Linearly disjoint submodules #
This file contains basics about linearly disjoint submodules.
Mathematical background #
We adapt the definitions in https://en.wikipedia.org/wiki/Linearly_disjoint.
Let R
be a commutative ring, S
be an R
-algebra (not necessarily commutative).
Let M
and N
be R
-submodules in S
(Submodule R S
).
M
andN
are linearly disjoint (Submodule.LinearDisjoint M N
or simplyM.LinearDisjoint N
), if the naturalR
-linear mapM ⊗[R] N →ₗ[R] S
(Submodule.mulMap M N
) induced by the multiplication inS
is injective.
The following is the first equivalent characterization of linear disjointness:
Submodule.LinearDisjoint.linearIndependent_left_of_flat
: ifM
andN
are linearly disjoint, ifN
is a flatR
-module, then for any family ofR
-linearly independent elements{ m_i }
ofM
, they are alsoN
-linearly independent, in the sense that theR
-linear map fromι →₀ N
toS
which maps{ n_i }
to the sum ofm_i * n_i
(Submodule.mulLeftMap N m
) is injective.Submodule.LinearDisjoint.of_basis_left
: conversely, if{ m_i }
is anR
-basis ofM
, which is alsoN
-linearly independent, thenM
andN
are linearly disjoint.
Dually, we have:
Submodule.LinearDisjoint.linearIndependent_right_of_flat
: ifM
andN
are linearly disjoint, ifM
is a flatR
-module, then for any family ofR
-linearly independent elements{ n_i }
ofN
, they are alsoM
-linearly independent, in the sense that theR
-linear map fromι →₀ M
toS
which maps{ m_i }
to the sum ofm_i * n_i
(Submodule.mulRightMap M n
) is injective.Submodule.LinearDisjoint.of_basis_right
: conversely, if{ n_i }
is anR
-basis ofN
, which is alsoM
-linearly independent, thenM
andN
are linearly disjoint.
The following is the second equivalent characterization of linear disjointness:
Submodule.LinearDisjoint.linearIndependent_mul_of_flat
: ifM
andN
are linearly disjoint, if one ofM
andN
is flat, then for any family ofR
-linearly independent elements{ m_i }
ofM
, and any family ofR
-linearly independent elements{ n_j }
ofN
, the family{ m_i * n_j }
inS
is alsoR
-linearly independent.Submodule.LinearDisjoint.of_basis_mul
: conversely, if{ m_i }
is anR
-basis ofM
, if{ n_i }
is anR
-basis ofN
, such that the family{ m_i * n_j }
inS
isR
-linearly independent, thenM
andN
are linearly disjoint.
Other main results #
Submodule.LinearDisjoint.symm_of_commute
,Submodule.linearDisjoint_comm_of_commute
: linear disjointness is symmetric under some commutative conditions.Submodule.LinearDisjoint.map
: linear disjointness is preserved by injective algebra homomorphisms.Submodule.linearDisjoint_op
: linear disjointness is preserved by taking multiplicative opposite.Submodule.LinearDisjoint.of_le_left_of_flat
,Submodule.LinearDisjoint.of_le_right_of_flat
,Submodule.LinearDisjoint.of_le_of_flat_left
,Submodule.LinearDisjoint.of_le_of_flat_right
: linear disjointness is preserved by taking submodules under some flatness conditions.Submodule.LinearDisjoint.of_linearDisjoint_fg_left
,Submodule.LinearDisjoint.of_linearDisjoint_fg_right
,Submodule.LinearDisjoint.of_linearDisjoint_fg
: conversely, if any finitely generated submodules ofM
andN
are linearly disjoint, thenM
andN
themselves are linearly disjoint.Submodule.LinearDisjoint.bot_left
,Submodule.LinearDisjoint.bot_right
: the zero module is linearly disjoint with any other submodules.Submodule.LinearDisjoint.one_left
,Submodule.LinearDisjoint.one_right
: the image ofR
inS
is linearly disjoint with any other submodules.Submodule.LinearDisjoint.of_left_le_one_of_flat
,Submodule.LinearDisjoint.of_right_le_one_of_flat
: if a submodule is contained in the image ofR
inS
, then it is linearly disjoint with any other submodules, under some flatness conditions.Submodule.LinearDisjoint.not_linearIndependent_pair_of_commute_of_flat
,Submodule.LinearDisjoint.rank_inf_le_one_of_commute_of_flat
: ifM
andN
are linearly disjoint, if one ofM
andN
is flat, then any two commutative elements contained in the intersection ofM
andN
are notR
-linearly independent (namely, their span is notR ^ 2
). In particular, if any two elements in the intersection ofM
andN
are commutative, then the rank of the intersection ofM
andN
is at most one.These results are stated using bundled version (i.e.
