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Simplify proof key lemma in chow

Large set of changes in order to simplify the proof of the key lemma
used in proofs of properties for intersecting with effective Cartier
divisors and capping with c_1 of invertible modules.

Probably the original proof was a bit too original. It is still there
in an appendix, but the current proof can be read within an hour by
those familiar with standard (Noetherian) commutative algebra. The
original proof required checking commutativity of many diagrams and
checking many signs.

We also added more explanation of what is happening in the introduction
as well as a comment on Milnor K-theory giving the reference to Kato's
paper and how it is more general.
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aisejohan committed Nov 12, 2017
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  1. +0 −138 algebra.tex
  2. +7,804 −6,809 chow.tex
  3. +55 −2 obsolete.tex
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@@ -28622,144 +28622,6 @@ \section{Orders of vanishing}
by definition the lemma is proved.
\end{proof}
\noindent
We can extend some of the results above to reduced $1$-dimensional
Noetherian local rings which are not domains by the following lemma.
\begin{lemma}
\label{lemma-not-domain}
Let $(R, \mathfrak m)$ be a reduced Noetherian local ring of dimension $1$
and let $x \in \mathfrak m$ be a nonzerodivisor. Let
$\mathfrak q_1, \ldots, \mathfrak q_r$ be the minimal primes of $R$.
Then
$$
\text{length}_R(R/(x)) = \sum\nolimits_i \text{ord}_{R/\mathfrak q_i}(x)
$$
\end{lemma}
\begin{proof}
Note that $R_i = R/\mathfrak q_i$ is a Noetherian $1$-dimensional local
domain. Denote $K_i$ the fraction field of $R_i$.
If $x$ is a unit in $R$, then both sides are zero. Hence we may assume
$x \in \mathfrak m$. Consider the map $\Psi : R \to \prod R_i$.
As $R$ is reduced this map is injective, see
Lemma \ref{lemma-Zariski-topology}.
By Lemma \ref{lemma-total-ring-fractions-no-embedded-points} we have
$Q(R) = \prod K_i$. Hence the finite $R$-module $\Coker(\Psi)$
is annihilated by a nonzerodivisor $y \in R$, hence has support
$\{\mathfrak m\}$, is annihilated by some power of $x$ and has finite
length over $R$, see Lemma \ref{lemma-support-point}.
Consider the short exact sequence
$$
0 \to R \to \prod R_i \to \Coker(\Psi) \to 0
$$
Applying multiplication by $x^n$ to this for $n \gg 0$ we obtain
from the snake lemma
$$
0 \to \Coker(\Psi) \to R/x^nR \to \prod R_i/x^nR_i \to
\Coker(\Psi) \to 0
$$
Thus we see that
$$
\text{length}_R(R/x^nR)
=
\text{length}_R(\prod R_i/x^nR_i) = \sum \text{length}_R(R_i/x^nR_i)
$$
by Lemma \ref{lemma-length-additive}.
By Lemma \ref{lemma-length-independent} we have
$\text{length}_R(R_i/x^nR_i) = \text{length}_{R_i}(R_i/x^nR_i)$.
Now the result follows from the additivity of
Lemma \ref{lemma-ord-additive} and the definition of the order
of vanishing along $R_i$.
\end{proof}
\begin{lemma}
\label{lemma-length-multiplication}
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$.
Let $M$ be a finite $R$-module. Let $x \in R$. Assume that
\begin{enumerate}
\item $\dim(\text{Supp}(M)) \leq 1$, and
\item $\dim(\text{Supp}(M/xM)) \leq 0$.
\end{enumerate}
Write
$\text{Supp}(M) = \{\mathfrak m, \mathfrak q_1, \ldots, \mathfrak q_t\}$.
Then
$$
\text{length}_R(M_x) - \text{length}_R({}_xM) =
\sum\nolimits_{i = 1, \ldots, t}
\text{ord}_{R/\mathfrak q_i}(x)
\text{length}_{R_{\mathfrak q_i}}(M_{\mathfrak q_i}).
$$
where $M_x = M/xM$ and ${}_xM = \Ker(x : M \to M)$.
\end{lemma}
\begin{proof}
We first make some preparatory remarks.
The result of the lemma holds if $M$ has finite length, i.e., if $t = 0$,
because in this case the exact sequence $0 \to {}_xM \to M \to M \to M_x \to 0$
and additivity of length shows that
$\text{length}_R(M_x) = \text{length}_R({}_xM)$.
Also, if we have a short exact sequence $0 \to M \to M' \to M'' \to 0$
of modules satisfying (1) and (2), then lemma for 2 out of 3
of these implies the lemma for the third by the snake lemma.
\medskip\noindent
Denote $M_i$ the image of $M$ in $M_{\mathfrak q_i}$, so
$\text{Supp}(M_i) = \{\mathfrak m, \mathfrak q_i\}$.
The kernel and cokernel of the map $M \to \bigoplus M_i$
have support $\{\mathfrak m\}$ and hence have finite length.
By our preparatory remarks, it follows that it suffices to
prove the lemma for each $M_i$. Thus we may assume that
$\text{Supp}(M) = \{\mathfrak m, \mathfrak q\}$.
In this case we can filter $M$ by powers of $\mathfrak q$.
Again additivity shows that it suffices to prove the lemma
in the case $M$ is annihilated by $\mathfrak q$.
In this case we can view $M$ as a $R/\mathfrak q$-module,
i.e., we may assume that $R$ is a Noetherian local domain
of dimension $1$ with fraction field $K$.
Dividing by the torsion submodule, i.e., by the
kernel of $M \to M \otimes_R K = V$ (the torsion has
finite length hence is handled by our preliminary remarks)
we may assume that $M \subset V$ is a lattice.
Then $\text{length}({}_xM) = 0$ and
$\text{length}_R(M_x) = d(M, xM)$. Since
$\text{length}_K(V) = \dim_K(V)$
we see that $\det(x : V \to V) = x^{\dim_K(V)}$ and
$\text{ord}_R(\det(x : V \to V)) = \dim_K(V) \text{ord}_R(x)$.
Thus the lemma follows from
Lemma \ref{lemma-order-vanishing-determinant}
in this particular case.
\end{proof}
\begin{lemma}
\label{lemma-additivity-divisors-restricted}
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$.
Let $I \subset R$ be an ideal and let $x \in R$.
Assume $x$ is a nonzerodivisor on $R/I$ and that $\dim(R/I) = 1$.
Then
$$
\text{length}_R(R/(x, I))
=
\sum\nolimits_i \text{length}_R(R/(x, \mathfrak q_i))
\text{length}_{R_{\mathfrak q_i}}((R/I)_{\mathfrak q_i})
$$
where $\mathfrak q_1, \ldots, \mathfrak q_n$ are the minimal
primes over $I$. More generally if $M$ is any finite Cohen-Macaulay
module of dimension $1$ over $R$ and $\dim(\text{Supp}(M/xM)) = 0$, then
$$
\text{length}_R(M/xM)
=
\sum\nolimits_i \text{length}_R(R/(x, \mathfrak q_i))
\text{length}_{R_{\mathfrak q_i}}(M_{\mathfrak q_i}).
$$
where $\mathfrak q_1, \ldots, \mathfrak q_t$ are the
minimal primes of the support of $M$.
\end{lemma}
\begin{proof}
These are special cases of Lemma \ref{lemma-length-multiplication}.
\end{proof}
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