Theory Field_ZF

(*
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    Copyright (C) 2005, 2006  Slawomir Kolodynski

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header {*\isaheader{Field\_ZF.thy}*}

theory Field_ZF imports Ring_ZF

begin

text{*This theory covers basic facts about fields.*}

section{*Definition and basic properties*}

text{*In this section we define what is a field and list the basic properties
  of fields. *}

text{*Field is a notrivial commutative ring such that all 
  non-zero elements have an inverse. We define the notion of being a field as
  a statement about three sets. The first set, denoted @{text "K"} is the 
  carrier of the field. The second set, denoted @{text "A"} represents the 
  additive operation on @{text "K"} (recall that in ZF set theory functions 
  are sets). The third set @{text "M"} represents the multiplicative operation 
  on @{text "K"}.*}

constdefs
  "IsAfield(K,A,M) ≡ 
  (IsAring(K,A,M) ∧ (M {is commutative on} K) ∧
  TheNeutralElement(K,A) ≠ TheNeutralElement(K,M) ∧ 
  (∀a∈K. a≠TheNeutralElement(K,A)-->
  (∃b∈K. M`⟨a,b⟩ = TheNeutralElement(K,M))))"

text{*The @{text "field0"} context extends the @{text "ring0"}
  context adding field-related assumptions and notation related to the 
  multiplicative inverse. *}

locale field0 = ring0 K +
  assumes mult_commute: "M {is commutative on} K"
  
  assumes not_triv: "\<zero> ≠ \<one>"

  assumes inv_exists: "∀a∈K. a≠\<zero> --> (∃b∈K. a·b = \<one>)"

  fixes non_zero ("K0")
  defines non_zero_def[simp]: "K0 ≡ K-{\<zero>}"

  fixes inv ("_¯ " [96] 97)
  defines inv_def[simp]: "a¯ ≡ GroupInv(K0,restrict(M,K0×K0))`(a)";

text{*The next lemma assures us that we are talking fields 
  in the @{text "field0"} context.*}

lemma (in field0) Field_ZF_1_L1: shows "IsAfield(K,A,M)"
  using ringAssum mult_commute not_triv inv_exists IsAfield_def
  by simp;

text{*We can use theorems proven in the @{text "field0"} context whenever we
  talk about a field.*}

lemma Field_ZF_1_L2: assumes "IsAfield(K,A,M)"
  shows "field0(K,A,M)"
  using prems IsAfield_def field0_axioms.intro ring0_def field0_def 
  by simp;

text{*Let's have an explicit statement that the multiplication
  in fields is commutative.*}

lemma (in field0) field_mult_comm: assumes "a∈K"  "b∈K"
  shows "a·b = b·a"
  using mult_commute prems IsCommutative_def by simp;

text{*Fields do not have zero divisors.*}

lemma (in field0) field_has_no_zero_divs: shows "HasNoZeroDivs(K,A,M)"
proof -
  { fix a b assume A1: "a∈K"  "b∈K" and A2: "a·b = \<zero>"  and A3: "b≠\<zero>"
    from inv_exists A1 A3 obtain c where I: "c∈K" and II: "b·c = \<one>"
      by auto;
    from A2 have "a·b·c = \<zero>·c" by simp;
    with A1 I have "a·(b·c) = \<zero>" 
      using Ring_ZF_1_L11 Ring_ZF_1_L6 by simp
    with A1 II have "a=\<zero> "using Ring_ZF_1_L3 by simp } 
  then have "∀a∈K.∀b∈K. a·b = \<zero> --> a=\<zero> ∨ b=\<zero>" by auto;
    then show ?thesis using HasNoZeroDivs_def by auto;
qed;

text{*$K_0$ (the set of nonzero field elements is closed with respect
  to multiplication.*}

lemma (in field0) Field_ZF_1_L2: "K0 {is closed under} M"
  using Ring_ZF_1_L4 field_has_no_zero_divs Ring_ZF_1_L12
    IsOpClosed_def by auto;

