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chow.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Chow Homology and Chern Classes}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we discuss Chow homology groups and the construction
of chern classes of vector bundles as elements of operational
Chow cohomology groups (everything with $\mathbf{Z}$-coefficients).
\medskip\noindent
In the first part of this chapter we work on determinants of finite
length modules, we define periodic complexes, their determinants,
and properties of these. All of this is done to give a direct proof
of the Key Lemma \ref{lemma-secondary-ramification}.
Presumably a more standard approach to this lemma would be to
use K-theory of local Noetherian rings.
\medskip\noindent
Next, we introduce the basic setup we work with in the rest of this
chapter in Section \ref{section-setup}. To make the material a little
bit more challenging we decided to treat a somewhat more general case
than is usually done. Namely we assume our schemes $X$ are locally of
finite type over a fixed locally Noetherian base scheme which is universally
catenary and is endowed with a dimension function. These assumption suffice
to be able to define the Chow homology groups $A_*(X)$ and the action of
capping with chern classes on them. This is an indication that we should
be able to define these also for algebraic stacks locally of finite type
over such a base.
\medskip\noindent
Next, we follow the first few chapters of \cite{F} in order to define
cycles, flat pullback, proper pushforward, and rational equivalence,
except that we have been less precise about the supports of the cycles
involved.
\medskip\noindent
We diverge from the presentation given in \cite{F} by using the
Key lemma mentioned above to prove a basic commutativity relation in
Section \ref{section-key}. Using this we prove that the operation
of intersecting with an invertible sheaf passes through rational
equivalence and is commutative, see Section \ref{section-commutativity}.
One more application of the Key
lemma proves that the Gysin map of an effective Cartier divisor
passes through rational equivalence, see Section \ref{section-gysin}.
Having proved this, it is straightforward to define chern
classes of vector bundles, prove additivity, prove the splitting principle,
introduce chern characters, Todd classes, and state the
Grothendieck-Riemann-Roch theorem.
\medskip\noindent
In the appendix we collect some hints to different approaches to this material.
\medskip\noindent
We will return to the Chow groups $A_*(X)$ for smooth projective varieties
over algebraically closed fields in the next chapter. Using a moving
lemma as in \cite{Samuel}, \cite{ChevalleyI}, and \cite{ChevalleyII}
and Serre's Tor-formula
(see \cite{Serre_local_algebra} or \cite{Serre_algebre_locale})
we will define a ring structure on $A_*(X)$. See
Intersection Theory, Section \ref{intersection-section-introduction} ff.
\section{Determinants of finite length modules}
\label{section-determinants-finite-length}
\noindent
The material in this section is related to the material in
the paper \cite{determinant} and to the material in the
thesis \cite{Joe}.
\medskip\noindent
Given any field $\kappa$ and any finite dimensional $\kappa$-vector space
$V$ we set $\det_\kappa(V) = \wedge^n(V)$ where $n = \dim_\kappa(V)$.
We will generalize this to finite length modules over local rings.
If the local ring contains a field, then the determinant constructed
below is a ``usual'' determinant, see
Remark \ref{remark-explain-determinant}.
\begin{definition}
\label{definition-determinant}
Let $R$ be a local ring with maximal ideal $\mathfrak m$ and
residue field $\kappa$. Let $M$ be a finite length $R$-module.
Say $l = \text{length}_R(M)$.
\begin{enumerate}
\item Given elements $x_1, \ldots, x_r \in M$ we denote
$\langle x_1, \ldots, x_r \rangle = Rx_1 + \ldots + Rx_r$ the
$R$-submodule of $M$ generated by $x_1, \ldots, x_r$.
\item We will say an $l$-tuple of elements
$(e_1, \ldots, e_l)$ of $M$ is {\it admissible} if
$\mathfrak m e_i \in \langle e_1, \ldots, e_{i - 1} \rangle$
for $i = 1, \ldots, l$.
\item A {\it symbol} $[e_1, \ldots, e_l]$ will mean
$(e_1, \ldots, e_l)$ is an admissible $l$-tuple.
\item An {\it admissible relation} between symbols is one of the following:
\begin{enumerate}
\item if $(e_1, \ldots, e_l)$ is an admissible sequence and
for some $1 \leq a \leq l$ we have
$e_a \in \langle e_1, \ldots, e_{a - 1}\rangle$, then
$[e_1, \ldots, e_l] = 0$,
\item if $(e_1, \ldots, e_l)$ is an admissible sequence and
for some $1 \leq a \leq l$ we have $e_a = \lambda e'_a + x$
with $\lambda \in R^*$, and
$x \in \langle e_1, \ldots, e_{a - 1}\rangle$, then
$$
[e_1, \ldots, e_l] =
\overline{\lambda} [e_1, \ldots, e_{a - 1}, e'_a, e_{a + 1}, \ldots, e_l]
$$
where $\overline{\lambda} \in \kappa^*$ is the image of $\lambda$ in
the residue field, and
\item if $(e_1, \ldots, e_l)$ is an admissible sequence and
$\mathfrak m e_a \subset \langle e_1, \ldots, e_{a - 2}\rangle$ then
$$
[e_1, \ldots, e_l] =
- [e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l].
