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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
We start this chapter by giving the shortest possible
algebraic proof of the Key Lemma \ref{lemma-milnor-gersten-low-degree}.
We first define the Herbrand quotient
(Section \ref{section-periodic-complexes})
and we compute it in some cases
(Section \ref{section-calculation}).
Next, we prove some simple algebra lemmas on
existence of suitable factorizations after modifications
(Section \ref{section-preparation-tame-symbol}).
Using these we construct/define the tame symbol in
Section \ref{section-tame-symbol}.
Only the most basic properties of the tame symbol
are needed to prove the Key Lemma, which we do
in Section \ref{section-key-lemma}.
\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 assumptions suffice
to be able to define the Chow homology groups $\CH_*(X)$ 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
There are two appendices. In Appendix A (Section \ref{section-appendix-A})
we discuss an alternative (longer) construction of the
tame symbol and corresponding proof of the Key Lemma.
Finally, in Appendix B (Section \ref{section-appendix-chow})
we briefly discuss the relationship with $K$-theory of coherent
sheaves and we discuss some blowup lemmas.
We suggest the reader look at their introductions for
more information.
\medskip\noindent
We will return to the Chow groups $\CH_*(X)$ 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 $\CH_*(X)$. See
Intersection Theory, Section \ref{intersection-section-introduction} ff.
\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
$$
\xymatrix{
\ldots \ar[r] &
M \ar[r]^\varphi &
M \ar[r]^\psi &
M \ar[r]^\varphi &
M \ar[r] & \ldots
}
$$
is a complex. Since this is a special case of a $2$-periodic complex
we have its {\it cohomology modules} $H^0(M, \varphi, \psi)$,
$H^1(M, \varphi, \psi)$ and a notion of exactness.
\end{enumerate}
\end{definition}
\noindent
In the following we will use any result proved for $2$-periodic
complexes without further mention for $(2, 1)$-periodic complexes.
It is clear that the collection of $2$-periodic complexes forms a
category with morphisms
$(f, g) : (M, N, \varphi, \psi) \to (M', N', \varphi', \psi')$
pairs of morphisms $f : M \to M'$ and $g : N \to N'$ such
that $\varphi' \circ f = g \circ \varphi$ and $\psi' \circ g = f \circ \psi$.
We obtain an abelian category, with kernels and cokernels as in
Homology, Lemma \ref{homology-lemma-cat-chain-abelian}.
\begin{definition}
\label{definition-periodic-length}
Let $(M, N, \varphi, \psi)$ be a $2$-periodic complex
over a ring $R$ whose cohomology modules have finite length.
In this case we define the {\it multiplicity} of $(M, N, \varphi, \psi)$
to be the integer
$$
e_R(M, N, \varphi, \psi) =
\text{length}_R(H^0(M, N, \varphi, \psi))
-
\text{length}_R(H^1(M, N, \varphi, \psi))
$$
In the case of a $(2, 1)$-periodic complex $(M, \varphi, \psi)$,
we denote this by $e_R(M, \varphi, \psi)$ and we will sometimes call this
the {\it (additive) Herbrand quotient}.
\end{definition}
\noindent
If the cohomology groups of $(M, \varphi, \psi)$
are finite abelian groups, then it is customary to call the
{\it (multiplicative) Herbrand quotient}
$$
q(M, \varphi, \psi) =
\frac{\# H^0(M, \varphi, \psi)}{\# H^1(M, \varphi, \psi)}
$$
In words: the multiplicative Herbrand quotient is the number of elements of
$H^0$ divided by the number of elements of $H^1$. If $R$ is local and if
the residue field of $R$ is finite with $q$ elements, then we see that
$$
q(M, \varphi, \psi) = q^{e_R(M, \varphi, \psi)}
$$
\medskip\noindent
An example of a $(2, 1)$-periodic complex over a ring $R$ is any triple of
the form $(M, 0, \psi)$ where $M$ is an $R$-module and $\psi$ is an
$R$-linear map. If the kernel and cokernel of $\psi$ have finite length,
then we obtain
\begin{equation}
\label{equation-multiplicity-coker-ker}
e_R(M, 0, \psi) = \text{length}_R(\Coker(\psi)) - \text{length}_R(\Ker(\psi))
\end{equation}
We state and prove the obligatory lemmas on these notations.
\begin{lemma}
\label{lemma-additivity-periodic-length}
Let $R$ be a ring. Suppose that we have a short exact sequence of
$2$-periodic complexes
$$
0 \to (M_1, N_1, \varphi_1, \psi_1)
\to (M_2, N_2, \varphi_2, \psi_2)
\to (M_3, N_3, \varphi_3, \psi_3)
\to 0
$$
If two out of three have cohomology modules of finite length so does
the third and we have
$$
e_R(M_2, N_2, \varphi_2, \psi_2) =
e_R(M_1, N_1, \varphi_1, \psi_1) +
e_R(M_3, N_3, \varphi_3, \psi_3).
$$
\end{lemma}
\begin{proof}
We abbreviate $A = (M_1, N_1, \varphi_1, \psi_1)$,
$B = (M_2, N_2, \varphi_2, \psi_2)$ and $C = (M_3, N_3, \varphi_3, \psi_3)$.
