Bourbaki

Lemma 7

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

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

Retain the notation of lemma 6.

a) One has the inequality HM′′(1)≥1−Tδ1−THMH_{M''}^{(1)} \geq \frac{1 - T^\delta}{1 - T} H_M.

b) In order that one have equality, it is necessary and sufficient that φ\varphi be injective.

c) If this is so, one has M′⊂⋂nFnM' \subset \bigcap_n F_n, and the exact sequence of A-modules

0⟶gr(M)→φgr(M)→vgr(M′′)⟶0, 0 \longrightarrow \mathrm{gr}(M) \xrightarrow{\varphi} \mathrm{gr}(M) \xrightarrow{v} \mathrm{gr}(M'') \longrightarrow 0,

where vv is the canonical mapping, is exact.

Assertions a) and b) follow from formula (18) of lemma 6, and from the definition HM(1)=(1−T)−1.HMH_M^{(1)} = (1 - T)^{-1}.H_M.

Suppose that φ\varphi is injective. By III, § 2, no 8, th. 1, (i), one has

Ker(f)⊂f−1(Fn+δ)=Fn \mathrm{Ker}(f) \subset f^{-1}(F_{n+\delta}) = F_n

for all nn, whence the first assertion of c). Moreover there is an exact sequence

0⟶M/M′→f′M⟶M/f(M)⟶0, 0 \longrightarrow M/M' \xrightarrow{f'} M \longrightarrow M/f(M) \longrightarrow 0,

where f′f' is deduced from ff by passing to the quotient. If φ\varphi is injective, one has as above f′−1(Fn)=Fn−δ/M′{f'}^{-1}(F_n) = F_{n-\delta}/M'. It follows that the filtration on M/M′M/M' deduced as inverse image by f′f' of the filtration FF on MM is the filtration n↦Fn−δ/M′n \mapsto F_{n-\delta}/M'; the associated graded is gr(M)(−δ)\mathrm{gr}(M)(-\delta) and there is an exact sequence of graded modules (III, § 2, no. 4, prop. 2)

0⟶gr(M)(−δ)→φ′gr(M)⟶gr(M′′)⟶0, 0 \longrightarrow \mathrm{gr}(M)(-\delta) \xrightarrow{\varphi'} \mathrm{gr}(M) \longrightarrow \mathrm{gr}(M'') \longrightarrow 0,

where φn′=φn−δ\varphi'_n = \varphi_{n-\delta} for all nn. This completes the proof of c).

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

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