Invariant basis number property

We say that a ring R satisfies the invariant basis number property if there is a well-defined notion of the rank of a finitely generated free (left) R-module. Since a finitely generated free module with a basis consisting of n elements is linearly equivalent to Fin n → R, it is sufficient that (Fin n → R) ≃ₗ[R] (Fin m → R) implies n = m.

It is also useful to consider two stronger conditions, namely the rank condition, that a surjective linear map (Fin n → R) →ₗ[R] (Fin m → R) implies m ≤ n and the strong rank condition, that an injective linear map (Fin n → R) →ₗ[R] (Fin m → R) implies n ≤ m.

The strong rank condition implies the rank condition, and the rank condition implies the invariant basis number property.

Main definitions

StrongRankCondition R is a type class stating that R satisfies the strong rank condition. RankCondition R is a type class stating that R satisfies the rank condition. InvariantBasisNumber R is a type class stating that R has the invariant basis number property.

Main results

We show that every nontrivial left-noetherian ring satisfies the strong rank condition, (and so in particular every division ring or field), and then use this to show every nontrivial commutative ring has the invariant basis number property.

More generally, every commutative ring satisfies the strong rank condition. This is proved in LinearAlgebra/FreeModule/StrongRankCondition. We keep invariantBasisNumber_of_nontrivial_of_commRing here since it imports fewer files.

Future work

So far, there is no API at all for the InvariantBasisNumber class. There are several natural ways to formulate that a module M is finitely generated and free, for example M ≃ₗ[R] (Fin n → R), M ≃ₗ[R] (ι → R), where ι is a fintype, or providing a basis indexed by a finite type. There should be lemmas applying the invariant basis number property to each situation.

The finite version of the invariant basis number property implies the infinite analogue, i.e., that (ι →₀ R) ≃ₗ[R] (ι' →₀ R) implies that Cardinal.mk ι = Cardinal.mk ι'. This fact (and its variants) should be formalized.

References

• https://en.wikipedia.org/wiki/Invariant_basis_number
• https://mathoverflow.net/a/2574/

Tags

free module, rank, invariant basis number, IBN

theorem StrongRankCondition_iff (R : Type u) [Semiring R] : StrongRankCondition R ↔ ∀ {n m : ℕ} (f : (Fin n → R) →ₗ[R] Fin m → R), Function.Injective ↑f → n ≤ m

class StrongRankCondition (R : Type u) [Semiring R] : Prop

• le_of_fin_injective : ∀ {n m : ℕ} (f : (Fin n → R) →ₗ[R] Fin m → R), Function.Injective ↑f → n ≤ m

  Any injective linear map from Rⁿ to Rᵐ guarantees n ≤ m.

We say that R satisfies the strong rank condition if (Fin n → R) →ₗ[R] (Fin m → R) injective implies n ≤ m.

theorem le_of_fin_injective (R : Type u) [Semiring R] [StrongRankCondition R] {n : ℕ} {m : ℕ} (f : (Fin n → R) →ₗ[R] Fin m → R) : Function.Injective ↑f → n ≤ m

theorem strongRankCondition_iff_succ (R : Type u) [Semiring R] : StrongRankCondition R ↔ ∀ (n : ℕ) (f : (Fin (n + 1) → R) →ₗ[R] Fin n → R), ¬Function.Injective ↑f

A ring satisfies the strong rank condition if and only if, for all n : ℕ, any linear map (Fin (n + 1) → R) →ₗ[R] (Fin n → R) is not injective.

theorem card_le_of_injective (R : Type u) [Semiring R] [StrongRankCondition R] {α : Type u_1} {β : Type u_2} [Fintype α] [Fintype β] (f : (α → R) →ₗ[R] β → R) (i : Function.Injective ↑f) : Fintype.card α ≤ Fintype.card β

theorem card_le_of_injective' (R : Type u) [Semiring R] [StrongRankCondition R] {α : Type u_1} {β : Type u_2} [Fintype α] [Fintype β] (f : (α →₀ R) →ₗ[R] β →₀ R) (i : Function.Injective ↑f) : Fintype.card α ≤ Fintype.card β

class RankCondition (R : Type u) [Semiring R] : Prop

• le_of_fin_surjective : ∀ {n m : ℕ} (f : (Fin n → R) →ₗ[R] Fin m → R), Function.Surjective ↑f → m ≤ n

  Any surjective linear map from Rⁿ to Rᵐ guarantees m ≤ n.