a : ↥(M ⊓ N)
). If you want a not bundled version (i.e.a : S
withha : a ∈ M ⊓ N
), you may useLinearIndependent.of_comp
andFinVec.map_eq
(inMathlib/Data/Fin/Tuple/Reflection.lean
), see the following code snippet:have h := H.not_linearIndependent_pair_of_commute_of_flat hf ⟨a, ha⟩ ⟨b, hb⟩ hc contrapose! h refine .of_comp (M ⊓ N).subtype ?_ convert h exact (FinVec.map_eq _ _).symm
Submodule.LinearDisjoint.rank_le_one_of_commute_of_flat_of_self
: ifM
and itself are linearly disjoint, ifM
is flat, if any two elements inM
are commutative, then the rank ofM
is at most one.
The results with name containing "of_commute" also have corresponding specialized versions
assuming S
is commutative.
Tags #
linearly disjoint, linearly independent, tensor product
Two submodules M
and N
in an algebra S
over R
are linearly disjoint if the natural map
M ⊗[R] N →ₗ[R] S
induced by multiplication in S
is injective.
- injective : Function.Injective ⇑(M.mulMap N)
Instances For
If M
and N
are linearly disjoint submodules, then there is the natural isomorphism
M ⊗[R] N ≃ₗ[R] M * N
induced by multiplication in S
.
Equations
- H.mulMap = (LinearEquiv.ofInjective (M.mulMap N) ⋯).trans (LinearEquiv.ofEq (LinearMap.range (M.mulMap N)) (M * N) ⋯)
Instances For
Linear disjointness is preserved by taking multiplicative opposite.
Alias of the forward direction of Submodule.linearDisjoint_op
.
Linear disjointness is preserved by taking multiplicative opposite.
Alias of the reverse direction of Submodule.linearDisjoint_op
.
Linear disjointness is preserved by taking multiplicative opposite.
Linear disjointness is symmetric if elements in the module commute.
Linear disjointness is symmetric if elements in the module commute.
Linear disjointness is preserved by injective algebra homomorphisms.
If { m_i }
is an R
-basis of M
, which is also N
-linearly independent
(in this result it is stated as Submodule.mulLeftMap
is injective),
then M
and N
are linearly disjoint.
If { n_i }
is an R
-basis of N
, which is also M
-linearly independent
(in this result it is stated as Submodule.mulRightMap
is injective),
then M
and N
are linearly disjoint.
If { m_i }
is an R
-basis of M
, if { n_i }
is an R
-basis of N
,
such that the family { m_i * n_j }
in S
is R
-linearly independent
(in this result it is stated as the relevant Finsupp.linearCombination
is injective),
then M
and N
are linearly disjoint.
The zero module is linearly disjoint with any other submodules.
The zero module is linearly disjoint with any other submodules.
The image of R
in S
is linearly disjoint with any other submodules.
The image of R
in S
is linearly disjoint with any other submodules.
If for any finitely generated submodules M'
of M
, M'
and N
are linearly disjoint,
then M
and N
themselves are linearly disjoint.
If for any finitely generated submodules N'
of N
, M
and N'
are linearly disjoint,
then M
and N
themselves are linearly disjoint.
If for any finitely generated submodules M'
and N'
of M
and N
, respectively,
M'
and N'
are linearly disjoint, then M
and N
themselves are linearly disjoint.
Linear disjointness is symmetric in a commutative ring.
Linear disjointness is symmetric in a commutative ring.
If M
and N
are linearly disjoint, if N
is a flat R
-module, then for any family of
R
-linearly independent elements { m_i }
of M
, they are also N
-linearly independent,
in the sense that the R
-linear map from ι →₀ N
to S
which maps { n_i }
to the sum of m_i * n_i
(Submodule.mulLeftMap N m
) has trivial kernel.
If { m_i }
is an R
-basis of M
, which is also N
-linearly independent,
then M
and N
are linearly disjoint.
