Lemma 42.68.18. Let $R$ be a local ring with residue field $\kappa $. Suppose that we have a short exact sequence of $(2, 1)$-periodic complexes
\[ 0 \to (M_1, \varphi _1, \psi _1) \to (M_2, \varphi _2, \psi _2) \to (M_3, \varphi _3, \psi _3) \to 0 \]
with all $M_ i$ of finite length, and each $(M_1, \varphi _1, \psi _1)$ exact. Then
\[ \det \nolimits _\kappa (M_2, \varphi _2, \psi _2) = \det \nolimits _\kappa (M_1, \varphi _1, \psi _1) \det \nolimits _\kappa (M_3, \varphi _3, \psi _3). \]
in $\kappa ^*$.
Proof.
Let us abbreviate $I_{\varphi , i} = \mathop{\mathrm{Im}}(\varphi _ i)$, $K_{\varphi , i} = \mathop{\mathrm{Ker}}(\varphi _ i)$, $I_{\psi , i} = \mathop{\mathrm{Im}}(\psi _ i)$, and $K_{\psi , i} = \mathop{\mathrm{Ker}}(\psi _ i)$. Observe that we have a commutative square
\[ \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & K_{\varphi , 1} \ar[r] \ar[d] & K_{\varphi , 2} \ar[r] \ar[d] & K_{\varphi , 3} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] \ar[d] & M_2 \ar[r] \ar[d] & M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{\varphi , 1} \ar[r] \ar[d] & I_{\varphi , 2} \ar[r] \ar[d] & I_{\varphi , 3} \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 & } \]
of finite length $R$-modules with exact rows and columns. The top row is exact since it can be identified with the sequence $I_{\psi , 1} \to I_{\psi , 2} \to I_{\psi , 3} \to 0$ of images, and similarly for the bottom row. There is a similar diagram involving the modules $I_{\psi , i}$ and $K_{\psi , i}$. By definition $\det _\kappa (M_2, \varphi _2, \psi _2)$ corresponds, up to a sign, to the composition of the left vertical maps in the following diagram
\[ \xymatrix{ \det _\kappa (M_1) \otimes \det _\kappa (M_3) \ar[r]^\gamma \ar[d]^{\gamma ^{-1} \otimes \gamma ^{-1}} & \det _\kappa (M_2) \ar[d]^{\gamma ^{-1}} \\ \det \nolimits _\kappa (K_{\varphi , 1}) \otimes \det \nolimits _\kappa (I_{\varphi , 1}) \otimes \det \nolimits _\kappa (K_{\varphi , 3}) \otimes \det \nolimits _\kappa (I_{\varphi , 3}) \ar[d]^{\sigma \otimes \sigma } \ar[r]^-{\gamma \otimes \gamma } & \det \nolimits _\kappa (K_{\varphi , 2}) \otimes \det \nolimits _\kappa (I_{\varphi , 2}) \ar[d]^\sigma \\ \det \nolimits _\kappa (K_{\psi , 1}) \otimes \det \nolimits _\kappa (I_{\psi , 1}) \otimes \det \nolimits _\kappa (K_{\psi , 3}) \otimes \det \nolimits _\kappa (I_{\psi , 3}) \ar[d]^{\gamma \otimes \gamma } \ar[r]^-{\gamma \otimes \gamma } & \det \nolimits _\kappa (K_{\psi , 2}) \otimes \det \nolimits _\kappa (I_{\psi , 2}) \ar[d]^\gamma \\ \det _\kappa (M_1) \otimes \det _\kappa (M_3) \ar[r]^\gamma & \det _\kappa (M_2) } \]
The top and bottom squares are commutative up to sign by applying Lemma 42.68.6 (2). The middle square is trivially commutative (we are just switching factors). Hence we see that $\det \nolimits _\kappa (M_2, \varphi _2, \psi _2) = \epsilon \det \nolimits _\kappa (M_1, \varphi _1, \psi _1) \det \nolimits _\kappa (M_3, \varphi _3, \psi _3) $ for some sign $\epsilon $. And the sign can be worked out, namely the outer rectangle in the diagram above commutes up to
\begin{eqnarray*} \epsilon & = & (-1)^{\text{length}(I_{\varphi , 1})\text{length}(K_{\varphi , 3}) + \text{length}(I_{\psi , 1})\text{length}(K_{\psi , 3})} \\ & = & (-1)^{\text{length}(I_{\varphi , 1})\text{length}(I_{\psi , 3}) + \text{length}(I_{\psi , 1})\text{length}(I_{\varphi , 3})} \end{eqnarray*}
(proof omitted). It follows easily from this that the signs work out as well.
$\square$
Comments (0)