text{*Any nonzero element has a right inverse that is nonzero.*}

lemma (in field0) Field_ZF_1_L3: assumes A1: "a∈K0"
  shows "∃b∈K0. a·b = \<one>"
proof -
  from inv_exists A1 obtain b where "b∈K" and "a·b = \<one>"
    by auto;
  with not_triv A1 show "∃b∈K0. a·b = \<one>"
    using Ring_ZF_1_L6 by auto;
qed;

text{*If we remove zero, the field with multiplication
  becomes a group and we can use all theorems proven in 
  @{text "group0"} context.*}

theorem (in field0) Field_ZF_1_L4: shows 
  "IsAgroup(K0,restrict(M,K0×K0))"
  "group0(K0,restrict(M,K0×K0))"
  "\<one> = TheNeutralElement(K0,restrict(M,K0×K0))"
proof-
  let ?f = "restrict(M,K0×K0)"
  have 
    "M {is associative on} K"
    "K0 ⊆ K"  "K0 {is closed under} M"
    using Field_ZF_1_L1 IsAfield_def IsAring_def IsAgroup_def 
      IsAmonoid_def Field_ZF_1_L2 by auto;
  then have "?f {is associative on} K0"
    using func_ZF_4_L3 by simp;
  moreover
  from not_triv have 
    I: "\<one>∈K0 ∧ (∀a∈K0. ?f`⟨\<one>,a⟩ = a ∧  ?f`⟨a,\<one>⟩ = a)"
    using Ring_ZF_1_L2 Ring_ZF_1_L3 by auto;
  then have "∃n∈K0. ∀a∈K0. ?f`⟨n,a⟩ = a ∧  ?f`⟨a,n⟩ = a"
    by blast;
  ultimately have II: "IsAmonoid(K0,?f)" using IsAmonoid_def
    by simp;
  then have "monoid0(K0,?f)" using monoid0_def by simp;
  moreover note I
  ultimately show "\<one> = TheNeutralElement(K0,?f)"
    by (rule monoid0.group0_1_L4);
  then have "∀a∈K0.∃b∈K0. ?f`⟨a,b⟩ =  TheNeutralElement(K0,?f)"
    using Field_ZF_1_L3 by auto;
  with II show "IsAgroup(K0,?f)" by (rule definition_of_group)
  then show "group0(K0,?f)" using group0_def by simp
qed;

text{*The inverse of a nonzero field element is nonzero.*}

lemma (in field0) Field_ZF_1_L5: assumes A1: "a∈K"  "a≠\<zero>"
  shows "a¯ ∈ K0"  "(a¯)² ∈ K0"   "a¯ ∈ K"  "a¯ ≠ \<zero>"
proof -
  from A1 have "a ∈ K0" by simp;
  then show "a¯ ∈ K0" using Field_ZF_1_L4 group0.inverse_in_group
    by auto;
  then show  "(a¯)² ∈ K0"  "a¯ ∈ K"  "a¯ ≠ \<zero>" 
    using Field_ZF_1_L2 IsOpClosed_def by auto
qed;

text{*The inverse is really the inverse.*}

lemma (in field0) Field_ZF_1_L6: assumes A1: "a∈K"  "a≠\<zero>"
  shows "a·a¯ = \<one>"  "a¯·a = \<one>"
proof -
  let ?f = "restrict(M,K0×K0)"
  from A1 have 
    "group0(K0,?f)"
    "a ∈ K0"
    using Field_ZF_1_L4 by auto;
  then have 
    "?f`⟨a,GroupInv(K0, ?f)`(a)⟩ = TheNeutralElement(K0,?f) ∧
    ?f`⟨GroupInv(K0,?f)`(a),a⟩ = TheNeutralElement(K0, ?f)"
    by (rule group0.group0_2_L6);
  with A1 show "a·a¯ = \<one>"  "a¯·a = \<one>"
    using Field_ZF_1_L5 Field_ZF_1_L4 by auto;
qed;

