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. $\square$

[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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