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. $\square$

[1] 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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