Bourbaki

Lemma 6

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

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

Let A be a ring, M an A-module and (Fn)(F_n) a decreasing filtration on M consisting of submodules; suppose that there exists an integer n1n_1 such that Fn1=MF_{n_1} = M. Let f be an endomorphism of M, M′M' its kernel and M′′M'' its cokernel. We equip M′M' with the filtration (Fn′)(F'_n) induced by (Fn)(F_n) and M′′M'' with the quotient filtration (Fn′′)(F''_n) of (Fn)(F_n). Suppose that Fn/Fn+1F_n/F_{n+1} is of finite length for every n∈Zn \in \mathbf{Z} and that there exists an integer δ\delta such that f(Fn)⊂Fn+δf(F_n) \subset F_{n+\delta}. Let φ\varphi be the graded endomorphism of degree δ\delta of the graded module gr(M)=⨁n∈ZFn/Fn+1\mathrm{gr}(M) = \bigoplus_{n \in \mathbf{Z}} F_n/F_{n+1} deduced from f. Between the following elements of Z((T))\mathbf{Z}((T))

HM=∑n∈Zlong⁡A(Fn/Fn+1).Tn H_M = \sum_{n \in \mathbf{Z}} \operatorname{long}_A(F_n/F_{n+1}).T^n
HM′=∑n∈Zlong⁡A(Fn′/Fn+1′).Tn H_{M'} = \sum_{n \in \mathbf{Z}} \operatorname{long}_A(F'_n/F'_{n+1}).T^n
HM′′=∑n∈Zlong⁡A(Fn′′/Fn+1′′).Tn H_{M''} = \sum_{n \in \mathbf{Z}} \operatorname{long}_A(F''_n/F''_{n+1}).T^n
PKer(φ)=∑n∈Zlong⁡A(Ker(φn)).Tn, P_{\mathrm{Ker}(\varphi)} = \sum_{n \in \mathbf{Z}} \operatorname{long}_A(\mathrm{Ker}(\varphi_n)).T^n ,

one has the inequalities

(17) HM′≤PKer(φ)H_{M'} \leq P_{\mathrm{Ker}(\varphi)}

(18) (1−Tδ).HM(1)+Tδ.PKer(φ)≤HM′′(1)≤(1−Tδ).HM(1)+Tδ.PKer(φ)(1)(1 - T^\delta).H_M^{(1)} + T^\delta.P_{\mathrm{Ker}(\varphi)} \leq H_{M''}^{(1)} \leq (1 - T^\delta).H_M^{(1)} + T^\delta.P_{\mathrm{Ker}(\varphi)}^{(1)}.

The sequence of submodules Gn=f−1(Fn+δ)G_n = f^{-1}(F_{n+\delta}) of MM is a decreasing filtration, and one has Fn⊂GnF_n \subset G_n for every integer nn. By definition, one has Ker(φn)=(Gn+1∩Fn)/Fn+1\mathrm{Ker}(\varphi_n) = (G_{n+1} \cap F_n)/F_{n+1}, whence

(19) PKer(φ)=∑n∈ZlongA((Gn+1∩Fn)/Fn+1).TnP_{\mathrm{Ker}(\varphi)} = \sum_{n \in \mathbf{Z}} \mathrm{long}_A((G_{n+1} \cap F_n)/F_{n+1}).T^n.

For every nn, the A-module (M′∩Fn)/(M′∩Fn+1)(M' \cap F_n)/(M' \cap F_{n+1}) identifies with a submodule of (Gn+1∩Fn)/Fn+1(G_{n+1} \cap F_n)/F_{n+1}, and inequality (17) follows at once from (19). By lemma 5, one has moreover

(20) PKer(φ)≤∑n∈ZlongA(Gn+1/Fn+1).Tn≤PKer(φ)(1)P_{\mathrm{Ker}(\varphi)} \leq \sum_{n \in \mathbf{Z}} \mathrm{long}_A(G_{n+1}/F_{n+1}).T^n \leq P_{\mathrm{Ker}(\varphi)}^{(1)}.

For every n∈Zn \in \mathbf{Z}, one has an exact sequence of A-modules

0⟶Gn+1/Fn+1⟶M/Fn+1→fnM/Fn+δ+1⟶M′′/Fn+δ+1′′⟶0, 0 \longrightarrow G_{n+1}/F_{n+1} \longrightarrow M/F_{n+1} \xrightarrow{f_n} M/F_{n+\delta+1} \longrightarrow M''/F''_{n+\delta+1} \longrightarrow 0,

where fnf_n is deduced from ff by passing to quotients. Consequently one has

longA(M′′/Fn+δ+1′′)=longA(M/Fn+δ+1)−longA(M/Fn+1)+longA(Gn+1/Fn+1)\mathrm{long}_A(M''/F''_{n+\delta+1}) = \mathrm{long}_A(M/F_{n+\delta+1}) - \mathrm{long}_A(M/F_{n+1}) + \mathrm{long}_A(G_{n+1}/F_{n+1}).

Multiplying by Tn+δT^{n+\delta} and summing over nn, one obtains

(21) HM′′(1)=(1−Tδ)HM(1)+Tδ.∑n∈ZlongA(Gn+1/Fn+1).Tn,H_{M''}^{(1)} = (1 - T^\delta)H_M^{(1)} + T^\delta.\sum_{n \in \mathbf{Z}} \mathrm{long}_A(G_{n+1}/F_{n+1}).T^n,

and inequality (18) follows at once from (20) and (21).

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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Translated with machine help, not yet reviewed by a person.