The Stacks project

Proof. Fix an admissible sequence $(f_1, \ldots , f_ l)$ of $M$ such that

for $i = 1, \ldots , l$. Such an admissible sequence exists exactly because $M$ has length $l$. We will show that any element of $\det _\kappa (M)$ is a $\kappa $-multiple of the symbol $[f_1, \ldots , f_ l]$. This will prove the lemma.

Let $(e_1, \ldots , e_ l)$ be an admissible sequence of $M$. It suffices to show that $[e_1, \ldots , e_ l]$ is a multiple of $[f_1, \ldots , f_ l]$. First assume that $\langle e_1, \ldots , e_ l\rangle \not= M$. Then there exists an $i \in [1, \ldots , l]$ such that $e_ i \in \langle e_1, \ldots , e_{i - 1}\rangle $. It immediately follows from the first admissible relation that $[e_1, \ldots , e_ n] = 0$ in $\det _\kappa (M)$. Hence we may assume that $\langle e_1, \ldots , e_ l\rangle = M$. In particular there exists a smallest index $i \in \{ 1, \ldots , l\} $ such that $f_1 \in \langle e_1, \ldots , e_ i\rangle $. This means that $e_ i = \lambda f_1 + x$ with $x \in \langle e_1, \ldots , e_{i - 1}\rangle $ and $\lambda \in R^*$. By the second admissible relation this means that $[e_1, \ldots , e_ l] = \overline{\lambda }[e_1, \ldots , e_{i - 1}, f_1, e_{i + 1}, \ldots , e_ l]$. Note that $\mathfrak m f_1 = 0$. Hence by applying the third admissible relation $i - 1$ times we see that

Note that it is also the case that $ \langle f_1, e_1, \ldots , e_{i - 1}, e_{i + 1}, \ldots , e_ l\rangle = M$. By induction suppose we have proven that our original symbol is equal to a scalar times

for some admissible sequence $(f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l)$ whose elements generate $M$, i.e., with $\langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l\rangle = M$. Then we find the smallest $i$ such that $f_{j + 1} \in \langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ i\rangle $ and we go through the same process as above to see that

Continuing in this vein we obtain the desired result. $\square$

Proof. It is clear that the rule described in the lemma gives a $\kappa $-linear map since all of the admissible relations are satisfied by the usual symbols $e_1 \wedge \ldots \wedge e_ l$. It is also clearly a surjective map. Since by Lemma 42.68.3 the left hand side has dimension at most one we see that the map is an isomorphism. $\square$

Proof. We know $\det _\kappa (M)$ has dimension at most $1$, and in fact that it is generated by $[f_1, \ldots , f_ l]$, by Lemma 42.68.3 and its proof. We will show by induction on $l = \text{length}(M)$ that it is nonzero. For $l = 1$ it follows from Lemma 42.68.4. Choose a nonzero element $f \in M$ with $\mathfrak m f = 0$. Set $\overline{M} = M /\langle f \rangle $, and denote the quotient map $x \mapsto \overline{x}$. We will define a surjective map

which will prove the lemma since by induction the determinant of $\overline{M}$ is nonzero.

We define $\psi $ on symbols as follows. Let $(e_1, \ldots , e_ l)$ be an admissible sequence. If $f \not\in \langle e_1, \ldots , e_ l \rangle $ then we simply set $\psi ([e_1, \ldots , e_ l]) = 0$. If $f \in \langle e_1, \ldots , e_ l \rangle $ then we choose an $i$ minimal such that $f \in \langle e_1, \ldots , e_ i \rangle $. We may write $e_ i = \lambda f + x$ for some unit $\lambda \in R$ and $x \in \langle e_1, \ldots , e_{i - 1} \rangle $. In this case we set

Note that it is indeed the case that $(\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l)$ is an admissible sequence in $\overline{M}$, so this makes sense. Let us show that extending this rule $\kappa $-linearly to linear combinations of symbols does indeed lead to a map on determinants. To do this we have to show that the admissible relations are mapped to zero.

