Maps out of a Koszul complex that are injective modulo an ideal #
Let M be a free module with basis (bᵢ) and let φ : M → R be a linear functional, with
values xᵢ = φ(bᵢ). This file proves the inductive step of the proof of Stacks, Lemma 10.110.3
(00OA). Suppose the xᵢ lie in an ideal I and are linearly independent in I/I², and compare
the Koszul complex with a complex whose differential d' lands in I times its target, as for a
minimal free resolution over a local ring with I the maximal ideal. If a chain map α is
injective modulo I in degree n, with free target, then it is injective modulo I in degree
n + 1: if α(z) ∈ I F' then α(d z) = d'(α z) ∈ I² F, so ∑ xᵢ α(ιᵢ z) ∈ I² F, where ιᵢ is
the contraction by the coordinate bᵢ*, whence every α(ιᵢ z) ∈ I F, every ιᵢ z ∈ I ⋀ⁿ M,
and finally z ∈ I ⋀ⁿ⁺¹ M by Module.Basis.mem_smul_top_iff_forall_koszulDifferential_coord_mem.
Iterating the step, a chain map from the Koszul complex into a complex of free modules whose
differentials land in I times their targets is injective modulo I in every degree as soon as
it is so in degree 0. In degree 0 this holds whenever the chain map lies over the augmentation
⋀⁰ M ≅ R → R ⧸ I.
Main results #
LinearMap.comap_smul_top_le_of_comp_eq_mkQ_zeroEquiv: a map out of⋀⁰ Mlifting the augmentation⋀⁰ M ≅ R → R ⧸ Iis injective moduloI.Module.Basis.comap_smul_top_le_of_comp_koszulDifferential: a map compatible with the Koszul differential and a differential landing inItimes its target is injective moduloIin degreen + 1if it is so in degreen.Module.Basis.comap_smul_top_le_of_hom_koszulComplex: a chain map from the Koszul complex into such a complex of free modules is injective moduloIin every degree if it is so in degree0.
References #
Degree 0 of the comparison with a Koszul complex. A map α : ⋀⁰ M → F lifting the
augmentation ⋀⁰ M ≅ R → R ⧸ I along some π : F → R ⧸ I is injective modulo I: α(z) ∈ I F
forces z ∈ I ⋀⁰ M, as π carries I F into I (R ⧸ I) = 0.
The inductive step of Stacks 00OA. Let φ(bᵢ) ∈ I be linearly independent in I/I²,
let d : F' → F be a map into a free module with image in I F, and let α, α' be maps out of
⋀ⁿ M and ⋀ⁿ⁺¹ M with d ∘ α' = α ∘ d_φ, where d_φ is the Koszul differential of φ. If
α is injective modulo I, i.e. α(z) ∈ I F forces z ∈ I ⋀ⁿ M, then so is α'.
Comparison of a Koszul complex with a minimal complex (Stacks 00OA). Let φ(bᵢ) ∈ I be
linearly independent in I/I², and let α : K(φ) → F be a chain map into a complex of free
modules whose differentials land in I times their targets. If α is injective modulo I in
degree 0, then it is injective modulo I in every degree: α(z) ∈ I Fₙ forces
z ∈ I ⋀ⁿ M.