The Stacks project

Proof. Let $\mathfrak q$ be a minimal prime of the support of $M$. Then $M_{\mathfrak q}$ is a finite length $A_{\mathfrak q}$-module, see Algebra, Lemma 10.62.3. Hence both $\varphi$ and $\psi$ induce isomorphisms $M_{\mathfrak q} \to M_{\mathfrak q}$. Thus the support of $M/\varphi \psi M$ is $\{ \mathfrak m_ A\}$ and hence it has finite length (see lemma cited above). Finally, the kernel of $\varphi$ on $M/\varphi \psi M$ is clearly $\psi M/\varphi \psi M$, and hence the kernel of $\varphi$ is the image of $\psi$ on $M/\varphi \psi M$. Similarly the other way since $M/\varphi \psi M = M/\psi \varphi M$ by assumption. $\square$

Proof. Follows from Lemma 42.68.27. $\square$

Proof. The first statement follows from Lemma 42.68.24 applied to $M/abcM$ and endomorphisms $\alpha , \beta , \gamma$ given by multiplication by $a, b, c$. The second comes from Lemma 42.68.15. $\square$

Proof. Immediate from Lemma 42.68.16. $\square$

Proof. Note that in this case $M/ubM = M/bM$ on which multiplication by $b$ is zero. Hence $d_ M(u, b) = \det _\kappa (u|_{M/bM})$ by Lemma 42.68.17. The lemma then follows from Lemma 42.68.10. $\square$

Proof. It is easy to see that this leads to a short exact sequence of exact $(2, 1)$-periodic complexes

Hence the lemma follows from Lemma 42.68.18. $\square$

Proof. If $a \in A^*$, then the equality follows from the equality $\text{length}(M/bM) = \text{length}(M'/bM')$ and Lemma 42.68.33. Similarly if $b$ is a unit the lemma holds as well (by the symmetry of Lemma 42.68.30). Hence we may assume that $a, b \in \mathfrak m_ A$. This in particular implies that $\mathfrak m$ is not an associated prime of $M$, and hence $\alpha : M \to M'$ is injective. This permits us to think of $M$ as a submodule of $M'$. By assumption $M'/M$ is a finite $A$-module with support $\{ \mathfrak m_ A\}$ and hence has finite length. Note that for any third module $M''$ with $M \subset M'' \subset M'$ the maps $M \to M''$ and $M'' \to M'$ satisfy the assumptions of the lemma as well. This reduces us, by induction on the length of $M'/M$, to the case where $\text{length}_ A(M'/M) = 1$. Finally, in this case consider the map

By construction the cokernel $Q$ of $\overline{\alpha }$ has length $1$. Since $a, b \in \mathfrak m_ A$, they act trivially on $Q$. It also follows that the kernel $K$ of $\overline{\alpha }$ has length $1$ and hence also $a$, $b$ act trivially on $K$. Hence we may apply Lemma 42.68.25. Thus it suffices to see that the two maps $\alpha _ i : Q \to K$ are the same. In fact, both maps are equal to the map $q = x' \bmod \mathop{\mathrm{Im}}(\overline{\alpha }) \mapsto abx' \in K$. We omit the verification. $\square$

Proof. Choose a filtration by $A$-submodules

such that each quotient $M_ j/M_{j - 1}$ is isomorphic to $A/\mathfrak p_ j$ for some prime ideal $\mathfrak p_ j$ of $A$. See Algebra, Lemma 10.62.1. For each $j$ we have either $\mathfrak p_ j = \mathfrak q_ i$ for some $i$, or $\mathfrak p_ j = \mathfrak m_ A$. Moreover, for a fixed $i$, the number of $j$ such that $\mathfrak p_ j = \mathfrak q_ i$ is equal to $\text{length}_{A_{\mathfrak q_ i}}(M_{\mathfrak q_ i})$ by Algebra, Lemma 10.62.5. Hence $d_{M_ j}(a, b)$ is defined for each $j$ and

by Lemma 42.68.34 in the first instance and Lemma 42.68.35 in the second. Hence the lemma. $\square$

Proof. By multiplicativity it suffices to prove this when $x, y \in A$. Let $t \in A$ be a uniformizer. Write $x = t^ bu$ and $y = t^ bv$ for some $a, b \geq 0$ and $u, v \in A^*$. Set $l = a + b$. Then $t^{l - 1}, \ldots , t^ b$ is an admissible sequence in $(x)/(xy)$ and $t^{l - 1}, \ldots , t^ a$ is an admissible sequence in $(y)/(xy)$. Hence by Remark 42.68.14 we see that $d_ A(x, y)$ is characterized by the equation

Hence by the admissible relations for the symbols $[x_1, \ldots , x_ l]$ we see that

as desired. $\square$

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

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

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

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

Hence we conclude that

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. $\square$

Proof. Write $x = a/b$ with $a, b \in A$. The hypothesis implies, since $1 - x = (b - a)/b$, that also $b - a \not= 0$. Hence we compute

Thus we have to show that $d_ A(a, b - a) d_ A(b, b) = d_ A(b, b - a) d_ A(a, b)$. This is Lemma 42.68.38. $\square$