$$
\end{enumerate}
\item
We define the {\it determinant of the finite length $R$-module $M$} to be
$$
\det\nolimits_\kappa(M) =
\left\{
\frac{\kappa\text{-vector space generated by symbols}}
{\kappa\text{-linear combinations of admissible relations}}
\right\}
$$
\end{enumerate}
\end{definition}
\noindent
We stress that always $l = \text{length}_R(M)$. We also stress that
it does not follow that the symbol $[e_1, \ldots, e_l]$ is
additive in the entries (this will typically not be the case).
Before we can show that the determinant $\det_\kappa(M)$ actually
has dimension $1$ we have to show that it has dimension at most $1$.
\begin{lemma}
\label{lemma-dimension-at-most-one}
With notations as above we have $\dim_\kappa(\det_\kappa(M)) \leq 1$.
\end{lemma}
\begin{proof}
Fix an admissible sequence $(f_1, \ldots, f_l)$ of $M$ such that
$$
\text{length}_R(\langle f_1, \ldots, f_i\rangle) = i
$$
for $i = 1, \ldots, l$. Such an admissible sequence exists exactly because
$M$ has length $l$. We will show that any element of
$\det_\kappa(M)$ is a $\kappa$-multiple of the symbol
$[f_1, \ldots, f_l]$. This will prove the lemma.
\medskip\noindent
Let $(e_1, \ldots, e_l)$ be an admissible sequence of $M$.
It suffices to show that $[e_1, \ldots, e_l]$ is a multiple
of $[f_1, \ldots, f_l]$. First assume that
$\langle e_1, \ldots, e_l\rangle \not = M$. Then there exists
an $i \in [1, \ldots, l]$ such that
$e_i \in \langle e_1, \ldots, e_{i - 1}\rangle$. It immediately
follows from the first admissible relation that
$[e_1, \ldots, e_n] = 0$ in $\det_\kappa(M)$.
Hence we may assume that $\langle e_1, \ldots, e_l\rangle = M$.
In particular there exists a smallest index $i \in \{1, \ldots, l\}$
such that $f_1 \in \langle e_1, \ldots, e_i\rangle$. This means
that $e_i = \lambda f_1 + x$ with
$x \in \langle e_1, \ldots, e_{i - 1}\rangle$ and $\lambda \in R^*$.
By the second admissible relation this means that
$[e_1, \ldots, e_l] =
\overline{\lambda}[e_1, \ldots, e_{i - 1}, f_1, e_{i + 1}, \ldots, e_l]$.
Note that $\mathfrak m f_1 = 0$. Hence by applying the third
admissible relation $i - 1$ times we see that
$$
[e_1, \ldots, e_l] =
(-1)^{i - 1}\overline{\lambda}
[f_1, e_1, \ldots, e_{i - 1}, e_{i + 1}, \ldots, e_l].
$$
Note that it is also the case that
$ \langle f_1, e_1, \ldots, e_{i - 1}, e_{i + 1}, \ldots, e_l\rangle = M$.
By induction suppose we have proven that our original
symbol is equal to a scalar times
$$
[f_1, \ldots, f_j, e_{j + 1}, \ldots, e_l]
$$
for some admissible sequence $(f_1, \ldots, f_j, e_{j + 1}, \ldots, e_l)$
whose elements generate $M$, i.e., \ with
$\langle f_1, \ldots, f_j, e_{j + 1}, \ldots, e_l\rangle = M$.
Then we find the smallest $i$ such that
$f_{j + 1} \in \langle f_1, \ldots, f_j, e_{j + 1}, \ldots, e_i\rangle$
and we go through the same process as above to see that
$$
[f_1, \ldots, f_j, e_{j + 1}, \ldots, e_l]
=
(\text{scalar}) [f_1, \ldots, f_j, f_{j + 1}, e_{j + 1},
\ldots, \hat{e_i}, \ldots, e_l]
$$
Continuing in this vein we obtain the desired result.
\end{proof}
\noindent
Before we show that $\det_\kappa(M)$ always has dimension $1$,
let us show that it agrees with the usual top exterior power in
the case the module is a vector space over $\kappa$.
\begin{lemma}
\label{lemma-compare-det}
Let $R$ be a local ring with maximal ideal $\mathfrak m$ and
residue field $\kappa$. Let $M$ be a finite length $R$-module
which is annihilated by $\mathfrak m$. Let $l = \dim_\kappa(M)$.
