The Stacks project

Proof. We choose a presentation

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

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

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

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

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

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

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

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

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

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

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. $\square$