Bourbaki

Example

Tag 05M0, permanent. Cite it as [Bourbaki, Tag 05M0].

Algebra, chapter II, § 1. Modules

It has been seen (no. 1) that a homothety hα:x↦αxh_\alpha : x \mapsto \alpha x on an AA-module EE is not necessarily a linear mapping. But if α\alpha is invertible, hαh_\alpha is a semi-linear mapping (which is moreover bijective) relative to the inner automorphism ξ↦αξα−1\xi \mapsto \alpha \xi \alpha^{-1} of AA, for α(λx)=(αλα−1)(αx)\alpha(\lambda x) = (\alpha \lambda \alpha^{-1})(\alpha x).

Let C be a third ring, ρ′:C→B\rho': C \to B a homomorphism and G a C-module. If v:G→Fv: G \to F is a semi-linear mapping relative to ρ′\rho', the composition w=u∘vw = u \circ v is a semi-linear mapping of G into E relative to the homomorphism ρ′′=ρ∘ρ′\rho'' = \rho \circ \rho'. If ρ\rho is an isomorphism and u:F→Eu: F \to E is a bijective semi-linear mapping relative to ρ\rho, the inverse mapping u′:E→Fu': E \to F is a semi-linear mapping relative to the inverse isomorphism ρ′:A→B\rho': A \to B of ρ\rho.

It is thus seen that, for the species of structure defined by giving on an ordered pair (A, E) of sets a ring structure on A and a left A-module structure on E, the dimorphisms (u,ϕ)(u, \phi) can be taken as morphisms (Set Theory, IV, § 2, no. 1); we shall always assume in what follows that this choice of morphisms has been made.

Remark (3). Let A1,A2A_1, A_2 be two rings, A=A1×A2A = A_1 \times A_2 their product and let e1=(1,0), e2=(0,1)e_1 = (1, 0),\ e_2 = (0, 1) in A, so that A1A_1 and A2A_2 are canonically identified with the two-sided ideals Ae1Ae_1 and Ae2Ae_2 of A. For every A-module E, e1Ee_1E and e2Ee_2E are sub-A-modules E1, E2E_1,\ E_2 of E, annihilated respectively by e2e_2 and e1e_1, so that, canonically identifying A/Ae2A/Ae_2 with A1A_1 and A/Ae1A/Ae_1 with A2A_2, E1E_1 (resp. E2E_2) is given an A1A_1-module (resp. A2A_2-module) structure. Moreover, E is the direct sum of E1E_1 and E2E_2, for every x∈Ex \in E can be written as x=e1x+e2xx = e_1x + e_2x and the relation e1x=e2ye_1x = e_2y implies e1x=e12x=e1e2y=0e_1x = e_1^2x = e_1e_2y = 0. Conversely, for every ordered pair consisting of an A1A_1-module F1F_1 and an A2A_2-module F2F_2, let E1E_1 be the A-module (p1)∗(F1)(p_1) * (F_1), E2E_2 the A-module (p2)∗(F2)(p_2) * (F_2), p1p_1 and p2p_2 being the projections of A onto A1A_1 and A2A_2 respectively; then in the A-module E=E1⊕E2, E1=e1E, E2=e2EE = E_1 \oplus E_2,\ E_1 = e_1E,\ E_2 = e_2E. The study of A-modules is thus reduced to that of A1A_1-modules and that of A2A_2-modules. In particular, every submodule M of E is of the form M1⊕M2M_1 \oplus M_2, where M1=e1MM_1 = e_1M and M2=e2MM_2 = e_2M.

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

Cited by

Nothing.

Cites

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