Then the map
$$
\det\nolimits_\kappa(M) \longrightarrow \wedge^l_\kappa(M),
\quad
[e_1, \ldots, e_l] \longmapsto e_1 \wedge \ldots \wedge e_l
$$
is an isomorphism.
\end{lemma}
\begin{proof}
It is clear that the rule described in the lemma gives a $\kappa$-linear
map since all of the admissible relations are satisfied by the usual
symbols $e_1 \wedge \ldots \wedge e_l$. It is also clearly a surjective
map. Since by Lemma \ref{lemma-dimension-at-most-one} the left hand side
has dimension at most one
we see that the map is an isomorphism.
\end{proof}
\begin{lemma}
\label{lemma-determinant-dimension-one}
Let $R$ be a local ring with maximal ideal $\mathfrak m$ and
residue field $\kappa$. Let $M$ be a finite length $R$-module.
The determinant $\det_\kappa(M)$ defined above is a $\kappa$-vector
space of dimension $1$. It is generated by the symbol
$[f_1, \ldots, f_l]$ for any admissible sequence such
that $\langle f_1, \ldots f_l \rangle = M$.
\end{lemma}
\begin{proof}
We know $\det_\kappa(M)$ has dimension at most $1$, and in fact that it
is generated by $[f_1, \ldots, f_l]$, by
Lemma \ref{lemma-dimension-at-most-one} and its proof.
We will show by induction on $l = \text{length}(M)$
that it is nonzero. For $l = 1$ it follows from Lemma \ref{lemma-compare-det}.
Choose a nonzero element $f \in M$
with $\mathfrak m f = 0$. Set $\overline{M} = M /\langle f \rangle$,
and denote the quotient map $x \mapsto \overline{x}$.
We will define a surjective map
$$
\psi : \det\nolimits_k(M) \to \det\nolimits_\kappa(\overline{M})
$$
which will prove the lemma since by induction the determinant of
$\overline{M}$ is nonzero.
\medskip\noindent
We define $\psi$ on symbols as follows.
Let $(e_1, \ldots, e_l)$ be an admissible sequence.
If $f \not \in \langle e_1, \ldots, e_l \rangle$ then
we simply set $\psi([e_1, \ldots, e_l]) = 0$.
If $f \in \langle e_1, \ldots, e_l \rangle$ then we choose
an $i$ minimal such that $f \in \langle e_1, \ldots, e_i \rangle$.
We may write $e_i = \lambda f + x$ for some unit $\lambda \in R$
and $x \in \langle e_1, \ldots, e_{i - 1} \rangle$.
In this case we set
$$
\psi([e_1, \ldots, e_l]) =
(-1)^i
\overline{\lambda}[\overline{e}_1, \ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1}, \ldots, \overline{e}_l].
$$
Note that it is indeed the case that
$(\overline{e}_1, \ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1}, \ldots, \overline{e}_l)$
is an admissible sequence in $\overline{M}$, so this makes sense.
Let us show that extending this rule $\kappa$-linearly to
linear combinations of symbols does indeed lead to a map on
determinants. To do this we have to show that the admissible
relations are mapped to zero.
\medskip\noindent
Type (a) relations. Suppose we have $(e_1, \ldots, e_l)$ an
admissible sequence and for some $1 \leq a \leq l$ we have
$e_a \in \langle e_1, \ldots, e_{a - 1}\rangle$.
Suppose that $f \in \langle e_1, \ldots, e_i\rangle$ with $i$ minimal.
Then $i \not = a$ and
$\overline{e}_a \in \langle \overline{e}_1, \ldots,
\hat{\overline{e}_i}, \ldots, \overline{e}_{a - 1}\rangle$ if $i < a$
or
$\overline{e}_a \in \langle \overline{e}_1, \ldots,
\overline{e}_{a - 1}\rangle$ if $i > a$.
Thus the same admissible relation for $\det_\kappa(\overline{M})$ forces
the symbol $[\overline{e}_1, \ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1}, \ldots, \overline{e}_l]$
to be zero as desired.
\medskip\noindent
Type (b) relations. Suppose we have $(e_1, \ldots, e_l)$ an
admissible sequence and for some $1 \leq a \leq l$ we have
$e_a = \lambda e'_a + x$ with $\lambda \in R^*$, and
$x \in \langle e_1, \ldots, e_{a - 1}\rangle$.
Suppose that $f \in \langle e_1, \ldots, e_i\rangle$ with $i$ minimal.
Say $e_i = \mu f + y$ with $y \in \langle e_1, \ldots, e_{i - 1}\rangle$.
If $i < a$ then the desired equality is
$$
(-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l]
=
(-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_{a - 1},
\overline{e}'_a,
\overline{e}_{a + 1},
\ldots,
\overline{e}_l]
$$
which follows from $\overline{e}_a = \lambda \overline{e}'_a + \overline{x}$
and the corresponding admissible relation for $\det_\kappa(\overline{M})$.