text{*A lemma with two field elements and cancelling.*}

lemma (in field0) Field_ZF_1_L7: assumes "a∈K" "b∈K" "b≠\<zero>"
  shows 
  "a·b·b¯ = a"
  "a·b¯·b = a"
  using prems Field_ZF_1_L5 Ring_ZF_1_L11 Field_ZF_1_L6 Ring_ZF_1_L3
  by auto;

section{*Equations and identities*}

text{*This section deals with more specialized identities that are true in 
  fields.*}

text{*$a/(a^2) = a$.*}

lemma (in field0) Field_ZF_2_L1: assumes A1: "a∈K"  "a≠\<zero>"
  shows "a·(a¯)² = a¯"
proof -
  have "a·(a¯)² = a·(a¯·a¯)" by simp;
  also from A1 have "… =  (a·a¯)·a¯" 
    using Field_ZF_1_L5 Ring_ZF_1_L11 
    by simp;
  also from A1 have "… = a¯" 
    using Field_ZF_1_L6 Field_ZF_1_L5 Ring_ZF_1_L3
    by simp;
  finally show "a·(a¯)² = a¯" by simp;
qed;

text{*If we multiply two different numbers by a nonzero number, the results 
  will be different.*}

lemma (in field0) Field_ZF_2_L2: 
  assumes "a∈K"  "b∈K"  "c∈K"  "a≠b"  "c≠\<zero>"
  shows "a·c¯ ≠ b·c¯"
  using prems field_has_no_zero_divs Field_ZF_1_L5 Ring_ZF_1_L12B
  by simp;

text{*We can put a nonzero factor on the other side of non-identity 
  (is this the best way to call it?) changing it to the inverse.*}

lemma (in field0) Field_ZF_2_L3:
  assumes A1: "a∈K"  "b∈K"  "b≠\<zero>"  "c∈K"   and A2: "a·b ≠ c"
  shows "a ≠ c·b¯"
proof -
  from A1 A2 have "a·b·b¯ ≠ c·b¯" 
    using  Ring_ZF_1_L4 Field_ZF_2_L2 by simp;
  with A1 show "a ≠ c·b¯" using Field_ZF_1_L7
    by simp;
qed;

text{*If if the inverse of $b$ is different than $a$, then the
  inverse of $a$ is different than $b$.*}

lemma (in field0) Field_ZF_2_L4:
  assumes "a∈K"  "a≠\<zero>" and "b¯ ≠ a"
  shows "a¯ ≠ b"
  using prems Field_ZF_1_L4 group0.group0_2_L11B
  by simp;

text{*An identity with two field elements, one and an inverse.*}

lemma (in field0) Field_ZF_2_L5:
  assumes "a∈K"  "b∈K" "b≠\<zero>"
  shows "(\<one> \<ra> a·b)·b¯ = a \<ra> b¯"
  using prems Ring_ZF_1_L4 Field_ZF_1_L5 Ring_ZF_1_L2 ring_oper_distr 
    Field_ZF_1_L7 Ring_ZF_1_L3 by simp;

text{*An identity with three field elements, inverse and cancelling.*}

lemma (in field0) Field_ZF_2_L6: assumes A1: "a∈K"  "b∈K"  "b≠\<zero>"  "c∈K"
  shows "a·b·(c·b¯) = a·c"
proof -
  from A1 have T: "a·b ∈ K"  "b¯ ∈ K"
    using Ring_ZF_1_L4 Field_ZF_1_L5 by auto;
  with mult_commute A1 have "a·b·(c·b¯) = a·b·(b¯·c)"
    using IsCommutative_def by simp;
  moreover
  from A1 T have "a·b ∈ K"  "b¯ ∈ K"  "c∈K"
    by auto;
  then have "a·b·b¯·c = a·b·(b¯·c)"
    by (rule Ring_ZF_1_L11);
  ultimately have "a·b·(c·b¯) = a·b·b¯·c" by simp;
  with A1 show "a·b·(c·b¯) = a·c" 
    using Field_ZF_1_L7 by simp;
qed;

end