Bourbaki

Proposition 10

Tag 00B4, permanent. Cite it as [Bourbaki, Tag 00B4].

Algebra, VIII, p. 163, § 9. Radical

Let M be an A-module, P a projective A-module, and u:P→Mu: P\rightarrow M a linear mapping. Let a\mathfrak{a} be a two-sided ideal of A. Suppose that the linear mapping u‾:P/aP→M/aM\overline{u}: P/\mathfrak{a}P\rightarrow M/\mathfrak{a}M deduced from uu by passing to the quotients is bijective and that one of the following conditions are satisfied:

(i) The A-modules M and P are finitely generated, and a\mathfrak{a} is contained in the radical of A.

(ii) The ideal a\mathfrak{a} is nilpotent.

Then (P,u)(P, u) is a projective cover of M.

Under the assumptions, the homomorphism uu is surjective (VIII, p. 159, Corollary 2) and its kernel N is contained in aP\mathfrak{a}P. Let P′P' be a proper submodule of P. By Nakayama’s lemma (VIII, p. 158, Theorem 2), we have P′+aP≠PP'+\mathfrak{a}P\not= P and therefore u(P′)≠Mu(P')\not= M. So (P,u)(P, u) is a projective cover of M.

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

Cites

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