### 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
$$\label{chow-equation-functor} \left\{ \begin{matrix} \text{finite length }R\text{-modules} \\ \text{with isomorphisms} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} 1\text{-dimensional }\kappa \text{-vector spaces} \\ \text{with isomorphisms} \end{matrix} \right\}$$

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

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