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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 $$\xymatrix{ \ldots \ar[r] & M \ar[r]^\varphi & M \ar[r]^\psi & M \ar[r]^\varphi & M \ar[r] & \ldots }$$ 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 (resp.\ $(2, 1)$-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 $\varphi' \circ f = f \circ \varphi$ and $\psi' \circ g = g \circ \psi$. In fact it is an abelian category, with kernels and cokernels as in Homology, Lemma \ref{homology-lemma-cat-chain-abelian}. Also, note that a special case are the $(2, 1)$-periodic complexes of the form $(M, 0, \psi)$. In this special case we have $$H^0(M, 0, \psi) = \Coker(\psi) , \quad\text{and}\quad H^1(M, 0, \psi) = \Ker(\psi).$$ \begin{definition} \label{definition-periodic-length} Let $R$ be a local ring. Let $(M, N, \varphi, \psi)$ be a $2$-periodic complex over $R$ whose cohomology groups have finite length over $R$. In this case we define the {\it multiplicity} of $(M, N, \varphi, \psi)$ to be the integer $$e_R(M, N, \varphi, \psi) = \text{length}_R(H^0(M, N, \varphi, \psi)) - \text{length}_R(H^1(M, N, \varphi, \psi))$$ We will sometimes (especially in the case of a $(2, 1)$-periodic complex with $\varphi = 0$) call this the {\it Herbrand quotient}\footnote{If the residue field of $R$ is finite with $q$ elements it is customary to call the Herbrand quotient $h(M, N, \varphi, \psi) = q^{e_R(M, N, \varphi, \psi)}$ which is equal to the number of elements of $H^0$ divided by the number of elements of $H^1$.}. \end{definition} \begin{lemma} \label{lemma-periodic-length} Let $R$ be a local ring. \begin{enumerate} \item If $(M, N, \varphi, \psi)$ is a $2$-periodic complex such that $M$, $N$ have finite length. Then $e_R(M, N, \varphi, \psi) = \text{length}_R(M) - \text{length}_R(N)$. \item If $(M, \varphi, \psi)$ is a $(2, 1)$-periodic complex such that $M$ has finite length. Then $e_R(M, \varphi, \psi) = 0$. \item Suppose that we have a short exact sequence of $(2, 1)$-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{enumerate} \end{lemma} \begin{proof} Proof of (3). 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. The proofs of (1) and (2) are omitted. \end{proof} \section{Periodic complexes and determinants} \label{section-periodic-complexes-determinants} \noindent Let $R$ be a local ring with residue field $\kappa$. Let $(M, \varphi, \psi)$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi, \psi)$ is exact. We are going to use the determinant construction to define an invariant of this situation. See Section \ref{section-determinants-finite-length}. Let us abbreviate $K_\varphi = \Ker(\varphi)$, $I_\varphi = \Im(\varphi)$, $K_\psi = \Ker(\psi)$, and $I_\psi = \Im(\psi)$. The short exact sequences $$0 \to K_\varphi \to M \to I_\varphi \to 0, \quad 0 \to K_\psi \to M \to I_\psi \to 0$$ give isomorphisms $$\gamma_\varphi : \det\nolimits_\kappa(K_\varphi) \otimes \det\nolimits_\kappa(I_\varphi) \longrightarrow \det\nolimits_\kappa(M), \quad \gamma_\psi : \det\nolimits_\kappa(K_\psi) \otimes \det\nolimits_\kappa(I_\psi) \longrightarrow \det\nolimits_\kappa(M),$$ see Lemma \ref{lemma-det-exact-sequences}. On the other hand the exactness of the complex gives equalities $K_\varphi = I_\psi$, and $K_\psi = I_\varphi$ and hence an isomorphism $$\sigma : \det\nolimits_\kappa(K_\varphi) \otimes \det\nolimits_\kappa(I_\varphi) \longrightarrow \det\nolimits_\kappa(K_\psi) \otimes \det\nolimits_\kappa(I_\psi)$$ by switching the factors. Using this notation we can define our invariant. \begin{definition} \label{definition-periodic-determinant} Let $R$ be a local ring with residue field $\kappa$. Let $(M, \varphi, \psi)$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi, \psi)$ is exact. The {\it determinant of $(M, \varphi, \psi)$} is the element $$\det\nolimits_\kappa(M, \varphi, \psi) \in \kappa^*$$ such that the composition $$\det\nolimits_\kappa(M) \xrightarrow{\gamma_\psi \circ \sigma \circ \gamma_\varphi^{-1}} \det\nolimits_\kappa(M)$$ is multiplication by $(-1)^{\text{length}_R(I_\varphi)\text{length}_R(I_\psi)} \det\nolimits_\kappa(M, \varphi, \psi)$. \end{definition} \begin{remark} \label{remark-more-elementary} Here is a more down to earth description of the determinant introduced above. Let $R$ be a local ring with residue field $\kappa$. Let $(M, \varphi, \psi)$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi, \psi)$ is exact. Let us abbreviate $I_\varphi = \Im(\varphi)$, $I_\psi = \Im(\psi)$ as above. Assume that $\text{length}_R(I_\varphi) = a$ and $\text{length}_R(I_\psi) = b$, so that $a + b = \text{length}_R(M)$ by exactness. Choose admissible sequences $x_1, \ldots, x_a \in I_\varphi$ and $y_1, \ldots, y_b \in I_\psi$ such that the symbol $[x_1, \ldots, x_a]$ generates $\det_\kappa(I_\varphi)$ and the symbol $[x_1, \ldots, x_b]$ generates $\det_\kappa(I_\psi)$. Choose $\tilde x_i \in M$ such that $\varphi(\tilde x_i) = x_i$. Choose $\tilde y_j \in M$ such that $\psi(\tilde y_j) = y_j$. Then $\det_\kappa(M, \varphi, \psi)$ is characterized by the equality $$[x_1, \ldots, x_a, \tilde y_1, \ldots, \tilde y_b] = (-1)^{ab} \det\nolimits_\kappa(M, \varphi, \psi) [y_1, \ldots, y_b, \tilde x_1, \ldots, \tilde x_a]$$ in $\det_\kappa(M)$. This also explains the sign. \end{remark} \begin{lemma} \label{lemma-periodic-determinant-shift} Let $R$ be a local ring with residue field $\kappa$. Let $(M, \varphi, \psi)$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi, \psi)$ is exact. Then $$\det\nolimits_\kappa(M, \varphi, \psi) \det\nolimits_\kappa(M, \psi, \varphi) = 1.$$ \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-periodic-determinant-sign} Let $R$ be a local ring with residue field $\kappa$. Let $(M, \varphi, \varphi)$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi, \varphi)$ is exact. Then $\text{length}_R(M) = 2 \text{length}_R(\Im(\varphi))$ and $$\det\nolimits_\kappa(M, \varphi, \varphi) = (-1)^{\text{length}_R(\Im(\varphi))} = (-1)^{\frac{1}{2}\text{length}_R(M)}$$ \end{lemma} \begin{proof} Follows directly from the sign rule in the definitions. \end{proof} \begin{lemma} \label{lemma-periodic-determinant-easy-case} Let $R$ be a local ring with residue field $\kappa$. Let $M$ be a finite length $R$-module. \begin{enumerate} \item if $\varphi : M \to M$ is an isomorphism then $\det_\kappa(M, \varphi, 0) = \det_\kappa(\varphi)$. \item if $\psi : M \to M$ is an isomorphism then $\det_\kappa(M, 0, \psi) = \det_\kappa(\psi)^{-1}$. \end{enumerate} \end{lemma} \begin{proof} Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition \ref{definition-periodic-determinant}, identify $K_\varphi = I_\psi = 0$, $I_\varphi = K_\psi = M$. With these identifications, the map $$\gamma_\varphi : \kappa \otimes \det\nolimits_\kappa(M) = \det\nolimits_\kappa(K_\varphi) \otimes \det\nolimits_\kappa(I_\varphi) \longrightarrow \det\nolimits_\kappa(M)$$ is identified with $\det_\kappa(\varphi^{-1})$. On the other hand the map $\gamma_\psi$ is identified with the identity map. Hence $\gamma_\psi \circ \sigma \circ \gamma_\varphi^{-1}$ is equal to $\det_\kappa(\varphi)$ in this case. Whence the result. We omit the proof of (2). \end{proof} \begin{lemma} \label{lemma-periodic-determinant} Let $R$ be a local ring with residue field $\kappa$. Suppose that we have a short exact sequence of $(2, 1)$-periodic complexes $$0 \to (M_1, \varphi_1, \psi_1) \to (M_2, \varphi_2, \psi_2) \to (M_3, \varphi_3, \psi_3) \to 0$$ with all $M_i$ of finite length, and each $(M_1, \varphi_1, \psi_1)$ exact. Then $$\det\nolimits_\kappa(M_2, \varphi_2, \psi_2) = \det\nolimits_\kappa(M_1, \varphi_1, \psi_1) \det\nolimits_\kappa(M_3, \varphi_3, \psi_3).$$ in $\kappa^*$. \end{lemma} \begin{proof} Let us abbreviate $I_{\varphi, i} = \Im(\varphi_i)$, $K_{\varphi, i} = \Ker(\varphi_i)$, $I_{\psi, i} = \Im(\psi_i)$, and $K_{\psi, i} = \Ker(\psi_i)$. Observe that we have a commutative square $$\xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & K_{\varphi, 1} \ar[r] \ar[d] & K_{\varphi, 2} \ar[r] \ar[d] & K_{\varphi, 3} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] \ar[d] & M_2 \ar[r] \ar[d] & M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{\varphi, 1} \ar[r] \ar[d] & I_{\varphi, 2} \ar[r] \ar[d] & I_{\varphi, 3} \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 & }$$ of finite length $R$-modules with exact rows and columns. The top row is exact since it can be identified with the sequence $I_{\psi, 1} \to I_{\psi, 2} \to I_{\psi, 3} \to 0$ of images, and similarly for the bottom row. There is a similar diagram involving the modules $I_{\psi, i}$ and $K_{\psi, i}$. By definition $\det_\kappa(M_2, \varphi_2, \psi_2)$ corresponds, up to a sign, to the composition of the left vertical maps in the following diagram $$\xymatrix{ \det_\kappa(M_1) \otimes \det_\kappa(M_3) \ar[r]^\gamma \ar[d]^{\gamma^{-1} \otimes \gamma^{-1}} & \det_\kappa(M_2) \ar[d]^{\gamma^{-1}} \\ \det\nolimits_\kappa(K_{\varphi, 1}) \otimes \det\nolimits_\kappa(I_{\varphi, 1}) \otimes \det\nolimits_\kappa(K_{\varphi, 3}) \otimes \det\nolimits_\kappa(I_{\varphi, 3}) \ar[d]^{\sigma \otimes \sigma} \ar[r]^-{\gamma \otimes \gamma} & \det\nolimits_\kappa(K_{\varphi, 2}) \otimes \det\nolimits_\kappa(I_{\varphi, 2}) \ar[d]^\sigma \\ \det\nolimits_\kappa(K_{\psi, 1}) \otimes \det\nolimits_\kappa(I_{\psi, 1}) \otimes \det\nolimits_\kappa(K_{\psi, 3}) \otimes \det\nolimits_\kappa(I_{\psi, 3}) \ar[d]^{\gamma \otimes \gamma} \ar[r]^-{\gamma \otimes \gamma} & \det\nolimits_\kappa(K_{\psi, 2}) \otimes \det\nolimits_\kappa(I_{\psi, 2}) \ar[d]^\gamma \\ \det_\kappa(M_1) \otimes \det_\kappa(M_3) \ar[r]^\gamma & \det_\kappa(M_2) }$$ The top and bottom squares are commutative up to sign by applying Lemma \ref{lemma-det-exact-sequences} (2). The middle square is trivially commutative (we are just switching factors). Hence we see that $\det\nolimits_\kappa(M_2, \varphi_2, \psi_2) = \epsilon \det\nolimits_\kappa(M_1, \varphi_1, \psi_1) \det\nolimits_\kappa(M_3, \varphi_3, \psi_3)$ for some sign $\epsilon$. And the sign can be worked out, namely the outer rectangle in the diagram above commutes up to \begin{eqnarray*} \epsilon & = & (-1)^{\text{length}(I_{\varphi, 1})\text{length}(K_{\varphi, 3}) + \text{length}(I_{\psi, 1})\text{length}(K_{\psi, 3})} \\ & = & (-1)^{\text{length}(I_{\varphi, 1})\text{length}(I_{\psi, 3}) + \text{length}(I_{\psi, 1})\text{length}(I_{\varphi, 3})} \end{eqnarray*} (proof omitted). It follows easily from this that the signs work out as well. \end{proof} \begin{example} \label{example-dual-numbers} Let $k$ be a field. Consider the ring $R = k[T]/(T^2)$ of dual numbers over $k$. Denote $t$ the class of $T$ in $R$. Let $M = R$ and $\varphi = ut$, $\psi = vt$ with $u, v \in k^*$. In this case $\det_k(M)$ has generator $e = [t, 1]$. We identify $I_\varphi = K_\varphi = I_\psi = K_\psi = (t)$. Then $\gamma_\varphi(t \otimes t) = u^{-1}[t, 1]$ (since $u^{-1} \in M$ is a lift of $t \in I_\varphi$) and $\gamma_\psi(t \otimes t) = v^{-1}[t, 1]$ (same reason). Hence we see that $\det_k(M, \varphi, \psi) = -u/v \in k^*$. \end{example} \begin{example} \label{example-Zp} Let $R = \mathbf{Z}_p$ and let $M = \mathbf{Z}_p/(p^l)$. Let $\varphi = p^b u$ and $\varphi = p^a v$ with $a, b \geq 0$, $a + b = l$ and $u, v \in \mathbf{Z}_p^*$. Then a computation as in Example \ref{example-dual-numbers} shows that \begin{eqnarray*} \det\nolimits_{\mathbf{F}_p}(\mathbf{Z}_p/(p^l), p^bu, p^av) & = & (-1)^{ab}u^a/v^b \bmod p \\ & = & (-1)^{\text{ord}_p(\alpha)\text{ord}_p(\beta)} \frac{\alpha^{\text{ord}_p(\beta)}}{\beta^{\text{ord}_p(\alpha)}} \bmod p \end{eqnarray*} with $\alpha = p^bu, \beta = p^av \in \mathbf{Z}_p$. See Lemma \ref{lemma-symbol-is-usual-tame-symbol} for a more general case (and a proof). \end{example} \begin{example} \label{example-generic-vector-space} Let $R = k$ be a field. Let $M = k^{\oplus a} \oplus k^{\oplus b}$ be $l = a + b$ dimensional. Let $\varphi$ and $\psi$ be the following diagonal matrices $$\varphi = \text{diag}(u_1, \ldots, u_a, 0, \ldots, 0), \quad \psi = \text{diag}(0, \ldots, 0, v_1, \ldots, v_b)$$ with $u_i, v_j \in k^*$. In this case we have $$\det\nolimits_k(M, \varphi, \psi) = \frac{u_1 \ldots u_a}{v_1 \ldots v_b}.$$ This can be seen by a direct computation or by computing in case $l = 1$ and using the additivity of Lemma \ref{lemma-periodic-determinant}. \end{example} \begin{example} \label{example-special-vector-space} Let $R = k$ be a field. Let $M = k^{\oplus a} \oplus k^{\oplus a}$ be $l = 2a$ dimensional. Let $\varphi$ and $\psi$ be the following block matrices $$\varphi = \left( \begin{matrix} 0 & U \\ 0 & 0 \end{matrix} \right), \quad \psi = \left( \begin{matrix} 0 & V \\ 0 & 0 \end{matrix} \right),$$ with $U, V \in \text{Mat}(a \times a, k)$ invertible. In this case we have $$\det\nolimits_k(M, \varphi, \psi) = (-1)^a\frac{\det(U)}{\det(V)}.$$ This can be seen by a direct computation. The case $a = 1$ is similar to the computation in Example \ref{example-dual-numbers}. \end{example} \begin{example} \label{example-a-la-oort} Let $R = k$ be a field. Let $M = k^{\oplus 4}$. Let $$\varphi = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ u_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & u_2 & 0 \end{matrix} \right) \quad \varphi = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & v_2 & 0 \\ 0 & 0 & 0 & 0 \\ v_1 & 0 & 0 & 0 \end{matrix} \right) \quad$$ with $u_1, u_2, v_1, v_2 \in k^*$. Then we have $$\det\nolimits_k(M, \varphi, \psi) = -\frac{u_1u_2}{v_1v_2}.$$ \end{example} \noindent Next we come to the analogue of the fact that the determinant of a composition of linear endomorphisms is the product of the determinants. To avoid very long formulae we write $I_\varphi = \Im(\varphi)$, and $K_\varphi = \Ker(\varphi)$ for any $R$-module map $\varphi : M \to M$. We also denote $\varphi\psi = \varphi \circ \psi$ for a pair of morphisms $\varphi, \psi : M \to M$. \begin{lemma} \label{lemma-multiplicativity-determinant} Let $R$ be a local ring with residue field $\kappa$. Let $M$ be a finite length $R$-module. Let $\alpha, \beta, \gamma$ be endomorphisms of $M$. Assume that \begin{enumerate} \item $I_\alpha = K_{\beta\gamma}$, and similarly for any permutation of $\alpha, \beta, \gamma$, \item $K_\alpha = I_{\beta\gamma}$, and similarly for any permutation of $\alpha, \beta, \gamma$. \end{enumerate} Then \begin{enumerate} \item The triple $(M, \alpha, \beta\gamma)$ is an exact $(2, 1)$-periodic complex. \item The triple $(I_\gamma, \alpha, \beta)$ is an exact $(2, 1)$-periodic complex. \item The triple $(M/K_\beta, \alpha, \gamma)$ is an exact $(2, 1)$-periodic complex. \item We have $$\det\nolimits_\kappa(M, \alpha, \beta\gamma) = \det\nolimits_\kappa(I_\gamma, \alpha, \beta) \det\nolimits_\kappa(M/K_\beta, \alpha, \gamma).$$ \end{enumerate} \end{lemma} \begin{proof} It is clear that the assumptions imply part (1) of the lemma. \medskip\noindent To see part (1) note that the assumptions imply that $I_{\gamma\alpha} = I_{\alpha\gamma}$, and similarly for kernels and any other pair of morphisms. Moreover, we see that $I_{\gamma\beta} =I_{\beta\gamma} = K_\alpha \subset I_\gamma$ and similarly for any other pair. In particular we get a short exact sequence $$0 \to I_{\beta\gamma} \to I_\gamma \xrightarrow{\alpha} I_{\alpha\gamma} \to 0$$ and similarly we get a short exact sequence $$0 \to I_{\alpha\gamma} \to I_\gamma \xrightarrow{\beta} I_{\beta\gamma} \to 0.$$ This proves $(I_\gamma, \alpha, \beta)$ is an exact $(2, 1)$-periodic complex. Hence part (2) of the lemma holds. \medskip\noindent To see that $\alpha$, $\gamma$ give well defined endomorphisms of $M/K_\beta$ we have to check that $\alpha(K_\beta) \subset K_\beta$ and $\gamma(K_\beta) \subset K_\beta$. This is true because $\alpha(K_\beta) = \alpha(I_{\gamma\alpha}) = I_{\alpha\gamma\alpha} \subset I_{\alpha\gamma} = K_\beta$, and similarly in the other case. The kernel of the map $\alpha : M/K_\beta \to M/K_\beta$ is $K_{\beta\alpha}/K_\beta = I_\gamma/K_\beta$. Similarly, the kernel of $\gamma : M/K_\beta \to M/K_\beta$ is equal to $I_\alpha/K_\beta$. Hence we conclude that (3) holds. \medskip\noindent We introduce $r = \text{length}_R(K_\alpha)$, $s = \text{length}_R(K_\beta)$ and $t = \text{length}_R(K_\gamma)$. By the exact sequences above and our hypotheses we have $\text{length}_R(I_\alpha) = s + t$, $\text{length}_R(I_\beta) = r + t$, $\text{length}_R(I_\gamma) = r + s$, and $\text{length}(M) = r + s + t$. Choose \begin{enumerate} \item an admissible sequence $x_1, \ldots, x_r \in K_\alpha$ generating $K_\alpha$ \item an admissible sequence $y_1, \ldots, y_s \in K_\beta$ generating $K_\beta$, \item an admissible sequence $z_1, \ldots, z_t \in K_\gamma$ generating $K_\gamma$, \item elements $\tilde x_i \in M$ such that $\beta\gamma\tilde x_i = x_i$, \item elements $\tilde y_i \in M$ such that $\alpha\gamma\tilde y_i = y_i$, \item elements $\tilde z_i \in M$ such that $\beta\alpha\tilde z_i = z_i$. \end{enumerate} With these choices the sequence $y_1, \ldots, y_s, \alpha\tilde z_1, \ldots, \alpha\tilde z_t$ is an admissible sequence in $I_\alpha$ generating it. Hence, by Remark \ref{remark-more-elementary} the determinant $D = \det_\kappa(M, \alpha, \beta\gamma)$ is the unique element of $\kappa^*$ such that \begin{align*} [y_1, \ldots, y_s, \alpha\tilde z_1, \ldots, \alpha\tilde z_s, \tilde x_1, \ldots, \tilde x_r] \\ = (-1)^{r(s + t)} D [x_1, \ldots, x_r, \gamma\tilde y_1, \ldots, \gamma\tilde y_s, \tilde z_1, \ldots, \tilde z_t] \end{align*} By the same remark, we see that $D_1 = \det_\kappa(M/K_\beta, \alpha, \gamma)$ is characterized by $$[y_1, \ldots, y_s, \alpha\tilde z_1, \ldots, \alpha\tilde z_t, \tilde x_1, \ldots, \tilde x_r] = (-1)^{rt} D_1 [y_1, \ldots, y_s, \gamma\tilde x_1, \ldots, \gamma\tilde x_r, \tilde z_1, \ldots, \tilde z_t]$$ By the same remark, we see that $D_2 = \det_\kappa(I_\gamma, \alpha, \beta)$ is characterized by $$[y_1, \ldots, y_s, \gamma\tilde x_1, \ldots, \gamma\tilde x_r, \tilde z_1, \ldots, \tilde z_t] = (-1)^{rs} D_2 [x_1, \ldots, x_r, \gamma\tilde y_1, \ldots, \gamma\tilde y_s, \tilde z_1, \ldots, \tilde z_t]$$ Combining the formulas above we see that $D = D_1 D_2$ as desired. \end{proof} \begin{lemma} \label{lemma-tricky} Let $R$ be a local ring with residue field $\kappa$. Let $\alpha : (M, \varphi, \psi) \to (M', \varphi', \psi')$ be a morphism of $(2, 1)$-periodic complexes over $R$. Assume \begin{enumerate} \item $M$, $M'$ have finite length, \item $(M, \varphi, \psi)$, $(M', \varphi', \psi')$ are exact, \item the maps $\varphi$, $\psi$ induce the zero map on $K = \Ker(\alpha)$, and \item the maps $\varphi$, $\psi$ induce the zero map on $Q = \Coker(\alpha)$. \end{enumerate} Denote $N = \alpha(M) \subset M'$. We obtain two short exact sequences of $(2, 1)$-periodic complexes $$\begin{matrix} 0 \to (N, \varphi', \psi') \to (M', \varphi', \psi') \to (Q, 0, 0) \to 0 \\ 0 \to (K, 0, 0) \to (M, \varphi, \psi) \to (N, \varphi', \psi') \to 0 \end{matrix}$$ which induce two isomorphisms $\alpha_i : Q \to K$, $i = 0, 1$. Then $$\det\nolimits_\kappa(M, \varphi, \psi) = \det\nolimits_\kappa(\alpha_0^{-1} \circ \alpha_1) \det\nolimits_\kappa(M', \varphi', \psi')$$ In particular, if $\alpha_0 = \alpha_1$, then $\det\nolimits_\kappa(M, \varphi, \psi) = \det\nolimits_\kappa(M', \varphi', \psi')$. \end{lemma} \begin{proof} There are (at least) two ways to prove this lemma. One is to produce an enormous commutative diagram using the properties of the determinants. The other is to use the characterization of the determinants in terms of admissible sequences of elements. It is the second approach that we will use. \medskip\noindent First let us explain precisely what the maps $\alpha_i$ are. Namely, $\alpha_0$ is the composition $$\alpha_0 : Q = H^0(Q, 0, 0) \to H^1(N, \varphi', \psi') \to H^2(K, 0, 0) = K$$ and $\alpha_1$ is the composition $$\alpha_1 : Q = H^1(Q, 0, 0) \to H^2(N, \varphi', \psi') \to H^3(K, 0, 0) = K$$ coming from the boundary maps of the short exact sequences of complexes displayed in the lemma. The fact that the complexes $(M, \varphi, \psi)$, $(M', \varphi', \psi')$ are exact implies these maps are isomorphisms. \medskip\noindent We will use the notation $I_\varphi = \Im(\varphi)$, $K_\varphi = \Ker(\varphi)$ and similarly for the other maps. Exactness for $M$ and $M'$ means that $K_\varphi = I_\psi$ and three similar equalities. We introduce $k = \text{length}_R(K)$, $a = \text{length}_R(I_\varphi)$, $b = \text{length}_R(I_\psi)$. Then we see that $\text{length}_R(M) = a + b$, and $\text{length}_R(N) = a + b - k$, $\text{length}_R(Q) = k$ and $\text{length}_R(M') = a + b$. The exact sequences below will show that also $\text{length}_R(I_{\varphi'}) = a$ and $\text{length}_R(I_{\psi'}) = b$. \medskip\noindent The assumption that $K \subset K_\varphi = I_\psi$ means that $\varphi$ factors through $N$ to give an exact sequence $$0 \to \alpha(I_\psi) \to N \xrightarrow{\varphi\alpha^{-1}} I_\psi \to 0.$$ Here $\varphi\alpha^{-1}(x') = y$ means $x' = \alpha(x)$ and $y = \varphi(x)$. Similarly, we have $$0 \to \alpha(I_\varphi) \to N \xrightarrow{\psi\alpha^{-1}} I_\varphi \to 0.$$ The assumption that $\psi'$ induces the zero map on $Q$ means that $I_{\psi'} = K_{\varphi'} \subset N$. This means the quotient $\varphi'(N) \subset I_{\varphi'}$ is identified with $Q$. Note that $\varphi'(N) = \alpha(I_\varphi)$. Hence we conclude there is an isomorphism $$\varphi' : Q \to I_{\varphi'}/\alpha(I_\varphi)$$ simply described by $\varphi'(x' \bmod N) = \varphi'(x') \bmod \alpha(I_\varphi)$. In exactly the same way we get $$\psi' : Q \to I_{\psi'}/\alpha(I_\psi)$$ Finally, note that $\alpha_0$ is the composition $$\xymatrix{ Q \ar[r]^-{\varphi'} & I_{\varphi'}/\alpha(I_\varphi) \ar[rrr]^-{\psi\alpha^{-1}|_{I_{\varphi'}/\alpha(I_\varphi)}} & & & K }$$ and similarly $\alpha_1 = \varphi\alpha^{-1}|_{I_{\psi'}/\alpha(I_\psi)} \circ \psi'$. \medskip\noindent To shorten the formulas below we are going to write $\alpha x$ instead of $\alpha(x)$ in the following. No confusion should result since all maps are indicated by Greek letters and elements by Roman letters. We are going to choose \begin{enumerate} \item an admissible sequence $z_1, \ldots, z_k \in K$ generating $K$, \item elements $z'_i \in M$ such that $\varphi z'_i = z_i$, \item elements $z''_i \in M$ such that $\psi z''_i = z_i$, \item elements $x_{k + 1}, \ldots, x_a \in I_\varphi$ such that $z_1, \ldots, z_k, x_{k + 1}, \ldots, x_a$ is an admissible sequence generating $I_\varphi$, \item elements $\tilde x_i \in M$ such that $\varphi \tilde x_i = x_i$, \item elements $y_{k + 1}, \ldots, y_b \in I_\psi$ such that $z_1, \ldots, z_k, y_{k + 1}, \ldots, y_b$ is an admissible sequence generating $I_\psi$, \item elements $\tilde y_i \in M$ such that $\psi \tilde y_i = y_i$, and \item elements $w_1, \ldots, w_k \in M'$ such that $w_1 \bmod N, \ldots, w_k \bmod N$ are an admissible sequence in $Q$ generating $Q$. \end{enumerate} By Remark \ref{remark-more-elementary} the element $D = \det_\kappa(M, \varphi, \psi) \in \kappa^*$ is characterized by \begin{eqnarray*} & & [z_1, \ldots, z_k, x_{k + 1}, \ldots, x_a, z''_1, \ldots, z''_k, \tilde y_{k + 1}, \ldots, \tilde y_b] \\ & = & (-1)^{ab} D [z_1, \ldots, z_k, y_{k + 1}, \ldots, y_b, z'_1, \ldots, z'_k, \tilde x_{k + 1}, \ldots, \tilde x_a] \end{eqnarray*} Note that by the discussion above $\alpha x_{k + 1}, \ldots, \alpha x_a, \varphi w_1, \ldots, \varphi w_k$ is an admissible sequence generating $I_{\varphi'}$ and $\alpha y_{k + 1}, \ldots, \alpha y_b, \psi w_1, \ldots, \psi w_k$ is an admissible sequence generating $I_{\psi'}$. Hence by Remark \ref{remark-more-elementary} the element $D' = \det_\kappa(M', \varphi', \psi') \in \kappa^*$ is characterized by \begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots, \alpha x_a, \varphi' w_1, \ldots, \varphi' w_k, \alpha \tilde y_{k + 1}, \ldots, \alpha \tilde y_b, w_1, \ldots, w_k] \\ & = & (-1)^{ab} D' [\alpha y_{k + 1}, \ldots, \alpha y_b, \psi' w_1, \ldots, \psi' w_k, \alpha \tilde x_{k + 1}, \ldots, \alpha \tilde x_a, w_1, \ldots, w_k] \end{eqnarray*} Note how in the first, resp.\ second displayed formula the the first, resp.\ last $k$ entries of the symbols on both sides are the same. Hence these formulas are really equivalent to the equalities \begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots, \alpha x_a, \alpha z''_1, \ldots, \alpha z''_k, \alpha \tilde y_{k + 1}, \ldots, \alpha \tilde y_b] \\ & = & (-1)^{ab} D [\alpha y_{k + 1}, \ldots, \alpha y_b, \alpha z'_1, \ldots, \alpha z'_k, \alpha \tilde x_{k + 1}, \ldots, \alpha \tilde x_a] \end{eqnarray*} and \begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots, \alpha x_a, \varphi' w_1, \ldots, \varphi' w_k, \alpha \tilde y_{k + 1}, \ldots, \alpha \tilde y_b] \\ & = & (-1)^{ab} D' [\alpha y_{k + 1}, \ldots, \alpha y_b, \psi' w_1, \ldots, \psi' w_k, \alpha \tilde x_{k + 1}, \ldots, \alpha \tilde x_a] \end{eqnarray*} in $\det_\kappa(N)$. Note that $\varphi' w_1, \ldots, \varphi' w_k$ and $\alpha z''_1, \ldots, z''_k$ are admissible sequences generating the module $I_{\varphi'}/\alpha(I_\varphi)$. Write $$[\varphi' w_1, \ldots, \varphi' w_k] = \lambda_0 [\alpha z''_1, \ldots, \alpha z''_k]$$ in $\det_\kappa(I_{\varphi'}/\alpha(I_\varphi))$ for some $\lambda_0 \in \kappa^*$. Similarly, write $$[\psi' w_1, \ldots, \psi' w_k] = \lambda_1 [\alpha z'_1, \ldots, \alpha z'_k]$$ in $\det_\kappa(I_{\psi'}/\alpha(I_\psi))$ for some $\lambda_1 \in \kappa^*$. On the one hand it is clear that $$\alpha_i([w_1, \ldots, w_k]) = \lambda_i[z_1, \ldots, z_k]$$ for $i = 0, 1$ by our description of $\alpha_i$ above, which means that $$\det\nolimits_\kappa(\alpha_0^{-1} \circ \alpha_1) = \lambda_1/\lambda_0$$ and on the other hand it is clear that \begin{eqnarray*} & & \lambda_0 [\alpha x_{k + 1}, \ldots, \alpha x_a, \alpha z''_1, \ldots, \alpha z''_k, \alpha \tilde y_{k + 1}, \ldots, \alpha \tilde y_b] \\ & = & [\alpha x_{k + 1}, \ldots, \alpha x_a, \varphi' w_1, \ldots, \varphi' w_k, \alpha \tilde y_{k + 1}, \ldots, \alpha \tilde y_b] \end{eqnarray*} and \begin{eqnarray*} & & \lambda_1[\alpha y_{k + 1}, \ldots, \alpha y_b, \alpha z'_1, \ldots, \alpha z'_k, \alpha \tilde x_{k + 1}, \ldots, \alpha \tilde x_a] \\ & = & [\alpha y_{k + 1}, \ldots, \alpha y_b, \psi' w_1, \ldots, \psi' w_k, \alpha \tilde x_{k + 1}, \ldots, \alpha \tilde x_a] \end{eqnarray*} which imply $\lambda_0 D = \lambda_1 D'$. The lemma follows. \end{proof} \section{Symbols} \label{section-symbols} \noindent The correct generality for this construction is perhaps the situation of the following lemma. \begin{lemma} \label{lemma-pre-symbol} Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module of dimension $1$. Assume $\varphi, \psi : M \to M$ are two injective $A$-module maps, and assume $\varphi(\psi(M)) = \psi(\varphi(M))$, for example if $\varphi$ and $\psi$ commute. Then $\text{length}_R(M/\varphi\psi M) < \infty$ and $(M/\varphi\psi M, \varphi, \psi)$ is an exact $(2, 1)$-periodic complex. \end{lemma} \begin{proof} Let $\mathfrak q$ be a minimal prime of the support of $M$. Then $M_{\mathfrak q}$ is a finite length $A_{\mathfrak q}$-module, see Algebra, Lemma \ref{algebra-lemma-support-point}. Hence both $\varphi$ and $\psi$ induce isomorphisms $M_{\mathfrak q} \to M_{\mathfrak q}$. Thus the support of $M/\varphi\psi M$ is $\{\mathfrak m_A\}$ and hence it has finite length (see lemma cited above). Finally, the kernel of $\varphi$ on $M/\varphi\psi M$ is clearly $\psi M/\varphi\psi M$, and hence the kernel of $\varphi$ is the image of $\psi$ on $M/\varphi\psi M$. Similarly the other way since $M/\varphi\psi M = M/\psi\varphi M$ by assumption. \end{proof} \begin{lemma} \label{lemma-symbol-defined} Let $A$ be a Noetherian local ring. Let $a, b \in A$. \begin{enumerate} \item If $M$ is a finite $A$-module of dimension $1$ such that $a, b$ are nonzerodivisors on $M$, then $\text{length}_A(M/abM) < \infty$ and $(M/abM, a, b)$ is a $(2, 1)$-periodic exact complex. \item If $a, b$ are nonzerodivisors and $\dim(A) = 1$ then $\text{length}_A(A/(ab)) < \infty$ and $(A/(ab), a, b)$ is a $(2, 1)$-periodic exact complex. \end{enumerate} In particular, in these cases $\det_\kappa(M/abM, a, b) \in \kappa^*$, resp.