Invariant basis number property #
Main definitions #
Let R
be a (not necessary commutative) ring.
InvariantBasisNumber R
is a type class stating that(Fin n → R) ≃ₗ[R] (Fin m → R)
impliesn = m
, a property known as the invariant basis number property.This assumption implies that there is a well-defined notion of the rank of a finitely generated free (left)
R
-module.
It is also useful to consider the following stronger conditions:
the rank condition, witnessed by the type class
RankCondition R
, states that the existence of a surjective linear map(Fin n → R) →ₗ[R] (Fin m → R)
impliesm ≤ n
the strong rank condition, witnessed by the type class
StrongRankCondition R
, states that the existence of an injective linear map(Fin n → R) →ₗ[R] (Fin m → R)
impliesn ≤ m
.OrzechProperty R
, defined inMathlib/RingTheory/OrzechProperty.lean
, states that for any finitely generatedR
-moduleM
, any surjective homomorphismf : N → M
from a submoduleN
ofM
toM
is injective.
Instances #
IsNoetherianRing.orzechProperty
(defined inMathlib/RingTheory/Noetherian.lean
) : any left-noetherian ring satisfies the Orzech property. This applies in particular to division rings.strongRankCondition_of_orzechProperty
: the Orzech property implies the strong rank condition (for non trivial rings).IsNoetherianRing.strongRankCondition
: every nontrivial left-noetherian ring satisfies the strong rank condition (and so in particular every division ring or field).rankCondition_of_strongRankCondition
: the strong rank condition implies the rank condition.invariantBasisNumber_of_rankCondition
: the rank condition implies the invariant basis number property.invariantBasisNumber_of_nontrivial_of_commRing
: a nontrivial commutative ring satisfies the invariant basis number property.
More generally, every commutative ring satisfies the Orzech property,
hence the strong rank condition, which is proved in Mathlib/RingTheory/FiniteType.lean
.
We keep invariantBasisNumber_of_nontrivial_of_commRing
here since it imports fewer files.
Counterexamples to converse results #
The following examples can be found in the book of Lam [lam_1999] (see also https://math.stackexchange.com/questions/4711904):
- Let
k
be a field, then the free (non-commutative) algebrak⟨x, y⟩
satisfies the rank condition but not the strong rank condition. - The free (non-commutative) algebra
ℚ⟨a, b, c, d⟩
quotient by the two-sided ideal(ac − 1, bd − 1, ab, cd)
satisfies the invariant basis number property but not the rank condition.
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/
- [Lam, T. Y. Lectures on Modules and Rings][lam_1999]
- [Orzech, Morris. Onto endomorphisms are isomorphisms][orzech1971]
- [Djoković, D. Ž. Epimorphisms of modules which must be isomorphisms][djokovic1973]
- [Ribenboim, Paulo. Épimorphismes de modules qui sont nécessairement des isomorphismes][ribenboim1971]
Tags #
free module, rank, Orzech property, (strong) rank condition, invariant basis number, IBN
We say that R
satisfies the strong rank condition if (Fin n → R) →ₗ[R] (Fin m → R)
injective
implies n ≤ m
.
Any injective linear map from
Rⁿ
toRᵐ
guaranteesn ≤ m
.
Instances
Any injective linear map from Rⁿ
to Rᵐ
guarantees n ≤ m
.
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.
Any nontrivial ring satisfying Orzech property also satisfies strong rank condition.
Equations
- ⋯ = ⋯
Any surjective linear map from Rⁿ
to Rᵐ
guarantees m ≤ n
.
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
- ⋯ = ⋯
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.
Any linear equiv between
Rⁿ
andRᵐ
guaranteesm = n
.
Instances
Equations
- ⋯ = ⋯
Any nontrivial noetherian ring satisfies the strong rank condition, since it satisfies Orzech property.
Equations
- ⋯ = ⋯
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:
- We construct the ring
R^n/I^n
, show that it is anR/I
-module and show that there is an isomorphism ofR/I
-modulesR^n/I^n ≃ (R/I)^n
. This isomorphism is calledIdeal.piQuotEquiv
and is located in the fileRingTheory/Ideals.lean
. - We construct an isomorphism of
R/I
-modulesR^n/I^n ≃ R^m/I^m
using the isomorphismR^n ≃ R^m
.
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
- ⋯ = ⋯