Corollary 1
Tag 0021, permanent. Cite it as [Bourbaki, Tag 0021].
Algebra, VIII, p. 28, § 2. The Structure of Modules of Finite Length
Let A be a ring, and let M be an A-module.
a) If the module M is Noetherian, then every surjective endomorphism of M is bijective.
b) If the module M is Artinian, then every injective endomorphism of M is bijective.
c) If the module M has finite length, then every injective or surjective endomorphism of M is bijective.
d) If the ring A is commutative and the A-module M is finitely generated, then every surjective endomorphism of M is bijective.
Let be an endomorphism of the A-module M. We use the notation introduced at the beginning of this subsection. If the endomorphism is surjective, then we have . Assertion a) then follows from Proposition 2, a) and Remark 3. Likewise, if the endomorphism is injective, we have . Assertion b) therefore follows from Proposition 2, b) and Remark 3. Assertion c) follows immediately from a) and b).
Now, suppose that the ring A is commutative, the A-module M is finitely generated, and the endomorphism is surjective. Let us prove that is injective. Let be an element of M such that . Choose a finite generating family of the A-module M and, for every , an element of M such that . There exist families , and of elements of A such that we have
for every . Let be a Noetherian subring of A containing the elements , and (VIII, p. 12, Corollary 3). Let be the -submodule of M generated by the family . We have , and for every ; hence defines, by restriction, a surjective endomorphism of the -module . Since the ring is Noetherian, the finitely generated -module is Noetherian (VIII, p. 7, Proposition 4 a)). By a), the endomorphism of is bijective. By construction, belongs to , and we have . We therefore have , which proves d).
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.