\ $\det_\kappa(A/(ab), a, b) \in \kappa^*$ are defined. \end{lemma} \begin{proof} Follows from Lemma \ref{lemma-pre-symbol}. \end{proof} \begin{definition} \label{definition-symbol-M} Let $A$ be a Noetherian local ring with residue field $\kappa$. Let $a, b \in A$. Let $M$ be a finite $A$-module of dimension $1$ such that $a, b$ are nonzerodivisors on $M$. We define the {\it symbol associated to $M, a, b$} to be the element $$d_M(a, b) = \det\nolimits_\kappa(M/abM, a, b) \in \kappa^*$$ \end{definition} \begin{lemma} \label{lemma-multiplicativity-symbol} Let $A$ be a Noetherian local ring. Let $a, b, c \in A$. Let $M$ be a finite $A$-module with $\dim(\text{Supp}(M)) = 1$. Assume $a, b, c$ are nonzerodivisors on $M$. Then $$d_M(a, bc) = d_M(a, b) d_M(a, c)$$ and $d_M(a, b)d_M(b, a) = 1$. \end{lemma} \begin{proof} The first statement follows from Lemma \ref{lemma-multiplicativity-determinant} applied to $M/abcM$ and endomorphisms $\alpha, \beta, \gamma$ given by multiplication by $a, b, c$. The second comes from Lemma \ref{lemma-periodic-determinant-shift}. \end{proof} \begin{definition} \label{definition-tame-symbol} Let $A$ be a Noetherian local domain of dimension $1$ with residue field $\kappa$. Let $K$ be the fraction field of $A$. We define the {\it tame symbol} of $A$ to be the map $$K^* \times K^* \longrightarrow \kappa^*, \quad (x, y) \longmapsto d_A(x, y)$$ where $d_A(x, y)$ is extended to $K^* \times K^*$ by the multiplicativity of Lemma \ref{lemma-multiplicativity-symbol}. \end{definition} \noindent It is clear that we may extend more generally $d_M(-, -)$ to certain rings of fractions of $A$ (even if $A$ is not a domain). \begin{lemma} \label{lemma-symbol-when-equal} Let $A$ be a Noetherian local ring and $M$ a finite $A$-module of dimension $1$. Let $a \in A$ be a nonzerodivisor on $M$. Then $d_M(a, a) = (-1)^{\text{length}_A(M/aM)}$. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-periodic-determinant-sign}. \end{proof} \begin{lemma} \label{lemma-symbol-when-one-is-a-unit} Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module of dimension $1$. Let $b \in A$ be a nonzerodivisor on $M$, and let $u \in A^*$. Then $$d_M(u, b) = u^{\text{length}_A(M/bM)} \bmod \mathfrak m_A.$$ In particular, if $M = A$, then $d_A(u, b) = u^{\text{ord}_A(b)} \bmod \mathfrak m_A$. \end{lemma} \begin{proof} Note that in this case $M/ubM = M/bM$ on which multiplication by $b$ is zero. Hence $d_M(u, b) = \det_\kappa(u|_{M/bM})$ by Lemma \ref{lemma-periodic-determinant-easy-case}. The lemma then follows from Lemma \ref{lemma-times-u-determinant}. \end{proof} \begin{lemma} \label{lemma-symbol-short-exact-sequence} Let $A$ be a Noetherian local ring. Let $a, b \in A$. Let $$0 \to M \to M' \to M'' \to 0$$ be a short exact sequence of $A$-modules of dimension $1$ such that $a, b$ are nonzerodivisors on all three $A$-modules. Then $$d_{M'}(a, b) = d_M(a, b) d_{M''}(a, b)$$ in $\kappa^*$. \end{lemma} \begin{proof} It is easy to see that this leads to a short exact sequence of exact $(2, 1)$-periodic complexes $$0 \to (M/abM, a, b) \to (M'/abM', a, b) \to (M''/abM'', a, b) \to 0$$ Hence the lemma follows from Lemma \ref{lemma-periodic-determinant}. \end{proof} \begin{lemma} \label{lemma-symbol-compare-modules} Let $A$ be a Noetherian local ring. Let $\alpha : M \to M'$ be a homomorphism of finite $A$-modules of dimension $1$. Let $a, b \in A$. Assume \begin{enumerate} \item $a$, $b$ are nonzerodivisors on both $M$ and $M'$, and \item $\dim(\Ker(\alpha)), \dim(\Coker(\alpha)) \leq 0$. \end{enumerate} Then $d_M(a, b) = d_{M'}(a, b)$. \end{lemma} \begin{proof} If $a \in A^*$, then the equality follows from the equality $\text{length}(M/bM) = \text{length}(M'/bM')$ and Lemma \ref{lemma-symbol-when-one-is-a-unit}. Similarly if $b$ is a unit the lemma holds as well (by the symmetry of Lemma \ref{lemma-multiplicativity-symbol}). Hence we may assume that $a, b \in \mathfrak m_A$. This in particular implies that $\mathfrak m$ is not an associated prime of $M$, and hence $\alpha : M \to M'$ is injective. This permits us to think of $M$ as a submodule of $M'$. By assumption $M'/M$ is a finite $A$-module with support $\{\mathfrak m_A\}$ and hence has finite length. Note that for any third module $M''$ with $M \subset M'' \subset M'$ the maps $M \to M''$ and $M'' \to M'$ satisfy the assumptions of the lemma as well. This reduces us, by induction on the length of $M'/M$, to the case where $\text{length}_A(M'/M) = 1$. Finally, in this case consider the map $$\overline{\alpha} : M/abM \longrightarrow M'/abM'.$$ By construction the cokernel $Q$ of $\overline{\alpha}$ has length $1$. Since $a, b \in \mathfrak m_A$, they act trivially on $Q$. It also follows that the kernel $K$ of $\overline{\alpha}$ has length $1$ and hence also $a$, $b$ act trivially on $K$. Hence we may apply Lemma \ref{lemma-tricky}. Thus it suffices to see that the two maps $\alpha_i : Q \to K$ are the same. In fact, both maps are equal to the map $q = x' \bmod \Im(\overline{\alpha}) \mapsto abx' \in K$. We omit the verification. \end{proof} \begin{lemma} \label{lemma-compute-symbol-M} Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module with $\dim(\text{Supp}(M)) = 1$. Let $a, b \in A$ nonzerodivisors on $M$. Let $\mathfrak q_1, \ldots, \mathfrak q_t$ be the minimal primes in the support of $M$. Then $$d_M(a, b) = \prod\nolimits_{i = 1, \ldots, t} d_{A/\mathfrak q_i}(a, b)^{ \text{length}_{A_{\mathfrak q_i}}(M_{\mathfrak q_i})}$$ as elements of $\kappa^*$. \end{lemma} \begin{proof} Choose a filtration by $A$-submodules $$0 = M_0 \subset M_1 \subset \ldots \subset M_n = M$$ such that each quotient $M_j/M_{j - 1}$ is isomorphic to $A/\mathfrak p_j$ for some prime ideal $\mathfrak p_j$ of $A$. See Algebra, Lemma \ref{algebra-lemma-filter-Noetherian-module}. For each $j$ we have either $\mathfrak p_j = \mathfrak q_i$ for some $i$, or $\mathfrak p_j = \mathfrak m_A$. Moreover, for a fixed $i$, the number of $j$ such that $\mathfrak p_j = \mathfrak q_i$ is equal to $\text{length}_{A_{\mathfrak q_i}}(M_{\mathfrak q_i})$ by Algebra, Lemma \ref{algebra-lemma-filter-minimal-primes-in-support}. Hence $d_{M_j}(a, b)$ is defined for each $j$ and $$d_{M_j}(a, b) = \left\{ \begin{matrix} d_{M_{j - 1}}(a, b) d_{A/\mathfrak q_i}(a, b) & \text{if} & \mathfrak p_j = \mathfrak q_i \\ d_{M_{j - 1}}(a, b) & \text{if} & \mathfrak p_j = \mathfrak m_A \end{matrix} \right.$$ by Lemma \ref{lemma-symbol-short-exact-sequence} in the first instance and Lemma \ref{lemma-symbol-compare-modules} in the second. Hence the lemma. \end{proof} \begin{lemma} \label{lemma-symbol-is-usual-tame-symbol} Let $A$ be a discrete valuation ring with fraction field $K$. For nonzero $x, y \in K$ we have $$d_A(x, y) = (-1)^{\text{ord}_A(x)\text{ord}_A(y)} \frac{x^{\text{ord}_A(y)}}{y^{\text{ord}_A(x)}} \bmod \mathfrak m_A,$$ in other words the symbol is equal to the usual tame symbol. \end{lemma} \begin{proof} By multiplicativity it suffices to prove this when $x, y \in A$. Let $t \in A$ be a uniformizer. Write $x = t^bu$ and $y = t^bv$ for some $a, b \geq 0$ and $u, v \in A^*$. Set $l = a + b$. Then $t^{l - 1}, \ldots, t^b$ is an admissible sequence in $(x)/(xy)$ and $t^{l - 1}, \ldots, t^a$ is an admissible sequence in $(y)/(xy)$. Hence by Remark \ref{remark-more-elementary} we see that $d_A(x, y)$ is characterized by the equation $$[t^{l - 1}, \ldots, t^b, v^{-1}t^{b - 1}, \ldots, v^{-1}] = (-1)^{ab} d_A(x, y) [t^{l - 1}, \ldots, t^a, u^{-1}t^{a - 1}, \ldots, u^{-1}].$$ Hence by the admissible relations for the symbols $[x_1, \ldots, x_l]$ we see that $$d_A(x, y) = (-1)^{ab} u^a/v^b \bmod \mathfrak m_A$$ as desired. \end{proof} \noindent We add the following lemma here. It is very similar to Algebra, Lemma \ref{algebra-lemma-nonregular-dimension-one}. \begin{lemma} \label{lemma-Noetherian-domain-dim-1-two-elements} Let $R$ be a local Noetherian domain of dimension $1$ with maximal ideal $\mathfrak m$. Let $a, b \in \mathfrak m$ be nonzero. There exists a finite ring extension $R \subset R'$ with same field of fractions, and $t, a', b' \in R'$ such that $a = ta'$ and $b = tb'$ and $R' = a'R' + b'R'$. \end{lemma} \begin{proof} 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-blow-up-integral-scheme} this is an integral scheme. The morphism $X \to \Spec(R)$ is projective by Divisors, Lemma \ref{divisors-lemma-blowing-up-projective}. By Algebra, Lemma \ref{algebra-lemma-finite-in-codim-1} and the fact that $X$ is quasi-compact we see that the fibre of $X \to \Spec(R)$ over $\mathfrak m$ is finite. By Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine} there exists an affine open $U \subset X$ containing this fibre. Hence $X = U$ because $X \to \Spec(R)$ is closed. In other words $X$ is affine, say $X = \Spec(R')$. By Morphisms, Lemma \ref{morphisms-lemma-locally-finite-type-characterize} we see that $R \to R'$ is of finite type. Since $X \to \Spec(R)$ is proper and affine it is integral (see Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}). Hence $R \to R'$ is of finite type and integral, hence finite (Algebra, Lemma \ref{algebra-lemma-characterize-finite-in-terms-of-integral}). 