We have a long exact cohomology sequence
$$
\ldots
\to H^1(C)
\to H^0(A)
\to H^0(B)
\to H^0(C)
\to H^1(A)
\to H^1(B)
\to H^1(C)
\to \ldots
$$
This gives a finite exact sequence
$$
0 \to I
\to H^0(A)
\to H^0(B)
\to H^0(C)
\to H^1(A)
\to H^1(B)
\to K \to 0
$$
with $0 \to K \to H^1(C) \to I \to 0$ a filtration. By additivity of
the length function (Algebra, Lemma \ref{algebra-lemma-length-additive})
we see the result.
\end{proof}
\begin{lemma}
\label{lemma-finite-periodic-length}
Let $R$ be a ring. If $(M, N, \varphi, \psi)$ is a $2$-periodic complex
such that $M$, $N$ have finite length, then
$e_R(M, N, \varphi, \psi) = \text{length}_R(M) - \text{length}_R(N)$.
In particular, if $(M, \varphi, \psi)$ is a $(2, 1)$-periodic complex
such that $M$ has finite length, then
$e_R(M, \varphi, \psi) = 0$.
\end{lemma}
\begin{proof}
Observe that on the category of $2$-periodic complexes with $M$, $N$
of finite length the quantity ``$\text{length}_R(M) - \text{length}_R(N)$''
is additive in short exact sequences (precise statement left to the reader).
Consider the short exact sequence
$$
0 \to (M, \Im(\varphi), \varphi, 0) \to
(M, N, \varphi, \psi) \to (0, N/\Im(\varphi), 0, 0) \to 0
$$
The initial remark combined with the additivity of
Lemma \ref{lemma-additivity-periodic-length}
reduces us to the cases (a) $M = 0$ and (b) $\varphi$ is surjective.
We leave those cases to the reader.
\end{proof}
\begin{lemma}
\label{lemma-compare-periodic-lengths}
Let $R$ be a ring. Let $f : (M, \varphi, \psi) \to (M', \varphi', \psi')$
be a map of $(2, 1)$-periodic complexes whose cohomology modules
have finite length. If $\Ker(f)$ and $\Coker(f)$ have finite length,
then $e_R(M, \varphi, \psi) = e_R(M', \varphi', \psi')$.
\end{lemma}
\begin{proof}
Apply the additivity of Lemma \ref{lemma-additivity-periodic-length}
and observe that $(\Ker(f), \varphi, \psi)$ and
$(\Coker(f), \varphi', \psi')$ have vanishing multiplicity by
Lemma \ref{lemma-finite-periodic-length}.
\end{proof}
\section{Calculation of some multiplicities}
\label{section-calculation}
\noindent
To prove equality of certain cycles later on we need to
compute some multiplicities. Our main tool, besides the
elementary lemmas on multiplicities given in the previous section,
will be Algebra, Lemma \ref{algebra-lemma-order-vanishing-determinant}.
\begin{lemma}
\label{lemma-length-multiplication}
Let $R$ be a Noetherian local ring.
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
$$
e_R(M, 0, x) =
\sum\nolimits_{i = 1, \ldots, t}
\text{ord}_{R/\mathfrak q_i}(x)
\text{length}_{R_{\mathfrak q_i}}(M_{\mathfrak q_i}).
$$
\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 both the left hand side and the right hand side are zero
in this case, see Lemma \ref{lemma-finite-periodic-length}.
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
additivity of length (Algebra, Lemma \ref{algebra-lemma-length-additive}) and
additivty of multiplicities (Lemma \ref{lemma-additivity-periodic-length}).
\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 have a finite filtration
$M \supset \mathfrak qM \supset \mathfrak q^2M \supset \ldots \supset
\mathfrak q^nM = 0$ by Algebra, Lemma
\ref{algebra-lemma-Noetherian-power-ideal-kills-module}.
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
(Algebra, Definition \ref{algebra-definition-lattice}).
Then $x : M \to M$ is injective and
$\text{length}_R(M/xM) = d(M, xM)$
(Algebra, Definition \ref{algebra-definition-distance}). 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 desired equality follows from
Algebra, Lemma \ref{algebra-lemma-order-vanishing-determinant}
in this case.
\end{proof}
\begin{lemma}
\label{lemma-additivity-divisors-restricted}
Let $R$ be a Noetherian local ring.
Let $x \in R$. If $M$ is a finite Cohen-Macaulay module over $R$
with $\dim(\text{Supp}(M)) = 1$ 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 $M$. If $I \subset R$ is an ideal
such that $x$ is a nonzerodivisor on $R/I$ and $\dim(R/I) = 1$, 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 $I$.
\end{lemma}
\begin{proof}
These are special cases of Lemma \ref{lemma-length-multiplication}.
\end{proof}
\noindent
Here is another case where we can determine the value of a multiplicity.
\begin{lemma}
\label{lemma-powers-period-length-zero}
Let $R$ be a ring. Let $M$ be an $R$-module.
