Lemma 16.9.3. Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation 16.9.1. Let $\mathfrak p = R \cap \mathfrak q$. Assume that $\mathfrak q$ is minimal over $\mathfrak h_ A$ and that $R_\mathfrak p \to A_\mathfrak p \to \Lambda _\mathfrak q \supset \mathfrak q\Lambda _\mathfrak q$ can be resolved. Then there exists a factorization $A \to C \to \Lambda $ with $C$ of finite presentation such that $H_{C/R} \Lambda \not\subset \mathfrak q$.
[Lemma 12.2, swan] or [Lemma 2, popescu-GND]
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
\begin{align*} F_ j(\lambda _1, \ldots , \lambda _{n + m + 1}) & = (\lambda _{n + m + 1})^{\deg (F_ j)} F_ j(\lambda _1, \ldots , \lambda _ n, \frac{\lambda _{n + 1}}{\lambda _{n + m + 1}}, \ldots , \frac{\lambda _{n + m}}{\lambda _{n + m + 1}}, 1) \\ & = (\lambda _{n + m + 1})^{\deg (F_ j)} f_ j(\lambda _1, \ldots , \lambda _ n, \frac{\lambda _{n + 1}}{\lambda _{n + m + 1}}, \ldots , \frac{\lambda _{n + m}}{\lambda _{n + m + 1}}) \\ & = 0 \end{align*}
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
\[ R[x_0, \ldots , x_{n + m + 1}]/ (g_1, \ldots , g_ t, x_0F_1(x_1, \ldots , x_{n + m + 1}), \ldots , x_0F_ c(x_1, \ldots , x_{n + m + 1})) \]
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
\[ R_{s_0s_1 \ldots s_ t}[x_0, x_1, \ldots , x_{n + m + 1}, 1/x_0x_{n + m + 1}]/ (f_1, \ldots , f_ c) \]
which is smooth over $R$ by construction. Since $b$ does not map to an element of $\mathfrak q$, we win. $\square$
Comments (4)
Comment #3986 by Kestutis Cesnavicius on
Comment #3987 by Kestutis Cesnavicius on
Comment #4064 by Kestutis Cesnavicius on
Comment #4111 by Johan on