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-symbol-is-steinberg-prepare} Let $A$ be a Noetherian local ring. Let $a, b \in A$. Let $M$ be a finite $A$-module of dimension $1$ on which each of $a$, $b$, $b - a$ are nonzerodivisors. Then $$d_M(a, b - a)d_M(b, b) = d_M(b, b - a)d_M(a, b)$$ in $\kappa^*$. \end{lemma} \begin{proof} By Lemma \ref{lemma-compute-symbol-M} it suffices to show the relation when $M = A/\mathfrak q$ for some prime $\mathfrak q \subset A$ with $\dim(A/\mathfrak q) = 1$. \medskip\noindent In case $M = A/\mathfrak q$ we may replace $A$ by $A/\mathfrak q$ and $a, b$ by their images in $A/\mathfrak q$. Hence we may assume $A = M$ and $A$ a local Noetherian domain of dimension $1$. The reason is that the residue field $\kappa$ of $A$ and $A/\mathfrak q$ are the same and that for any $A/\mathfrak q$-module $M$ the determinant taken over $A$ or over $A/\mathfrak q$ are canonically identified. See Lemma \ref{lemma-determinant-quotient-ring}. \medskip\noindent It suffices to show the relation when both $a, b$ are in the maximal ideal. Namely, the case where one or both are units follows from Lemmas \ref{lemma-symbol-when-one-is-a-unit} and \ref{lemma-symbol-when-equal}. \medskip\noindent Choose an extension $A \subset A'$ and factorizations $a = ta'$, $b = tb'$ as in Lemma \ref{lemma-Noetherian-domain-dim-1-two-elements}. Note that also $b - a = t(b' - a')$ and that $A' = (a', b') = (a', b' - a') = (b' - a', b')$. Here and in the following we think of $A'$ as an $A$-module and $a, b, a', b', t$ as $A$-module endomorphisms of $A'$. We will use the notation $d^A_{A'}(a', b')$ and so on to indicate $$d^A_{A'}(a', b') = \det\nolimits_\kappa(A'/a'b'A', a', b')$$ which is defined by Lemma \ref{lemma-pre-symbol}. The upper index ${}^A$ is used to distinguish this from the already defined symbol $d_{A'}(a', b')$ which is different (for example because it has values in the residue field of $A'$ which may be different from $\kappa$). By Lemma \ref{lemma-symbol-compare-modules} we see that $d_A(a, b) = d^A_{A'}(a, b)$, and similarly for the other combinations. Using this and multiplicativity we see that it suffices to prove $$d^A_{A'}(a', b' - a') d^A_{A'}(b', b') = d^A_{A'}(b', b' - a') d^A_{A'}(a', b')$$ Now, since $(a', b') = A'$ and so on we have $$\begin{matrix} A'/(a'(b' - a')) & \cong & A'/(a') \oplus A'/(b' - a') \\ A'/(b'(b' - a')) & \cong & A'/(b') \oplus A'/(b' - a') \\ A'/(a'b') & \cong & A'/(a') \oplus A'/(b') \end{matrix}$$ Moreover, note that multiplication by $b' - a'$ on $A/(a')$ is equal to multiplication by $b'$, and that multiplication by $b' - a'$ on $A/(b')$ is equal to multiplication by $-a'$. Using Lemmas \ref{lemma-periodic-determinant-easy-case} and \ref{lemma-periodic-determinant} we conclude $$\begin{matrix} d^A_{A'}(a', b' - a') & = & \det\nolimits_\kappa(b'|_{A'/(a')})^{-1} \det\nolimits_\kappa(a'|_{A'/(b' - a')}) \\ d^A_{A'}(b', b' - a') & = & \det\nolimits_\kappa(-a'|_{A'/(b')})^{-1} \det\nolimits_\kappa(b'|_{A'/(b' - a')}) \\ d^A_{A'}(a', b') & = & \det\nolimits_\kappa(b'|_{A'/(a')})^{-1} \det\nolimits_\kappa(a'|_{A'/(b')}) \end{matrix}$$ Hence we conclude that $$(-1)^{\text{length}_A(A'/(b'))} d^A_{A'}(a', b' - a') = d^A_{A'}(b', b' - a') d^A_{A'}(a', b')$$ the sign coming from the $-a'$ in the second equality above. On the other hand, by Lemma \ref{lemma-periodic-determinant-sign} we have $d^A_{A'}(b', b') = (-1)^{\text{length}_A(A'/(b'))}$ and the lemma is proved. \end{proof} \noindent The tame symbol is a Steinberg symbol. \begin{lemma} \label{lemma-symbol-is-steinberg} Let $A$ be a Noetherian local domain of dimension $1$. Let $K = f.f.(A)$. For $x \in K \setminus \{0, 1\}$ we have $$d_A(x, 1 -x) = 1$$ \end{lemma} \begin{proof} Write $x = a/b$ with $a, b \in A$. The hypothesis implies, since $1 - x = (b - a)/b$, that also $b - a \not = 0$. Hence we compute $$d_A(x, 1 - x) = d_A(a, b - a)d_A(a, b)^{-1}d_A(b, b - a)^{-1}d_A(b, b)$$ Thus we have to show that $d_A(a, b - a) d_A(b, b) = d_A(b, b - a) d_A(a, b)$. This is Lemma \ref{lemma-symbol-is-steinberg-prepare}. \end{proof} \section{Lengths and determinants} \label{section-length-determinant} \noindent In this section we use the determinant to compare lattices. The key lemma is the following. \begin{lemma} \label{lemma-key-lemma} Let $R$ be a noetherian local ring. Let $\mathfrak q \subset R$ be a prime with $\dim(R/\mathfrak q) = 1$. Let $\varphi : M \to N$ be a homomorphism of finite $R$-modules. Assume there exist $x_1, \ldots, x_l \in M$ and $y_1, \ldots, y_l \in M$ with the following properties \begin{enumerate} \item $M = \langle x_1, \ldots, x_l\rangle$, \item $\langle x_1, \ldots, x_i\rangle / \langle x_1, \ldots, x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots, l$, \item $N = \langle y_1, \ldots, y_l\rangle$, and \item $\langle y_1, \ldots, y_i\rangle / \langle y_1, \ldots, y_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots, l$. \end{enumerate} Then $\varphi$ is injective if and only if $\varphi_{\mathfrak q}$ is an isomorphism, and in this case we have $$\text{length}_R(\Coker(\varphi)) = \text{ord}_{R/\mathfrak q}(f)$$ where $f \in \kappa(\mathfrak q)$ is the element such that $$[\varphi(x_1), \ldots, \varphi(x_l)] = f [y_1, \ldots, y_l]$$ in $\det_{\kappa(\mathfrak q)}(N_{\mathfrak q})$. \end{lemma} \begin{proof} First, note that the lemma holds in case $l = 1$. Namely, in this case $x_1$ is a basis of $M$ over $R/\mathfrak q$ and $y_1$ is a basis of $N$ over $R/\mathfrak q$ and we have $\varphi(x_1) = fy_1$ for some $f \in R$. Thus $\varphi$ is injective if and only if $f \not \in \mathfrak q$. Moreover, $\Coker(\varphi) = R/(f, \mathfrak q)$ and hence the lemma holds by definition of $\text{ord}_{R/q}(f)$ (see Algebra, Definition \ref{algebra-definition-ord}). \medskip\noindent In fact, suppose more generally that $\varphi(x_i) = f_iy_i$ for some $f_i \in R$, $f_i \not \in \mathfrak q$. Then the induced maps $$\langle x_1, \ldots, x_i\rangle / \langle x_1, \ldots, x_{i - 1}\rangle \longrightarrow \langle y_1, \ldots, y_i\rangle / \langle y_1, \ldots, y_{i - 1}\rangle$$ are all injective and have cokernels isomorphic to $R/(f_i, \mathfrak q)$. Hence we see that $$\text{length}_R(\Coker(\varphi)) = \sum \text{ord}_{R/\mathfrak q}(f_i).$$ On the other hand it is clear that $$[\varphi(x_1), \ldots, \varphi(x_l)] = f_1 \ldots f_l [y_1, \ldots, y_l]$$ in this case from the admissible relation (b) for symbols. Hence we see the result holds in this case also. \medskip\noindent We prove the general case by induction on $l$. Assume $l > 1$. Let $i \in \{1, \ldots, l\}$ be minimal such that $\varphi(x_1) \in \langle y_1, \ldots, y_i\rangle$. We will argue by induction on $i$. If $i = 1$, then we get a commutative diagram $$\xymatrix{ 0 \ar[r] & \langle x_1 \rangle \ar[r] \ar[d] & \langle x_1, \ldots, x_l \rangle \ar[r] \ar[d] & \langle x_1, \ldots, x_l \rangle / \langle x_1 \rangle \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \langle y_1 \rangle \ar[r] & \langle y_1, \ldots, y_l \rangle \ar[r] & \langle y_1, \ldots, y_l \rangle / \langle y_1 \rangle \ar[r] & 0 }$$ and the lemma follows from the snake lemma and induction on $l$. Assume now that $i > 1$. Write $\varphi(x_1) = a_1 y_1 + \ldots + a_{i - 1} y_{i - 1} + a y_i$ with $a_j, a \in R$ and $a \not \in \mathfrak q$ (since otherwise $i$ was not minimal). Set $$x'_j = \left\{ \begin{matrix} x_j & \text{if} & j = 1 \\ ax_j & \text{if} & j \geq 2 \end{matrix} \right. \quad\text{and}\quad y'_j = \left\{ \begin{matrix} y_j & \text{if} & j < i \\ ay_j & \text{if} & j \geq i \end{matrix} \right.$$ Let $M' = \langle x'_1, \ldots, x'_l \rangle$ and $N' = \langle y'_1, \ldots, y'_l \rangle$. Since $\varphi(x'_1) = a_1 y'_1 + \ldots + a_{i - 1} y'_{i - 1} + y'_i$ by construction and since for $j > 1$ we have $\varphi(x'_j) = a\varphi(x_i) \in \langle y'_1, \ldots, y'_l\rangle$ we get a commutative diagram of $R$-modules and maps $$\xymatrix{ M' \ar[d] \ar[r]_{\varphi'} & N' \ar[d] \\ M \ar[r]^\varphi & N }$$ By the result of the second paragraph of the proof we know that $\text{length}_R(M/M') = (l - 1)\text{ord}_{R/\mathfrak q}(a)$ and similarly $\text{length}_R(M/M') = (l - i + 1)\text{ord}_{R/\mathfrak q}(a)$. By a diagram chase this implies that $$\text{length}_R(\Coker(\varphi')) = \text{length}_R(\Coker(\varphi)) + i\ \text{ord}_{R/\mathfrak q}(a).$$ On the other hand, it is clear that writing $$[\varphi(x_1), \ldots, \varphi(x_l)] = f [y_1, \ldots, y_l], \quad [\varphi'(x'_1), \ldots, \varphi(x'_l)] = f' [y'_1, \ldots, y'_l]$$ we have $f' = a^if$. Hence it suffices to prove the lemma for the case that $\varphi(x_1) = a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_i$, i.e., in the case that $a = 1$. Next, recall that $$[y_1, \ldots, y_l] = [y_1, \ldots, y_{i - 1}, a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_i, y_{i + 1}, \ldots, y_l]$$ by the admissible relations for symbols. The sequence $y_1, \ldots, y_{i - 1}, a_1y_1 + \ldots + a_{i - 1}y_{i - 1} + y_i, y_{i + 1}, \ldots, y_l$ satisfies the conditions (3), (4) of the lemma also. Hence, we may actually assume that $\varphi(x_1) = y_i$. In this case, note that we have $\mathfrak q x_1 = 0$ which implies also $\mathfrak q y_i = 0$. We have $$[y_1, \ldots, y_l] = - [y_1, \ldots, y_{i - 2}, y_i, y_{i - 1}, y_{i + 1}, \ldots, y_l]$$ by the third of the admissible relations defining $\det_{\kappa(\mathfrak q)}(N_{\mathfrak q})$. Hence we may replace $y_1, \ldots, y_l$ by the sequence $y'_1, \ldots, y'_l = y_1, \ldots, y_{i - 2}, y_i, y_{i - 1}, y_{i + 1}, \ldots, y_l$ (which also satisfies conditions (3) and (4) of the lemma). Clearly this decreases the invariant $i$ by $1$ and we win by induction on $i$. \end{proof} \noindent To use the previous lemma we show that often sequences