Lemma 42.68.38. Let $A$ be a Noetherian local ring. Let $a, b \in A$. Let $M$ be a finite $A$-module of dimension $1$ on which each of $a$, $b$, $b - a$ are nonzerodivisors. Then

$d_ M(a, b - a)d_ M(b, b) = d_ M(b, b - a)d_ M(a, b)$

in $\kappa ^*$.

Proof. By Lemma 42.68.36 it suffices to show the relation when $M = A/\mathfrak q$ for some prime $\mathfrak q \subset A$ with $\dim (A/\mathfrak q) = 1$.

In case $M = A/\mathfrak q$ we may replace $A$ by $A/\mathfrak q$ and $a, b$ by their images in $A/\mathfrak q$. Hence we may assume $A = M$ and $A$ a local Noetherian domain of dimension $1$. The reason is that the residue field $\kappa$ of $A$ and $A/\mathfrak q$ are the same and that for any $A/\mathfrak q$-module $M$ the determinant taken over $A$ or over $A/\mathfrak q$ are canonically identified. See Lemma 42.68.8.

It suffices to show the relation when both $a, b$ are in the maximal ideal. Namely, the case where one or both are units follows from Lemmas 42.68.33 and 42.68.32.

Choose an extension $A \subset A'$ and factorizations $a = ta'$, $b = tb'$ as in Lemma 42.4.2. Note that also $b - a = t(b' - a')$ and that $A' = (a', b') = (a', b' - a') = (b' - a', b')$. Here and in the following we think of $A'$ as an $A$-module and $a, b, a', b', t$ as $A$-module endomorphisms of $A'$. We will use the notation $d^ A_{A'}(a', b')$ and so on to indicate

$d^ A_{A'}(a', b') = \det \nolimits _\kappa (A'/a'b'A', a', b')$

which is defined by Lemma 42.68.27. The upper index ${}^ A$ is used to distinguish this from the already defined symbol $d_{A'}(a', b')$ which is different (for example because it has values in the residue field of $A'$ which may be different from $\kappa$). By Lemma 42.68.35 we see that $d_ A(a, b) = d^ A_{A'}(a, b)$, and similarly for the other combinations. Using this and multiplicativity we see that it suffices to prove

$d^ A_{A'}(a', b' - a') d^ A_{A'}(b', b') = d^ A_{A'}(b', b' - a') d^ A_{A'}(a', b')$

Now, since $(a', b') = A'$ and so on we have

$\begin{matrix} A'/(a'(b' - a')) & \cong & A'/(a') \oplus A'/(b' - a') \\ A'/(b'(b' - a')) & \cong & A'/(b') \oplus A'/(b' - a') \\ A'/(a'b') & \cong & A'/(a') \oplus A'/(b') \end{matrix}$

Moreover, note that multiplication by $b' - a'$ on $A/(a')$ is equal to multiplication by $b'$, and that multiplication by $b' - a'$ on $A/(b')$ is equal to multiplication by $-a'$. Using Lemmas 42.68.17 and 42.68.18 we conclude

$\begin{matrix} d^ A_{A'}(a', b' - a') & = & \det \nolimits _\kappa (b'|_{A'/(a')})^{-1} \det \nolimits _\kappa (a'|_{A'/(b' - a')}) \\ d^ A_{A'}(b', b' - a') & = & \det \nolimits _\kappa (-a'|_{A'/(b')})^{-1} \det \nolimits _\kappa (b'|_{A'/(b' - a')}) \\ d^ A_{A'}(a', b') & = & \det \nolimits _\kappa (b'|_{A'/(a')})^{-1} \det \nolimits _\kappa (a'|_{A'/(b')}) \end{matrix}$

Hence we conclude that

$(-1)^{\text{length}_ A(A'/(b'))} d^ A_{A'}(a', b' - a') = d^ A_{A'}(b', b' - a') d^ A_{A'}(a', b')$

the sign coming from the $-a'$ in the second equality above. On the other hand, by Lemma 42.68.16 we have $d^ A_{A'}(b', b') = (-1)^{\text{length}_ A(A'/(b'))}$ and the lemma is proved. $\square$

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