## 42.68 Appendix A: Alternative approach to key lemma

In this appendix we first define determinants $\det _\kappa (M)$ of finite length modules $M$ over local rings $(R, \mathfrak m, \kappa )$, see Subsection 42.68.1. The determinant $\det _\kappa (M)$ is a $1$-dimensional $\kappa$-vector space. We use this in Subsection 42.68.12 to define the determinant $\det _\kappa (M, \varphi , \psi ) \in \kappa ^*$ of an exact $(2, 1)$-periodic complex $(M, \varphi , \psi )$ with $M$ of finite length. In Subsection 42.68.26 we use these determinants to construct a tame symbol $d_ R(a, b) = \det _\kappa (R/ab, a, b)$ for a pair of nonzerodivisors $a, b \in R$ when $R$ is Noetherian of dimension $1$. Although there is no doubt that

$d_ R(a, b) = \partial _ R(a, b)$

where $\partial _ R$ is as in Section 42.5, we have not (yet) added the verification. The advantage of the tame symbol as constructed in this appendix is that it extends (for example) to pairs of injective endomorphisms $\varphi , \psi$ of a finite $R$-module $M$ of dimension $1$ such that $\varphi (\psi (M)) = \psi (\varphi (M))$. In Subsection 42.68.40 we relate Herbrand quotients and determinants. An easy to state version of main the main result (Proposition 42.68.43) is the formula

$-e_ R(M, \varphi , \psi ) = \text{ord}_ R(\det \nolimits _ K(M_ K, \varphi , \psi ))$

when $(M, \varphi , \psi )$ is a $(2, 1)$-periodic complex whose Herbrand quotient $e_ R$ (Definition 42.2.2) is defined over a $1$-dimensonal Noetherian local domain $R$ with fraction field $K$. We use this proposition to give an alternative proof of the key lemma (Lemma 42.6.3) for the tame symbol constructed in this appendix, see Lemma 42.68.46.

### 42.68.1 Determinants of finite length modules

The material in this section is related to the material in the paper and to the material in the thesis [Joe].

Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $\varphi : M \to M$ be an $R$-linear endomorphism of a finite length $R$-module $M$. In More on Algebra, Section 15.120 we have already defined the determinant $\det _\kappa (\varphi )$ (and the trace and the characteristic polynomial) of $\varphi$ relative to $\kappa$. In this section, we will construct a canonical $1$-dimensional $\kappa$-vector space $\det _\kappa (M)$ such that $\det _\kappa (\varphi : M \to M) : \det _\kappa (M) \to \det _\kappa (M)$ is equal to multiplication by $\det _\kappa (\varphi )$. If $M$ is annihilated by $\mathfrak m$, then $M$ can be viewed as a finite dimension $\kappa$-vector space and then we have $\det _\kappa (M) = \wedge ^ n_\kappa (M)$ where $n = \dim _\kappa (M)$. Our construction will generalize this to all finite length modules over $R$ and if $R$ contains its residue field, then the determinant $\det _\kappa (M)$ will be given by the usual determinant in a suitable sense, see Remark 42.68.9.

Definition 42.68.2. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. Let $M$ be a finite length $R$-module. Say $l = \text{length}_ R(M)$.

1. 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 $x_1, \ldots , x_ r$.

2. We will say an $l$-tuple of elements $(e_1, \ldots , e_ l)$ of $M$ is admissible if $\mathfrak m e_ i \subset \langle e_1, \ldots , e_{i - 1} \rangle$ for $i = 1, \ldots , l$.

3. A symbol $[e_1, \ldots , e_ l]$ will mean $(e_1, \ldots , e_ l)$ is an admissible $l$-tuple.

4. An admissible relation between symbols is one of the following:

1. 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 $[e_1, \ldots , e_ l] = 0$,

2. 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

3. 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

$[e_1, \ldots , e_ l] = - [e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l].$
5. We define the determinant of the finite length $R$-module $M$ to be

$\det \nolimits _\kappa (M) = \left\{ \frac{\kappa \text{-vector space generated by symbols}}{\kappa \text{-linear combinations of admissible relations}} \right\}$

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 $1$.

Lemma 42.68.3. With notations as above we have $\dim _\kappa (\det _\kappa (M)) \leq 1$.

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.

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. $\square$

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 $\kappa$.

Lemma 42.68.4. 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.

