Proposition 5

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

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

Let 0 \to M' \to M \to M'' \to 0 be an exact sequence of finitely generated modules over a noetherian ring A, and q an ideal of A. a) In order that M/qM be of finite length, it is necessary and sufficient that the same be true of M'/qM' and M''/qM''. b) Suppose M/qM of finite length. Then one is in one of the following three cases: 1) d_q(M) = d_q(M') > d_q(M'') \text{ and } e_q(M) = e_q(M'), 2) d_q(M) = d_q(M'') > d_q(M') \text{ and } e_q(M) = e_q(M''), 3) d_q(M) = d_q(M') = d_q(M'') \text{ and } e_q(M) = e_q(M') + e_q(M''). a) One has Supp(M) = Supp(M') \cup Supp(M'') and the assertion results from Remark 2. b) Let us endow M with a q-good filtration F (for example the q-adic filtration), M'' with the quotient filtration F'', and M' with the induced filtration F'. The filtrations F' and F'' are q-good (III, § 3, No. 1, Prop. 1). Then for each n one has an exact sequence of A-modules

0 \to F'_n/F'_{n+1} \to F_n/F_{n+1} \to F''_n/F''_{n+1} \to 0

(III, § 2, No. 4, Prop. 2), so that one has H_{M,F} = H_{M',F'} + H_{M'',F''}, or again

(1 - T)^{-d_q(M)}R = (1 - T)^{-d_q(M')}R' + (1 - T)^{-d_q(M'')}R''

with R, R', R'' \in \mathbf{Z}[T, T^{-1}], R(1) = e_q(M), R'(1) = e_q(M'), R''(1) = e_q(M''). Assertion b) results from this at once.

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

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Cites

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