Theory Polynomials.More_Modules
theory More_Modules
  imports HOL.Modules
begin
text ‹More facts about modules.›
section ‹Modules over Commutative Rings›
context module
begin
lemma scale_minus_both [simp]: "(- a) *s (- x) = a *s x"
  by simp
subsection ‹Submodules Spanned by Sets of Module-Elements›
lemma span_insertI:
  assumes "p ∈ span B"
  shows "p ∈ span (insert r B)"
proof -
  have "B ⊆ insert r B" by blast
  hence "span B ⊆ span (insert r B)" by (rule span_mono)
  with assms show ?thesis ..
qed
lemma span_insertD:
  assumes "p ∈ span (insert r B)" and "r ∈ span B"
  shows "p ∈ span B"
  using assms(1)
proof (induct p rule: span_induct_alt)
  case base
  show "0 ∈ span B" by (fact span_zero)
next
  case step: (step q b a)
  from step(1) have "b = r ∨ b ∈ B" by simp
  thus "q *s b + a ∈ span B"
  proof
    assume eq: "b = r"
    from step(2) assms(2) show ?thesis unfolding eq by (intro span_add span_scale)
  next
    assume "b ∈ B"
    hence "b ∈ span B" using span_superset ..
    with step(2) show ?thesis by (intro span_add span_scale)
  qed
qed
lemma span_insert_idI:
  assumes "r ∈ span B"
  shows "span (insert r B) = span B"
proof (intro subset_antisym subsetI)
  fix p
  assume "p ∈ span (insert r B)"
  from this assms show "p ∈ span B" by (rule span_insertD)
next
  fix p
  assume "p ∈ span B"
  thus "p ∈ span (insert r B)" by (rule span_insertI)
qed
lemma span_insert_zero: "span (insert 0 B) = span B"
  using span_zero by (rule span_insert_idI)
lemma span_Diff_zero: "span (B - {0}) = span B"
  by (metis span_insert_zero insert_Diff_single)
lemma span_insert_subset:
  assumes "span A ⊆ span B" and "r ∈ span B"
  shows "span (insert r A) ⊆ span B"
proof
  fix p
  assume "p ∈ span (insert r A)"
  thus "p ∈ span B"
  proof (induct p rule: span_induct_alt)
    case base
    show ?case by (fact span_zero)
  next
    case step: (step q b a)
    show ?case
    proof (intro span_add span_scale)
      from ‹b ∈ insert r A› show "b ∈ span B"
      proof
        assume "b = r"
        thus "b ∈ span B" using assms(2) by simp
      next
        assume "b ∈ A"
        hence "b ∈ span A" using span_superset ..
        thus "b ∈ span B" using assms(1) ..
      qed
    qed fact
  qed
qed
lemma replace_span:
  assumes "q ∈ span B"
  shows "span (insert q (B - {p})) ⊆ span B"
  by (rule span_insert_subset, rule span_mono, fact Diff_subset, fact)
lemma sum_in_spanI: "(∑b∈B. q b *s b) ∈ span B"
  by (auto simp: intro: span_sum span_scale dest: span_base)
lemma span_closed_sum_list: "(⋀x. x ∈ set xs ⟹ x ∈ span B) ⟹ sum_list xs ∈ span B"
  by (induct xs) (auto intro: span_zero span_add)
lemma spanE:
  assumes "p ∈ span B"
  obtains A q where "finite A" and "A ⊆ B" and "p = (∑b∈A. (q b) *s b)"
  using assms by (auto simp: span_explicit)
lemma span_finite_subset:
  assumes "p ∈ span B"
  obtains A where "finite A" and "A ⊆ B" and "p ∈ span A"
proof -
  from assms obtain A q where "finite A" and "A ⊆ B" and p: "p = (∑a∈A. q a *s a)"
    by (rule spanE)
  note this(1, 2)
  moreover have "p ∈ span A" unfolding p by (rule sum_in_spanI)
  ultimately show ?thesis ..
qed
lemma span_finiteE:
  assumes "finite B" and "p ∈ span B"
  obtains q where "p = (∑b∈B. (q b) *s b)"
  using assms by (auto simp: span_finite)
lemma span_subset_spanI:
  assumes "A ⊆ span B"
  shows "span A ⊆ span B"
  using assms subspace_span by (rule span_minimal)
lemma span_insert_cong:
  assumes "span A = span B"
  shows "span (insert p A) = span (insert p B)" (is "?l = ?r")
proof
  have 1: "span (insert p C1) ⊆ span (insert p C2)" if "span C1 = span C2" for C1 C2
  proof (rule span_subset_spanI)
    show "insert p C1 ⊆ span (insert p C2)"
    proof (rule insert_subsetI)
      show "p ∈ span (insert p C2)" by (rule span_base) simp
    next
      have "C1 ⊆ span C1" by (rule span_superset)
      also from that have "… = span C2" .
      also have "… ⊆ span (insert p C2)" by (rule span_mono) blast
      finally show "C1 ⊆ span (insert p C2)" .
