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
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.
$\square$
\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*}
\[ R[x_0, \ldots , x_{n + m + 1}]/ (g_1, \ldots , g_ t, x_0F_1(x_1, \ldots , x_{n + m + 1}), \ldots , x_0F_ c(x_1, \ldots , x_{n + m + 1})) \]
\[ R_{s_0s_1 \ldots s_ t}[x_0, x_1, \ldots , x_{n + m + 1}, 1/x_0x_{n + m + 1}]/ (f_1, \ldots , f_ c) \]
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
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