Corollary

Tag 00CZ, permanent. Cite it as [Bourbaki, Tag 00CZ].

Algebra, VIII, p. 184, § 11. Grothendieck Groups

Let $(E_i)_{0 \leq i \leq n}$ be a finite sequence of modules of type $\mathscr{C}$ . If there exists an exact sequence

0 \longrightarrow E_0 \xrightarrow{u_0} E_1 \xrightarrow{u_1} \cdots \xrightarrow{u_{n-1}} E_{n-1} \xrightarrow{u_n} E_n \longrightarrow 0,

then we have

(4)

\sum_{i=0}^n (-1)^i \varphi(E_i) = 0.

Let us prove the corollary by induction on $n$ ; the cases $n= 0$ and $n= 1$ are trivial. So let $n\geqslant 1$ , and let

0\longrightarrow E_0\longrightarrow^{u_0}E_1\longrightarrow \cdots \longrightarrow^{u_1}E_{n-1}\longrightarrow^{u_{n-1}}E_n\overset{u}{-}-\rightarrow^{^n}E_{n+1}\longrightarrow 0

be an exact sequence of modules of type $\mathscr{C}$ . By Proposition 1, the kernel F of $u_n$ is a module of type $\mathscr{C}$ , and we have

\varphi (F) =\varphi (E_n)-\varphi (E_{n+1}) \tag{5}

Moreover, we have an exact sequence

0\longrightarrow E_0\longrightarrow E_1\longrightarrow E_2\longrightarrow \cdots \longrightarrow E_{n-1}\longrightarrow F\longrightarrow 0

and the induction hypothesis gives the relation

\sum^{n-1}_{i=0}(-1)^i\varphi (E_i) + (-1)^n\varphi (F) = 0 \tag{6}

We then immediately deduce from (5) and (6) that $\sum^{n+1}_{i=0}(-1)^i\varphi (E_i) = 0$ , and the corollary follows.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Proposition 1