ldots , x_ n, z_1, \ldots , z_ m]/ (f_ j - \pi z_ j, p_{k, \ell }). \]

Here $j \in \{ 1, \ldots , m\} $, $k \in \{ 1, \ldots , r\} $, $\ell \in \{ 1, \ldots , m\} $, and

\[ p_{k, \ell } = a_ k z_\ell - \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j z_ j - \pi g_{k, \ell }. \]

Note that for $\ell \in E_ k$ we have $p_{k, \ell } = 0$ by our choices above.

The map $R \to D$ is the given one. Say $\bar C \to \Lambda /\pi ^2\Lambda $ maps $x_ i$ to the class of $\lambda _ i$ modulo $\pi ^2$. For an element $f \in R[x_1, \ldots , x_ n]$ we denote $f(\lambda ) \in \Lambda $ the result of substituting $\lambda _ i$ for $x_ i$. Then we know that $f_ j(\lambda ) = \pi ^2 \mu _ j$ for some $\mu _ j \in \Lambda $. Define $D \to \Lambda $ by the rules $x_ i \mapsto \lambda _ i$ and $z_ j \mapsto \pi \mu _ j$. This is well defined because

\begin{align*} p_{k, \ell } & \mapsto a_ k(\lambda ) \pi \mu _\ell - \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j(\lambda ) \pi \mu _ j - \pi g_{k, \ell }(\lambda ) \\ & = \pi \left(a_ k(\lambda ) \mu _\ell - \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j(\lambda ) \mu _ j - g_{k, \ell }(\lambda )\right) \end{align*}

Substituting $x_ i = \lambda _ i$ in (4) above we see that the expression inside the brackets is annihilated by $\pi ^2$, hence it is annihilated by $\pi $ as we have assumed $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$. The map $\bar C \to D/\pi D$ is determined by $x_ i \mapsto x_ i$ (clearly well defined). Thus we are done if we can prove (b), (c), and (d).

Using (4) we obtain the following key equality

\begin{align*} \pi p_{k, \ell } & = \pi a_ k z_\ell - \sum \nolimits _{j \in E_ k} \pi h_{k, \ell }^ jz_ j - \pi ^2 g_{k, \ell } \\ & = - a_ k (f_\ell - \pi z_\ell ) + a_ k f_\ell + \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j (f_ j - \pi z_ j) - \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j f_ j - \pi ^2 g_{k, \ell } \\ & = -a_ k(f_\ell - \pi z_\ell ) + \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j(f_ j - \pi z_ j) \end{align*}

The end result is an element of the ideal generated by $f_ j - \pi z_ j$. In particular, we see that $D[1/\pi ]$ is isomorphic to $R[1/\pi ][x_1, \ldots , x_ n, z_1, \ldots , z_ m]/(f_ j - \pi z_ j)$ which is isomorphic to $R[1/\pi ][x_1, \ldots , x_ n]$ hence smooth over $R$. This proves (b).

For fixed $k \in \{ 1, \ldots , r\} $ consider the ring

\[ D_ k = R[x_1, \ldots , x_ n, z_1, \ldots , z_ m]/ (f_ j - \pi z_ j, j \in E_ k, p_{k, \ell }) \]

The number of equations is $m = |E_ k| + (m - |E_ k|)$ as $p_{k, \ell }$ is zero if $\ell \in E_ k$. Also, note that

\begin{align*} (D_ k/\pi D_ k)_{a_ k} & = R/\pi R[x_1, \ldots , x_ n, 1/a_ k, z_1, \ldots , z_ m]/ (f_ j, j \in E_ k, p_{k, \ell }) \\ & = (\bar C/\pi \bar C)_{a_ k}[z_1, \ldots , z_ m]/ (a_ kz_\ell - \sum \nolimits _{j \in E_ k} h_{k, \ell }^ j z_ j) \\ & \cong (\bar C/\pi \bar C)_{a_ k}[z_ j, j \in E_ k] \end{align*}

In particular $(D_ k/\pi D_ k)_{a_ k}$ is smooth over $(\bar C/\pi \bar C)_{a_ k}$. By our choice of $a_ k$ we have that $(\bar C/\pi \bar C)_{a_ k}$ is smooth over $R/\pi R$ of relative dimension $n - |E_ k|$, see (2). Hence for a prime $\mathfrak q_ k \subset D_ k$ containing $\pi $ and lying over $\mathop{\mathrm{Spec}}(\bar C_{a_ k})$ the fibre ring of $R \to D_ k$ is smooth at $\mathfrak q_ k$ of dimension $n$. Thus $R \to D_ k$ is syntomic at $\mathfrak q_ k$ by our count of the number of equations above, see Algebra, Lemma 10.136.10. Hence $R \to D_ k$ is smooth at $\mathfrak q_ k$, see Algebra, Lemma 10.137.17.

