Remark 1

Tag 0COL, permanent. Cite it as [Bourbaki, Tag 0COL].

Commutative Algebra, chapter VIII, § 4. Séries de Hilbert-Samuel

With the notation of a), let us write $R = \sum_{i \in \mathbf{Z}} a_i T^i$ , and suppose $d > 0$ . By No. 1, the relation $H_{M,F} = (1 - T)^{-d} R$ may also be written

\text{(11)} \quad \operatorname{long}_A(F_n/F_{n+1}) = \sum_{i \in \mathbf{Z}} a_i \binom{n - i + d - 1}{d - 1} = \sum_{i \leq n} a_i \binom{n - i + d - 1}{d - 1}.

Analogously, since

H^{(1)}_{M,F} = (1 - T)^{-d - 1} R

, one has

\text{(12)} \quad \operatorname{long}_A(M/F_{n+1}) = \sum_{i \in \mathbf{Z}} a_i \binom{n - i + d}{d} = \sum_{i \leq n} a_i \binom{n - i + d}{d}.

Let $A$ be a noetherian ring, $q$ an ideal of $A$ , $M$ a finitely generated $A$ -module such that $M/qM$ is of finite length. If $M \neq qM$ , there exist by Theorem 2, b) integers $d_q(M) \geq 0$ and $e_q(M) > 0$ such that, for every q-good filtration $F$ on $M$ , there exists $R \in \mathbf{Z}[T, T^{-1}]$ with

H_{M,F} = (1 - T)^{-d_q(M)} R , \quad R(1) = e_q(M) .

M = qM

, one puts by convention

d_q(M) = -\infty

e_q(M) = 0

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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