Let $\varphi : M \to M$ be an endomorphism and $n > 0$
such that $\varphi^n = 0$ and such that $\Ker(\varphi)/\Im(\varphi^{n - 1})$
has finite length as an $R$-module.
Then
$$
e_R(M, \varphi^i, \varphi^{n - i}) = 0
$$
for $i = 0, \ldots, n$.
\end{lemma}
\begin{proof}
The cases $i = 0, n$ are trivial as $\varphi^0 = \text{id}_M$ by convention.
Let us think of $M$ as an $R[t]$-module where multiplication by $t$
is given by $\varphi$. Let us write
$K_i = \Ker(t^i : M \to M)$ and
$$
a_i = \text{length}_R(K_i/t^{n - i}M),\quad
b_i = \text{length}_R(K_i/tK_{i + 1}),\quad
c_i = \text{length}_R(K_1/t^iK_{i + 1})
$$
Boundary values are $a_0 = a_n = b_0 = c_0 = 0$.
The $c_i$ are integers for $i < n$ as $K_1/t^iK_{i + 1}$
is a quotient of $K_1/t^{n - 1}M$ which is assumed to have finite length.
We will use frequently that $K_i \cap t^jM = t^jK_{i + j}$.
For $0 < i < n - 1$ we have an exact sequence
$$
0 \to
K_1/t^{n - i - 1}K_{n - i} \to
K_{i + 1}/t^{n - i - 1}M \xrightarrow{t} K_i/t^{n - i}M
\to K_i/tK_{i + 1} \to 0
$$
By induction on $i$ we conclude that $a_i$ and $b_i$ are
integers for $i < n$ and that
$$
c_{n - i - 1} - a_{i + 1} + a_i - b_i = 0
$$
For $0 < i < n - 1$ there is a short exact sequence
$$
0 \to
K_i/tK_{i + 1} \to
K_{i + 1}/tK_{i + 2} \xrightarrow{t^i}
K_1/t^{i + 1}K_{i + 2} \to
K_1/t^iK_{i + 1} \to 0
$$
which gives
$$
b_i - b_{i + 1} + c_{i + 1} - c_i = 0
$$
Since $b_0 = c_0$ we conclude that $b_i = c_i$ for $i < n$.
Then we see that
$$
a_2 = a_1 + b_{n - 2} - b_1,\quad
a_3 = a_2 + b_{n - 3} - b_2,\quad \ldots
$$
It is straightforward to see that this implies $a_i = a_{n - i}$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-multiply-period-length}
Let $(R, \mathfrak m)$ be a Noetherian local ring. Let
$(M, \varphi, \psi)$ be a $(2, 1)$-periodic complex over $R$
with $M$ finite and with cohomology groups of finite length over $R$.
Let $x \in R$ be such that $\dim(\text{Supp}(M/xM)) \leq 0$. Then
$$
e_R(M, x\varphi, \psi) = e_R(M, \varphi, \psi) - e_R(\Im(\varphi), 0, x)
$$
and
$$
e_R(M, \varphi, x\psi) = e_R(M, \varphi, \psi) + e_R(\Im(\psi), 0, x)
$$
\end{lemma}
\begin{proof}
We will only prove the first formula as the second is proved
in exactly the same manner.
Let $M' = M[x^\infty]$ be the $x$-power torsion submodule of $M$.
Consider the short exact sequence $0 \to M' \to M \to M'' \to 0$.
Then $M''$ is $x$-power torsion free (More on Algebra, Lemma
\ref{more-algebra-lemma-divide-by-torsion}).
Since $\varphi$, $\psi$ map $M'$ into $M'$
we obtain a short exact sequence
$$
0 \to (M', \varphi', \psi') \to (M, \varphi, \psi) \to
(M'', \varphi'', \psi'') \to 0
$$
of $(2, 1)$-periodic complexes. Also, we get a short exact sequence
$0 \to M' \cap \Im(\varphi) \to \Im(\varphi) \to \Im(\varphi'') \to 0$.
We have
$e_R(M', \varphi, \psi) = e_R(M', x\varphi, \psi) =
e_R(M' \cap \Im(\varphi), 0, x) = 0$
by Lemma \ref{lemma-compare-periodic-lengths}.
By additivity (Lemma \ref{lemma-additivity-periodic-length})
we see that it suffices to prove the lemma for $(M'', \varphi'', \psi'')$.
This reduces us to the case discussed in the next paragraph.
\medskip\noindent
Assume $x : M \to M$ is injective.
In this case $\Ker(x\varphi) = \Ker(\varphi)$.
On the other hand we have a short exact sequence
$$
0 \to \Im(\varphi)/x\Im(\varphi) \to
\Ker(\psi)/\Im(x\varphi) \to \Ker(\psi)/\Im(\varphi) \to 0
$$
This together with (\ref{equation-multiplicity-coker-ker}) proves the formula.