If $i > a$ then the desired equality is
$$
(-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l]
=
(-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 1},
\overline{e}'_a,
\overline{e}_{a + 1},
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l]
$$
which follows from $\overline{e}_a = \lambda \overline{e}'_a + \overline{x}$
and the corresponding admissible relation for $\det_\kappa(\overline{M})$.
The interesting case is when $i = a$. In this case we have
$e_a = \lambda e'_a + x = \mu f + y$. Hence also
$e'_a = \lambda^{-1}(\mu f + y - x)$. Thus we see that
$$
\psi([e_1, \ldots, e_l])
= (-1)^i \overline{\mu}
[\overline{e}_1, \ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1}, \ldots, \overline{e}_l]
=
\psi(
\overline{\lambda}
[e_1, \ldots, e_{a - 1}, e'_a, e_{a + 1}, \ldots, e_l]
)
$$
as desired.
\medskip\noindent
Type (c) relations. Suppose that $(e_1, \ldots, e_l)$
is an admissible sequence and
$\mathfrak m e_a \subset \langle e_1, \ldots, e_{a - 2}\rangle$.
Suppose that $f \in \langle e_1, \ldots, e_i\rangle$ with $i$ minimal.
Say $e_i = \lambda f + x$ with $x \in \langle e_1, \ldots, e_{i - 1}\rangle$.
We distinguish $4$ cases:
\medskip\noindent
Case 1: $i < a - 1$. The desired equality is
\begin{align*}
& (-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l] \\
& =
(-1)^{i + 1}
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_{a - 2},
\overline{e}_a,
\overline{e}_{a - 1},
\overline{e}_{a + 1},
\ldots,
\overline{e}_l]
\end{align*}
which follows from the type (c) admissible relation for
$\det_\kappa(\overline{M})$.
\medskip\noindent
Case 2: $i > a$. The desired equality is
\begin{align*}
& (-1)^i
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l] \\
& =
(-1)^{i + 1}
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 2},
\overline{e}_a,
\overline{e}_{a - 1},
\overline{e}_{a + 1},
\ldots,
\overline{e}_{i - 1},
\overline{e}_{i + 1},
\ldots,
\overline{e}_l]
\end{align*}
which follows from the type (c) admissible relation for
$\det_\kappa(\overline{M})$.
\medskip\noindent
Case 3: $i = a$. We write $e_a = \lambda f + \mu e_{a - 1} + y$
with $y \in \langle e_1, \ldots, e_{a - 2}\rangle$. Then
$$
\psi([e_1, \ldots, e_l]) =
(-1)^a
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 1},
\overline{e}_{a + 1},
\ldots,
\overline{e}_l]
$$
by definition. If $\overline{\mu}$ is nonzero, then we have
$e_{a - 1} = - \mu^{-1} \lambda f + \mu^{-1}e_a - \mu^{-1} y$
and we obtain
$$
\psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) =
(-1)^a
\overline{\mu^{-1}\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 2},
\overline{e}_a,
\overline{e}_{a + 1},
\ldots,
\overline{e}_l]
$$
by definition. Since in $\overline{M}$ we have
$\overline{e}_a = \mu \overline{e}_{a - 1} + \overline{y}$ we see
the two outcomes are equal by relation (a) for $\det_\kappa(\overline{M})$.
If on the other hand $\overline{\mu}$ is zero, then we can write
$e_a = \lambda f + y$ with $y \in \langle e_1, \ldots, e_{a - 2}\rangle$
and we have
$$
\psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) =
(-1)^a
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 1},
\overline{e}_{a + 1},
\ldots,
\overline{e}_l]
$$
which is equal to $\psi([e_1, \ldots, e_l])$.
\medskip\noindent
Case 4: $i = a - 1$. Here we have
$$
\psi([e_1, \ldots, e_l]) =
(-1)^{a - 1}
\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 2},
\overline{e}_a,
\ldots,
\overline{e}_l]
$$
by definition. If $f \not \in \langle e_1, \ldots, e_{a - 2}, e_a \rangle$
then
$$
\psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) =
(-1)^{a + 1}\overline{\lambda}
[\overline{e}_1,
\ldots,
\overline{e}_{a - 2},
\overline{e}_a,
\ldots,
\overline{e}_l]
$$
Since $(-1)^{a - 1} = (-1)^{a + 1}$ the two expressions are the same.
Finally, assume $f \in \langle e_1, \ldots, e_{a - 2}, e_a \rangle$.
In this case we see that $e_{a - 1} = \lambda f + x$ with
$x \in \langle e_1, \ldots, e_{a - 2}\rangle$ and
$e_a = \mu f + y$ with $y \in \langle e_1, \ldots, e_{a - 2}\rangle$
for units $\lambda, \mu \in R$.
We conclude that both
$e_a \in \langle e_1, \ldots, e_{a - 1} \rangle$ and
$e_{a - 1} \in \langle e_1, \ldots, e_{a - 2}, e_a\rangle$.