Type (a) relations. Suppose we have $(e_1, \ldots , e_ l)$ an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a \in \langle e_1, \ldots , e_{a - 1}\rangle $. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle $ with $i$ minimal. Then $i \not= a$ and $\overline{e}_ a \in \langle \overline{e}_1, \ldots , \hat{\overline{e}_ i}, \ldots , \overline{e}_{a - 1}\rangle $ if $i < a$ or $\overline{e}_ a \in \langle \overline{e}_1, \ldots , \overline{e}_{a - 1}\rangle $ if $i > a$. Thus the same admissible relation for $\det _\kappa (\overline{M})$ forces the symbol $[\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l]$ to be zero as desired.

Type (b) relations. Suppose we have $(e_1, \ldots , e_ l)$ an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a = \lambda e'_ a + x$ with $\lambda \in R^*$, and $x \in \langle e_1, \ldots , e_{a - 1}\rangle $. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle $ with $i$ minimal. Say $e_ i = \mu f + y$ with $y \in \langle e_1, \ldots , e_{i - 1}\rangle $. If $i < a$ then the desired equality is

which follows from $\overline{e}_ a = \lambda \overline{e}'_ a + \overline{x}$ and the corresponding admissible relation for $\det _\kappa (\overline{M})$. If $i > a$ then the desired equality is

which follows from $\overline{e}_ a = \lambda \overline{e}'_ a + \overline{x}$ and the corresponding admissible relation for $\det _\kappa (\overline{M})$. The interesting case is when $i = a$. In this case we have $e_ a = \lambda e'_ a + x = \mu f + y$. Hence also $e'_ a = \lambda ^{-1}(\mu f + y - x)$. Thus we see that

as desired.

Type (c) relations. Suppose that $(e_1, \ldots , e_ l)$ is an admissible sequence and $\mathfrak m e_ a \subset \langle e_1, \ldots , e_{a - 2}\rangle $. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle $ with $i$ minimal. Say $e_ i = \lambda f + x$ with $x \in \langle e_1, \ldots , e_{i - 1}\rangle $. We distinguish $4$ cases:

Case 1: $i < a - 1$. The desired equality is

which follows from the type (c) admissible relation for $\det _\kappa (\overline{M})$.

Case 2: $i > a$. The desired equality is

which follows from the type (c) admissible relation for $\det _\kappa (\overline{M})$.

Case 3: $i = a$. We write $e_ a = \lambda f + \mu e_{a - 1} + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle $. Then

by definition. If $\overline{\mu }$ is nonzero, then we have $e_{a - 1} = - \mu ^{-1} \lambda f + \mu ^{-1}e_ a - \mu ^{-1} y$ and we obtain

by definition. Since in $\overline{M}$ we have $\overline{e}_ a = \mu \overline{e}_{a - 1} + \overline{y}$ we see the two outcomes are equal by relation (a) for $\det _\kappa (\overline{M})$. If on the other hand $\overline{\mu }$ is zero, then we can write $e_ a = \lambda f + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle $ and we have

which is equal to $\psi ([e_1, \ldots , e_ l])$.

Case 4: $i = a - 1$. Here we have

by definition. If $f \not\in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle $ then

Since $(-1)^{a - 1} = (-1)^{a + 1}$ the two expressions are the same. Finally, assume $f \in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle $. In this case we see that $e_{a - 1} = \lambda f + x$ with $x \in \langle e_1, \ldots , e_{a - 2}\rangle $ and $e_ a = \mu f + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle $ for units $\lambda , \mu \in R$. We conclude that both $e_ a \in \langle e_1, \ldots , e_{a - 1} \rangle $ and $e_{a - 1} \in \langle e_1, \ldots , e_{a - 2}, e_ a\rangle $. In this case a relation of type (a) applies to both $[e_1, \ldots , e_ l]$ and $[e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l]$ and the compatibility of $\psi $ with these shown above to see that both

are zero, as desired.