\end{proof}
\section{Preparation for tame symbols}
\label{section-preparation-tame-symbol}
\noindent
In this section we put some lemma that will help us define the
tame symbol in the next section.
\begin{lemma}
\label{lemma-glue-at-max}
Let $A$ be a Noetherian ring. Let $\mathfrak m_1, \ldots, \mathfrak m_r$
be pairwise distinct maximal ideals of $A$. For $i = 1, \ldots, r$ let
$\varphi_i : A_{\mathfrak m_i} \to B_i$ be a ring map whose
kernel and cokernel are annihilated by a power
of $\mathfrak m_i$. Then there exists a ring map $\varphi : A \to B$ such
that
\begin{enumerate}
\item the localization of $\varphi$ at $\mathfrak m_i$ is
isomorphic to $\varphi_i$, and
\item $\Ker(\varphi)$ and $\Coker(\varphi)$ are annihilated
by a power of $\mathfrak m_1 \cap \ldots \cap \mathfrak m_r$.
\end{enumerate}
Moreover, if each $\varphi_i$ is finite, injective, or
surjective then so is $\varphi$.
\end{lemma}
\begin{proof}
Set $I = \mathfrak m_1 \cap \ldots \cap \mathfrak m_r$. Set
$A_i = A_{\mathfrak m_i}$ and $A' = \prod A_i$.
Then $IA' = \prod \mathfrak m_i A_i$ and $A \to A'$
is a flat ring map such that $A/I \cong A'/IA'$.
Thus we may use More on Algebra, Lemma
\ref{more-algebra-lemma-application-formal-glueing}
to see that there exists an $A$-module map $\varphi : A \to B$
with $\varphi_i$ isomorphic to the localization of $\varphi$
at $\mathfrak m_i$. Then we can use the discussion in
More on Algebra, Remark \ref{more-algebra-remark-formal-glueing-algebras}
to endow $B$ with an $A$-algebra structure
matching the given $A$-algebra structure on $B_i$.
The final statement of the lemma follows easily from
the fact that $\Ker(\varphi)_{\mathfrak m_i} \cong \Ker(\varphi_i)$
and $\Coker(\varphi)_{\mathfrak m_i} \cong \Coker(\varphi_i)$.
\end{proof}
\noindent
The following lemma is very similar to
Algebra, Lemma \ref{algebra-lemma-nonregular-dimension-one}.
\begin{lemma}
\label{lemma-Noetherian-domain-dim-1-two-elements}
Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $1$.
Let $a, b \in R$ be nonzerodivisors.
There exists a finite ring extension $R \subset R'$
with $R'/R$ annihilated by a power of $\mathfrak m$
and nonzerodivisors $t, a', b' \in R'$ such that
$a = ta'$ and $b = tb'$ and $R' = a'R' + b'R'$.
\end{lemma}
\begin{proof}
If $a$ or $b$ is a unit, then the lemma is true with $R = R'$.
Thus we may assume $a, b \in \mathfrak m$.
Set $I = (a, b)$. The idea is to blow up $R$ in $I$.
Instead of doing the algebraic argument we work geometrically.
Let $X = \text{Proj}(\bigoplus_{d \geq 0} I^d)$.
By Divisors, Lemma
\ref{divisors-lemma-blowing-up-gives-effective-Cartier-divisor}
the morphism $X \to \Spec(R)$ is an isomorphism over
the punctured spectrum $U = \Spec(R) \setminus \{\mathfrak m\}$.
Thus we may and do view $U$ as an open subscheme of $X$.
The morphism $X \to \Spec(R)$ is projective by
Divisors, Lemma \ref{divisors-lemma-blowing-up-projective}.
Also, every generic point of $X$ lies in $U$, for example
by Divisors, Lemma \ref{divisors-lemma-blow-up-and-irreducible-components}.
It follows from Varieties, Lemma \ref{varieties-lemma-finite-in-codim-1}
that $X \to \Spec(R)$ is finite. Thus $X = \Spec(R')$ is
affine and $R \to R'$ is finite. We have $R_a \cong R'_a$ as $U = D(a)$.
Hence a power of $a$ annihilates the finite $R$-module $R'/R$.
As $\mathfrak m = \sqrt{(a)}$ we see that $R'/R$ is annihilated
by a power of $\mathfrak m$. By
Divisors, Lemma \ref{divisors-lemma-blowing-up-gives-effective-Cartier-divisor}
we see that $IR'$ is a locally principal ideal.
Since $R'$ is semi-local we see that $IR'$ is principal,
see Algebra, Lemma \ref{algebra-lemma-locally-free-semi-local-free},
say $IR' = (t)$. Then we have $a = a't$ and $b = b't$ and everything is
clear.
\end{proof}
\begin{lemma}
\label{lemma-not-infinitely-divisible}
Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $1$.
Let $a, b \in R$ be nonzerodivisors with $a \in \mathfrak m$.
There exists an integer $n = n(R, a, b)$ such that for a finite ring
extension $R \subset R'$ if $b = a^m c$ for some $c \in R'$, then $m \leq n$.
\end{lemma}
\begin{proof}
Choose a minimal prime $\mathfrak q \subset R$. Observe that
$\dim(R/\mathfrak q) = 1$, in particular $R/\mathfrak q$ is not a field.
