# 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

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`

to`Rᵐ`

guarantees`n ≤ 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 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`

and`Rᵐ`

guarantees`m = n`

.

## Instances

## Equations

- ⋯ = ⋯

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

- ⋯ = ⋯

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 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`

. - We construct an isomorphism of
`R/I`

-modules`R^n/I^n ≃ R^m/I^m`

using the isomorphism`R^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

- ⋯ = ⋯