## 43.15 Algebraic multiplicities

Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $M$ be a finite $A$-module and let $I \subset A$ be an ideal of definition (Algebra, Definition 10.59.1). Recall that the function

$\chi _{I, M}(n) = \text{length}_ A(M/I^ nM) = \sum \nolimits _{p = 0, \ldots , n - 1} \text{length}_ A(I^ pM/I^{p + 1}M)$

is a numerical polynomial (Algebra, Proposition 10.59.5). The degree of this polynomial is equal to $\dim (\text{Supp}(M))$ by Algebra, Lemma 10.62.6.

Definition 43.15.1. In the situation above, assume $d \geq \dim (\text{Supp}(M))$. In this case, if $d > \dim (\text{Supp}(M))$, then we set $e_ I(M, d) = 0$ and if $d = \dim (\text{Supp}(M))$, then we set $e_ I(M, d)$ equal to $d!$ times the leading coefficient of the numerical polynomial $\chi _{I, M}$. Thus in both cases we have

$\chi _{I, M}(n) \sim e_ I(M, d) \frac{n^ d}{d!} + \text{lower order terms}$

The multiplicity of $M$ for the ideal of definition $I$ is $e_ I(M) = e_ I(M, \dim (\text{Supp}(M)))$.

We have the following properties of these multiplicities.

Lemma 43.15.2. Let $A$ be a Noetherian local ring. Let $I \subset A$ be an ideal of definition. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of finite $A$-modules. Let $d \geq \dim (\text{Supp}(M))$. Then

$e_ I(M, d) = e_ I(M', d) + e_ I(M'', d)$

Proof. Immediate from the definitions and Algebra, Lemma 10.59.10. $\square$

Lemma 43.15.3. Let $A$ be a Noetherian local ring. Let $I \subset A$ be an ideal of definition. Let $M$ be a finite $A$-module. Let $d \geq \dim (\text{Supp}(M))$. Then

$e_ I(M, d) = \sum \text{length}_{A_\mathfrak p}(M_\mathfrak p) e_ I(A/\mathfrak p, d)$

where the sum is over primes $\mathfrak p \subset A$ with $\dim (A/\mathfrak p) = d$.

Proof. Both the left and side and the right hand side are additive in short exact sequences of modules of dimension $\leq d$, see Lemma 43.15.2 and Algebra, Lemma 10.52.3. Hence by Algebra, Lemma 10.62.1 it suffices to prove this when $M = A/\mathfrak q$ for some prime $\mathfrak q$ of $A$ with $\dim (A/\mathfrak q) \leq d$. This case is obvious. $\square$

Lemma 43.15.4. Let $P$ be a polynomial of degree $r$ with leading coefficient $a$. Then

$r! a = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i{r \choose i} P(t - i)$

for any $t$.

Proof. Let us write $\Delta$ the operator which to a polynomial $P$ associates the polynomial $\Delta (P) = P(t) - P(t - 1)$. We claim that

$\Delta ^ r(P) = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} P(t - i)$

This is true for $r = 0, 1$ by inspection. Assume it is true for $r$. Then we compute

\begin{align*} \Delta ^{r + 1}(P) & = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} \Delta (P)(t - i) \\ & = \sum \nolimits _{n = -r, \ldots , 0} (-1)^ i {r \choose i} (P(t - i) - P(t - i - 1)) \end{align*}

Thus the claim follows from the equality

${r + 1 \choose i} = {r \choose i} + {r \choose i - 1}$

The lemma follows from the fact that $\Delta (P)$ is of degree $r - 1$ with leading coefficient $ra$ if the degree of $P$ is $r$. $\square$

An important fact is that one can compute the multiplicity in terms of the Koszul complex. Recall that if $R$ is a ring and $f_1, \ldots , f_ r \in R$, then $K_\bullet (f_1, \ldots , f_ r)$ denotes the Koszul complex, see More on Algebra, Section 15.28.

Theorem 43.15.5. Let $A$ be a Noetherian local ring. Let $I = (f_1, \ldots , f_ r) \subset A$ be an ideal of definition. Let $M$ be a finite $A$-module. Then

$e_ I(M, r) = \sum (-1)^ i\text{length}_ A H_ i(K_\bullet (f_1, \ldots , f_ r) \otimes _ A M)$

Proof. Let us change the Koszul complex $K_\bullet (f_1, \ldots , f_ r)$ into a cochain complex $K^\bullet$ by setting $K^ n = K_{-n}(f_1, \ldots , f_ r)$. Then $K^\bullet$ is sitting in degrees $-r, \ldots , 0$ and $H^ i(K^\bullet \otimes _ A M) = H_{-i}(K_\bullet (f_1, \ldots , f_ r) \otimes _ A M)$. The statement of the theorem makes sense as the modules $H^ i(K^\bullet \otimes M)$ are annihilated by $f_1, \ldots , f_ r$ (More on Algebra, Lemma 15.28.6) hence have finite length. Define a filtration on the complex $K^\bullet$ by setting

$F^ p(K^ n \otimes _ A M) = I^{\max (0, p + n)}(K^ n \otimes _ A M),\quad p \in \mathbf{Z}$