of elements with the required properties exist. \begin{lemma} \label{lemma-good-sequence-exists} Let $R$ be a local Noetherian ring. Let $\mathfrak q \subset R$ be a prime ideal. Let $M$ be a finite $R$-module such that $\mathfrak q$ is one of the minimal primes of the support of $M$. Then there exist $x_1, \ldots, x_l \in M$ such that \begin{enumerate} \item the support of $M / \langle x_1, \ldots, x_l\rangle$ does not contain $\mathfrak q$, and \item $\langle x_1, \ldots, x_i\rangle / \langle x_1, \ldots, x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots, l$. \end{enumerate} Moreover, in this case $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$. \end{lemma} \begin{proof} The condition that $\mathfrak q$ is a minimal prime in the support of $M$ implies that $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$ is finite (see Algebra, Lemma \ref{algebra-lemma-support-point}). Hence we can find $y_1, \ldots, y_l \in M_{\mathfrak q}$ such that $\langle y_1, \ldots, y_i\rangle / \langle y_1, \ldots, y_{i - 1}\rangle \cong \kappa(\mathfrak q)$ for $i = 1, \ldots, l$. We can find $f_i \in R$, $f_i \not \in \mathfrak q$ such that $f_i y_i$ is the image of some element $z_i \in M$. Moreover, as $R$ is Noetherian we can write $\mathfrak q = (g_1, \ldots, g_t)$ for some $g_j \in R$. By assumption $g_j y_i \in \langle y_1, \ldots, y_{i - 1} \rangle$ inside the module $M_{\mathfrak q}$. By our choice of $z_i$ we can find some further elements $f_{ji} \in R$, $f_{ij} \not \in \mathfrak q$ such that $f_{ij} g_j z_i \in \langle z_1, \ldots, z_{i - 1} \rangle$ (equality in the module $M$). The lemma follows by taking $$x_1 = f_{11}f_{12}\ldots f_{1t}z_1, \quad x_2 = f_{11}f_{12}\ldots f_{1t}f_{21}f_{22}\ldots f_{2t}z_2,$$ and so on. Namely, since all the elements $f_i, f_{ij}$ are invertible in $R_{\mathfrak q}$ we still have that $R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_i / R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_{i - 1} \cong \kappa(\mathfrak q)$ for $i = 1, \ldots, l$. By construction, $\mathfrak q x_i \in \langle x_1, \ldots, x_{i - 1}\rangle$. Thus $\langle x_1, \ldots, x_i\rangle / \langle x_1, \ldots, x_{i - 1}\rangle$ is an $R$-module generated by one element, annihilated $\mathfrak q$ such that localizing at $\mathfrak q$ gives a $q$-dimensional vector space over $\kappa(\mathfrak q)$. Hence it is isomorphic to $R/\mathfrak q$. \end{proof} \noindent Here is the main result of this section. We will see below the various different consequences of this proposition. The reader is encouraged to first prove the easier Lemma \ref{lemma-application-herbrand-quotient} his/herself. \begin{proposition} \label{proposition-length-determinant-periodic-complex} Let $R$ be a local Noetherian ring with residue field $\kappa$. Suppose that $(M, \varphi, \psi)$ is a $(2, 1)$-periodic complex over $R$. Assume \begin{enumerate} \item $M$ is a finite $R$-module, \item the cohomology modules of $(M, \varphi, \psi)$ are of finite length, and \item $\dim(\text{Supp}(M)) = 1$. \end{enumerate} Let $\mathfrak q_i$, $i = 1, \ldots, t$ be the minimal primes of the support of $M$. Then we have\footnote{ Obviously we could get rid of the minus sign by redefining $\det_\kappa(M, \varphi, \psi)$ as the inverse of its current value, see Definition \ref{definition-periodic-determinant}.} $$- e_R(M, \varphi, \psi) = \sum\nolimits_{i = 1, \ldots, t} \text{ord}_{R/\mathfrak q_i}\left( \det\nolimits_{\kappa(\mathfrak q_i)} (M_{\mathfrak q_i}, \varphi_{\mathfrak q_i}, \psi_{\mathfrak q_i}) \right)$$ \end{proposition} \begin{proof} We first reduce to the case $t = 1$ in the following way. Note that $\text{Supp}(M) = \{\mathfrak m, \mathfrak q_1, \ldots, \mathfrak q_t\}$, where $\mathfrak m \subset R$ is the maximal ideal. Let $M_i$ denote the image of $M \to M_{\mathfrak q_i}$, so $\text{Supp}(M_i) = \{\mathfrak m, \mathfrak q_i\}$. The map $\varphi$ (resp.\ $\psi$) induces an $R$-module map $\varphi_i : M_i \to M_i$ (resp.\ $\psi_i : M_i \to M_i$). Thus we get a morphism of $(2, 1)$-periodic complexes $$(M, \varphi, \psi) \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, t} (M_i, \varphi_i, \psi_i).$$ The kernel and cokernel of this map have support equal to $\{\mathfrak m\}$ (or are zero). Hence by Lemma \ref{lemma-periodic-length} these $(2, 1)$-periodic complexes have multiplicity $0$. In other words we have $$e_R(M, \varphi, \psi) = \sum\nolimits_{i = 1, \ldots, t} e_R(M_i, \varphi_i, \psi_i)$$ On the other hand we clearly have $M_{\mathfrak q_i} = M_{i, \mathfrak q_i}$, and hence the terms of the right hand side of the formula of the lemma are equal to the expressions $$\text{ord}_{R/\mathfrak q_i}\left( \det\nolimits_{\kappa(\mathfrak q_i)} (M_{i, \mathfrak q_i}, \varphi_{i, \mathfrak q_i}, \psi_{i, \mathfrak q_i}) \right)$$ In other words, if we can prove the lemma for each of the modules $M_i$, then the lemma holds. This reduces us to the case $t = 1$. \medskip\noindent Assume we have a $(2, 1)$-periodic complex $(M, \varphi, \psi)$ over a Noetherian local ring with $M$ a finite $R$-module, $\text{Supp}(M) = \{\mathfrak m, \mathfrak q\}$, and finite length cohomology modules. The proof in this case follows from Lemma \ref{lemma-key-lemma} and careful bookkeeping. Denote $K_\varphi = \Ker(\varphi)$, $I_\varphi = \Im(\varphi)$, $K_\psi = \Ker(\psi)$, and $I_\psi = \Im(\psi)$. Since $R$ is Noetherian these are all finite $R$-modules. Set $$a = \text{length}_{R_{\mathfrak q}}(I_{\varphi, \mathfrak q}) = \text{length}_{R_{\mathfrak q}}(K_{\psi, \mathfrak q}), \quad b = \text{length}_{R_{\mathfrak q}}(I_{\psi, \mathfrak q}) = \text{length}_{R_{\mathfrak q}}(K_{\varphi, \mathfrak q}).$$ Equalities because the complex becomes exact after localizing at $\mathfrak q$. Note that $l = \text{length}_{R_{\mathfrak q}}(M_{\mathfrak q})$ is equal to $l = a + b$. \medskip\noindent We are going to use Lemma \ref{lemma-good-sequence-exists} to choose sequences of elements in finite $R$-modules $N$ with support contained in $\{\mathfrak m, \mathfrak q\}$. In this case $N_{\mathfrak q}$ has finite length, say $n \in \mathbf{N}$. Let us call a sequence $w_1, \ldots, w_n \in N$ with properties (1) and (2) of Lemma \ref{lemma-good-sequence-exists} a good sequence''. Note that the quotient $N/\langle w_1, \ldots, w_n \rangle$ of $N$ by the submodule generated by a good sequence has support (contained in) $\{\mathfrak m\}$ and hence has finite length (Algebra, Lemma \ref{algebra-lemma-support-point}). Moreover, the symbol $[w_1, \ldots, w_n] \in \det_{\kappa(\mathfrak q)}(N_{\mathfrak q})$ is a generator, see Lemma \ref{lemma-determinant-dimension-one}. \medskip\noindent Having said this we choose good sequences $$\begin{matrix} x_1, \ldots, x_b & \text{in} & K_\varphi, & t_1, \ldots, t_a & \text{in} & K_\psi, \\ y_1, \ldots, y_a & \text{in} & I_\varphi \cap \langle t_1, \ldots t_a\rangle, & s_1, \ldots, s_b & \text{in} & I_\psi \cap \langle x_1, \ldots, x_b\rangle. \end{matrix}$$ We will adjust our choices a little bit as follows. Choose lifts $\tilde y_i \in M$ of $y_i \in I_\varphi$ and $\tilde s_i \in M$ of $s_i \in I_\psi$. It may not be the case that $\mathfrak q \tilde y_1 \subset \langle x_1, \ldots, x_b\rangle$ and it may not be the case that $\mathfrak q \tilde s_1 \subset \langle t_1, \ldots, t_a\rangle$. However, using that $\mathfrak q$ is finitely generated (as in the proof of Lemma \ref{lemma-good-sequence-exists}) we can find a $d \in R$, $d \not \in \mathfrak q$ such that $\mathfrak q d\tilde y_1 \subset \langle x_1, \ldots, x_b\rangle$ and $\mathfrak q d\tilde s_1 \subset \langle t_1, \ldots, t_a\rangle$. Thus after replacing $y_i$ by $dy_i$, $\tilde y_i$ by $d\tilde y_i$, $s_i$ by $ds_i$ and $\tilde s_i$ by $d\tilde s_i$ we see that we may assume also that $x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_b$ and $t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b$ are good sequences in $M$. \medskip\noindent Finally, we choose a good sequence $z_1, \ldots, z_l$ in the finite $R$-module $$\langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a \rangle \cap \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b \rangle.