In this case a relation of type (a) applies to both
$[e_1, \ldots, e_l]$ and
$[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]$
and the compatibility of $\psi$ with these shown above to see that both
$$
\psi([e_1, \ldots, e_l])
\quad\text{and}\quad
\psi([e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l])
$$
are zero, as desired.
\medskip\noindent
At this point we have shown that $\psi$ is well defined, and all that remains
is to show that it is surjective. To see this let
$(\overline{f}_2, \ldots, \overline{f}_l)$ be an admissible sequence
in $\overline{M}$. We can choose lifts $f_2, \ldots, f_l \in M$, and
then $(f, f_2, \ldots, f_l)$ is an admissible sequence in $M$.
Since $\psi([f, f_2, \ldots, f_l]) = [f_2, \ldots, f_l]$ we win.
\end{proof}
\noindent
Let $R$ be a local ring with maximal ideal $\mathfrak m$ and
residue field $\kappa$. Note that if $\varphi : M \to N$ is an
isomorphism of finite length $R$-modules, then we get an
isomorphism
$$
\det\nolimits_\kappa(\varphi) :
\det\nolimits_\kappa(M)
\to
\det\nolimits_\kappa(N)
$$
simply by the rule
$$
\det\nolimits_\kappa(\varphi)([e_1, \ldots, e_l])
=
[\varphi(e_1), \ldots, \varphi(e_l)]
$$
for any symbol $[e_1, \ldots, e_l]$ for $M$.
Hence we see that $\det\nolimits_\kappa$ is a functor
\begin{equation}
\label{equation-functor}
\left\{
\begin{matrix}
\text{finite length }R\text{-modules}\\
\text{with isomorphisms}
\end{matrix}
\right\}
\longrightarrow
\left\{
\begin{matrix}
1\text{-dimensional }\kappa\text{-vector spaces}\\
\text{with isomorphisms}
\end{matrix}
\right\}
\end{equation}
This is typical for a ``determinant functor''
(see \cite{Knudsen}), as is the following additivity
property.
\begin{lemma}
\label{lemma-det-exact-sequences}
Let $(R, \mathfrak m, \kappa)$ be a local ring.
For every short exact sequence
$$
0 \to K \to L \to M \to 0
$$
of finite length $R$-modules there exists a canonical isomorphism
$$
\gamma_{K \to L \to M} :
\det\nolimits_\kappa(K) \otimes_\kappa \det\nolimits_\kappa(M)
\longrightarrow
\det\nolimits_\kappa(L)
$$
defined by the rule on nonzero symbols
$$
[e_1, \ldots, e_k]
\otimes
[\overline{f}_1, \ldots, \overline{f}_m]
\longrightarrow
[e_1, \ldots, e_k, f_1, \ldots, f_m]
$$
with the following properties:
\begin{enumerate}
\item For every isomorphism of short exact sequences, i.e., for
every commutative diagram
$$
\xymatrix{
0 \ar[r] &
K \ar[r] \ar[d]^u &
L \ar[r] \ar[d]^v &
M \ar[r] \ar[d]^w &
0 \\
0 \ar[r] &
K' \ar[r] &
L' \ar[r] &
M' \ar[r] &
0
}
$$
with short exact rows and isomorphisms $u, v, w$ we have
$$
\gamma_{K' \to L' \to M'} \circ
(\det\nolimits_\kappa(u) \otimes \det\nolimits_\kappa(w))
=
\det\nolimits_\kappa(v) \circ
\gamma_{K \to L \to M},
$$
\item for every commutative square of finite length $R$-modules
with exact rows and columns
$$
\xymatrix{
& 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\
0 \ar[r] & A \ar[r] \ar[d] & B \ar[r] \ar[d] & C \ar[r] \ar[d] & 0 \\
0 \ar[r] & D \ar[r] \ar[d] & E \ar[r] \ar[d] & F \ar[r] \ar[d] & 0 \\
0 \ar[r] & G \ar[r] \ar[d] & H \ar[r] \ar[d] & I \ar[r] \ar[d] & 0 \\
& 0 & 0 & 0 &
}
$$
the following diagram is commutative
$$
\xymatrix{
\det\nolimits_\kappa(A) \otimes
\det\nolimits_\kappa(C) \otimes
\det\nolimits_\kappa(G) \otimes
\det\nolimits_\kappa(I)
\ar[dd]_{\epsilon}
\ar[rrr]_-{\gamma_{A \to B \to C} \otimes \gamma_{G \to H \to I}}
& & &
\det\nolimits_\kappa(B) \otimes
\det\nolimits_\kappa(H)
\ar[d]^{\gamma_{B \to E \to H}}
\\
& & & \det\nolimits_\kappa(E)
\\
\det\nolimits_\kappa(A) \otimes
\det\nolimits_\kappa(G) \otimes
\det\nolimits_\kappa(C) \otimes
\det\nolimits_\kappa(I)
\ar[rrr]^-{\gamma_{A \to D \to G} \otimes \gamma_{C \to F \to I}}
& & &
\det\nolimits_\kappa(D) \otimes
\det\nolimits_\kappa(F)
\ar[u]_{\gamma_{D \to E \to F}}
}
$$
where $\epsilon$ is the switch of the factors in the tensor product
times $(-1)^{cg}$ with $c = \text{length}_R(C)$ and $g = \text{length}_R(G)$,
and
\item the map $\gamma_{K \to L \to M}$ agrees with the usual isomorphism
if $0 \to K \to L \to M \to 0$ is actually a short exact sequence
of $\kappa$-vector spaces.
