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.
$\square$
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