Since $f_ i I^ p \subset I^{p + 1}$ this is a filtration by subcomplexes. Thus we have a filtered complex and we obtain a spectral sequence, see Homology, Section 12.24. We have

$E_0 = \bigoplus \nolimits _{p, q} E_0^{p, q} = \bigoplus \nolimits _{p, q} \text{gr}^ p(K^{p + q} \otimes _ A M) = \text{Gr}_ I(K^\bullet \otimes _ A M)$

Since $K^ n$ is finite free we have

$\text{Gr}_ I(K^\bullet \otimes _ A M) = \text{Gr}_ I(K^\bullet ) \otimes _{\text{Gr}_ I(A)} \text{Gr}_ I(M)$

Note that $\text{Gr}_ I(K^\bullet )$ is the Koszul complex over $\text{Gr}_ I(A)$ on the elements $\overline{f}_1, \ldots , \overline{f}_ r \in I/I^2$. A simple calculation (omitted) shows that the differential $d_0$ on $E_0$ agrees with the differential coming from the Koszul complex. Since $\text{Gr}_ I(M)$ is a finite $\text{Gr}_ I(A)$-module and since $\text{Gr}_ I(A)$ is Noetherian (as a quotient of $A/I[x_1, \ldots , x_ r]$ with $x_ i \mapsto \overline{f}_ i$), the cohomology module $E_1 = \bigoplus E_1^{p, q}$ is a finite $\text{Gr}_ I(A)$-module. However, as above $E_1$ is annihilated by $\overline{f}_1, \ldots , \overline{f}_ r$. We conclude $E_1$ has finite length. In particular we find that $\text{Gr}^ p_ F(K^\bullet \otimes M)$ is acyclic for $p \gg 0$.

Next, we check that the spectral sequence above converges using Homology, Lemma 12.24.10. The required equalities follow easily from the Artin-Rees lemma in the form stated in Algebra, Lemma 10.51.3. Thus we see that

\begin{align*} \sum (-1)^ i\text{length}_ A(H^ i(K^\bullet \otimes _ A M)) & = \sum (-1)^{p + q} \text{length}_ A(E_\infty ^{p, q}) \\ & = \sum (-1)^{p + q} \text{length}_ A(E_1^{p, q}) \end{align*}

because as we've seen above the length of $E_1$ is finite (of course this uses additivity of lengths). Pick $t$ so large that $\text{Gr}^ p_ F(K^\bullet \otimes M)$ is acyclic for $p \geq t$ (see above). Using additivity again we see that

$\sum (-1)^{p + q} \text{length}_ A(E_1^{p, q}) = \sum \nolimits _ n \sum \nolimits _{p \leq t} (-1)^ n \text{length}_ A(\text{gr}^ p(K^ n \otimes _ A M))$

This is equal to

$\sum \nolimits _{n = -r, \ldots , 0} (-1)^ n{r \choose |n|} \chi _{I, M}(t + n)$

by our choice of filtration above and the definition of $\chi _{I, M}$ in Algebra, Section 10.59. The lemma follows from Lemma 43.15.4 and the definition of $e_ I(M, r)$. $\square$

Remark 43.15.6 (Trivial generalization). Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $M$ be a finite $A$-module. Let $I \subset A$ be an ideal. The following are equivalent

1. $I' = I + \text{Ann}(M)$ is an ideal of definition (Algebra, Definition 10.59.1),

2. the image $\overline{I}$ of $I$ in $\overline{A} = A/\text{Ann}(M)$ is an ideal of definition,

3. $\text{Supp}(M/IM) \subset \{ \mathfrak m\}$,

4. $\dim (\text{Supp}(M/IM)) \leq 0$, and

5. $\text{length}_ A(M/IM) < \infty$.

This follows from Algebra, Lemma 10.62.3 (details omitted). If this is the case we have $M/I^ nM = M/(I')^ nM$ for all $n$ and $M/I^ nM = M/\overline{I}^ nM$ for all $n$ if $M$ is viewed as an $\overline{A}$-module. Thus we can define

$\chi _{I, M}(n) = \text{length}_ A(M/I^ nM) = \sum \nolimits _{p = 0, \ldots , n - 1} \text{length}_ A(I^ pM/I^{p + 1}M)$

and we get

$\chi _{I, M}(n) = \chi _{I', M}(n) = \chi _{\overline{I}, M}(n)$

for all $n$ by the equalities above. All the results of Algebra, Section 10.59 and all the results in this section, have analogues in this setting. In particular we can define multiplicities $e_ I(M, d)$ for $d \geq \dim (\text{Supp}(M))$ and we have

$\chi _{I, M}(n) \sim e_ I(M, d) \frac{n^ d}{d!} + \text{lower order terms}$

as in the case where $I$ is an ideal of definition.

Comment #8836 by Ben Moonen on

Something's wrong with the phrasing of Definition 43.15.1 (tag 0AZV). I think this needs to be rewritten along the following lines: 'In the situation above, [...] if d > dim(Supp(M)), and equal to d! times [...] if d = dim(Supp(M)). For d = dim(Supp(M)) we therefore have [...]'

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