Lemmas about rank and finrank in rings satisfying strong rank condition. #
Main statements #
For modules over rings satisfying the rank condition
Basis.le_span: the cardinality of a basis is bounded by the cardinality of any spanning set
For modules over rings satisfying the strong rank condition
linearIndependent_le_span: For any linearly independent familyv : ι → Mand any finite spanning setw : Set M, the cardinality ofιis bounded by the cardinality ofw.linearIndependent_le_basis: Ifbis a basis for a moduleM, andsis a linearly independent set, then the cardinality ofsis bounded by the cardinality ofb.
For modules over rings with invariant basis number (including all commutative rings and all Noetherian rings)
mk_eq_mk_of_basis: the dimension theorem, any two bases of the same vector space have the same cardinality.
Additional definition #
Algebra.IsQuadraticExtension: An extension of ringsR ⊆ Sis quadratic ifSis a freeR-algebra of rank2.
The dimension theorem: if v and v' are two bases, their index types
have the same cardinalities.
Given two bases indexed by ι and ι' of an R-module, where R satisfies the invariant
basis number property, an equiv ι ≃ ι'.
Equations
- v.indexEquiv v' = ⋯.some
Instances For
An auxiliary lemma for Basis.le_span.
If R satisfies the rank condition,
then for any finite basis b : Basis ι R M,
and any finite spanning set w : Set M,
the cardinality of ι is bounded by the cardinality of w.
Another auxiliary lemma for Basis.le_span, which does not require assuming the basis is finite,
but still assumes we have a finite spanning set.
If R satisfies the rank condition,
then the cardinality of any basis is bounded by the cardinality of any spanning set.
If R satisfies the strong rank condition,
then any linearly independent family v : ι → M
contained in the span of some finite w : Set M,
is itself finite.
If R satisfies the strong rank condition,
then for any linearly independent family v : ι → M
contained in the span of some finite w : Set M,
the cardinality of ι is bounded by the cardinality of w.
If R satisfies the strong rank condition,
then for any linearly independent family v : ι → M
and any finite spanning set w : Set M,
the cardinality of ι is bounded by the cardinality of w.
A version of linearIndependent_le_span for Finset.
An auxiliary lemma for linearIndependent_le_basis:
we handle the case where the basis b is infinite.
Over any ring R satisfying the strong rank condition,
if b is a basis for a module M,
and s is a linearly independent set,
then the cardinality of s is bounded by the cardinality of b.
StrongRankCondition implies that if there is an injective linear map (α →₀ R) →ₗ[R] β →₀ R,
then the cardinal of α is smaller than or equal to the cardinal of β.
If R satisfies the strong rank condition, then for any linearly independent family v : ι → M
and spanning set w : Set M, the cardinality of ι is bounded by the cardinality of w.
Let R satisfy the strong rank condition. If m elements of a free rank n R-module are
linearly independent, then m ≤ n.
Over any ring R satisfying the strong rank condition,
if b is an infinite basis for a module M,
then every maximal linearly independent set has the same cardinality as b.
This proof (along with some of the lemmas above) comes from Les familles libres maximales d'un module ont-elles le meme cardinal?
If a vector space has a finite basis, then its dimension (seen as a cardinal) is equal to the cardinality of the basis.
An induction (and recursion) principle for proving results about all submodules of a fixed
finite free module M. A property is true for all submodules of M if it satisfies the following
"inductive step": the property is true for a submodule N if it's true for all submodules N'
of N with the property that there exists 0 ≠ x ∈ N such that the sum N' + Rx is direct.
Equations
- Submodule.inductionOnRank b P ih N = Submodule.inductionOnRankAux b P ih (Fintype.card ι) N ⋯
Instances For
If S a module-finite free R-algebra, then the R-rank of a nonzero R-free
ideal I of S is the same as the rank of S.
If a vector space (or module) has a finite basis, then its dimension (or rank) is equal to the cardinality of the basis.
If a free module is of finite rank, then the cardinality of any basis is equal to its
finrank.
If a vector space (or module) has a finite basis, then its dimension (or rank) is equal to the
cardinality of the basis. This lemma uses a Finset instead of indexed types.
A ring satisfying StrongRankCondition (such as a DivisionRing) is one-dimensional as a
module over itself.
Given a basis of a ring over itself indexed by a type ι, then ι is Unique.
Equations
- Module.Basis.unique R b = ⋯.some
Instances For
The rank of a finite module is finite.
Equations
If M is finite, finrank M = rank M.
If M is finite, then finrank N = rank N for all N : Submodule M. Note that
such an N need not be finitely generated.
Also see Module.finrank_top_le_finrank_of_isScalarTower_of_free
for a version with different typeclass constraints.
Also see Module.finrank_bot_le_finrank_of_isScalarTower_of_free
for a version with different typeclass constraints.
This is a version of exists_linearIndependent
with an upper estimate on the size of the finite set we choose.
This is a version of mem_span_set with an estimate on the number of terms in the sum.
An extension of rings R ⊆ S is quadratic if S is a free R-algebra of rank 2.
- exists_basis : Nonempty ((I : Type u_3) × Basis I R S)