Bourbaki

Proposition 12

Tag 05NR, permanent. Cite it as [Bourbaki, Tag 05NR].

Algebra, chapter II, § 2. Modules of linear mappings. Duality

Let E be an A-module and (at)t∈T(a_t)_{t \in T} a generating system of E. The following conditions are equivalent: (a) E is a projective A-module. (b) There exists a family (at∗)t∈T(a_t^*)_{t \in T} of linear forms on E such that, for all x∈Ex \in E, the family (⟨x,at∗⟩)t∈T(\langle x, a_t^* \rangle)_{t \in T} has finite support and

x=∑t∈T⟨x,at∗⟩at. x = \sum_{t \in T} \langle x, a_t^* \rangle a_t.

There exists a surjective homomorphism u:L→Eu : L \to E, where L=As(T)L = A_s^{(T)}, such that if (et)t∈T(e_t)_{t \in T} is the canonical basis of L then u(et)=atu(e_t) = a_t (§ 1, no. 11, Proposition 17); for E to be projective, it is necessary and sufficient that there exist a linear mapping v:E→Lv : E \to L such that u∘v=1Eu \circ v = 1_E (no. 2, Proposition 4 and § 1, no. 9, Proposition 15). If such a mapping exists and we write tv(et∗)=at∗{}^tv(e_t^*) = a_t^*, then ⟨x,at∗⟩=⟨x,tv(et∗)⟩=⟨v(x),et∗⟩\langle x, a_t^* \rangle = \langle x, {}^tv(e_t^*) \rangle = \langle v(x), e_t^* \rangle, hence the family (⟨x,at∗⟩)(\langle x, a_t^* \rangle) has finite support and x=u(∑t∈T⟨(x),et∗⟩et)=∑t∈T⟨x,at∗⟩atx = u \left( \sum_{t \in T} \langle (x), e_t^* \rangle e_t \right) = \sum_{t \in T} \langle x, a_t^* \rangle a_t for all x∈Ex \in E. Conversely, if condition (b) of the statement is fulfilled, the sum ∑t∈T⟨x,at∗⟩et\sum_{t \in T} \langle x, a_t^* \rangle e_t is defined for all x∈Ex \in E and x→∑t∈T⟨x,at∗⟩etx \to \sum_{t \in T} \langle x, a_t^* \rangle e_t is a linear mapping v:E→Lv : E \to L such that u∘v=1Eu \circ v = 1_E.

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

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Cites

Transcribed from the printed edition, 1998, Springer. Not translated.