The Stacks project

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. $\square$