Linearly disjoint of submodules #
This file contains basics about the linearly disjoint of 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).
MandNare linearly disjoint (Submodule.LinearDisjoint M Nor simplyM.LinearDisjoint N), if the naturalR-linear mapM ⊗[R] N →ₗ[R] S(Submodule.mulMap M N) induced by the multiplication inSis injective.
The following is the first equivalent characterization of linearly disjointness:
Submodule.LinearDisjoint.linearIndependent_left_of_flat: ifMandNare linearly disjoint, ifNis 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 NtoSwhich 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, thenMandNare linearly disjoint.
Dually, we have:
Submodule.LinearDisjoint.linearIndependent_right_of_flat: ifMandNare linearly disjoint, ifMis 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 MtoSwhich 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, thenMandNare linearly disjoint.
The following is the second equivalent characterization of linearly disjointness:
Submodule.LinearDisjoint.linearIndependent_mul_of_flat: ifMandNare linearly disjoint, if one ofMandNis 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 }inSis 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 }inSisR-linearly independent, thenMandNare linearly disjoint.
Other main results #
Submodule.LinearDisjoint.symm_of_commute,Submodule.linearDisjoint_symm_of_commute: linearly disjoint is symmetric under some commutative conditions.Submodule.linearDisjoint_op: linearly disjoint 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: linearly disjoint 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 ofMandNare linearly disjoint, thenMandNthemselves 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 ofRinSis 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 ofRinS, 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: ifMandNare linearly disjoint, if one ofMandNis flat, then any two commutative elements contained in the intersection ofMandNare notR-linearly independent (namely, their span is notR ^ 2). In particular, if any two elements in the intersection ofMandNare commutative, then the rank of the intersection ofMandNis 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 : Swithha : a ∈ M ⊓ N), you may useLinearIndependent.of_compandFinVec.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 _ _).symmSubmodule.LinearDisjoint.rank_le_one_of_commute_of_flat_of_self: ifMand itself are linearly disjoint, ifMis flat, if any two elements inMare commutative, then the rank ofMis at most one.
The results with name containing "of_commute" also have corresponding specified 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
Linearly disjoint is preserved by taking multiplicative opposite.
Alias of the forward direction of Submodule.linearDisjoint_op.
Linearly disjoint is preserved by taking multiplicative opposite.
Alias of the reverse direction of Submodule.linearDisjoint_op.
Linearly disjoint is preserved by taking multiplicative opposite.
Linearly disjoint is symmetric if elements in the module commute.
Linearly disjoint is symmetric if elements in the module commute.
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.total 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.
Linearly disjoint is symmetric in a commutative ring.
Linearly disjoint 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_i } 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.