    qed
  qed
  from assms show "?l ⊆ ?r" by (rule 1)
  from assms[symmetric] show "?r ⊆ ?l" by (rule 1)
qed
lemma span_induct' [consumes 1, case_names base step]:
  assumes "p ∈ span B" and "P 0"
    and "⋀a q p. a ∈ span B ⟹ P a ⟹ p ∈ B ⟹ q ≠ 0 ⟹ P (a + q *s p)"
  shows "P p"
  using assms(1, 1)
proof (induct p rule: span_induct_alt)
  case base
  from assms(2) show ?case .
next
  case (step q b a)
  from step.hyps(1) have "b ∈ span B" by (rule span_base)
  hence "q *s b ∈ span B" by (rule span_scale)
  with step.prems have "a ∈ span B" by (simp only: span_add_eq)
  hence "P a" by (rule step.hyps)
  show ?case
  proof (cases "q = 0")
    case True
    from ‹P a› show ?thesis by (simp add: True)
  next
    case False
    with ‹a ∈ span B› ‹P a› step.hyps(1) have "P (a + q *s b)" by (rule assms(3))
    thus ?thesis by (simp only: add.commute)
  qed
qed
lemma span_INT_subset: "span (⋂a∈A. f a) ⊆ (⋂a∈A. span (f a))" (is "?l ⊆ ?r")
proof
  fix p
  assume "p ∈ ?l"
  show "p ∈ ?r"
  proof
    fix a
    assume "a ∈ A"
    from ‹p ∈ ?l› show "p ∈ span (f a)"
    proof (induct p rule: span_induct')
      case base
      show ?case by (fact span_zero)
    next
      case (step p q b)
      from step(3) ‹a ∈ A› have "b ∈ f a" ..
      hence "b ∈ span (f a)" by (rule span_base)
      with step(2) show ?case by (intro span_add span_scale)
    qed
  qed
qed
lemma span_INT: "span (⋂a∈A. span (f a)) = (⋂a∈A. span (f a))" (is "?l = ?r")
proof
  have "?l ⊆ (⋂a∈A. span (span (f a)))" by (rule span_INT_subset)
  also have "... = ?r" by (simp add: span_span)
  finally show "?l ⊆ ?r" .
qed (fact span_superset)
lemma span_Int_subset: "span (A ∩ B) ⊆ span A ∩ span B"
proof -
  have "span (A ∩ B) = span (⋂x∈{A, B}. x)" by simp
  also have "… ⊆ (⋂x∈{A, B}. span x)" by (fact span_INT_subset)
  also have "… = span A ∩ span B" by simp
  finally show ?thesis .
qed
lemma span_Int: "span (span A ∩ span B) = span A ∩ span B"
proof -
  have "span (span A ∩ span B) = span (⋂x∈{A, B}. span x)" by simp
  also have "… = (⋂x∈{A, B}. span x)" by (fact span_INT)
  also have "… = span A ∩ span B" by simp
  finally show ?thesis .
qed
lemma span_image_scale_eq_image_scale: "span ((*s) q ` F) = (*s) q ` span F" (is "?A = ?B")
proof (intro subset_antisym subsetI)
  fix p
  assume "p ∈ ?A"
  thus "p ∈ ?B"
  proof (induct p rule: span_induct')
    case base
    from span_zero show ?case by (rule rev_image_eqI) simp
  next
    case (step p r a)
    from step.hyps(2) obtain p' where "p' ∈ span F" and p: "p = q *s p'" ..
    from step.hyps(3) obtain a' where "a' ∈ F" and a: "a = q *s a'" ..
    from this(1) have "a' ∈ span F" by (rule span_base)
    hence "r *s a' ∈ span F" by (rule span_scale)
    with ‹p' ∈ span F› have "p' + r *s a' ∈ span F" by (rule span_add)
    hence "q *s (p' + r *s a') ∈ ?B" by (rule imageI)
    also have "q *s (p' + r *s a') = p + r *s a" by (simp add: a p algebra_simps)
    finally show ?case .
  qed
next
  fix p
  assume "p ∈ ?B"
  then obtain p' where "p' ∈ span F" and "p = q *s p'" ..
  from this(1) show "p ∈ ?A" unfolding ‹p = q *s p'›
  proof (induct p' rule: span_induct')
    case base
    show ?case by (simp add: span_zero)
  next
    case (step p r a)
    from step.hyps(3) have "q *s a ∈ (*s) q ` F" by (rule imageI)
    hence "q *s a ∈ ?A" by (rule span_base)
    hence "r *s (q *s a) ∈ ?A" by (rule span_scale)
    with step.hyps(2) have "q *s p + r *s (q *s a) ∈ ?A" by (rule span_add)
    also have "q *s p + r *s (q *s a) = q *s (p + r *s a)" by (simp add: algebra_simps)
    finally show ?case .
  qed
qed
end 
section ‹Ideals over Commutative Rings›
lemma module_times: "module (*)"
  by (standard, simp_all add: algebra_simps)