\end{enumerate}
\end{lemma}
\begin{proof}
The significance of taking nonzero symbols in the explicit description
of the map $\gamma_{K \to L \to M}$ is simply that if $(e_1, \ldots, e_l)$
is an admissible sequence in $K$, and
$(\overline{f}_1, \ldots, \overline{f}_m)$ is an admissible sequence in
$M$, then it is not guaranteed that $(e_1, \ldots, e_l, f_1, \ldots, f_m)$
is an admissible sequence in $L$ (where of course $f_i \in L$ signifies
a lift of $\overline{f}_i$). However, if the symbol
$[e_1, \ldots, e_l]$ is nonzero in $\det_\kappa(K)$, then
necessarily $K = \langle e_1, \ldots, e_k\rangle$ (see
proof of Lemma \ref{lemma-dimension-at-most-one}), and
in this case it is true that $(e_1, \ldots, e_k, f_1, \ldots, f_m)$
is an admissible sequence.
Moreover, by the admissible relations of type (b) for $\det_\kappa(L)$
we see that the value of $[e_1, \ldots, e_k, f_1, \ldots, f_m]$ in
$\det_\kappa(L)$ is independent of the choice of the lifts
$f_i$ in this case also. Given this remark, it is clear
that an admissible relation for $e_1, \ldots, e_k$ in $K$
translates into an admissible relation among
$e_1, \ldots, e_k, f_1, \ldots, f_m$ in $L$, and
similarly for an admissible relation among the
$\overline{f}_1, \ldots, \overline{f}_m$.
Thus $\gamma$ defines a linear map of vector spaces as claimed in the lemma.
\medskip\noindent
By Lemma \ref{lemma-determinant-dimension-one} we know
$\det_\kappa(L)$ is generated by any single
symbol $[x_1, \ldots, x_{k + m}]$ such that
$(x_1, \ldots, x_{k + m})$ is an admissible sequence
with $L = \langle x_1, \ldots, x_{k + m}\rangle$. Hence it is
clear that the map $\gamma_{K \to L \to M}$ is surjective and
hence an isomorphism.
\medskip\noindent
Property (1) holds because
\begin{eqnarray*}
& & \det\nolimits_\kappa(v)([e_1, \ldots, e_k, f_1, \ldots, f_m]) \\
& = &
[v(e_1), \ldots, v(e_k), v(f_1), \ldots, v(f_m)] \\
& = &
\gamma_{K' \to L' \to M'}([u(e_1), \ldots, u(e_k)]
\otimes [w(f_1), \ldots, w(f_m)]).
\end{eqnarray*}
Property (2) means that given a symbol
$[\alpha_1, \ldots, \alpha_a]$ generating $\det_\kappa(A)$,
a symbol $[\gamma_1, \ldots, \gamma_c]$ generating $\det_\kappa(C)$,
a symbol $[\zeta_1, \ldots, \zeta_g]$ generating $\det_\kappa(G)$, and
a symbol $[\iota_1, \ldots, \iota_i]$ generating $\det_\kappa(I)$
we have
\begin{eqnarray*}
& & [\alpha_1, \ldots, \alpha_a, \tilde\gamma_1, \ldots, \tilde\gamma_c,
\tilde\zeta_1, \ldots, \tilde\zeta_g, \tilde\iota_1, \ldots, \tilde\iota_i] \\
& = &
(-1)^{cg} [\alpha_1, \ldots, \alpha_a, \tilde\zeta_1, \ldots, \tilde\zeta_g,
\tilde\gamma_1, \ldots, \tilde\gamma_c, \tilde\iota_1, \ldots, \tilde\iota_i]
\end{eqnarray*}
(for suitable lifts $\tilde{x}$ in $E$) in $\det_\kappa(E)$.
This holds because we may use the admissible relations of type (c)
$cg$ times in the following order: move the
$\tilde\zeta_1$ past the elements
$\tilde\gamma_c, \ldots, \tilde\gamma_1$
(allowed since $\mathfrak m\tilde\zeta_1 \subset A$),
then move $\tilde\zeta_2$ past the elements
$\tilde\gamma_c, \ldots, \tilde\gamma_1$
(allowed since $\mathfrak m\tilde\zeta_2 \subset A + R\tilde\zeta_1$),
and so on.