At this point we have shown that $\psi $ is well defined, and all that remains is to show that it is surjective. To see this let $(\overline{f}_2, \ldots , \overline{f}_ l)$ be an admissible sequence in $\overline{M}$. We can choose lifts $f_2, \ldots , f_ l \in M$, and then $(f, f_2, \ldots , f_ l)$ is an admissible sequence in $M$. Since $\psi ([f, f_2, \ldots , f_ l]) = [f_2, \ldots , f_ l]$ we win. $\square$

Proof. The significance of taking nonzero symbols in the explicit description of the map $\gamma _{K \to L \to M}$ is simply that if $(e_1, \ldots , e_ l)$ is an admissible sequence in $K$, and $(\overline{f}_1, \ldots , \overline{f}_ m)$ is an admissible sequence in $M$, then it is not guaranteed that $(e_1, \ldots , e_ l, f_1, \ldots , f_ m)$ is an admissible sequence in $L$ (where of course $f_ i \in L$ signifies a lift of $\overline{f}_ i$). However, if the symbol $[e_1, \ldots , e_ l]$ is nonzero in $\det _\kappa (K)$, then necessarily $K = \langle e_1, \ldots , e_ k\rangle $ (see proof of Lemma 42.68.3), and in this case it is true that $(e_1, \ldots , e_ k, f_1, \ldots , f_ m)$ is an admissible sequence. Moreover, by the admissible relations of type (b) for $\det _\kappa (L)$ we see that the value of $[e_1, \ldots , e_ k, f_1, \ldots , f_ m]$ in $\det _\kappa (L)$ is independent of the choice of the lifts $f_ i$ in this case also. Given this remark, it is clear that an admissible relation for $e_1, \ldots , e_ k$ in $K$ translates into an admissible relation among $e_1, \ldots , e_ k, f_1, \ldots , f_ m$ in $L$, and similarly for an admissible relation among the $\overline{f}_1, \ldots , \overline{f}_ m$. Thus $\gamma $ defines a linear map of vector spaces as claimed in the lemma.

By Lemma 42.68.5 we know $\det _\kappa (L)$ is generated by any single symbol $[x_1, \ldots , x_{k + m}]$ such that $(x_1, \ldots , x_{k + m})$ is an admissible sequence with $L = \langle x_1, \ldots , x_{k + m}\rangle $. Hence it is clear that the map $\gamma _{K \to L \to M}$ is surjective and hence an isomorphism.

Property (1) holds because

Property (2) means that given a symbol $[\alpha _1, \ldots , \alpha _ a]$ generating $\det _\kappa (A)$, a symbol $[\gamma _1, \ldots , \gamma _ c]$ generating $\det _\kappa (C)$, a symbol $[\zeta _1, \ldots , \zeta _ g]$ generating $\det _\kappa (G)$, and a symbol $[\iota _1, \ldots , \iota _ i]$ generating $\det _\kappa (I)$ we have

(for suitable lifts $\tilde{x}$ in $E$) in $\det _\kappa (E)$. This holds because we may use the admissible relations of type (c) $cg$ times in the following order: move the $\tilde\zeta _1$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _1 \subset A$), then move $\tilde\zeta _2$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _2 \subset A + R\tilde\zeta _1$), and so on.