$$ Note that this is also a good sequence in $M$. \medskip\noindent Since $I_{\varphi, \mathfrak q} = K_{\psi, \mathfrak q}$ there is a unique element $h \in \kappa(\mathfrak q)$ such that $[y_1, \ldots, y_a] = h [t_1, \ldots, t_a]$ inside $\det_{\kappa(\mathfrak q)}(K_{\psi, \mathfrak q})$. Similarly, as $I_{\psi, \mathfrak q} = K_{\varphi, \mathfrak q}$ there is a unique element $h \in \kappa(\mathfrak q)$ such that $[s_1, \ldots, s_b] = g [x_1, \ldots, x_b]$ inside $\det_{\kappa(\mathfrak q)}(K_{\varphi, \mathfrak q})$. We can also do this with the three good sequences we have in $M$. All in all we get the following identities \begin{align*} [y_1, \ldots, y_a] & = h [t_1, \ldots, t_a] \\ [s_1, \ldots, s_b] & = g [x_1, \ldots, x_b] \\ [z_1, \ldots, z_l] & = f_\varphi [x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a] \\ [z_1, \ldots, z_l] & = f_\psi [t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b] \end{align*} for some $g, h, f_\varphi, f_\psi \in \kappa(\mathfrak q)$. \medskip\noindent Having set up all this notation let us compute $\det_{\kappa(\mathfrak q)}(M, \varphi, \psi)$. Namely, consider the element $[z_1, \ldots, z_l]$. Under the map $\gamma_\psi \circ \sigma \circ \gamma_\varphi^{-1}$ of Definition \ref{definition-periodic-determinant} we have \begin{eqnarray*} [z_1, \ldots, z_l] & = & f_\varphi [x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a] \\ & \mapsto & f_\varphi [x_1, \ldots, x_b] \otimes [y_1, \ldots, y_a] \\ & \mapsto & f_\varphi h/g [t_1, \ldots, t_a] \otimes [s_1, \ldots, s_b] \\ & \mapsto & f_\varphi h/g [t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b] \\ & = & f_\varphi h/f_\psi g [z_1, \ldots, z_l] \end{eqnarray*} This means that $\det_{\kappa(\mathfrak q)} (M_{\mathfrak q}, \varphi_{\mathfrak q}, \psi_{\mathfrak q})$ is equal to $f_\varphi h/f_\psi g$ up to a sign. \medskip\noindent We abbreviate the following quantities \begin{eqnarray*} k_\varphi & = & \text{length}_R(K_\varphi/\langle x_1, \ldots, x_b\rangle) \\ k_\psi & = & \text{length}_R(K_\psi/\langle t_1, \ldots, t_a\rangle) \\ i_\varphi & = & \text{length}_R(I_\varphi/\langle y_1, \ldots, y_a\rangle) \\ i_\psi & = & \text{length}_R(I_\psi/\langle s_1, \ldots, s_a\rangle) \\ m_\varphi & = & \text{length}_R(M/ \langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a\rangle) \\ m_\psi & = & \text{length}_R(M/ \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b\rangle) \\ \delta_\varphi & = & \text{length}_R( \langle x_1, \ldots, x_b, \tilde y_1, \ldots, \tilde y_a\rangle \langle z_1, \ldots, z_l\rangle) \\ \delta_\psi & = & \text{length}_R( \langle t_1, \ldots, t_a, \tilde s_1, \ldots, \tilde s_b\rangle \langle z_1, \ldots, z_l\rangle) \end{eqnarray*} Using the exact sequences $0 \to K_\varphi \to M \to I_\varphi \to 0$ we get $m_\varphi = k_\varphi + i_\varphi$. Similarly we have $m_\psi = k_\psi + i_\psi$. We have $\delta_\varphi + m_\varphi = \delta_\psi + m_\psi$ since this is equal to the colength of $\langle z_1, \ldots, z_l \rangle$ in $M$. Finally, we have $$\delta_\varphi = \text{ord}_{R/\mathfrak q}(f_\varphi), \quad \delta_\psi = \text{ord}_{R/\mathfrak q}(f_\psi)$$ by our first application of the key Lemma \ref{lemma-key-lemma}. \medskip\noindent Next, let us compute the multiplicity of the periodic complex \begin{eqnarray*} e_R(M, \varphi, \psi) & = & \text{length}_R(K_\varphi/I_\psi) - \text{length}_R(K_\psi/I_\varphi) \\ & = & \text{length}_R( \langle x_1, \ldots, x_b\rangle/ \langle s_1, \ldots, s_b\rangle) + k_\varphi - i_\psi \\ & & - \text{length}_R( \langle t_1, \ldots, t_a\rangle/ \langle y_1, \ldots, y_a\rangle) - k_\psi + i_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + k_\varphi - i_\psi - k_\psi + i_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + m_\varphi - m_\psi \\ & = & \text{ord}_{R/\mathfrak q}(g/h) + \delta_\psi - \delta_\varphi \\ & = & \text{ord}_{R/\mathfrak q}(f_\psi g/f_\varphi h) \end{eqnarray*} where we used the key Lemma \ref{lemma-key-lemma} twice in the third equality. By our computation of $\det_{\kappa(\mathfrak q)} (M_{\mathfrak q}, \varphi_{\mathfrak q}, \psi_{\mathfrak q})$ this proves the proposition. \end{proof} \noindent In most applications the following lemma suffices. \begin{lemma} \label{lemma-application-herbrand-quotient} Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a finite $R$-module, and let $\psi : M \to M$ be an $R$-module map. Assume that \begin{enumerate} \item $\Ker(\psi)$ and $\Coker(\psi)$ have finite length, and \item $\dim(\text{Supp}(M)) \leq 1$. \end{enumerate} Write $\text{Supp}(M) = \{\mathfrak m, \mathfrak q_1, \ldots, \mathfrak q_t\}$ and denote $f_i \in \kappa(\mathfrak q_i)^*$ the element such that $\det_{\kappa(\mathfrak q_i)}(\psi_{\mathfrak q_i}) : \det_{\kappa(\mathfrak q_i)}(M_{\mathfrak q_i}) \to \det_{\kappa(\mathfrak q_i)}(M_{\mathfrak q_i})$ is multiplication by $f_i$. Then we have $$\text{length}_R(\Coker(\psi)) - \text{length}_R(\Ker(\psi)) = \sum\nolimits_{i = 1, \ldots, t} \text{ord}_{R/\mathfrak q_i}(f_i).$$ \end{lemma} \begin{proof} Recall that $H^0(M, 0, \psi) = \Coker(\psi)$ and $H^1(M, 0, \psi) = \Ker(\psi)$, see remarks above Definition \ref{definition-periodic-length}. The lemma follows by combining Proposition \ref{proposition-length-determinant-periodic-complex} with Lemma \ref{lemma-periodic-determinant-easy-case}. \medskip\noindent Alternative proof. Reduce to the case $\text{Supp}(M) = \{\mathfrak m, \mathfrak q\}$ as in the proof of Proposition \ref{proposition-length-determinant-periodic-complex}. Then directly combine Lemmas \ref{lemma-key-lemma} and \ref{lemma-good-sequence-exists} to prove this specific case of Proposition \ref{proposition-length-determinant-periodic-complex}. There is much less bookkeeping in this case, and the reader is encouraged to work this out. Details omitted. \end{proof} \section{Application to tame symbol} \label{section-application-tame-symbol} \noindent In this section we apply the results above to show the following key lemma. This lemma is a low degree case of the statement that there is a complex for Milnor K-theory similar to the Gersten-Quillen complex in Quillen's K-theory. See \cite{Kato-Milnor-K}. \begin{lemma}[Key Lemma] \label{lemma-secondary-ramification} \begin{reference} When $A$ is an excellent ring this is \cite[Proposition 1]{Kato-Milnor-K}. \end{reference} Let $A$ be a $2$-dimensional Noetherian local domain. Let $K = f.f.(A)$. Let $f, g \in K^*$. Let $\mathfrak q_1, \ldots, \mathfrak q_t$ be the height $1$ primes $\mathfrak q$ of $A$ such that either $f$ or $g$ is not an element of $A^*_{\mathfrak q}$. Then we have $$\sum\nolimits_{i = 1, \ldots, t} \text{ord}_{A/\mathfrak q_i}(d_{A_{\mathfrak q_i}}(f, g)) = 0$$ We can also write this as $$\sum\nolimits_{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(d_{A_{\mathfrak q}}(f, g)) = 0$$ since at any height one prime $\mathfrak q$ of $A$ where $f, g \in A^*_{\mathfrak q}$ we have $d_{A_{\mathfrak q}}(f, g) = 1$ by Lemma \ref{lemma-symbol-when-one-is-a-unit}. \end{lemma} \begin{proof} Since the tame symbols $d_{A_{\mathfrak q}}(f, g)$ are additive (Lemma \ref{lemma-multiplicativity-symbol}) and the order functions $\text{ord}_{A/\mathfrak q}$ are additive (Algebra, Lemma \ref{algebra-lemma-ord-additive}) it suffices to prove the formula when $f = a \in A$ and $g = b \in A$. In this case we see that we have to show $$\sum\nolimits_{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(\det\nolimits_\kappa(A_{\mathfrak q}/(ab), a, b)) = 0$$ By Proposition \ref{proposition-length-determinant-periodic-complex} this is equivalent to showing that $$e_A(A/(ab), a, b) = 0.$$ Since the complex $A/(ab) \xrightarrow{a} A/(ab) \xrightarrow{b} A/(ab) \xrightarrow{a} A/(ab)$ is exact we win. \end{proof} \section{Setup} \label{section-setup} \noindent We will throughout work over a locally Noetherian universally catenary base $S$ endowed with a dimension function $\delta$. Although it is likely possible to generalize (parts of) the discussion in the chapter, it seems that this is a good first approximation. We usually do not assume our schemes are separated or quasi-compact. Many interesting algebraic stacks are non-separated and/or non-quasi-compact and this is a good case study to see how to develop a reasonable theory for those as well. In order to reference these hypotheses we give it a number. \begin{situation} \label{situation-setup} Here $S$ is a locally Noetherian, and universally catenary scheme. Moreover, we assume $S$ is endowed with a dimension function $\delta : S \longrightarrow \mathbf{Z}$. \end{situation} \noindent See Morphisms, Definition \ref{morphisms-definition-universally-catenary} for the notion of a universally catenary scheme, and see Topology, Definition \ref{topology-definition-dimension-function} for the notion of a dimension function. Recall that any locally Noetherian catenary scheme locally has a dimension function, see Properties, Lemma \ref{properties-lemma-catenary-dimension-function}. Moreover, there are lots of schemes which are universally catenary, see Morphisms, Lemma \ref{morphisms-lemma-ubiquity-uc}. \medskip\noindent Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Any scheme $X$ locally of finite type over $S$ is locally Noetherian and catenary. In fact, $X$ has a canonical dimension function $$\delta = \delta_{X/S} : X \longrightarrow \mathbf{Z}$$ associated to $(f : X \to S, \delta)$ given by the rule $\delta_{X/S}(x) = \delta(f(x)) + \text{trdeg}_{\kappa(f(x))}\kappa(x)$. See Morphisms, Lemma \ref{morphisms-lemma-dimension-function-propagates}. Moreover, if $h : X \to Y$ is a morphism of schemes locally of finite type over $S$, and $x \in X$, $y = h(x)$, then obviously $\delta_{X/S}(x) = \delta_{Y/S}(y) + \text{trdeg}_{\kappa(y)}\kappa(x)$. We will freely use this function and its properties in the following. \medskip\noindent Here are the basic examples of setups as above. In fact, the main interest lies in the case where the base is the spectrum of a field, or the case where the base is the spectrum of a Dedekind ring (e.g.