## 42.3 Calculation of some multiplicities

To prove equality of certain cycles later on we need to compute some multiplicities. Our main tool, besides the elementary lemmas on multiplicities given in the previous section, will be Algebra, Lemma 10.121.7.

Lemma 42.3.1. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Let $x \in R$. Assume that

1. $\dim (\text{Supp}(M)) \leq 1$, and

2. $\dim (\text{Supp}(M/xM)) \leq 0$.

Write $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ t\}$. Then

$e_ R(M, 0, x) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}(x) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}).$

Proof. We first make some preparatory remarks. The result of the lemma holds if $M$ has finite length, i.e., if $t = 0$, because both the left hand side and the right hand side are zero in this case, see Lemma 42.2.4. Also, if we have a short exact sequence $0 \to M \to M' \to M'' \to 0$ of modules satisfying (1) and (2), then lemma for 2 out of 3 of these implies the lemma for the third by the additivity of length (Algebra, Lemma 10.52.3) and additivty of multiplicities (Lemma 42.2.3).

Denote $M_ i$ the image of $M$ in $M_{\mathfrak q_ i}$, so $\text{Supp}(M_ i) = \{ \mathfrak m, \mathfrak q_ i\}$. The kernel and cokernel of the map $M \to \bigoplus M_ i$ have support $\{ \mathfrak m\}$ and hence have finite length. By our preparatory remarks, it follows that it suffices to prove the lemma for each $M_ i$. Thus we may assume that $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\}$. In this case we have a finite filtration $M \supset \mathfrak qM \supset \mathfrak q^2M \supset \ldots \supset \mathfrak q^ nM = 0$ by Algebra, Lemma 10.62.4. Again additivity shows that it suffices to prove the lemma in the case $M$ is annihilated by $\mathfrak q$. In this case we can view $M$ as a $R/\mathfrak q$-module, i.e., we may assume that $R$ is a Noetherian local domain of dimension $1$ with fraction field $K$. Dividing by the torsion submodule, i.e., by the kernel of $M \to M \otimes _ R K = V$ (the torsion has finite length hence is handled by our preliminary remarks) we may assume that $M \subset V$ is a lattice (Algebra, Definition 10.121.3). Then $x : M \to M$ is injective and $\text{length}_ R(M/xM) = d(M, xM)$ (Algebra, Definition 10.121.5). Since $\text{length}_ K(V) = \dim _ K(V)$ we see that $\det (x : V \to V) = x^{\dim _ K(V)}$ and $\text{ord}_ R(\det (x : V \to V)) = \dim _ K(V) \text{ord}_ R(x)$. Thus the desired equality follows from Algebra, Lemma 10.121.7 in this case. $\square$

Lemma 42.3.2. Let $R$ be a Noetherian local ring. Let $x \in R$. If $M$ is a finite Cohen-Macaulay module over $R$ with $\dim (\text{Supp}(M)) = 1$ and $\dim (\text{Supp}(M/xM)) = 0$, then

$\text{length}_ R(M/xM) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}).$

where $\mathfrak q_1, \ldots , \mathfrak q_ t$ are the minimal primes of the support of $M$. If $I \subset R$ is an ideal such that $x$ is a nonzerodivisor on $R/I$ and $\dim (R/I) = 1$, then

$\text{length}_ R(R/(x, I)) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}((R/I)_{\mathfrak q_ i})$

where $\mathfrak q_1, \ldots , \mathfrak q_ n$ are the minimal primes over $I$.

Proof. These are special cases of Lemma 42.3.1. $\square$

Here is another case where we can determine the value of a multiplicity.

Lemma 42.3.3. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\varphi : M \to M$ be an endomorphism and $n > 0$ such that $\varphi ^ n = 0$ and such that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\varphi ^{n - 1})$ has finite length as an $R$-module. Then

$e_ R(M, \varphi ^ i, \varphi ^{n - i}) = 0$

for $i = 0, \ldots , n$.

