Bourbaki

Corollary 1

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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 uu be an endomorphism of the A-module M. We use the notation introduced at the beginning of this subsection. If the endomorphism uu is surjective, then we have I∞=MI_{\infty}= M. Assertion a) then follows from Proposition 2, a) and Remark 3. Likewise, if the endomorphism uu is injective, we have N∞=0N_{\infty}= 0. 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 uu is surjective. Let us prove that uu is injective. Let xx be an element of M such that u(x)=0u(x) = 0. Choose a finite generating family (xi)i∈I(x_i)_{i\in I} of the A-module M and, for every i∈Ii\in I, an element yiy_i of M such that u(yi)=xiu(y_i) =x_i. There exist families (ai)i∈I,(bij)(i,j)∈I×I(a_i)_{i\in I}, (b_{ij})_{(i,j)\in I\times I}, and (cij)(i,j)∈I×I(c_{ij})_{(i,j)\in I\times I} of elements of A such that we have

x=∑i∈Iaixi,yj=∑i∈Ibijxi,u(xj)=∑i∈Icijxi x=\sum_{i\in I}a_ix_i,y_j=\sum_{i\in I}b_{ij}x_i,u(x_j) =\sum_{i\in I}c_{ij}x_i

for every j∈Ij\in I. Let A′A' be a Noetherian subring of A containing the elements ai,bija_i,b_{ij}, and cijc_{ij} (VIII, p. 12, Corollary 3). Let M′M' be the A′A'-submodule of M generated by the family (xi)i∈I(x_i)_{i\in I}. We have u(xj)∈M′,yj∈M′u(x_j)\in M',y_j\in M', and u(yj)=xju(y_j) =x_j for every j∈Ij\in I; hence uu defines, by restriction, a surjective endomorphism u′u' of the A′A'-module M′M'. Since the ring A′A' is Noetherian, the finitely generated A′A'-module M′M' is Noetherian (VIII, p. 7, Proposition 4 a)). By a), the endomorphism u′u' of M′M' is bijective. By construction, xx belongs to M′M', and we have u′(x)=u(x)=0u'(x) =u(x) = 0. We therefore have x=0x= 0, 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.

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