l3lab/ntp-mathlib huggingface dataset:
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  

Proof state:
case intro.h
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
⊢ (trace R M) (φ z ∘ₗ φ x) = 0

Next tactic:
suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    exact NoZeroSMulDivisors.algebraMap_injective R A this
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    

Proof state:
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
this : (algebraMap R A) ((trace R M) (φ z ∘ₗ φ x)) = 0
⊢ (trace R M) (φ z ∘ₗ φ x) = 0

Next tactic:
have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    

Proof state:
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
this : (algebraMap R A) ((trace R M) (φ z ∘ₗ φ x)) = 0
_i : NoZeroSMulDivisors R A
⊢ (trace R M) (φ z ∘ₗ φ x) = 0

Next tactic:
rw [← map_zero (algebraMap R A)] at this
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    

Proof state:
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
this : (algebraMap R A) ((trace R M) (φ z ∘ₗ φ x)) = (algebraMap R A) 0
_i : NoZeroSMulDivisors R A
⊢ (trace R M) (φ z ∘ₗ φ x) = 0

Next tactic:
exact NoZeroSMulDivisors.algebraMap_injective R A this
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    exact NoZeroSMulDivisors.algebraMap_injective R A this
  

Proof state:
case intro.h
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
⊢ (algebraMap R A) ((trace R M) (φ z ∘ₗ φ x)) = 0

Next tactic:
rw [← LinearMap.trace_baseChange, LinearMap.baseChange_comp, ← toEndomorphism_baseChange,
    ← toEndomorphism_baseChange]
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    exact NoZeroSMulDivisors.algebraMap_injective R A this
  rw [← LinearMap.trace_baseChange, LinearMap.baseChange_comp, ← toEndomorphism_baseChange,
    ← toEndomorphism_baseChange]
  

Proof state:
case intro.h
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
⊢ (trace A (A ⊗[R] M))
      ((toEndomorphism A (A ⊗[R] L) (A ⊗[R] M)) (1 ⊗ₜ[R] z) ∘ₗ (toEndomorphism A (A ⊗[R] L) (A ⊗[R] M)) (1 ⊗ₜ[R] x)) =
    0

Next tactic:
replace hz : 1 ⊗ₜ z ∈ lowerCentralSeries A (A ⊗[R] L) (A ⊗[R] L) 1 := by
    simp only [lowerCentralSeries_succ, lowerCentralSeries_zero] at hz ⊢
    rw [← LieSubmodule.baseChange_top, ← LieSubmodule.lie_baseChange]
    exact Submodule.tmul_mem_baseChange_of_mem 1 hz
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    exact NoZeroSMulDivisors.algebraMap_injective R A this
  rw [← LinearMap.trace_baseChange, LinearMap.baseChange_comp, ← toEndomorphism_baseChange,
    ← toEndomorphism_baseChange]
  replace hz : 1 ⊗ₜ z ∈ lowerCentralSeries A (A ⊗[R] L) (A ⊗[R] L) 1 := by
    

Proof state:
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hz : z ∈ lowerCentralSeries R L L 1
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
⊢ 1 ⊗ₜ[R] z ∈ lowerCentralSeries A (A ⊗[R] L) (A ⊗[R] L) 1

Next tactic:
simp only [lowerCentralSeries_succ, lowerCentralSeries_zero] at hz ⊢
lemma lowerCentralSeries_one_inf_center_le_ker_traceForm :
    lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L ≤ LinearMap.ker (traceForm R L M) := by
  /- Sketch of proof (due to Zassenhaus):

  Let `z ∈ lowerCentralSeries R L L 1 ⊓ LieAlgebra.center R L` and `x : L`. We must show that
  `trace (φ x ∘ φ z) = 0` where `φ z : End R M` indicates the action of `z` on `M` (and likewise
  for `φ x`).

  Because `z` belongs to the indicated intersection, it has two key properties:
  (a) the trace of the action of `z` vanishes on any Lie module of `L`
      (see `LieModule.trace_toEndomorphism_eq_zero_of_mem_lcs`),
  (b) `z` commutes with all elements of `L`.

  If `φ x` were triangularizable, we could write `M` as a direct sum of generalized eigenspaces of
  `φ x`. Because `L` is nilpotent these are all Lie submodules, thus Lie modules in their own right,
  and thus by (a) above we learn that `trace (φ z) = 0` restricted to each generalized eigenspace.
  Because `z` commutes with `x`, this forces `trace (φ x ∘ φ z) = 0` on each generalized eigenspace,
  and so by summing the traces on each generalized eigenspace we learn the total trace is zero, as
  required (see `LinearMap.trace_comp_eq_zero_of_commute_of_trace_restrict_eq_zero`).

  To cater for the fact that `φ x` may not be triangularizable, we first extend the scalars from `R`
  to `AlgebraicClosure (FractionRing R)` and argue using the action of `A ⊗ L` on `A ⊗ M`. -/
  rintro z ⟨hz : z ∈ lowerCentralSeries R L L 1, hzc : z ∈ LieAlgebra.center R L⟩
  ext x
  rw [traceForm_apply_apply, LinearMap.zero_apply]
  let A := AlgebraicClosure (FractionRing R)
  suffices algebraMap R A (trace R _ ((φ z).comp (φ x))) = 0 by
    have _i : NoZeroSMulDivisors R A := NoZeroSMulDivisors.trans R (FractionRing R) A
    rw [← map_zero (algebraMap R A)] at this
    exact NoZeroSMulDivisors.algebraMap_injective R A this
  rw [← LinearMap.trace_baseChange, LinearMap.baseChange_comp, ← toEndomorphism_baseChange,
    ← toEndomorphism_baseChange]
  replace hz : 1 ⊗ₜ z ∈ lowerCentralSeries A (A ⊗[R] L) (A ⊗[R] L) 1 := by
    simp only [lowerCentralSeries_succ, lowerCentralSeries_zero] at hz ⊢
    

Proof state:
R : Type u_1
K : Type u_2
L : Type u_3
M : Type u_4
inst✝¹⁶ : CommRing R
inst✝¹⁵ : LieRing L
inst✝¹⁴ : LieAlgebra R L
inst✝¹³ : AddCommGroup M
inst✝¹² : Module R M
inst✝¹¹ : LieRingModule L M
inst✝¹⁰ : LieModule R L M
inst✝⁹ : Module.Free R M
inst✝⁸ : Module.Finite R M
inst✝⁷ : Field K
inst✝⁶ : LieAlgebra K L
inst✝⁵ : Module K M
inst✝⁴ : LieModule K L M
inst✝³ : FiniteDimensional K M
inst✝² : LieAlgebra.IsNilpotent R L
inst✝¹ : IsDomain R
inst✝ : IsPrincipalIdealRing R
z : L
hzc : z ∈ LieAlgebra.center R L
x : L
A : Type u_1 := AlgebraicClosure (FractionRing R)
hz : z ∈ ⁅⊤, ⊤⁆
⊢ 1 ⊗ₜ[R] z ∈ ⁅⊤, ⊤⁆

Next tactic:
rw [← LieSubmodule.baseChange_top, ← LieSubmodule.lie_baseChange]