We can choose a discrete valuation ring $A$ dominating $R/\mathfrak q$
with the same fraction field, see
Algebra, Lemma \ref{algebra-lemma-dominate-by-dimension-1}. Observe that
$a$ and $b$ map to nonzero elements of $A$ as nonzerodivisors in $R$
are not contained in $\mathfrak q$. Let $v$ be the discrete valuation on $A$.
Then $v(a) > 0$ as $a \in \mathfrak m$.
We claim $n = v(b)/v(a)$ works.
\medskip\noindent
Let $R \subset R'$ be given. Set $A' = A \otimes_R R'$.
Since $\Spec(R') \to \Spec(R)$ is surjective
(Algebra, Lemma \ref{algebra-lemma-integral-overring-surjective})
also $\Spec(A') \to \Spec(A)$ is surjective
(Algebra, Lemma \ref{algebra-lemma-surjective-spec-radical-ideal}).
Pick a prime $\mathfrak q' \subset A'$ lying over $(0) \subset A$.
Then $A \subset A'' = A'/\mathfrak q'$ is a finite extension of rings
(again inducing a surjection on spectra).
Pick a maximal ideal $\mathfrak m'' \subset A''$
lying over the maximal ideal of $A$ and a discrete valuation ring
$A'''$ dominating $A''_{\mathfrak m''}$ (see lemma cited above).
Then $A \to A'''$ is an extension of discrete valuation rings
and we have $b = a^m c$ in $A'''$. Thus $v'''(b) \geq mv'''(a)$.
Since $v''' = ev$ where $e$ is the ramification index
of $A'''/A$, we find that $m \leq n$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-prepare-tame-symbol}
Let $(A, \mathfrak m)$ be a Noetherian local ring of dimension $1$.
Let $r \geq 2$ and let $a_1, \ldots, a_r \in A$ be nonzerodivisors
not all units.
Then there exist
\begin{enumerate}
\item a finite ring extension $A \subset B$ with
$B/A$ annihilated by a power of $\mathfrak m$,
\item for each maximal ideal $\mathfrak m_j \subset B$
a nonzerodivisor $\pi_j \in B_j = B_{\mathfrak m_j}$, and
\item factorizations $a_i = u_{i, j} \pi_j^{e_{i, j}}$ in $B_j$
with $u_{i, j} \in B_j$ units and $e_{i, j} \geq 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
Since at least one $a_i$ is not a unit we find that $\mathfrak m$
is not an associated prime of $A$. Moreover, for any $A \subset B$
as in the statement $\mathfrak m$ is not an associated prime of $B$
and $\mathfrak m_j$ is not an associate prime of $B_j$.
Keeping this in mind will help check the arguments below.
\medskip\noindent
First, we claim that it suffices to prove the lemma for $r = 2$.
We will argue this by induction on $r$; we suggest the reader
skip the proof. Suppose we are given $A \subset B$ and $\pi_j$ in
$B_j = B_{\mathfrak m_j}$ and factorizations
$a_i = u_{i, j} \pi_j^{e_{i, j}}$ for $i = 1, \ldots, r - 1$ in $B_j$
with $u_{i, j} \in B_j$ units and $e_{i, j} \geq 0$.
Then by the case $r = 2$ for $\pi_j$ and $a_r$ in $B_j$
we can find extensions $B_j \subset C_j$ and for every maximal ideal
$\mathfrak m_{j, k}$ of $C_j$ a nonzerodivisor
$\pi_{j, k} \in C_{j, k} = (C_j)_{\mathfrak m_{j, k}}$
and factorizations
$$
\pi_j = v_{j, k} \pi_{j, k}^{f_{j, k}}
\quad\text{and}\quad
a_r = w_{j, k} \pi_{j, k}^{g_{j, k}}
$$
as in the lemma. There exists a unique finite extension $B \subset C$
with $C/B$ annihilated by a power of $\mathfrak m$ such
that $C_j \cong C_{\mathfrak m_j}$ for all $j$, see
Lemma \ref{lemma-glue-at-max}.
The maximal ideals of $C$ correspond $1$-to-$1$
to the maximal ideals $\mathfrak m_{j, k}$ in the localizations
and in these localizations we have
$$
a_i = u_{i, j} \pi_j^{e_{i, j}} =
u_{i, j} v_{j, k}^{e_{i, j}} \pi_{j, k}^{e_{i, j}f_{j, k}}
$$
for $i \leq r - 1$. Since $a_r$ factors correctly too the
proof of the induction step is complete.
\medskip\noindent
Proof of the case $r = 2$. We will use induction on
$$
\ell = \min(\text{length}_A(A/a_1A),\ \text{length}_A(A/a_2A)).
$$
If $\ell = 0$, then either $a_1$ or $a_2$ is a unit and
the lemma holds with $A = B$. Thus we may and do assume $\ell > 0$.
\medskip\noindent
Suppose we have a finite extension of rings $A \subset A'$ such that
$A'/A$ is annihilated by a power of $\mathfrak m$ and such that
$\mathfrak m$ is not an associated prime of $A'$.
