**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

an admissible sequence $x_1, \ldots , x_ r \in K_\alpha $ generating $K_\alpha $

an admissible sequence $y_1, \ldots , y_ s \in K_\beta $ generating $K_\beta $,

an admissible sequence $z_1, \ldots , z_ t \in K_\gamma $ generating $K_\gamma $,

elements $\tilde x_ i \in M$ such that $\beta \gamma \tilde x_ i = x_ i$,

elements $\tilde y_ i \in M$ such that $\alpha \gamma \tilde y_ i = y_ i$,

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.
$\square$

## Comments (0)