Lemma 42.68.17. Let $R$ be a local ring with residue field $\kappa$. Let $M$ be a finite length $R$-module.

1. if $\varphi : M \to M$ is an isomorphism then $\det _\kappa (M, \varphi , 0) = \det _\kappa (\varphi )$.

2. if $\psi : M \to M$ is an isomorphism then $\det _\kappa (M, 0, \psi ) = \det _\kappa (\psi )^{-1}$.

Proof. Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition 42.68.13, identify $K_\varphi = I_\psi = 0$, $I_\varphi = K_\psi = M$. With these identifications, the map

$\gamma _\varphi : \kappa \otimes \det \nolimits _\kappa (M) = \det \nolimits _\kappa (K_\varphi ) \otimes \det \nolimits _\kappa (I_\varphi ) \longrightarrow \det \nolimits _\kappa (M)$

is identified with $\det _\kappa (\varphi ^{-1})$. On the other hand the map $\gamma _\psi$ is identified with the identity map. Hence $\gamma _\psi \circ \sigma \circ \gamma _\varphi ^{-1}$ is equal to $\det _\kappa (\varphi )$ in this case. Whence the result. We omit the proof of (2). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).