If M
and N
are linearly disjoint, if M
is a flat R
-module, then for any family of
R
-linearly independent elements { n_i }
of N
, they are also M
-linearly independent,
in the sense that the R
-linear map from ι →₀ M
to S
which maps { m_i }
to the sum of m_i * n_i
(Submodule.mulRightMap M n
) has trivial kernel.
If { n_i }
is an R
-basis of N
, which is also M
-linearly independent,
then M
and N
are linearly disjoint.
If M
and N
are linearly disjoint, if M
is flat, then for any family of
R
-linearly independent elements { m_i }
of M
, and any family of
R
-linearly independent elements { n_j }
of N
, the family { m_i * n_j }
in S
is
also R
-linearly independent.
If M
and N
are linearly disjoint, if N
is flat, then for any family of
R
-linearly independent elements { m_i }
of M
, and any family of
R
-linearly independent elements { n_j }
of N
, the family { m_i * n_j }
in S
is
also R
-linearly independent.
If M
and N
are linearly disjoint, if one of M
and N
is flat, then for any family of
R
-linearly independent elements { m_i }
of M
, and any family of
R
-linearly independent elements { n_j }
of N
, the family { m_i * n_j }
in S
is
also R
-linearly independent.
If { m_i }
is an R
-basis of M
, if { n_j }
is an R
-basis of N
,
such that the family { m_i * n_j }
in S
is R
-linearly independent,
then M
and N
are linearly disjoint.
If M
and N
are linearly disjoint, if N
is flat, then for any submodule M'
of M
,
M'
and N
are also linearly disjoint.
If M
and N
are linearly disjoint, if M
is flat, then for any submodule N'
of N
,
M
and N'
are also linearly disjoint.
If M
and N
are linearly disjoint, M'
and N'
are submodules of M
and N
,
respectively, such that N
and M'
are flat, then M'
and N'
are also linearly disjoint.
If M
and N
are linearly disjoint, M'
and N'
are submodules of M
and N
,
respectively, such that M
and N'
are flat, then M'
and N'
are also linearly disjoint.
If N
is flat, M
is contained in i(R)
, where i : R → S
is the structure map,
then M
and N
are linearly disjoint.
If M
is flat, N
is contained in i(R)
, where i : R → S
is the structure map,
then M
and N
are linearly disjoint.
If M
and N
are linearly disjoint, if M
is flat, then any two commutative
elements of ↥(M ⊓ N)
are not R
-linearly independent (namely, their span is not R ^ 2
).
If M
and N
are linearly disjoint, if N
is flat, then any two commutative
elements of ↥(M ⊓ N)
are not R
-linearly independent (namely, their span is not R ^ 2
).
If M
and N
are linearly disjoint, if one of M
and N
is flat, then any two commutative
elements of ↥(M ⊓ N)
are not R
-linearly independent (namely, their span is not R ^ 2
).
If M
and N
are linearly disjoint, if one of M
and N
is flat,
if any two elements of ↥(M ⊓ N)
are commutative, then the rank of ↥(M ⊓ N)
is at most one.
If M
and N
are linearly disjoint, if M
is flat,
if any two elements of ↥(M ⊓ N)
are commutative, then the rank of ↥(M ⊓ N)
is at most one.
If M
and N
are linearly disjoint, if N
is flat,
if any two elements of ↥(M ⊓ N)
are commutative, then the rank of ↥(M ⊓ N)
is at most one.
If M
and itself are linearly disjoint, if M
is flat,
if any two elements of M
are commutative, then the rank of M
is at most one.
The Submodule.LinearDisjoint.not_linearIndependent_pair_of_commute_of_flat_left
for commutative rings.
The Submodule.LinearDisjoint.not_linearIndependent_pair_of_commute_of_flat_right
for commutative rings.
The Submodule.LinearDisjoint.not_linearIndependent_pair_of_commute_of_flat
for commutative rings.
The Submodule.LinearDisjoint.rank_inf_le_one_of_commute_of_flat
for commutative rings.
The Submodule.LinearDisjoint.rank_inf_le_one_of_commute_of_flat_left
for commutative rings.
The Submodule.LinearDisjoint.rank_inf_le_one_of_commute_of_flat_right
for commutative rings.
The Submodule.LinearDisjoint.rank_le_one_of_commute_of_flat_of_self
for commutative rings.