Bourbaki

Proposition 2

Tag 00D0, permanent. Cite it as [Bourbaki, Tag 00D0].

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

Suppose that the set C\mathscr{C} is hereditary. Let E be an A-module, and let M and N be submodules of E.

a) If the modules M and N are of type C\mathscr{C}, then so are the modules M∩NM\cap N and M + N, and we have

φ(M+N)+φ(M∩N)=φ(M)+φ(N) \varphi (M + N) +\varphi (M\cap N) =\varphi (M) +\varphi (N)

b) If the modules E/ME/M and E/NE/N are of type C\mathscr{C}, then so are the modules E/(M∩N)E/(M\cap N) and E/(M+N)E/(M + N), and we have

φ(E/(M+N))+φ(E/(M∩N))=φ(E/M)+φ(E/N) \varphi (E/(M + N)) +\varphi (E/(M\cap N)) =\varphi (E/M) +\varphi (E/N)

Assertions a) and b) follow from the existence of the exact sequences

0→M∩N→M⊕N→M+N→0 0\rightarrow M\cap N\rightarrow M\oplus N\rightarrow M + N\rightarrow 0

and

0→E/(M∩N)→(E/M)⊕(E/N)→E/(M+N)→0 0\rightarrow E/(M\cap N)\rightarrow (E/M)\oplus (E/N)\rightarrow E/(M + N)\rightarrow 0

(II, §1, No. 7, p. 207, Proposition 10).

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

  • II, §1, No. 7, p. 207, Proposition 10 not in this corpus
Transcribed from the printed edition, 2023, Springer Nature. Not translated.