Let $\mathfrak m_1, \ldots, \mathfrak m_r \subset A'$
be the maximal ideals and set $A'_i = A'_{\mathfrak m_i}$.
If we can solve the problem for $a_1, a_2$ in each $A'_i$,
then we can apply Lemma \ref{lemma-glue-at-max}
to produce a solution for $a_1, a_2$ in $A$.
Choose $x \in \{a_1, a_2\}$ such that $\ell = \text{length}_A(A/xA)$.
By Lemma \ref{lemma-compare-periodic-lengths}
and (\ref{equation-multiplicity-coker-ker})
we have $\text{length}_A(A/xA) = \text{length}_A(A'/xA')$.
On the other hand, we have
$$
\text{length}_A(A'/xA') =
\sum [\kappa(\mathfrak m_i) : \kappa(\mathfrak m)]
\text{length}_{A'_i}(A'_i/xA'_i)
$$
by Algebra, Lemma \ref{algebra-lemma-pushdown-module}.
Since $x \in \mathfrak m$ we see that each term on the right hand side
is positive. We conclude that the induction hypothesis applies
to $a_1, a_2$ in each $A'_i$ if $r > 1$ or if $r = 1$ and
$[\kappa(\mathfrak m_1) : \kappa(\mathfrak m)] > 1$.
We conclude that we may assume each $A'$ as above is local with
the same residue field as $A$.
\medskip\noindent
Applying the discussion of the previous paragraph,
we may replace $A$ by the ring constructed in
Lemma \ref{lemma-Noetherian-domain-dim-1-two-elements}
for $a_1, a_2 \in A$. Then since $A$ is local we find,
after possibly switching $a_1$ and $a_2$, that $a_2 \in (a_1)$.
Write $a_2 = a_1^m c$ with $m > 0$ maximal. In fact, by
Lemma \ref{lemma-not-infinitely-divisible}
we may assume $m$ is maximal even after replacing $A$
by any finite extension $A \subset A'$ as in the previous paragraph.
If $c$ is a unit, then we are done. If not, then we replace
$A$ by the ring constructed in
Lemma \ref{lemma-Noetherian-domain-dim-1-two-elements}
for $a_1, c \in A$. Then either (1) $c = a_1 c'$ or
(2) $a_1 = c a'_1$. The first case cannot happen since
it would give $a_2 = a_1^{m + 1} c'$ contradicting the
maximality of $m$. In the second case we get
$a_1 = c a'_1$ and $a_2 = c^{m + 1} (a'_1)^m$.
Then it suffices to prove the lemma for $A$ and $c, a'_1$.
If $a'_1$ is a unit we're done and if not, then
$\text{length}_A(A/cA) < \ell$ because $cA$ is a strictly
bigger ideal than $a_1A$. Thus we win by induction hypothesis.
\end{proof}
\section{Tame symbols}
\label{section-tame-symbol}
\noindent
Consider a Noetherian local ring $(A, \mathfrak m)$ of dimension $1$.
We denote $Q(A)$ the total ring of fractions of $A$, see
Algebra, Example \ref{algebra-example-localize-at-prime}.
The {\it tame symbol} will be a map
$$
\partial_A(-, -) : Q(A)^* \times Q(A)^* \longrightarrow \kappa(\mathfrak m)^*
$$
satisfying the following properties:
\begin{enumerate}
\item $\partial_A(f, gh) = \partial_A(f, g) \partial_A(f, h)$
\label{item-bilinear}
for $f, g, h \in Q(A)^*$,
\item $\partial_A(f, g) \partial_A(g, f) = 1$
\label{item-skew}
for $f, g \in Q(A)^*$,
\item $\partial_A(f, 1 - f) = 1$
\label{item-1-x}
for $f \in Q(A)^*$ such that $1 - f \in Q(A)^*$,
\item $\partial_A(aa', b) = \partial_A(a, b)\partial_A(a', b)$
\label{item-bilinear-better}
and $\partial_A(a, bb') = \partial_A(a, b)\partial_A(a, b')$
for $a, a', b, b' \in A$ nonzerodivisors,
\item $\partial_A(b, b) = (-1)^m$
\label{item-skew-better}
with $m = \text{length}_A(A/bA)$
for $b \in A$ a nonzerodivisor,
\item $\partial_A(u, b) = u^m \bmod \mathfrak m$
\label{item-normalization}
with $m = \text{length}_A(A/bA)$ for $u \in A$ a unit and
$b \in A$ a nonzerodivisor, and
\item
\label{item-1-x-better}
$\partial_A(a, b - a)\partial_A(b, b) = \partial_A(b, b - a)\partial_A(a, b)$
for $a, b \in A$ such that $a, b, b - a$ are nonzerodivisors.
