English
Let R be a ring, M an R-module, N a submodule of finite index, and I an ideal of finite index. If N has finite index, then I·N has finite index in M, given N is finitely generated.
Русский
Пусть R — кольцо, M — R-модуль, N — подмодуль с конечным индексом, и I — идеал с конечным индексом. Тогда I·N имеет конечный индекс в M, если N конечно порожден.
LaTeX
$$$$\text{Finite}(M/(I\cdot N)).$$$$
Lean4
/-- Let `N` be a finite index f.g. `R`-submodule, and `I` be a finite index ideal.
Then `I • N` also has finite index. -/
theorem finite_quotient_smul [Finite (R ⧸ I)] [Finite (M ⧸ N)] (hN : N.FG) : Finite (M ⧸ I • N) :=
by
suffices (I • N).toAddSubgroup.FiniteIndex by exact (I • N).toAddSubgroup.finite_quotient_of_finiteIndex
suffices Nat.card (N ⧸ (I • N).comap N.subtype) ≠ 0 by
constructor
rw [← AddSubgroup.relIndex_mul_index (H := (I • N).toAddSubgroup) (K := N.toAddSubgroup) Submodule.smul_le_right]
have inst : Finite (M ⧸ N.toAddSubgroup) := ‹_›
exact mul_ne_zero this AddSubgroup.index_ne_zero_of_finite
let e : (N ⧸ (I • N).comap N.subtype) ≃ₗ[R] (R ⧸ I) ⊗[R] N :=
Submodule.quotEquivOfEq _ (I • (⊤ : Submodule R N))
(Submodule.map_injective_of_injective N.injective_subtype (by simp [Submodule.smul_le_right])) ≪≫ₗ
(quotTensorEquivQuotSMul N I).symm
rw [Nat.card_congr e.toEquiv]
have : Module.Finite R N := Module.Finite.iff_fg.mpr hN
have : Finite ((R ⧸ I) ⊗[R] N) := Module.finite_of_finite (R ⧸ I)
exact Nat.card_pos.ne'