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 42.68.3 the left hand side has dimension at most one we see that the map is an isomorphism. $\square$

Lemma 42.68.5. 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 $\langle f_1, \ldots f_ l \rangle = M$.

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 42.68.3 and its proof. We will show by induction on $l = \text{length}(M)$ that it is nonzero. For $l = 1$ it follows from Lemma 42.68.4. 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.

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.

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.

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.

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:

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 $\det _\kappa (\overline{M})$.

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 $\det _\kappa (\overline{M})$.

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 $\psi ([e_1, \ldots , e_ l])$.

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.

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. $\square$

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

42.68.5.1
\begin{equation} \label{chow-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 [Knudsen]), as is the following additivity property.

Lemma 42.68.6. 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:

1. 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

$\gamma _{K' \to L' \to M'} \circ (\det \nolimits _\kappa (u) \otimes \det \nolimits _\kappa (w)) = \det \nolimits _\kappa (v) \circ \gamma _{K \to L \to M},$
2. 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

3. 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.

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 42.68.3), 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.

By Lemma 42.68.5 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.

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.

Part (3) of the lemma is obvious. This finishes the proof. $\square$

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 12.19.1)

$0 = F^ m \subset F^{m - 1} \subset \ldots \subset F^{n + 1} \subset F^ n = M$

then there is a well defined and canonical isomorphism

$\gamma _{(M, F)} : \det \nolimits _\kappa (F^{m - 1}/F^ m) \otimes _\kappa \ldots \otimes _ k \det \nolimits _\kappa (F^ n/F^{n + 1}) \longrightarrow \det \nolimits _\kappa (M)$

To construct it we use isomorphisms of Lemma 42.68.6 coming from the short exact sequences $0 \to F^{i - 1}/F^ i \to M/F^ i \to M/F^{i - 1} \to 0$. Part (2) of Lemma 42.68.6 with $G = 0$ shows we obtain the same isomorphism if we use the short exact sequences $0 \to F^ i \to F^{i - 1} \to F^{i - 1}/F^ i \to 0$.

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.

Lemma 42.68.7. 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

1. its restriction to the subcategory of modules annihilated by $\mathfrak m$ is isomorphic to the usual determinant functor (see Lemma 42.68.4), and

2. (1), (2) and (3) of Lemma 42.68.6 hold.

Proof. Omitted. $\square$

Lemma 42.68.8. 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 42.68.6 defined for $\det _{R, \kappa }$ and $\det _{R', \kappa }$.

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 10.52.12. 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 42.68.6. $\square$

Remark 42.68.9. 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 42.68.8 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 42.68.6 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.

For $I$ we can take the annihilator (Algebra, Definition 10.40.3) 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 10.160.8) which is a field by our assumption on $R$.

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 42.68.11 for a random example.

Lemma 42.68.10. 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 $u$.

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 42.68.6 (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. $\square$

Example 42.68.11. 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 $f = (1, 0) \in M$.

### 42.68.12 Periodic complexes and determinants

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 Subsection 42.68.1. Let us abbreviate $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$, $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\psi = \mathop{\mathrm{Ker}}(\psi )$, and $I_\psi = \mathop{\mathrm{Im}}(\psi )$. 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 42.68.6. 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.

Definition 42.68.13. 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 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 $(-1)^{\text{length}_ R(I_\varphi )\text{length}_ R(I_\psi )} \det \nolimits _\kappa (M, \varphi , \psi )$.

Remark 42.68.14. 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 $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $I_\psi = \mathop{\mathrm{Im}}(\psi )$ 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.

Lemma 42.68.15. 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

$\det \nolimits _\kappa (M, \varphi , \psi ) \det \nolimits _\kappa (M, \psi , \varphi ) = 1.$

Proof. Omitted. $\square$

Lemma 42.68.16. 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 $\text{length}_ R(M) = 2 \text{length}_ R(\mathop{\mathrm{Im}}(\varphi ))$ and

$\det \nolimits _\kappa (M, \varphi , \varphi ) = (-1)^{\text{length}_ R(\mathop{\mathrm{Im}}(\varphi ))} = (-1)^{\frac{1}{2}\text{length}_ R(M)}$

Proof. Follows directly from the sign rule in the definitions. $\square$

Lemma 42.68.17. Let $R$ be a local ring with residue field $\kappa$. Let $M$ be a finite length $R$-module.