\ $\mathbf{Z}$, or a discrete valuation ring). \begin{example} \label{example-field} Here $S = \Spec(k)$ and $k$ is a field. We set $\delta(pt) = 0$ where $pt$ indicates the unique point of $S$. The pair $(S, \delta)$ is an example of a situation as in Situation \ref{situation-setup} by Morphisms, Lemma \ref{morphisms-lemma-ubiquity-uc}. \end{example} \begin{example} \label{example-domain-dimension-1} Here $S = \Spec(A)$, where $A$ is a Noetherian domain of dimension $1$. For example we could consider $A = \mathbf{Z}$. We set $\delta(\mathfrak p) = 0$ if $\mathfrak p$ is a maximal ideal and $\delta(\mathfrak p) = 1$ if $\mathfrak p = (0)$ corresponds to the generic point. This is an example of Situation \ref{situation-setup} by Morphisms, Lemma \ref{morphisms-lemma-ubiquity-uc}. \end{example} \noindent In good cases $\delta$ corresponds to the dimension function. \begin{lemma} \label{lemma-delta-is-dimension} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Assume in addition $S$ is a Jacobson scheme, and $\delta(s) = 0$ for every closed point $s$ of $S$. Let $X$ be locally of finite type over $S$. Let $Z \subset X$ be an integral closed subscheme and let $\xi \in Z$ be its generic point. The following integers are the same: \begin{enumerate} \item $\delta_{X/S}(\xi)$, \item $\dim(Z)$, and \item $\dim(\mathcal{O}_{Z, z})$ where $z$ is a closed point of $Z$. \end{enumerate} \end{lemma} \begin{proof} Let $X \to S$, $\xi \in Z \subset X$ be as in the lemma. Since $X$ is locally of finite type over $S$ we see that $X$ is Jacobson, see Morphisms, Lemma \ref{morphisms-lemma-Jacobson-universally-Jacobson}. Hence closed points of $X$ are dense in every closed subset of $Z$ and map to closed points of $S$. Hence given any chain of irreducible closed subsets of $Z$ we can end it with a closed point of $Z$. It follows that $\dim(Z) = \sup_z(\dim(\mathcal{O}_{Z, z})$ (see Properties, Lemma \ref{properties-lemma-codimension-local-ring}) where $z \in Z$ runs over the closed points of $Z$. Note that $\dim(\mathcal{O}_{Z, z}) = \delta(\xi) - \delta(z))$ by the properties of a dimension function. For each closed $z \in Z$ the field extension $\kappa(z) \supset \kappa(f(z))$ is finite, see Morphisms, Lemma \ref{morphisms-lemma-jacobson-finite-type-points}. Hence $\delta_{X/S}(z) = \delta(f(z)) = 0$ for $z \in Z$ closed. It follows that all three integers are equal. \end{proof} \noindent In the situation of the lemma above the value of $\delta$ at the generic point of a closed irreducible subset is the dimension of the irreducible closed subset. However, in general we cannot expect the equality to hold. For example if $S = \Spec(\mathbf{C}[[t]])$ and $X = \Spec(\mathbf{C}((t)))$ then we would get $\delta(x) = 1$ for the unique point of $X$, but $\dim(X) = 0$. Still we want to think of $\delta_{X/S}$ as giving the dimension of the irreducible closed subschemes. Thus we introduce the following terminology. \begin{definition} \label{definition-delta-dimension} Let $(S, \delta)$ as in Situation \ref{situation-setup}. For any scheme $X$ locally of finite type over $S$ and any irreducible closed subset $Z \subset X$ we define $$\dim_\delta(Z) = \delta(\xi)$$ where $\xi \in Z$ is the generic point of $Z$. We will call this the {\it $\delta$-dimension of $Z$}. If $Z$ is a closed subscheme of $X$, then we define $\dim_\delta(Z)$ as the supremum of the $\delta$-dimensions of its irreducible components. \end{definition} \section{Cycles} \label{section-cycles} \noindent Since we are not assuming our schemes are quasi-compact we have to be a little careful when defining cycles. We have to allow infinite sums because a rational function may have infinitely many poles for example. In any case, if $X$ is quasi-compact then a cycle is a finite sum as usual. \begin{definition} \label{definition-cycles} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let $k \in \mathbf{Z}$. \begin{enumerate} \item A {\it cycle on $X$} is a formal sum $$\alpha = \sum n_Z [Z]$$ where the sum is over integral closed subschemes $Z \subset X$, each $n_Z \in \mathbf{Z}$, and the collection $\{Z; n_Z \not = 0\}$ is locally finite (Topology, Definition \ref{topology-definition-locally-finite}). \item A {\it $k$-cycle}, on $X$ is a cycle $$\alpha = \sum n_Z [Z]$$ where $n_Z \not = 0 \Rightarrow \dim_\delta(Z) = k$. \item The abelian group of all $k$-cycles on $X$ is denoted $Z_k(X)$. \end{enumerate} \end{definition} \noindent In other words, a $k$-cycle on $X$ is a locally finite formal $\mathbf{Z}$-linear combination of integral closed subschemes of $\delta$-dimension $k$. Addition of $k$-cycles $\alpha = \sum n_Z[Z]$ and $\beta = \sum m_Z[Z]$ is given by $$\alpha + \beta = \sum (n_Z + m_Z)[Z],$$ i.e., by adding the coefficients. \section{Cycle associated to a closed subscheme} \label{section-cycle-of-closed-subscheme} \begin{lemma} \label{lemma-multiplicity-finite} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let $Z \subset X$ be a closed subscheme. \begin{enumerate} \item Let $Z' \subset Z$ be an irreducible component and let $\xi \in Z'$ be its generic point. Then $$\text{length}_{\mathcal{O}_{X, \xi}} \mathcal{O}_{Z, \xi} < \infty$$ \item If $\dim_\delta(Z) \leq k$ and $\xi \in Z$ with $\delta(\xi) = k$, then $\xi$ is a generic point of an irreducible component of $Z$. \end{enumerate} \end{lemma} \begin{proof} Let $Z' \subset Z$, $\xi \in Z'$ be as in (1). Then $\dim(\mathcal{O}_{Z, \xi}) = 0$ (for example by Properties, Lemma \ref{properties-lemma-codimension-local-ring}). Hence $\mathcal{O}_{Z, \xi}$ is Noetherian local ring of dimension zero, and hence has finite length over itself (see Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}). Hence, it also has finite length over $\mathcal{O}_{X, \xi}$, see Algebra, Lemma \ref{algebra-lemma-pushdown-module}. \medskip\noindent Assume $\xi \in Z$ and $\delta(\xi) = k$. Consider the closure $Z' = \overline{\{\xi\}}$. It is an irreducible closed subscheme with $\dim_\delta(Z') = k$ by definition. Since $\dim_\delta(Z) = k$ it must be an irreducible component of $Z$. Hence we see (2) holds. \end{proof} \begin{definition} \label{definition-cycle-associated-to-closed-subscheme} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let $Z \subset X$ be a closed subscheme. \begin{enumerate} \item For any irreducible component $Z' \subset Z$ with generic point $\xi$ the integer $m_{Z', Z} = \text{length}_{\mathcal{O}_{X, \xi}} \mathcal{O}_{Z, \xi}$ (Lemma \ref{lemma-multiplicity-finite}) is called the {\it multiplicity of $Z'$ in $Z$}. \item Assume $\dim_\delta(Z) \leq k$. The {\it $k$-cycle associated to $Z$} is $$[Z]_k = \sum m_{Z', Z}[Z']$$ where the sum is over the irreducible components of $Z$ of $\delta$-dimension $k$. (This is a $k$-cycle by Divisors, Lemma \ref{divisors-lemma-components-locally-finite}.) \end{enumerate} \end{definition} \noindent It is important to note that we only define $[Z]_k$ if the $\delta$-dimension of $Z$ does not exceed $k$. In other words, by convention, if we write $[Z]_k$ then this implies that $\dim_\delta(Z) \leq k$. \section{Cycle associated to a coherent sheaf} \label{section-cycle-of-coherent-sheaf} \begin{lemma} \label{lemma-length-finite} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_X$-module. \begin{enumerate} \item The collection of irreducible components of the support of $\mathcal{F}$ is locally finite. \item Let $Z' \subset \text{Supp}(\mathcal{F})$ be an irreducible component and let $\xi \in Z'$ be its generic point. Then $$\text{length}_{\mathcal{O}_{X, \xi}} \mathcal{F}_\xi < \infty$$ \item If $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$ and $\xi \in Z$ with $\delta(\xi) = k$, then $\xi$ is a generic point of an irreducible component of $\text{Supp}(\mathcal{F})$. \end{enumerate} \end{lemma} \begin{proof} By Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-support-closed} the support $Z$ of $\mathcal{F}$ is a closed subset of $X$. We may think of $Z$ as a reduced closed subscheme of $X$ (Schemes, Lemma \ref{schemes-lemma-reduced-closed-subscheme}). Hence (1) follows from Divisors, Lemma \ref{divisors-lemma-components-locally-finite} applied to $Z$ and (3) follows from Lemma \ref{lemma-multiplicity-finite} applied to $Z$. \medskip\noindent Let $\xi \in Z'$ be as in (2). In this case for any specialization $\xi' \leadsto \xi$ in $X$ we have $\mathcal{F}_{\xi'} = 0$. Recall that the non-maximal primes of $\mathcal{O}_{X, \xi}$ correspond to the points of $X$ specializing to $\xi$ (Schemes, Lemma \ref{schemes-lemma-specialize-points}). Hence $\mathcal{F}_\xi$ is a finite $\mathcal{O}_{X, \xi}$-module whose support is $\{\mathfrak m_\xi\}$. Hence it has finite length by Algebra, Lemma \ref{algebra-lemma-support-point}. \end{proof} \begin{definition} \label{definition-cycle-associated-to-coherent-sheaf} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_X$-module. \begin{enumerate} \item For any irreducible component $Z' \subset \text{Supp}(\mathcal{F})$ with generic point $\xi$ the integer $m_{Z', \mathcal{F}} = \text{length}_{\mathcal{O}_{X, \xi}} \mathcal{F}_\xi$ (Lemma \ref{lemma-length-finite}) is called the {\it multiplicity of $Z'$ in $\mathcal{F}$}. \item Assume $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$. The {\it $k$-cycle associated to $\mathcal{F}$} is $$[\mathcal{F}]_k = \sum m_{Z', \mathcal{F}}[Z']$$ where the sum is over the irreducible components of $\text{Supp}(\mathcal{F})$ of $\delta$-dimension $k$. (This is a $k$-cycle by Lemma \ref{lemma-length-finite}.) \end{enumerate} \end{definition} \noindent It is important to note that we only define $[\mathcal{F}]_k$ if $\mathcal{F}$ is coherent and the $\delta$-dimension of $\text{Supp}(\mathcal{F})$ does not exceed $k$. In other words, by convention, if we write $[\mathcal{F}]_k$ then this implies that $\mathcal{F}$ is coherent on $X$ and $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$. \begin{lemma} \label{lemma-cycle-closed-coherent} Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $X$ be locally of finite type over $S$. Let \$Z \subset