Bourbaki

Lemma 2

Tag 002A, permanent. Cite it as [Bourbaki, Tag 002A].

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

Let M be an A-module that is the direct sum of a family (Mi)i∈I(M_i)_{i\in I} of primordial submodules, and let uu be an endomorphism of M. Set v=1M−uv= 1_M-u and MJ=⨁i∈JMiM_J=\bigoplus_{i\in J}M_i for every subset J of I. Then one of the following two properties holds:

a) There exists an index i∈Ii\in I such that uu induces an isomorphism from MiM_i to a direct factor submodule of M.

b) For every finite subset J of I,v,v induces an isomorphism from MJM_J to a submodule supplementary to MI−JM_{I-J}.

If property b) holds, then vv is injective.

Suppose that property a) does not hold, and let us establish property b) by induction on the cardinal of J. There is nothing to prove if J=∅J =\emptyset. Therefore, assume that J is nonempty, choose an element ii of J, and set J′=J−{i}J'= J-\{i\}. By the induction hypothesis, vv induces an isomorphism from MJ′M_{J'} to a submodule of M supplementary to MI−J′=MI−J⊕MiM_{I-J'}= M_{I-J}\oplus M_i; consequently, the submodule M′′=v(MJ′)⊕MI−JM''=v(M_{J'})\oplus M_{I-J} is supplementary to MiM_i. By Lemma 1 and the assumption on uu, the endomorphism vv induces an isomorphism from MiM_i to a submodule of M supplementary to M′′M''; consequently, vv induces an isomorphism from MJ=Mi⊕MJ′M_J= M_i\oplus M_{J'} to a submodule supplementary to MI−JM_{I-J}.

The last assertion follows from the fact that M is the union of the submodules MJM_J, where J runs through the finite subsets of I.

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.