We say that R satisfies the rank condition if (Fin n → R) →ₗ[R] (Fin m → R) surjective implies m ≤ n.

theorem le_of_fin_surjective (R : Type u) [Semiring R] [RankCondition R] {n : ℕ} {m : ℕ} (f : (Fin n → R) →ₗ[R] Fin m → R) : Function.Surjective ↑f → m ≤ n

theorem card_le_of_surjective (R : Type u) [Semiring R] [RankCondition R] {α : Type u_1} {β : Type u_2} [Fintype α] [Fintype β] (f : (α → R) →ₗ[R] β → R) (i : Function.Surjective ↑f) : Fintype.card β ≤ Fintype.card α

theorem card_le_of_surjective' (R : Type u) [Semiring R] [RankCondition R] {α : Type u_1} {β : Type u_2} [Fintype α] [Fintype β] (f : (α →₀ R) →ₗ[R] β →₀ R) (i : Function.Surjective ↑f) : Fintype.card β ≤ Fintype.card α

instance rankCondition_of_strongRankCondition (R : Type u) [Semiring R] [StrongRankCondition R] : RankCondition R

By the universal property for free modules, any surjective map (Fin n → R) →ₗ[R] (Fin m → R) has an injective splitting (Fin m → R) →ₗ[R] (Fin n → R) from which the strong rank condition gives the necessary inequality for the rank condition.

Equations

• rankCondition_of_strongRankCondition = rankCondition_of_strongRankCondition.proof_1

class InvariantBasisNumber (R : Type u) [Semiring R] : Prop

• eq_of_fin_equiv : ∀ {n m : ℕ}, ((Fin n → R) ≃ₗ[R] Fin m → R) → n = m

  Any linear equiv between Rⁿ and Rᵐ guarantees m = n.

We say that R has the invariant basis number property if (Fin n → R) ≃ₗ[R] (Fin m → R) implies n = m. This gives rise to a well-defined notion of rank of a finitely generated free module.

instance invariantBasisNumber_of_rankCondition (R : Type u) [Semiring R] [RankCondition R] : InvariantBasisNumber R

Equations

• invariantBasisNumber_of_rankCondition = invariantBasisNumber_of_rankCondition.proof_1

theorem eq_of_fin_equiv (R : Type u) [Semiring R] [InvariantBasisNumber R] {n : ℕ} {m : ℕ} : ((Fin n → R) ≃ₗ[R] Fin m → R) → n = m

theorem card_eq_of_linearEquiv (R : Type u) [Semiring R] [InvariantBasisNumber R] {α : Type u_1} {β : Type u_2} [Fintype α] [Fintype β] (f : (α → R) ≃ₗ[R] β → R) : Fintype.card α = Fintype.card β

theorem nontrivial_of_invariantBasisNumber (R : Type u) [Semiring R] [InvariantBasisNumber R] : Nontrivial R

instance IsNoetherianRing.strongRankCondition (R : Type u) [Ring R] [Nontrivial R] [IsNoetherianRing R] : StrongRankCondition R

Any nontrivial noetherian ring satisfies the strong rank condition.

An injective map ((Fin n ⊕ Fin (1 + m)) → R) →ₗ[R] (Fin n → R) for some left-noetherian R would force Fin (1 + m) → R ≃ₗ PUnit (via IsNoetherian.equivPUnitOfProdInjective), which is not the case!

Equations

• IsNoetherianRing.strongRankCondition = IsNoetherianRing.strongRankCondition.proof_1

We want to show that nontrivial commutative rings have invariant basis number. The idea is to take a maximal ideal I of R and use an isomorphism R^n ≃ R^m of R modules to produce an isomorphism (R/I)^n ≃ (R/I)^m of R/I-modules, which will imply n = m since R/I is a field and we know that fields have invariant basis number.

We construct the isomorphism in two steps:

1. We construct the ring R^n/I^n, show that it is an R/I-module and show that there is an isomorphism of R/I-modules R^n/I^n ≃ (R/I)^n. This isomorphism is called Ideal.piQuotEquiv and is located in the file RingTheory/Ideals.lean.
2. We construct an isomorphism of R/I-modules R^n/I^n ≃ R^m/I^m using the isomorphism R^n ≃ R^m.

instance invariantBasisNumber_of_nontrivial_of_commRing {R : Type u} [CommRing R] [Nontrivial R] : InvariantBasisNumber R

Nontrivial commutative rings have the invariant basis number property.

In fact, any nontrivial commutative ring satisfies the strong rank condition, see commRing_strongRankCondition. We prove this instance separately to avoid dependency on LinearAlgebra.Charpoly.Basic.

Equations

• @invariantBasisNumber_of_nontrivial_of_commRing = @invariantBasisNumber_of_nontrivial_of_commRing.proof_1