Part (3) of the lemma is obvious. This finishes the proof. $\square$

Proof. Omitted. $\square$

Proof. If the length of $M$ as an $R$-module is $l$, then the length of $M$ as an $R'$-module (i.e., $M_{R'}$) is $l$ as well, see Algebra, Lemma 10.52.12. Note that an admissible sequence $x_1, \ldots , x_ l$ of $M$ over $R$ is an admissible sequence of $M$ over $R'$ as $\mathfrak m'$ maps into $\mathfrak m$. The isomorphism is obtained by mapping the symbol $[x_1, \ldots , x_ l] \in \det \nolimits _{R, \kappa }(M)$ to the corresponding symbol $[x_1, \ldots , x_ l] \in \det \nolimits _{R', \kappa }(M)$. It is immediate to verify that this is functorial for isomorphisms and compatible with the isomorphisms $\gamma $ of Lemma 42.68.6. $\square$

Proof. Denote $f_ M \in \kappa ^*$ the element such that $\det \nolimits _\kappa (u_ M) = f_ M \text{id}_{\det \nolimits _\kappa (M)}$. Suppose that $0 \to K \to L \to M \to 0$ is a short exact sequence of finite $R$-modules. Then we see that $u_ k$, $u_ L$, $u_ M$ give an isomorphism of short exact sequences. Hence by Lemma 42.68.6 (1) we conclude that $f_ K f_ M = f_ L$. This means that by induction on length it suffices to prove the lemma in the case of length $1$ where it is trivial. $\square$

Proof. Omitted. $\square$

Proof. Follows directly from the sign rule in the definitions. $\square$

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

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). $\square$

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

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

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

(proof omitted). It follows easily from this that the signs work out as well. $\square$

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

and similarly we get a short exact sequence

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

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

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

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

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

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

6. 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

By the same remark, we see that $D_1 = \det _\kappa (M/K_\beta , \alpha , \gamma )$ is characterized by

By the same remark, we see that $D_2 = \det _\kappa (I_\gamma , \alpha , \beta )$ is characterized by

Combining the formulas above we see that $D = D_1 D_2$ as desired. $\square$

Proof. There are (at least) two ways to prove this lemma. One is to produce an enormous commutative diagram using the properties of the determinants. The other is to use the characterization of the determinants in terms of admissible sequences of elements. It is the second approach that we will use.

First let us explain precisely what the maps $\alpha _ i$ are. Namely, $\alpha _0$ is the composition

and $\alpha _1$ is the composition

coming from the boundary maps of the short exact sequences of complexes displayed in the lemma. The fact that the complexes $(M, \varphi , \psi )$, $(M', \varphi ', \psi ')$ are exact implies these maps are isomorphisms.

We will use the notation $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$ and similarly for the other maps. Exactness for $M$ and $M'$ means that $K_\varphi = I_\psi $ and three similar equalities. We introduce $k = \text{length}_ R(K)$, $a = \text{length}_ R(I_\varphi )$, $b = \text{length}_ R(I_\psi )$. Then we see that $\text{length}_ R(M) = a + b$, and $\text{length}_ R(N) = a + b - k$, $\text{length}_ R(Q) = k$ and $\text{length}_ R(M') = a + b$. The exact sequences below will show that also $\text{length}_ R(I_{\varphi '}) = a$ and $\text{length}_ R(I_{\psi '}) = b$.

The assumption that $K \subset K_\varphi = I_\psi $ means that $\varphi $ factors through $N$ to give an exact sequence

Here $\varphi \alpha ^{-1}(x') = y$ means $x' = \alpha (x)$ and $y = \varphi (x)$. Similarly, we have

The assumption that $\psi '$ induces the zero map on $Q$ means that $I_{\psi '} = K_{\varphi '} \subset N$. This means the quotient $\varphi '(N) \subset I_{\varphi '}$ is identified with $Q$. Note that $\varphi '(N) = \alpha (I_\varphi )$. Hence we conclude there is an isomorphism

simply described by $\varphi '(x' \bmod N) = \varphi '(x') \bmod \alpha (I_\varphi )$. In exactly the same way we get

Finally, note that $\alpha _0$ is the composition

and similarly $\alpha _1 = \varphi \alpha ^{-1}|_{I_{\psi '}/\alpha (I_\psi )} \circ \psi '$.

