The Stacks project

Proof. We are going to prove this by induction on $r$.

The case $r = 1$. Here the assumption is that there exists a factorization $A/\pi _1^8 \to \bar C \to \Lambda /\pi _1^8$ which resolves the situation modulo $\pi _1^8$. Conditions (1) and (2) are the assumptions needed to apply Lemma 16.7.3. Thus we can “lift” the resolution $\bar C$ to a resolution of $R \to A \to \Lambda \supset \mathfrak q$.

The case $r > 1$. In this case we apply the induction hypothesis for $r - 1$ to the situation $R/\pi _1^8 \to A/\pi _1^8 \to \Lambda /\pi _1^8 \supset \mathfrak q/\pi _1^8\Lambda$. Note that property (2) is preserved by Lemma 16.2.7. $\square$

Proof. Let $A_\mathfrak p \to C \to \Lambda _\mathfrak q$ be a resolution of $R_\mathfrak p \to A_\mathfrak p \to \Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q$. By our assumption that $\mathfrak q$ is minimal over $\mathfrak h_ A$ this means that $H_{C/R_\mathfrak p} \Lambda _\mathfrak q = \Lambda _\mathfrak q$. By Lemma 16.2.8 we may assume that $C$ is smooth over $R_\mathfrak p$. By Lemma 16.3.4 we may assume that $C$ is standard smooth over $R_\mathfrak p$. Write $A = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ t)$ and say $A \to \Lambda$ is given by $x_ i \mapsto \lambda _ i$. Write $C = R_\mathfrak p[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$ for some $c \geq n$ such that $A \to C$ maps $x_ i$ to $x_ i$ and such that $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is invertible in $C$, see Lemma 16.3.6. After clearing denominators we may assume $f_1, \ldots , f_ c$ are elements of $R[x_1, \ldots , x_{n + m}]$. Of course $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is not invertible in $R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$ but it becomes invertible after inverting some element $s_0 \in R$, $s_0 \not\in \mathfrak p$. As $g_ j$ maps to zero under $R[x_1, \ldots , x_ n] \to A \to C$ we can find $s_ j \in R$, $s_ j \not\in \mathfrak p$ such that $s_ j g_ j$ is zero in $R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$. Write $f_ j = F_ j(x_1, \ldots , x_{n + m}, 1)$ for some polynomial $F_ j \in R[x_1, \ldots , x_ n, X_{n + 1}, \ldots , X_{n + m + 1}]$ homogeneous in $X_{n + 1}, \ldots , X_{n + m + 1}$. Pick $\lambda _{n + i} \in \Lambda$, $i = 1, \ldots , m + 1$ with $\lambda _{n + m + 1} \not\in \mathfrak q$ such that $x_{n + i}$ maps to $\lambda _{n + i}/\lambda _{n + m + 1}$ in $\Lambda _\mathfrak q$. Then

in $\Lambda _\mathfrak q$. Thus we can find $\lambda _0 \in \Lambda$, $\lambda _0 \not\in \mathfrak q$ such that $\lambda _0 F_ j(\lambda _1, \ldots , \lambda _{n + m + 1}) = 0$ in $\Lambda$. Now we set $B$ equal to

which we map to $\Lambda$ by mapping $x_ i$ to $\lambda _ i$. Let $b$ be the image of $x_0 x_{n + m + 1} s_0 s_1 \ldots s_ t$ in $B$. Then $B_ b$ is isomorphic to

which is smooth over $R$ by construction. Since $b$ does not map to an element of $\mathfrak q$, we win. $\square$

Proof. By (3) the ring $\Lambda _\mathfrak q$ is Artinian local hence $\mathfrak q\Lambda _\mathfrak q$ is nilpotent. Thus $(\mathfrak h_ A)^ N \Lambda _\mathfrak q = 0$ for some $N > 0$. Thus there exists a $\lambda \in \Lambda$, $\lambda \not\in \mathfrak q$ such that $\lambda (\mathfrak h_ A)^ N = 0$ in $\Lambda$. Say $H_{A/R} = (a_1, \ldots , a_ r)$ so that $\lambda a_ i^ N = 0$ in $\Lambda$. By Lemma 16.9.3 we can find a factorization $A \to C \to \Lambda$ with $C$ of finite presentation such that $\mathfrak h_ C \not\subset \mathfrak q$. Write $C = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Set

where $t_{ij}$ is a set of $rm$ variables. Note that there is a map $B \to C[y_ i, z]/(y_ iz)$ given by setting $t_{ij}$ equal to zero. The map $B \to \Lambda$ is the composition $B \to C[y_ i, z]/(y_ iz) \to \Lambda$ where $C[y_ i, z]/(y_ iz) \to \Lambda$ is the given map $C \to \Lambda$, maps $z$ to $\lambda$, and maps $y_ i$ to the image of $a_ i^ N$ in $\Lambda$.

We claim that $B$ is a solution for $R \to A \to \Lambda \supset \mathfrak q$. First note that $B_ z$ is isomorphic to $C[y_1, \ldots , y_ r, z, z^{-1}]$ and hence is smooth. On the other hand, $B_{y_\ell } \cong A[x_ i, y_ i, y_\ell ^{-1}, t_{ij}, i \not= \ell ]$ which is smooth over $A$. Thus we see that $z$ and $a_\ell y_\ell$ (compositions of smooth maps are smooth) are all elements of $H_{B/R}$. This proves the lemma. $\square$