Corollary 3

Tag 05L6, permanent. Cite it as [Bourbaki, Tag 05L6].

Algebra, chapter II, § 1. Modules

Let $(M_i)_{0 \leq i \leq n}$ be a finite family of $A$ -modules of finite length. If there exists an exact sequence of linear mappings

0 \longrightarrow M_0 \xrightarrow{u_0} M_1 \xrightarrow{u_1} M_2 \longrightarrow \cdots \longrightarrow M_{n-1} \xrightarrow{u_{n-1}} M_n \longrightarrow 0

then

\sum_{k=0}^n (-1)^k \operatorname{long}(M_k) = 0.

The corollary is obvious for

n = 1

and is just Proposition 16 for

n = 2

; we argue by induction on

n

. If

M'_{n-1} = \operatorname{Im}(u_{n-2})

, then, by the induction hypothesis,

\sum_{k=0}^{n-2} (-1)^k \operatorname{long}(M_k) + (-1)^{n-1} \operatorname{long}(M'_{n-1}) = 0.

On the other hand, the exact sequence

0 \to M'_{n-1} \to M_{n-1} \to M_n \to 0

gives

\operatorname{long}(M'_{n-1}) + \operatorname{long}(M_n) = \operatorname{long}(M_{n-1}),

whence relation (38).

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

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