To shorten the formulas below we are going to write $\alpha x$ instead of $\alpha (x)$ in the following. No confusion should result since all maps are indicated by Greek letters and elements by Roman letters. We are going to choose

1. an admissible sequence $z_1, \ldots , z_ k \in K$ generating $K$,

2. elements $z'_ i \in M$ such that $\varphi z'_ i = z_ i$,

3. elements $z''_ i \in M$ such that $\psi z''_ i = z_ i$,

4. elements $x_{k + 1}, \ldots , x_ a \in I_\varphi $ such that $z_1, \ldots , z_ k, x_{k + 1}, \ldots , x_ a$ is an admissible sequence generating $I_\varphi $,

5. elements $\tilde x_ i \in M$ such that $\varphi \tilde x_ i = x_ i$,

6. elements $y_{k + 1}, \ldots , y_ b \in I_\psi $ such that $z_1, \ldots , z_ k, y_{k + 1}, \ldots , y_ b$ is an admissible sequence generating $I_\psi $,

7. elements $\tilde y_ i \in M$ such that $\psi \tilde y_ i = y_ i$, and

8. elements $w_1, \ldots , w_ k \in M'$ such that $w_1 \bmod N, \ldots , w_ k \bmod N$ are an admissible sequence in $Q$ generating $Q$.

By Remark 42.68.14 the element $D = \det _\kappa (M, \varphi , \psi ) \in \kappa ^*$ is characterized by

Note that by the discussion above $\alpha x_{k + 1}, \ldots , \alpha x_ a, \varphi w_1, \ldots , \varphi w_ k$ is an admissible sequence generating $I_{\varphi '}$ and $\alpha y_{k + 1}, \ldots , \alpha y_ b, \psi w_1, \ldots , \psi w_ k$ is an admissible sequence generating $I_{\psi '}$. Hence by Remark 42.68.14 the element $D' = \det _\kappa (M', \varphi ', \psi ') \in \kappa ^*$ is characterized by

Note how in the first, resp. second displayed formula the first, resp. last $k$ entries of the symbols on both sides are the same. Hence these formulas are really equivalent to the equalities

and

in $\det _\kappa (N)$. Note that $\varphi ' w_1, \ldots , \varphi ' w_ k$ and $\alpha z''_1, \ldots , z''_ k$ are admissible sequences generating the module $I_{\varphi '}/\alpha (I_\varphi )$. Write

in $\det _\kappa (I_{\varphi '}/\alpha (I_\varphi ))$ for some $\lambda _0 \in \kappa ^*$. Similarly, write

in $\det _\kappa (I_{\psi '}/\alpha (I_\psi ))$ for some $\lambda _1 \in \kappa ^*$. On the one hand it is clear that

for $i = 0, 1$ by our description of $\alpha _ i$ above, which means that

and on the other hand it is clear that

and

which imply $\lambda _0 D = \lambda _1 D'$. The lemma follows. $\square$

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. $\square$

Proof. First, note that the lemma holds in case $l = 1$. Namely, in this case $x_1$ is a basis of $M$ over $R/\mathfrak q$ and $y_1$ is a basis of $N$ over $R/\mathfrak q$ and we have $\varphi (x_1) = fy_1$ for some $f \in R$. Thus $\varphi $ is injective if and only if $f \not\in \mathfrak q$. Moreover, $\mathop{\mathrm{Coker}}(\varphi ) = R/(f, \mathfrak q)$ and hence the lemma holds by definition of $\text{ord}_{R/q}(f)$ (see Algebra, Definition 10.121.2).

In fact, suppose more generally that $\varphi (x_ i) = f_ iy_ i$ for some $f_ i \in R$, $f_ i \not\in \mathfrak q$. Then the induced maps

are all injective and have cokernels isomorphic to $R/(f_ i, \mathfrak q)$. Hence we see that

On the other hand it is clear that

in this case from the admissible relation (b) for symbols. Hence we see the result holds in this case also.

