Bourbaki

Lemma 1

Tag 0029, permanent. Cite it as [Bourbaki, Tag 0029].

Algebra, VIII, p. 32, § 2. The Structure of Modules of Finite Length

Let M be an A-module, M′M' a primordial submodule of M, and M′′M'' a submodule of M supplementary to M′M'. Let uu be an endomorphism of M. Then uu or 1M−u1_M-u induces an isomorphism from M′M' to a submodule of M supplementary to M′′M''.

Let pp be the projection from M onto M′M' with kernel M′′M'', and let vv be the restriction of p∘up\circ u to M′M'. First, suppose that vv is an automorphism of M′M'. Since vv is injective, the restriction of uu to M′M' is injective, and we have u(M′)∩M′′=0u(M')\cap M''= 0. Since vv is surjective, we have u(M′)⊕M′′=Mu(M')\oplus M''= M. Consequently, uu induces an isomorphism from M′M' to a submodule supplementary to M′′M'' in M. Now, suppose that vv is not an automorphism of M′M'. Then 1M′−v1_{M'}-v is an automorphism of M′M' because M′M' is primordial. Now, 1M′−v1_{M'}-v is the restriction of p∘(1M−u)p\circ (1_M-u) to M′M'. The previous reasoning proves that 1M−u1_M-u induces an isomorphism from M′M' to a submodule of M supplementary to M′′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.

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Cites

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