\medskip\noindent
Part (3) of the lemma is obvious.
This finishes the proof.
\end{proof}
\noindent
We can use the maps $\gamma$ of the lemma to define more general maps
$\gamma$ as follows. Suppose that $(R, \mathfrak m, \kappa)$ is a
local ring. Let $M$ be a finite length $R$-module and suppose we
are given a finite filtration (see
Homology, Definition \ref{homology-definition-filtered})
$$
M = F^n \supset F^{n + 1} \supset \ldots \supset F^{m - 1} \supset F^m = 0.
$$
Then there is a canonical isomorphism
$$
\gamma_{(M, F)} :
\bigotimes\nolimits_i \det\nolimits_\kappa(F^i/F^{i + 1})
\longrightarrow
\det\nolimits_\kappa(M)
$$
well defined up to sign(!). One can make the sign explicit either by
giving a well defined order of the terms in the tensor product (starting with
higher indices unfortunately), and by thinking of the target category for
the functor $\det_\kappa$ as the category of
$1$-dimensional super vector spaces. See \cite[Section 1]{determinant}.
\medskip\noindent
Here is another typical result for determinant functors.
It is not hard to show. The tricky part is usually to show the
existence of a determinant functor.
\begin{lemma}
\label{lemma-uniqueness-det}
Let $(R, \mathfrak m, \kappa)$ be any local ring.
The functor
$$
\det\nolimits_\kappa :
\left\{
\begin{matrix}
\text{finite length }R\text{-modules} \\
\text{with isomorphisms}
\end{matrix}
\right\}
\longrightarrow
\left\{
\begin{matrix}
1\text{-dimensional }\kappa\text{-vector spaces} \\
\text{with isomorphisms}
\end{matrix}
\right\}
$$
endowed with the maps $\gamma_{K \to L \to M}$ is characterized by
the following properties
\begin{enumerate}
\item its restriction to the subcategory of modules annihilated
by $\mathfrak m$ is isomorphic to the usual determinant functor
(see Lemma \ref{lemma-compare-det}), and
\item (1), (2) and (3) of Lemma \ref{lemma-det-exact-sequences}
hold.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-determinant-quotient-ring}
Let $(R', \mathfrak m') \to (R, \mathfrak m)$ be a local ring
homomorphism which induces an isomorphism on residue fields $\kappa$.
Then for every finite length $R$-module the restriction $M_{R'}$
is a finite length $R'$-module and there is a canonical isomorphism
$$
\det\nolimits_{R, \kappa}(M)
\longrightarrow
\det\nolimits_{R', \kappa}(M_{R'})
$$
This isomorphism is functorial in $M$ and compatible with the
isomorphisms $\gamma_{K \to L \to M}$ of Lemma \ref{lemma-det-exact-sequences}
defined for $\det_{R, \kappa}$ and $\det_{R', \kappa}$.
\end{lemma}
\begin{proof}
If the length of $M$ as an $R$-module is $l$, then the length
of $M$ as an $R'$-module (i.e., $M_{R'}$) is $l$ as well, see
Algebra, Lemma \ref{algebra-lemma-pushdown-module}.
Note that an admissible sequence $x_1, \ldots, x_l$ of $M$
over $R$ is an admissible sequence of $M$ over $R'$ as $\mathfrak m'$
maps into $\mathfrak m$.
The isomorphism is obtained by mapping the symbol
$[x_1, \ldots, x_l] \in \det\nolimits_{R, \kappa}(M)$
to the corresponding symbol
$[x_1, \ldots, x_l] \in \det\nolimits_{R', \kappa}(M)$.
It is immediate to verify that this is functorial for
isomorphisms and compatible with the isomorphisms
$\gamma$ of Lemma \ref{lemma-det-exact-sequences}.
\end{proof}
\begin{remark}
\label{remark-explain-determinant}
Let $(R, \mathfrak m, \kappa)$ be a local ring and assume either
the characteristic of $\kappa$ is zero or it is $p$ and $p R = 0$.
Let $M_1, \ldots, M_n$ be finite length $R$-modules.
We will show below that there exists an
ideal $I \subset \mathfrak m$ annihilating $M_i$ for $i = 1, \ldots, n$
and a section $\sigma : \kappa \to R/I$ of the canonical surjection
$R/I \to \kappa$. The restriction $M_{i, \kappa}$ of $M_i$ via $\sigma$
is a $\kappa$-vector space of dimension $l_i = \text{length}_R(M_i)$ and
using Lemma \ref{lemma-determinant-quotient-ring} we see that
$$
\det\nolimits_\kappa(M_i) = \wedge_\kappa^{l_i}(M_{i, \kappa})
$$
These isomorphisms are compatible with the isomorphisms
$\gamma_{K \to M \to L}$ of Lemma \ref{lemma-det-exact-sequences}
for short exact sequences of finite length $R$-modules annihilated
by $I$. The conclusion is that verifying a property of
$\det_\kappa$ often reduces to verifying corresponding properties
of the usual determinant on the category finite dimensional vector
spaces.