We prove the general case by induction on $l$. Assume $l > 1$. Let $i \in \{ 1, \ldots , l\} $ be minimal such that $\varphi (x_1) \in \langle y_1, \ldots , y_ i\rangle $. We will argue by induction on $i$. If $i = 1$, then we get a commutative diagram

and the lemma follows from the snake lemma and induction on $l$. Assume now that $i > 1$. Write $\varphi (x_1) = a_1 y_1 + \ldots + a_{i - 1} y_{i - 1} + a y_ i$ with $a_ j, a \in R$ and $a \not\in \mathfrak q$ (since otherwise $i$ was not minimal). Set

Let $M' = \langle x'_1, \ldots , x'_ l \rangle $ and $N' = \langle y'_1, \ldots , y'_ l \rangle $. Since $\varphi (x'_1) = a_1 y'_1 + \ldots + a_{i - 1} y'_{i - 1} + y'_ i$ by construction and since for $j > 1$ we have $\varphi (x'_ j) = a\varphi (x_ i) \in \langle y'_1, \ldots , y'_ l\rangle $ we get a commutative diagram of $R$-modules and maps

By the result of the second paragraph of the proof we know that $\text{length}_ R(M/M') = (l - 1)\text{ord}_{R/\mathfrak q}(a)$ and similarly $\text{length}_ R(M/M') = (l - i + 1)\text{ord}_{R/\mathfrak q}(a)$. By a diagram chase this implies that

On the other hand, it is clear that writing

we have $f' = a^ if$. Hence it suffices to prove the lemma for the case that $\varphi (x_1) = a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_ i$, i.e., in the case that $a = 1$. Next, recall that

by the admissible relations for symbols. The sequence $y_1, \ldots , y_{i - 1}, a_1y_1 + \ldots + a_{i - 1}y_{i - 1} + y_ i, y_{i + 1}, \ldots , y_ l$ satisfies the conditions (3), (4) of the lemma also. Hence, we may actually assume that $\varphi (x_1) = y_ i$. In this case, note that we have $\mathfrak q x_1 = 0$ which implies also $\mathfrak q y_ i = 0$. We have

by the third of the admissible relations defining $\det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$. Hence we may replace $y_1, \ldots , y_ l$ by the sequence $y'_1, \ldots , y'_ l = y_1, \ldots , y_{i - 2}, y_ i, y_{i - 1}, y_{i + 1}, \ldots , y_ l$ (which also satisfies conditions (3) and (4) of the lemma). Clearly this decreases the invariant $i$ by $1$ and we win by induction on $i$. $\square$

Proof. The condition that $\mathfrak q$ is a minimal prime in the support of $M$ implies that $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$ is finite (see Algebra, Lemma 10.62.3). Hence we can find $y_1, \ldots , y_ l \in M_{\mathfrak q}$ such that $\langle y_1, \ldots , y_ i\rangle / \langle y_1, \ldots , y_{i - 1}\rangle \cong \kappa (\mathfrak q)$ for $i = 1, \ldots , l$. We can find $f_ i \in R$, $f_ i \not\in \mathfrak q$ such that $f_ i y_ i$ is the image of some element $z_ i \in M$. Moreover, as $R$ is Noetherian we can write $\mathfrak q = (g_1, \ldots , g_ t)$ for some $g_ j \in R$. By assumption $g_ j y_ i \in \langle y_1, \ldots , y_{i - 1} \rangle $ inside the module $M_{\mathfrak q}$. By our choice of $z_ i$ we can find some further elements $f_{ji} \in R$, $f_{ij} \not\in \mathfrak q$ such that $f_{ij} g_ j z_ i \in \langle z_1, \ldots , z_{i - 1} \rangle $ (equality in the module $M$). The lemma follows by taking

and so on. Namely, since all the elements $f_ i, f_{ij}$ are invertible in $R_{\mathfrak q}$ we still have that $R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_ i / R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_{i - 1} \cong \kappa (\mathfrak q)$ for $i = 1, \ldots , l$. By construction, $\mathfrak q x_ i \in \langle x_1, \ldots , x_{i - 1}\rangle $. Thus $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle $ is an $R$-module generated by one element, annihilated $\mathfrak q$ such that localizing at $\mathfrak q$ gives a $q$-dimensional vector space over $\kappa (\mathfrak q)$. Hence it is isomorphic to $R/\mathfrak q$. $\square$

