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. $\square$

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