1. if $\varphi : M \to M$ is an isomorphism then $\det _\kappa (M, \varphi , 0) = \det _\kappa (\varphi )$.

2. if $\psi : M \to M$ is an isomorphism then $\det _\kappa (M, 0, \psi ) = \det _\kappa (\psi )^{-1}$.

Proof. Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition 42.68.13, 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). $\square$

Lemma 42.68.18. 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 $\kappa ^*$.

Proof. Let us abbreviate $I_{\varphi , i} = \mathop{\mathrm{Im}}(\varphi _ i)$, $K_{\varphi , i} = \mathop{\mathrm{Ker}}(\varphi _ i)$, $I_{\psi , i} = \mathop{\mathrm{Im}}(\psi _ i)$, and $K_{\psi , i} = \mathop{\mathrm{Ker}}(\psi _ i)$. 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 42.68.6 (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. $\square$

Example 42.68.19. 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 $\det _ k(M, \varphi , \psi ) = -u/v \in k^*$.

Example 42.68.20. 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 42.68.19 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 42.68.37 for a more general case (and a proof).

Example 42.68.21. 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 42.68.18.

Example 42.68.22. 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 42.68.19.

Example 42.68.23. 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

$\det \nolimits _ k(M, \varphi , \psi ) = -\frac{u_1u_2}{v_1v_2}.$

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 $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, and $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$ for any $R$-module map $\varphi : M \to M$. We also denote $\varphi \psi = \varphi \circ \psi$ for a pair of morphisms $\varphi , \psi : M \to M$.

Lemma 42.68.24. 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

1. $I_\alpha = K_{\beta \gamma }$, and similarly for any permutation of $\alpha , \beta , \gamma$,

2. $K_\alpha = I_{\beta \gamma }$, and similarly for any permutation of $\alpha , \beta , \gamma$.

Then

1. The triple $(M, \alpha , \beta \gamma )$ is an exact $(2, 1)$-periodic complex.

2. The triple $(I_\gamma , \alpha , \beta )$ is an exact $(2, 1)$-periodic complex.

3. The triple $(M/K_\beta , \alpha , \gamma )$ is an exact $(2, 1)$-periodic complex.

4. We have

$\det \nolimits _\kappa (M, \alpha , \beta \gamma ) = \det \nolimits _\kappa (I_\gamma , \alpha , \beta ) \det \nolimits _\kappa (M/K_\beta , \alpha , \gamma ).$

Proof. It is clear that the assumptions imply part (1) of the lemma.

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.

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.

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

1. an admissible sequence $x_1, \ldots , x_ r \in K_\alpha$ generating $K_\alpha$

2. an admissible sequence $y_1, \ldots , y_ s \in K_\beta$ generating $K_\beta$,

3. an admissible sequence $z_1, \ldots , z_ t \in K_\gamma$ generating $K_\gamma$,

4. elements $\tilde x_ i \in M$ such that $\beta \gamma \tilde x_ i = x_ i$,

5. elements $\tilde y_ i \in M$ such that $\alpha \gamma \tilde y_ i = y_ i$,

6. elements $\tilde z_ i \in M$ such that $\beta \alpha \tilde z_ i = z_ i$.

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 42.68.14 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. $\square$

Lemma 42.68.25. 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

1. $M$, $M'$ have finite length,

2. $(M, \varphi , \psi )$, $(M', \varphi ', \psi ')$ are exact,

3. the maps $\varphi$, $\psi$ induce the zero map on $K = \mathop{\mathrm{Ker}}(\alpha )$, and

4. the maps $\varphi$, $\psi$ induce the zero map on $Q = \mathop{\mathrm{Coker}}(\alpha )$.

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 $\det \nolimits _\kappa (M, \varphi , \psi ) = \det \nolimits _\kappa (M', \varphi ', \psi ')$.

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.

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.

We will use the notation $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$ 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 $\text{length}_ R(I_{\psi '}) = b$.

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 $\alpha _1 = \varphi \alpha ^{-1}|_{I_{\psi '}/\alpha (I_\psi )} \circ \psi '$.

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

1. an admissible sequence $z_1, \ldots , z_ k \in K$ generating $K$,

2. elements $z'_ i \in M$ such that $\varphi z'_ i = z_ i$,

3. elements $z''_ i \in M$ such that $\psi z''_ i = z_ i$,

4. 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 $I_\varphi$,

5. elements $\tilde x_ i \in M$ such that $\varphi \tilde x_ i = x_ i$,

6. 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 $I_\psi$,

7. elements $\tilde y_ i \in M$ such that $\psi \tilde y_ i = y_ i$, and

8. 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 $Q$.

