rank — Mathlib

English

Let E be a finite-dimensional space over a field K with a lattice L inside E. Then the ℤ-rank of L equals the K-dimension of E; i.e., the number of independent generators of L over ℤ matches the dimension of E over K.

Русский

Пусть E — конечномерное пространство над полем K, внутри которого находится решётка L. Тогда ℤ-ранг L равен размерности E над K.

LaTeX

$$$\\operatorname{finrank}_{\\mathbb{Z}} L = \\operatorname{finrank}_{K} E$$$

Lean4

theorem rank [hs : IsZLattice K L] : finrank ℤ L = finrank K E := by
  classical
  have : Module.Finite ℤ L := module_finite K L
  have : Module ℚ E := Module.compHom E (algebraMap ℚ K)
  let b₀ := Module.Free.chooseBasis ℤ L
  let b := Subtype.val ∘ b₀
  have : LinearIndependent ℤ b :=
    LinearIndependent.map' b₀.linearIndependent (L.subtype)
      (ker_subtype _)
        -- We prove some assertions that will be useful later on

  have h_spanL : span ℤ (Set.range b) = L :=
    by
    convert congrArg (map (Submodule.subtype L)) b₀.span_eq
    · rw [map_span, Set.range_comp]
      rfl
    · exact (map_subtype_top _).symm
  have h_spanE : span K (Set.range b) = ⊤ :=
    by
    rw [← span_span_of_tower (R := ℤ), h_spanL]
    exact hs.span_top
  have h_card : Fintype.card (Module.Free.ChooseBasisIndex ℤ L) = (Set.range b).toFinset.card :=
    by
    rw [Set.toFinset_range, Finset.univ.card_image_of_injective]
    · rfl
    · exact Subtype.coe_injective.comp (Basis.injective _)
  rw [finrank_eq_card_chooseBasisIndex]
    -- We prove that `finrank ℤ L ≤ finrank K E` and `finrank K E ≤ finrank ℤ L`

  refine le_antisymm ?_ ?_
  · -- To prove that `finrank ℤ L ≤ finrank K E`, we proceed by contradiction and prove that, in
        -- this case, there is a ℤ-relation between the vectors of `b`

    obtain ⟨t, ⟨ht_inc, ⟨ht_span, ht_lin⟩⟩⟩ :=
      exists_linearIndependent K
        (Set.range b)
          -- `e` is a `K`-basis of `E` formed of vectors of `b`

    let e : Basis t K E := Basis.mk ht_lin (by simp [ht_span, h_spanE])
    have : Fintype t := Set.Finite.fintype ((Set.range b).toFinite.subset ht_inc)
    have h : LinearIndepOn ℤ id (Set.range b) := by
      rwa [linearIndepOn_id_range_iff (Subtype.coe_injective.comp b₀.injective)]
    contrapose! h
    obtain ⟨v, hv⟩ : (Set.range b \ Set.range e).Nonempty :=
      by
      rw [Basis.coe_mk, Subtype.range_coe_subtype, Set.setOf_mem_eq, ← Set.toFinset_nonempty]
      contrapose h
      rw [Finset.not_nonempty_iff_eq_empty, Set.toFinset_diff, Finset.sdiff_eq_empty_iff_subset] at h
      replace h := Finset.card_le_card h
      rwa [not_lt, h_card, ← topEquiv.finrank_eq, ← h_spanE, ← ht_span, finrank_span_set_eq_card ht_lin]
        -- Assume that `e ∪ {v}` is not `ℤ`-linear independent then we get the contradiction

    suffices ¬LinearIndepOn ℤ id (insert v (Set.range e))
      by
      contrapose! this
      refine this.mono ?_
      exact
        Set.insert_subset (Set.mem_of_mem_diff hv)
          (by simp [e, ht_inc])
            -- We prove finally that `e ∪ {v}` is not ℤ-linear independent or, equivalently,
                -- not ℚ-linear independent by showing that `v ∈ span ℚ e`.

    rw [LinearIndepOn, LinearIndependent.iff_fractionRing ℤ ℚ, ← LinearIndepOn,
      linearIndepOn_id_insert (Set.notMem_of_mem_diff hv), not_and, not_not]
    intro
      _
        -- But that follows from the fact that there exist `n, m : ℕ`, `n ≠ m`
            -- such that `(n - m) • v ∈ span ℤ e` which is true since `n ↦ ZSpan.fract e (n • v)`
            -- takes value into the finite set `fundamentalDomain e ∩ L`

    have h_mapsto :
      Set.MapsTo (fun n : ℤ => fract e (n • v)) Set.univ (Metric.closedBall 0 (∑ i, ‖e i‖) ∩ (L : Set E)) :=
      by
      rw [Set.mapsTo_inter, Set.mapsTo_univ_iff, Set.mapsTo_univ_iff]
      refine ⟨fun _ ↦ mem_closedBall_zero_iff.mpr (norm_fract_le e _), fun _ => ?_⟩
      · rw [← h_spanL]
        refine sub_mem ?_ ?_
        · exact zsmul_mem (subset_span (Set.diff_subset hv)) _
        · exact span_mono (by simp [e, ht_inc]) (coe_mem _)
    have h_finite : Set.Finite (Metric.closedBall 0 (∑ i, ‖e i‖) ∩ (L : Set E)) :=
      by
      change ((_ : Set E) ∩ L.toAddSubgroup).Finite
      have : DiscreteTopology L.toAddSubgroup := (inferInstance : DiscreteTopology L)
      exact Metric.finite_isBounded_inter_isClosed Metric.isBounded_closedBall inferInstance
    obtain ⟨n, -, m, -, h_neq, h_eq⟩ := Set.Infinite.exists_ne_map_eq_of_mapsTo Set.infinite_univ h_mapsto h_finite
    have h_nz : (-n + m : ℚ) ≠ 0 := by rwa [Ne, add_eq_zero_iff_eq_neg.not, neg_inj, Rat.coe_int_inj, ← Ne]
    apply (smul_mem_iff _ h_nz).mp
    refine span_subset_span ℤ ℚ _ ?_
    rwa [add_smul, neg_smul, SetLike.mem_coe, ← fract_eq_fract, Int.cast_smul_eq_zsmul ℚ, Int.cast_smul_eq_zsmul ℚ]
  · -- To prove that `finrank K E ≤ finrank ℤ L`, we use the fact `b` generates `E` over `K`
        -- and thus `finrank K E ≤ card b = finrank ℤ L`

    rw [← topEquiv.finrank_eq, ← h_spanE]
    convert finrank_span_le_card (R := K) (Set.range b)

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