Rank of free modules #
Main result #
LinearEquiv.nonempty_equiv_iff_lift_rank_eq
: Two free modules are isomorphic iff they have the same dimension.FiniteDimensional.finBasis
: An arbitrary basis of a finite free module indexed byFin n
givenfinrank R M = n
.
Tower law: if A
is a K
-module and K
is an extension of F
then
$\operatorname{rank}_F(A) = \operatorname{rank}_F(K) * \operatorname{rank}_K(A)$.
The universe polymorphic version of rank_mul_rank
below.
Tower law: if A
is a K
-module and K
is an extension of F
then
$\operatorname{rank}_F(A) = \operatorname{rank}_F(K) * \operatorname{rank}_K(A)$.
This is a simpler version of lift_rank_mul_lift_rank
with K
and A
in the same universe.
Tower law: if A
is a K
-module and K
is an extension of F
then
$\operatorname{rank}_F(A) = \operatorname{rank}_F(K) * \operatorname{rank}_K(A)$.
The rank of a free module M
over R
is the cardinality of ChooseBasisIndex R M
.
The finrank of a free module M
over R
is the cardinality of ChooseBasisIndex R M
.
The rank of a free module M
over an infinite scalar ring R
is the cardinality of M
whenever #R < #M
.
Two vector spaces are isomorphic if they have the same dimension.
Two vector spaces are isomorphic if they have the same dimension.
Two vector spaces are isomorphic if they have the same dimension.
Equations
- LinearEquiv.ofLiftRankEq M M' cond = Classical.choice ⋯
Instances For
Two vector spaces are isomorphic if they have the same dimension.
Equations
- LinearEquiv.ofRankEq M M₁ cond = Classical.choice ⋯
Instances For
Two vector spaces are isomorphic if and only if they have the same dimension.
Two vector spaces are isomorphic if and only if they have the same dimension.
Two finite and free modules are isomorphic if they have the same (finite) rank.
Two finite and free modules are isomorphic if and only if they have the same (finite) rank.
Two finite and free modules are isomorphic if they have the same (finite) rank.
Equations
- LinearEquiv.ofFinrankEq M M' cond = Classical.choice ⋯
Instances For
See rank_lt_aleph0
for the inverse direction without Module.Free R M
.
A finite rank free module has a basis indexed by Fin (finrank R M)
.
Equations
Instances For
A rank n
free module has a basis indexed by Fin n
.
Equations
- FiniteDimensional.finBasisOfFinrankEq R M hn = Basis.reindex (FiniteDimensional.finBasis R M) (Fin.castIso hn).toEquiv
Instances For
A free module with rank 1 has a basis with one element.