\medskip\noindent
For $I$ we can take the annihilator
(Algebra, Definition \ref{algebra-definition-annihilator})
of the module $M = \bigoplus M_i$. In this case we see that
$R/I \subset \text{End}_R(M)$ hence has finite length.
Thus $R/I$ is an Artinian local ring with residue field $\kappa$.
Since an Artinian local ring is complete we see that $R/I$
has a coefficient ring by the Cohen structure theorem
(Algebra, Theorem \ref{algebra-theorem-cohen-structure-theorem})
which is a field by our assumption on $R$.
\end{remark}
\noindent
Here is a case where we can compute the determinant of a linear map.
In fact there is nothing mysterious about this in any case, see
Example \ref{example-determinant-map} for a random example.
\begin{lemma}
\label{lemma-times-u-determinant}
Let $R$ be a local ring with residue field $\kappa$.
Let $u \in R^*$ be a unit.
Let $M$ be a module of finite length over $R$.
Denote $u_M : M \to M$ the map multiplication by $u$.
Then
$$
\det\nolimits_\kappa(u_M) :
\det\nolimits_\kappa(M)
\longrightarrow
\det\nolimits_\kappa(M)
$$
is multiplication by $\overline{u}^l$ where $l = \text{length}_R(M)$
and $\overline{u} \in \kappa^*$ is the image of $u$.
\end{lemma}
\begin{proof}
Denote $f_M \in \kappa^*$ the element such that
$\det\nolimits_\kappa(u_M) = f_M \text{id}_{\det\nolimits_\kappa(M)}$.
Suppose that $0 \to K \to L \to M \to 0$ is a short
exact sequence of finite $R$-modules. Then we see that
$u_k$, $u_L$, $u_M$ give an isomorphism of short exact sequences.
Hence by Lemma \ref{lemma-det-exact-sequences} (1) we conclude that
$f_K f_M = f_L$.
This means that by induction on length it suffices to prove the
lemma in the case of length $1$ where it is trivial.
\end{proof}
\begin{example}
\label{example-determinant-map}
Consider the local ring $R = \mathbf{Z}_p$.
Set $M = \mathbf{Z}_p/(p^2) \oplus \mathbf{Z}_p/(p^3)$.
Let $u : M \to M$ be the map given by the matrix
$$
u =
\left(
\begin{matrix}
a & b \\
pc & d
\end{matrix}
\right)
$$
where $a, b, c, d \in \mathbf{Z}_p$, and $a, d \in \mathbf{Z}_p^*$.
In this case $\det_\kappa(u)$ equals multiplication by
$a^2d^3 \bmod p \in \mathbf{F}_p^*$. This can easily be seen
by consider the effect of $u$ on the symbol
$[p^2e, pe, pf, e, f]$ where $e = (0 , 1) \in M$ and
$f = (1, 0) \in M$.
\end{example}
\section{Periodic complexes and Herbrand quotients}
\label{section-periodic-complexes}
\noindent
Of course there is a very general notion of periodic complexes.
We can require periodicity of the maps, or periodicity of the objects.
We will add these here as needed. For the moment we only need
the following cases.
\begin{definition}
\label{definition-periodic-complex}
Let $R$ be a ring.
\begin{enumerate}
\item A {\it $2$-periodic complex} over $R$ is given
by a quadruple $(M, N, \varphi, \psi)$ consisting of
$R$-modules $M$, $N$ and $R$-module maps $\varphi : M \to N$,
$\psi : N \to M$ such that
$$
\xymatrix{
\ldots \ar[r] &
M \ar[r]^\varphi &
N \ar[r]^\psi &
M \ar[r]^\varphi &
N \ar[r] & \ldots
}
$$
is a complex. In this setting we define the {\it cohomology modules}
of the complex to be the $R$-modules
$$
H^0(M, N, \varphi, \psi) = \Ker(\varphi)/\Im(\psi)
, \quad\text{and}\quad
H^1(M, N, \varphi, \psi) = \Ker(\psi)/\Im(\varphi).
$$
We say the $2$-periodic complex is {\it exact} if the cohomology
groups are zero.
\item A {\it $(2, 1)$-periodic complex} over $R$ is given
by a triple $(M, \varphi, \psi)$ consisting of an $R$-module $M$ and
$R$-module maps $\varphi : M \to M$, $\psi : M \to M$
such that
$$
\xymatrix{
\ldots \ar[r] &
M \ar[r]^\varphi &
M \ar[r]^\psi &
M \ar[r]^\varphi &
M \ar[r] & \ldots
}
$$