\end{enumerate}
Since it is easier to work with elements of $A$ we will
often think of $\partial_A$ as a map defined on pairs of
nonzerodivisors of $A$ satisfying (\ref{item-bilinear-better}),
(\ref{item-skew-better}), (\ref{item-normalization}),
(\ref{item-1-x-better}). It is an exercise to see that
setting
$$
\partial_A(\frac{a}{b}, \frac{c}{d}) =
\partial_A(a, c) \partial_A(a, d)^{-1} \partial_A(b, c)^{-1} \partial_A(b, d)
$$
we get a well defined map $Q(A)^* \times Q(A)^* \to \kappa(\mathfrak m)^*$
satisfying (\ref{item-bilinear}), (\ref{item-skew}), (\ref{item-1-x})
as well as the other properties.
\medskip\noindent
We do not claim there is a unique map with these properties.
Instead, we will give a recipe for constructing such a map.
Namely, given $a_1, a_2 \in A$ nonzerodivisors, we choose
a ring extension $A \subset B$ and local factorizations
as in Lemma \ref{lemma-prepare-tame-symbol}.
Then we define
\begin{equation}
\label{equation-tame-symbol}
\partial_A(a_1, a_2) = \prod\nolimits_j
\text{Norm}_{\kappa(\mathfrak m_j)/\kappa(\mathfrak m)}
((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}}
\bmod \mathfrak m_j)^{m_j}
\end{equation}
where $m_j = \text{length}_{B_j}(B_j/\pi_j B_j)$ and the product
is taken over the maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_r$ of $B$.
\begin{lemma}
\label{lemma-well-defined-tame-symbol}
The formula (\ref{equation-tame-symbol}) determines a
well defined element of $\kappa(\mathfrak m)^*$. In other words, the
right hand side does not depend on the choice of the
local factorizations or the choice of $B$.
\end{lemma}
\begin{proof}
Independence of choice of factorizations. Suppose we have
a Noetherian $1$-dimensional local ring $B$, elements $a_1, a_2 \in B$,
and nonzerodivisors $\pi, \theta$ such that we can write
$$
a_1 = u_1 \pi^{e_1} = v_1 \theta^{f_1},\quad
a_2 = u_2 \pi^{e_2} = v_2 \theta^{f_2}
$$
with $e_i, f_i \geq 0$ integers and $u_i, v_i$ units in $B$.
Observe that this implies
$$
a_1^{e_2} = u_1^{e_2}u_2^{-e_1}a_2^{e_1},\quad
a_1^{f_2} = v_1^{f_2}v_2^{-f_1}a_2^{f_1}
$$
On the other hand, setting
$m = \text{length}_B(B/\pi B)$ and $k = \text{length}_B(B/\theta B)$
we find $e_2 m = \text{length}_B(B/a_2 B) = f_2 k$.
Expanding $a_1^{e_2m} = a_1^{f_2 k}$ using the above we find
$$
(u_1^{e_2}u_2^{-e_1})^m = (v_1^{f_2}v_2^{-f_1})^k
$$
This proves the desired equality up to signs. To see the signs
work out we have to show $me_1e_2$ is even if and only if
$kf_1f_2$ is even. This follows as both $me_2 = kf_2$ and
$me_1 = kf_1$ (same argument as above).
\medskip\noindent
Independence of choice of $B$. Suppose given two extensions
$A \subset B$ and $A \subset B'$ as in Lemma \ref{lemma-prepare-tame-symbol}.
Then
$$
C = (B \otimes_A B')/(\mathfrak m\text{-power torsion})
$$
will be a third one. Thus we may assume we have
$A \subset B \subset C$ and factorizations over the
local rings of $B$ and we have to show that using
the same factorizations over the local rings of $C$
gives the same element of $\kappa(\mathfrak m)$.
By transitivity of norms
(Fields, Lemma \ref{fields-lemma-trace-and-norm-tower})
this comes down to the following problem:
if $B$ is a Noetherian local ring of dimension $1$
and $\pi \in B$ is a nonzerodivisor, then
$$
\lambda^m = \prod \text{Norm}_{\kappa_k/\kappa}(\lambda)^{m_k}
$$
Here we have used the following notation:
(1) $\kappa$ is the residue field of $B$,
(2) $\lambda$ is an element of $\kappa$,
(3) $\mathfrak m_k \subset C$ are the maximal ideals of $C$,
(4) $\kappa_k = \kappa(\mathfrak m_k)$ is the residue field of
$C_k = C_{\mathfrak m_k}$,
(5) $m = \text{length}_B(B/\pi B)$, and
(6) $m_k = \text{length}_{C_k}(C_k/\pi C_k)$.
The displayed equality holds because
$\text{Norm}_{\kappa_k/\kappa}(\lambda) = \lambda^{[\kappa_k : \kappa]}$
as $\lambda \in \kappa$ and because $m = \sum m_k[\kappa_k:\kappa]$.
First, we have $m = \text{length}_B(B/xB) = \text{length}_B(C/\pi C)$
by Lemma \ref{lemma-compare-periodic-lengths}
and (\ref{equation-multiplicity-coker-ker}).
Finally, we have $\text{length}_B(C/\pi C) = \sum m_k[\kappa_k:\kappa]$
by Algebra, Lemma \ref{algebra-lemma-pushdown-module}.