By Remark 42.68.14 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 42.68.14 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 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. $\square$

### 42.68.26 Symbols

The correct generality for this construction is perhaps the situation of the following lemma.

Lemma 42.68.27. 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.

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 10.62.3. 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. $\square$

Lemma 42.68.28. Let $A$ be a Noetherian local ring. Let $a, b \in A$.

1. 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.

2. 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.

In particular, in these cases $\det _\kappa (M/abM, a, b) \in \kappa ^*$, resp. $\det _\kappa (A/(ab), a, b) \in \kappa ^*$ are defined.

Proof. Follows from Lemma 42.68.27. $\square$

Definition 42.68.29. 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 symbol associated to $M, a, b$ to be the element

$d_ M(a, b) = \det \nolimits _\kappa (M/abM, a, b) \in \kappa ^*$

Lemma 42.68.30. 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 $d_ M(a, b)d_ M(b, a) = 1$.

Proof. The first statement follows from Lemma 42.68.24 applied to $M/abcM$ and endomorphisms $\alpha , \beta , \gamma$ given by multiplication by $a, b, c$. The second comes from Lemma 42.68.15. $\square$

Definition 42.68.31. 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 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 42.68.30.

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).

Lemma 42.68.32. 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 $d_ M(a, a) = (-1)^{\text{length}_ A(M/aM)}$.

Proof. Immediate from Lemma 42.68.16. $\square$

Lemma 42.68.33. 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 $d_ A(u, b) = u^{\text{ord}_ A(b)} \bmod \mathfrak m_ A$.

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 42.68.17. The lemma then follows from Lemma 42.68.10. $\square$

Lemma 42.68.34. 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 $\kappa ^*$.

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 42.68.18. $\square$

Lemma 42.68.35. 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

1. $a$, $b$ are nonzerodivisors on both $M$ and $M'$, and

2. $\dim (\mathop{\mathrm{Ker}}(\alpha )), \dim (\mathop{\mathrm{Coker}}(\alpha )) \leq 0$.

Then $d_ M(a, b) = d_{M'}(a, b)$.

Proof. If $a \in A^*$, then the equality follows from the equality $\text{length}(M/bM) = \text{length}(M'/bM')$ and Lemma 42.68.33. Similarly if $b$ is a unit the lemma holds as well (by the symmetry of Lemma 42.68.30). 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 42.68.25. 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 $q = x' \bmod \mathop{\mathrm{Im}}(\overline{\alpha }) \mapsto abx' \in K$. We omit the verification. $\square$

Lemma 42.68.36. 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 $\kappa ^*$.

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 10.62.1. 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 10.62.5. 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 42.68.34 in the first instance and Lemma 42.68.35 in the second. Hence the lemma. $\square$

Lemma 42.68.37. 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.

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 42.68.14 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. $\square$

Lemma 42.68.38. 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 $\kappa ^*$.

Proof. By Lemma 42.68.36 it suffices to show the relation when $M = A/\mathfrak q$ for some prime $\mathfrak q \subset A$ with $\dim (A/\mathfrak q) = 1$.

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 42.68.8.

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 42.68.33 and 42.68.32.

Choose an extension $A \subset A'$ and factorizations $a = ta'$, $b = tb'$ as in Lemma 42.4.2. 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 42.68.27. 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 42.68.35 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 42.68.17 and 42.68.18 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 42.68.16 we have $d^ A_{A'}(b', b') = (-1)^{\text{length}_ A(A'/(b'))}$ and the lemma is proved. $\square$

The tame symbol is a Steinberg symbol.

Lemma 42.68.39. Let $A$ be a Noetherian local domain of dimension $1$ with fraction field $K$. For $x \in K \setminus \{ 0, 1\}$ we have

$d_ A(x, 1 -x) = 1$

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 42.68.38. $\square$

### 42.68.40 Lengths and determinants

In this section we use the determinant to compare lattices. The key lemma is the following.

Lemma 42.68.41. 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

1. $M = \langle x_1, \ldots , x_ l\rangle$,

2. $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$,

3. $N = \langle y_1, \ldots , y_ l\rangle$, and

4. $\langle y_1, \ldots , y_ i\rangle / \langle y_1, \ldots , y_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$.

