The Stacks project

[Lemma 12.2, swan] or [Lemma 2, popescu-GND]

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 $H_{C/R} \Lambda \not\subset \mathfrak q$.

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

In the third sentence of the proof it should be " is smooth over ".

Comment #3987 by Kestutis Cesnavicius on

I think the definition of should instead be . Also, in the description of one should have instead of .

In the statement, perhaps it would make sense to rename to in the view of the notation used in the proof? Also, in the statement it would be nice to mention that (or ) is of finite presentation over .

Sorry for a double comment.

Comment #4064 by Kestutis Cesnavicius on

One could insert references to this lemma: Lemma 12.2 in Swan "Néron-Popescu desingularization" or Lemma 2 in Popescu "General Néron desingularization."

Also, I think it would be better to write instead of .

Comment #4111 by on

Thanks for these comments. I fixed the typos in this commit. I didn't change into because this is used all the time: we can view as an -module and then we can localize it at , so the notation makes sense.

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