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 → M
and any finite spanning setw : Set M
, the cardinality ofι
is bounded by the cardinality ofw
.linearIndependent_le_basis
: Ifb
is a basis for a moduleM
, ands
is a linearly independent set, then the cardinality ofs
is 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.
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
.
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?][lazarus1973]
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
- 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.
Alias of LinearMap.finrank_le_of_isSMulRegular
.