Then $\varphi$ is injective if and only if $\varphi _{\mathfrak q}$ is an isomorphism, and in this case we have

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) = \text{ord}_{R/\mathfrak q}(f)$

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 $\det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$.

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, $\mathop{\mathrm{Coker}}(\varphi ) = R/(f, \mathfrak q)$ and hence the lemma holds by definition of $\text{ord}_{R/q}(f)$ (see Algebra, Definition 10.121.2).

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

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) = \sum \text{ord}_{R/\mathfrak q}(f_ i).$

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.

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

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi ')) = \text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) + i\ \text{ord}_{R/\mathfrak q}(a).$

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 $i$. $\square$

To use the previous lemma we show that often sequences of elements with the required properties exist.

Lemma 42.68.42. 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

1. the support of $M / \langle x_1, \ldots , x_ l\rangle$ does not contain $\mathfrak q$, and

2. $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$.

Moreover, in this case $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$.

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 10.62.3). 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 $R/\mathfrak q$. $\square$

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 42.68.44 his/herself.

Proposition 42.68.43. 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

1. $M$ is a finite $R$-module,

2. the cohomology modules of $(M, \varphi , \psi )$ are of finite length, and

3. $\dim (\text{Supp}(M)) = 1$.

Let $\mathfrak q_ i$, $i = 1, \ldots , t$ be the minimal primes of the support of $M$. Then we have1

$- e_ R(M, \varphi , \psi ) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}\left( \det \nolimits _{\kappa (\mathfrak q_ i)} (M_{\mathfrak q_ i}, \varphi _{\mathfrak q_ i}, \psi _{\mathfrak q_ i}) \right)$

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 contained in $\{ \mathfrak m\}$. Hence by Lemma 42.2.5 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 $t = 1$.

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 42.68.41 and careful bookkeeping. Denote $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$, $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\psi = \mathop{\mathrm{Ker}}(\psi )$, and $I_\psi = \mathop{\mathrm{Im}}(\psi )$. 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 $l = a + b$.

We are going to use Lemma 42.68.42 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 42.68.42 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 10.62.3). Moreover, the symbol $[w_1, \ldots , w_ n] \in \det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$ is a generator, see Lemma 42.68.5.

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 42.68.42) 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 $M$.

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 $M$.

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 $g, h, f_\varphi , f_\psi \in \kappa (\mathfrak q)$.

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 42.68.13 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.

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 42.68.41.

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 42.68.41 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. $\square$

In most applications the following lemma suffices.

Lemma 42.68.44. 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

1. $\mathop{\mathrm{Ker}}(\psi )$ and $\mathop{\mathrm{Coker}}(\psi )$ have finite length, and

2. $\dim (\text{Supp}(M)) \leq 1$.

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

$\text{length}_ R(\mathop{\mathrm{Coker}}(\psi )) - \text{length}_ R(\mathop{\mathrm{Ker}}(\psi )) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}(f_ i).$

Proof. Recall that $H^0(M, 0, \psi ) = \mathop{\mathrm{Coker}}(\psi )$ and $H^1(M, 0, \psi ) = \mathop{\mathrm{Ker}}(\psi )$, see remarks above Definition 42.2.2. The lemma follows by combining Proposition 42.68.43 with Lemma 42.68.17.

Alternative proof. Reduce to the case $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\}$ as in the proof of Proposition 42.68.43. Then directly combine Lemmas 42.68.41 and 42.68.42 to prove this specific case of Proposition 42.68.43. There is much less bookkeeping in this case, and the reader is encouraged to work this out. Details omitted. $\square$

### 42.68.45 Application to the key lemma

In this section we apply the results above to show the analogue of the key lemma (Lemma 42.6.3) with the tame symbol $d_ A$ constructed above. Please see Remark 42.6.4 for the relationship with Milnor $K$-theory.

Lemma 42.68.46 (Key Lemma). Let $A$ be a $2$-dimensional Noetherian local domain with fraction field $K$. 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 42.68.33.

Proof. Since the tame symbols $d_{A_{\mathfrak q}}(f, g)$ are additive (Lemma 42.68.30) and the order functions $\text{ord}_{A/\mathfrak q}$ are additive (Algebra, Lemma 10.121.1) 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 42.68.43 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. $\square$

 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 42.68.13.

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