\end{proof}
\begin{lemma}
\label{lemma-tame-symbol}
The tame symbol (\ref{equation-tame-symbol}) satisfies
(\ref{item-bilinear-better}), (\ref{item-skew-better}),
(\ref{item-normalization}), (\ref{item-1-x-better}) and hence
gives a map $\partial_A : Q(A)^* \times Q(A)^* \to \kappa(\mathfrak m)^*$
satisfying (\ref{item-bilinear}), (\ref{item-skew}), (\ref{item-1-x}).
\end{lemma}
\begin{proof}
Let us prove (\ref{item-bilinear-better}).
Let $a_1, a_2, a_3 \in A$ be nonzerodivisors.
Choose $A \subset B$ as in Lemma \ref{lemma-prepare-tame-symbol}
for $a_1, a_2, a_3$. Then the equality
$$
\partial_A(a_1a_2, a_3) = \partial_A(a_1, a_3) \partial_A(a_2, a_3)
$$
follows from the equality
$$
(-1)^{(e_{1, j} + e_{2, j})e_{3, j}}
(u_{1, j}u_{2, j})^{e_{3, j}}u_{3, j}^{-e_{1, j} - e_{2, j}} =
(-1)^{e_{1, j}e_{3, j}}
u_{1, j}^{e_{3, j}}u_{3, j}^{-e_{1, j}}
(-1)^{e_{2, j}e_{3, j}}
u_{2, j}^{e_{3, j}}u_{3, j}^{-e_{2, j}}
$$
in $B_j$. Properties (\ref{item-skew-better}) and
(\ref{item-normalization}) are equally immediate.
\medskip\noindent
Let us prove (\ref{item-1-x-better}). Let $a_1, a_2, a_1 - a_2 \in A$
be nonzerodivisors and set $a_3 = a_1 - a_2$.
Choose $A \subset B$ as in Lemma \ref{lemma-prepare-tame-symbol}
for $a_1, a_2, a_3$. Then it suffices to show
$$
(-1)^{e_{1, j}e_{2, j} + e_{1, j}e_{3, j} + e_{2, j}e_{3, j} + e_{2, j}}
u_{1, j}^{e_{2, j} - e_{3, j}}
u_{2, j}^{e_{3, j} - e_{1, j}}
u_{3, j}^{e_{1, j} - e_{2, j}} \bmod \mathfrak m_j = 1
$$
This is clear if $e_{1, j} = e_{2, j} = e_{3, j}$.
Say $e_{1, j} > e_{2, j}$. Then we see that $e_{3, j} = e_{2, j}$
because $a_3 = a_1 - a_2$ and we see that $u_{3, j}$
has the same residue class as $-u_{2, j}$. Hence
the formula is true -- the signs work out as well
and this verification is the reason for the choice of signs
in (\ref{equation-tame-symbol}).
The other cases are handled in exactly the same manner.
\end{proof}
\begin{lemma}
\label{lemma-norm-down-tame-symbol}
Let $(A, \mathfrak m)$ be a Noetherian local ring of dimension $1$.
Let $A \subset B$ be a finite ring extension with $B/A$
annihilated by a power of $\mathfrak m$ and $\mathfrak m$ not
an associated prime of $B$.
For $a, b \in A$ nonzerodivisors we have
$$
\partial_A(a, b) = \prod
\text{Norm}_{\kappa(\mathfrak m_j)/\kappa(\mathfrak m)}(\partial_{B_j}(a, b))
$$
where the product is over the maximal ideals $\mathfrak m_j$ of $B$
and $B_j = B_{\mathfrak m_j}$.
\end{lemma}
\begin{proof}
Choose $B_j \subset C_j$ as in
Lemma \ref{lemma-prepare-tame-symbol} for $a, b$.
By Lemma \ref{lemma-glue-at-max} we can choose a finite ring
extension $B \subset C$ with $C_j \cong C_{\mathfrak m_j}$ for all $j$.
Let $\mathfrak m_{j, k} \subset C$ be the maximal ideals of $C$
lying over $\mathfrak m_j$. Let
$$
a = u_{j, k}\pi_{j, k}^{f_{j, k}},\quad
b = v_{j, k}\pi_{j, k}^{g_{j, k}}
$$
be the local factorizations which exist by our choice of
$C_j \cong C_{\mathfrak m_j}$. By definition we have
$$
\partial_A(a, b) =
\prod\nolimits_{j, k}
\text{Norm}_{\kappa(\mathfrak m_{j, k})/\kappa(\mathfrak m)}
((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}}
\bmod \mathfrak m_{j, k})^{m_{j, k}}
$$
and
$$
\partial_{B_j}(a, b) =
\prod\nolimits_k
\text{Norm}_{\kappa(\mathfrak m_{j, k})/\kappa(\mathfrak m_j)}
((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}}
\bmod \mathfrak m_{j, k})^{m_{j, k}}
$$
The result follows by transitivity of norms
for $\kappa(\mathfrak m_{j, k})/\kappa(\mathfrak m_j)/\kappa(\mathfrak m)$, see
Fields, Lemma \ref{fields-lemma-trace-and-norm-tower}.
\end{proof}
\begin{lemma}
\label{lemma-tame-symbol-formally-smooth}