Proof. We first reduce to the case $t = 1$ in the following way. Note that $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ t\} $, where $\mathfrak m \subset R$ is the maximal ideal. Let $M_ i$ denote the image of $M \to M_{\mathfrak q_ i}$, so $\text{Supp}(M_ i) = \{ \mathfrak m, \mathfrak q_ i\} $. The map $\varphi $ (resp. $\psi $) induces an $R$-module map $\varphi _ i : M_ i \to M_ i$ (resp. $\psi _ i : M_ i \to M_ i$). Thus we get a morphism of $(2, 1)$-periodic complexes

The kernel and cokernel of this map have support contained in $\{ \mathfrak m\} $. Hence by Lemma 42.2.5 we have

On the other hand we clearly have $M_{\mathfrak q_ i} = M_{i, \mathfrak q_ i}$, and hence the terms of the right hand side of the formula of the lemma are equal to the expressions

In other words, if we can prove the lemma for each of the modules $M_ i$, then the lemma holds. This reduces us to the case $t = 1$.

Assume we have a $(2, 1)$-periodic complex $(M, \varphi , \psi )$ over a Noetherian local ring with $M$ a finite $R$-module, $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\} $, and finite length cohomology modules. The proof in this case follows from Lemma 42.68.41 and careful bookkeeping. Denote $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$, $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\psi = \mathop{\mathrm{Ker}}(\psi )$, and $I_\psi = \mathop{\mathrm{Im}}(\psi )$. Since $R$ is Noetherian these are all finite $R$-modules. Set

Equalities because the complex becomes exact after localizing at $\mathfrak q$. Note that $l = \text{length}_{R_{\mathfrak q}}(M_{\mathfrak q})$ is equal to $l = a + b$.

We are going to use Lemma 42.68.42 to choose sequences of elements in finite $R$-modules $N$ with support contained in $\{ \mathfrak m, \mathfrak q\} $. In this case $N_{\mathfrak q}$ has finite length, say $n \in \mathbf{N}$. Let us call a sequence $w_1, \ldots , w_ n \in N$ with properties (1) and (2) of Lemma 42.68.42 a “good sequence”. Note that the quotient $N/\langle w_1, \ldots , w_ n \rangle $ of $N$ by the submodule generated by a good sequence has support (contained in) $\{ \mathfrak m\} $ and hence has finite length (Algebra, Lemma 10.62.3). Moreover, the symbol $[w_1, \ldots , w_ n] \in \det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$ is a generator, see Lemma 42.68.5.

Having said this we choose good sequences

We will adjust our choices a little bit as follows. Choose lifts $\tilde y_ i \in M$ of $y_ i \in I_\varphi $ and $\tilde s_ i \in M$ of $s_ i \in I_\psi $. It may not be the case that $\mathfrak q \tilde y_1 \subset \langle x_1, \ldots , x_ b\rangle $ and it may not be the case that $\mathfrak q \tilde s_1 \subset \langle t_1, \ldots , t_ a\rangle $. However, using that $\mathfrak q$ is finitely generated (as in the proof of Lemma 42.68.42) we can find a $d \in R$, $d \not\in \mathfrak q$ such that $\mathfrak q d\tilde y_1 \subset \langle x_1, \ldots , x_ b\rangle $ and $\mathfrak q d\tilde s_1 \subset \langle t_1, \ldots , t_ a\rangle $. Thus after replacing $y_ i$ by $dy_ i$, $\tilde y_ i$ by $d\tilde y_ i$, $s_ i$ by $ds_ i$ and $\tilde s_ i$ by $d\tilde s_ i$ we see that we may assume also that $x_1, \ldots , x_ b, \tilde y_1, \ldots , \tilde y_ b$ and $t_1, \ldots , t_ a, \tilde s_1, \ldots , \tilde s_ b$ are good sequences in $M$.

