Bourbaki

Proposition 3

Tag 00D1, permanent. Cite it as [Bourbaki, Tag 00D1].

Algebra, VIII, p. 185, § 11. Grothendieck Groups

Suppose that C\mathscr{C} is hereditary. Let E be a module of type C\mathscr{C} and (Ei)0⩽i⩽n(E_i)_{0\leqslant i\leqslant n} a composition series of E (I, §4, No. 7, p. 41). For 1⩽i⩽n1\leqslant i\leqslant n, the module Ei−1/EiE_{i-1}/E_i is of type C\mathscr{C}, and we have

φ(E)=∑i=1nφ(Ei−1/Ei) \varphi (E) =\sum_{i=1}^n\varphi (E_{i-1}/E_i)

Since C\mathscr{C} is hereditary, the modules E0/E1E_0/E_1 and E1E_1 are of type C\mathscr{C}, and we have φ(E)=φ(E0/E1)+φ(E1)\varphi (E) =\varphi (E_0/E_1) +\varphi (E_1). Since the sequence (Ei+1)0⩽i⩽n−1(E_{i+1})_{0\leqslant i\leqslant n-1} is a composition series of E1E_1, Proposition 3 follows by induction on nn.

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.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.