Bourbaki

Theorem 1

Tag 0735, permanent. Cite it as [Bourbaki, Tag 0735].

Algebra, chapter VII, § 3. Free modules over a principal ideal domain

— Let AA be a ring such that every left ideal of AA is projective (II, p. 231, Def. 1) as an AA-module. Then every submodule MM of a free left AA-module LL is a direct sum of modules isomorphic to ideals of AA.

Let (ei)i∈I(e_i)_{i \in I} be a basis for LL, and let pip_i be the coordinate functions corresponding to this basis. Choose a well-ordering (Set Theory, III, p. 153) of II and let LαL_\alpha denote the submodule generated by the eie_i, for α≤i\alpha \leq i; put Mα=M∩LαM_\alpha = M \cap L_\alpha. The coordinate function pip_i maps MαM_\alpha onto an ideal aαa_\alpha of AA; since aαa_\alpha is a projective AA-module, there exists (II, p. 231, Prop. 4) a submodule NαN_\alpha of MM, such that the map x↦pi(x)x \mapsto p_i(x) from NαN_\alpha into aαa_\alpha is bijective. Let Mα′M'_\alpha be the submodule of LL generated by the NαN_\alpha, for α≤i\alpha \leq i; we will show that Mα′=MαM'_\alpha = M_\alpha for all α\alpha, which will imply that MM is generated by the family (Nα)α<i(N_\alpha)_{\alpha < i}. In fact, suppose Mλ′=MλM'_\lambda = M_\lambda for all λ<i\lambda < i; then pλ(x)∈aλp_\lambda(x) \in a_\lambda, for all x∈Mλx \in M_\lambda; thus there exists y∈Nλy \in N_\lambda such that x−yx - y is a linear combination of finitely many elements eie_i, with α<λ\alpha < \lambda; in other words x−yx - y is an element of MαM_\alpha, for some λ<i\lambda < i; the inductive hypothesis shows that x−y∈Mλ′⊂Mα′x - y \in M'_\lambda \subset M'_\alpha, that is that x∈Mα′x \in M'_\alpha, and so Mα′=MαM'_\alpha = M_\alpha. It remains to show that the sum of the NαN_\alpha is direct; now suppose there exists a linear relation ∑aα=0\sum a_\alpha = 0, with aα∈Nαa_\alpha \in N_\alpha, where the aαa_\alpha (all but finitely many of which are zero) are not all zero. Let pp be the greatest index α\alpha such that aα≠0a_\alpha \neq 0; since pμ(aλ)=0p_\mu(a_\lambda) = 0 for λ<μ\lambda < \mu, we have pμ(aμ)=pμ(∑aα)=0p_\mu(a_\mu) = p_\mu\left( \sum a_\alpha \right) = 0, so aμ=0a_\mu = 0, contradicting the choice of pp.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Transcribed from the printed edition, 2003, Springer. Not translated.