Finally, we choose a good sequence $z_1, \ldots , z_ l$ in the finite $R$-module

Note that this is also a good sequence in $M$.

Since $I_{\varphi , \mathfrak q} = K_{\psi , \mathfrak q}$ there is a unique element $h \in \kappa (\mathfrak q)$ such that $[y_1, \ldots , y_ a] = h [t_1, \ldots , t_ a]$ inside $\det _{\kappa (\mathfrak q)}(K_{\psi , \mathfrak q})$. Similarly, as $I_{\psi , \mathfrak q} = K_{\varphi , \mathfrak q}$ there is a unique element $h \in \kappa (\mathfrak q)$ such that $[s_1, \ldots , s_ b] = g [x_1, \ldots , x_ b]$ inside $\det _{\kappa (\mathfrak q)}(K_{\varphi , \mathfrak q})$. We can also do this with the three good sequences we have in $M$. All in all we get the following identities

for some $g, h, f_\varphi , f_\psi \in \kappa (\mathfrak q)$.

Having set up all this notation let us compute $\det _{\kappa (\mathfrak q)}(M, \varphi , \psi )$. Namely, consider the element $[z_1, \ldots , z_ l]$. Under the map $\gamma _\psi \circ \sigma \circ \gamma _\varphi ^{-1}$ of Definition 42.68.13 we have

This means that $\det _{\kappa (\mathfrak q)} (M_{\mathfrak q}, \varphi _{\mathfrak q}, \psi _{\mathfrak q})$ is equal to $f_\varphi h/f_\psi g$ up to a sign.

We abbreviate the following quantities

Using the exact sequences $0 \to K_\varphi \to M \to I_\varphi \to 0$ we get $m_\varphi = k_\varphi + i_\varphi $. Similarly we have $m_\psi = k_\psi + i_\psi $. We have $\delta _\varphi + m_\varphi = \delta _\psi + m_\psi $ since this is equal to the colength of $\langle z_1, \ldots , z_ l \rangle $ in $M$. Finally, we have

by our first application of the key Lemma 42.68.41.

Next, let us compute the multiplicity of the periodic complex

where we used the key Lemma 42.68.41 twice in the third equality. By our computation of $\det _{\kappa (\mathfrak q)} (M_{\mathfrak q}, \varphi _{\mathfrak q}, \psi _{\mathfrak q})$ this proves the proposition. $\square$

Proof. Recall that $H^0(M, 0, \psi ) = \mathop{\mathrm{Coker}}(\psi )$ and $H^1(M, 0, \psi ) = \mathop{\mathrm{Ker}}(\psi )$, see remarks above Definition 42.2.2. The lemma follows by combining Proposition 42.68.43 with Lemma 42.68.17.

Alternative proof. Reduce to the case $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\} $ as in the proof of Proposition 42.68.43. Then directly combine Lemmas 42.68.41 and 42.68.42 to prove this specific case of Proposition 42.68.43. There is much less bookkeeping in this case, and the reader is encouraged to work this out. Details omitted. $\square$

Proof. Since the tame symbols $d_{A_{\mathfrak q}}(f, g)$ are additive (Lemma 42.68.30) and the order functions $\text{ord}_{A/\mathfrak q}$ are additive (Algebra, Lemma 10.121.1) it suffices to prove the formula when $f = a \in A$ and $g = b \in A$. In this case we see that we have to show

By Proposition 42.68.43 this is equivalent to showing that

Since the complex $A/(ab) \xrightarrow {a} A/(ab) \xrightarrow {b} A/(ab) \xrightarrow {a} A/(ab)$ is exact we win. $\square$