To finish the proof, let $\mathfrak q \subset D$ be a prime containing $\pi $ lying over a prime where $R/\pi ^2 R \to \bar C$ is smooth. Then $a_ k \not\in \mathfrak q$ for some $k$ by (1). We will show that the surjection $D_ k \to D$ induces an isomorphism on local rings at $\mathfrak q$. Since we know that the ring maps $\bar C/\pi \bar C \to D_ k/\pi D_ k$ and $R \to D_ k$ are smooth at the corresponding prime $\mathfrak q_ k$ by the preceding paragraph this will prove (c) and (d) and thus finish the proof.

First, note that for any $\ell $ the equation $\pi p_{k, \ell } = -a_ k(f_\ell - \pi z_\ell ) + \sum _{j \in E_ k} h_{k, \ell }^ j (f_ j - \pi z_ j)$ proved above shows that $f_\ell - \pi z_\ell $ maps to zero in $(D_ k)_{a_ k}$ and in particular in $(D_ k)_{\mathfrak q_ k}$. The relations (4) imply that $a_ k f_\ell = \sum _{j \in E_ k} h_{k, \ell }^ j f_ j$ in $I/I^2$. Since $(\bar I_ k/\bar I_ k^2)_{a_ k}$ is free on $f_ j$, $j \in E_ k$ we see that

\[ a_{k'} h_{k, \ell }^ j - \sum \nolimits _{j' \in E_{k'}} h_{k', \ell }^{j'} h_{k, j'}^ j \]

is zero in $\bar C_{a_ k}$ for every $k, k', \ell $ and $j \in E_ k$. Hence we can find a large integer $N$ such that

\[ a_ k^ N\left( a_{k'} h_{k, \ell }^ j - \sum \nolimits _{j' \in E_{k'}} h_{k', \ell }^{j'} h_{k, j'}^ j \right) \]

is in $I_ k + \pi ^2R[x_1, \ldots , x_ n]$. Computing modulo $\pi $ we have

\begin{align*} & a_ kp_{k', \ell } - a_{k'}p_{k, \ell } + \sum h_{k', \ell }^{j'} p_{k, j'} \\ & = - a_ k \sum h_{k', \ell }^{j'} z_{j'} + a_{k'} \sum h_{k, \ell }^ j z_ j + \sum h_{k', \ell }^{j'} a_ k z_{j'} - \sum \sum h_{k', \ell }^{j'} h_{k, j'}^ j z_ j \\ & = \sum \left( a_{k'} h_{k, \ell }^ j - \sum h_{k', \ell }^{j'} h_{k, j'}^ j \right) z_ j \end{align*}

with Einstein summation convention. Combining with the above we see $a_ k^{N + 1} p_{k', \ell }$ is contained in the ideal generated by $I_ k$ and $\pi $ in $R[x_1, \ldots , x_ n, z_1, \ldots , z_ m]$. Thus $p_{k', \ell }$ maps into $\pi (D_ k)_{a_ k}$. On the other hand, the equation

\[ \pi p_{k', \ell } = -a_{k'} (f_\ell - \pi z_\ell ) + \sum \nolimits _{j' \in E_{k'}} h_{k', \ell }^{j'}(f_{j'} - \pi z_{j'}) \]

shows that $\pi p_{k', \ell }$ is zero in $(D_ k)_{a_ k}$. Since we have assumed that $\text{Ann}_ R(\pi ) = \text{Ann}_ R(\pi ^2)$ and since $(D_ k)_{\mathfrak q_ k}$ is smooth hence flat over $R$ we see that $\text{Ann}_{(D_ k)_{\mathfrak q_ k}}(\pi ) = \text{Ann}_{(D_ k)_{\mathfrak q_ k}}(\pi ^2)$. We conclude that $p_{k', \ell }$ maps to zero as well, hence $D_{\mathfrak q} = (D_ k)_{\mathfrak q_ k}$ and we win. $\square$

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