Proof. The cases $i = 0, n$ are trivial as $\varphi ^0 = \text{id}_ M$ by convention. Let us think of $M$ as an $R[t]$-module where multiplication by $t$ is given by $\varphi$. Let us write $K_ i = \mathop{\mathrm{Ker}}(t^ i : M \to M)$ and

$a_ i = \text{length}_ R(K_ i/t^{n - i}M),\quad b_ i = \text{length}_ R(K_ i/tK_{i + 1}),\quad c_ i = \text{length}_ R(K_1/t^ iK_{i + 1})$

Boundary values are $a_0 = a_ n = b_0 = c_0 = 0$. The $c_ i$ are integers for $i < n$ as $K_1/t^ iK_{i + 1}$ is a quotient of $K_1/t^{n - 1}M$ which is assumed to have finite length. We will use frequently that $K_ i \cap t^ jM = t^ jK_{i + j}$. For $0 < i < n - 1$ we have an exact sequence

$0 \to K_1/t^{n - i - 1}K_{n - i} \to K_{i + 1}/t^{n - i - 1}M \xrightarrow {t} K_ i/t^{n - i}M \to K_ i/tK_{i + 1} \to 0$

By induction on $i$ we conclude that $a_ i$ and $b_ i$ are integers for $i < n$ and that

$c_{n - i - 1} - a_{i + 1} + a_ i - b_ i = 0$

For $0 < i < n - 1$ there is a short exact sequence

$0 \to K_ i/tK_{i + 1} \to K_{i + 1}/tK_{i + 2} \xrightarrow {t^ i} K_1/t^{i + 1}K_{i + 2} \to K_1/t^ iK_{i + 1} \to 0$

which gives

$b_ i - b_{i + 1} + c_{i + 1} - c_ i = 0$

Since $b_0 = c_0$ we conclude that $b_ i = c_ i$ for $i < n$. Then we see that

$a_2 = a_1 + b_{n - 2} - b_1,\quad a_3 = a_2 + b_{n - 3} - b_2,\quad \ldots$

It is straighforward to see that this implies $a_ i = a_{n - i}$ as desired. $\square$

Lemma 42.3.4. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$ with $M$ finite and with cohomology groups of finite length over $R$. Let $x \in R$ be such that $\dim (\text{Supp}(M/xM)) \leq 0$. Then

$e_ R(M, x\varphi , \psi ) = e_ R(M, \varphi , \psi ) - e_ R(\mathop{\mathrm{Im}}(\varphi ), 0, x)$

and

$e_ R(M, \varphi , x\psi ) = e_ R(M, \varphi , \psi ) + e_ R(\mathop{\mathrm{Im}}(\psi ), 0, x)$

Proof. We will only prove the first formula as the second is proved in exactly the same manner. Let $M' = M[x^\infty ]$ be the $x$-power torsion submodule of $M$. Consider the short exact sequence $0 \to M' \to M \to M'' \to 0$. Then $M''$ is $x$-power torsion free (More on Algebra, Lemma 15.88.4). Since $\varphi$, $\psi$ map $M'$ into $M'$ we obtain a short exact sequence

$0 \to (M', \varphi ', \psi ') \to (M, \varphi , \psi ) \to (M'', \varphi '', \psi '') \to 0$

of $(2, 1)$-periodic complexes. Also, we get a short exact sequence $0 \to M' \cap \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi '') \to 0$. We have $e_ R(M', \varphi , \psi ) = e_ R(M', x\varphi , \psi ) = e_ R(M' \cap \mathop{\mathrm{Im}}(\varphi ), 0, x) = 0$ by Lemma 42.2.5. By additivity (Lemma 42.2.3) we see that it suffices to prove the lemma for $(M'', \varphi '', \psi '')$. This reduces us to the case discussed in the next paragraph.

Assume $x : M \to M$ is injective. In this case $\mathop{\mathrm{Ker}}(x\varphi ) = \mathop{\mathrm{Ker}}(\varphi )$. On the other hand we have a short exact sequence

$0 \to \mathop{\mathrm{Im}}(\varphi )/x\mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(x\varphi ) \to \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi ) \to 0$

This together with (42.2.2.1) proves the formula. $\square$

Comment #6725 by jok on

In lemma 42. 3. 1, the second sentence, what is t=0?

Comment #6728 by on

@#6725. The integer $t$ denotes the number of dimension $1$ primes in the support of $M$. So if $t = 0$, then $M$ is $0$ or has support of dimension $0$.

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