{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-field-01",
  "mko_id": "mko-finite-field",
  "title": {
    "zh-Hant": "完整表認證 F₃ 與有限整環引理",
    "en": "Certifying F3 and the finite-domain lemma"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-field-01",
    "source": {
      "mko_id": "mko-finite-field",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "a0f82af13b19a69bc7a3bab914cede8b8bae9db78543343f63b0b4086a684c5e",
      "english_sha256": "6afed76061d7bb9aae31a76426475e249103042ef12a966308ab75b556c16593"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "F₃ 是域，2/2=1；有限整環的非零乘法映射給逆元。",
      "en": "F3 is a field and 2/2=1; nonzero multiplication maps supply inverses in a finite domain."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "F₃={0,1,2}，運算為整數加乘後模3。另令 D 為有限交換含一整環且1≠0。",
          "en": "F3={0,1,2}, with integer addition/multiplication reduced modulo 3. Also let D be a finite commutative unital integral domain with 1 distinct from 0."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "F_3=\\{0,1,2\\}",
        "after_tex": "\\begin{array}{c|ccc}+&0&1&2\\\\\\hline 0&0&1&2\\\\1&1&2&0\\\\2&2&0&1\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "九個和全部列出，結果均在載體。加法零是0，每列給唯一和。",
          "en": "All nine sums are listed and lie in the carrier. Zero is the additive identity, and each ordered pair has exactly one sum."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "反查各列有三格，0列保持全部元素，不是抽樣表。",
            "en": "Each row has three entries and the zero row fixes every element; this is not a sample."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "\\begin{array}{c|ccc}+&0&1&2\\\\\\hline 0&0&1&2\\\\1&1&2&0\\\\2&2&0&1\\\\\\end{array}",
        "after_tex": "-0=0,\\quad -1=2,\\quad -2=1",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由表讀得每個加法逆元；結合律與交換律由整數加法取餘保留。",
          "en": "The table supplies every additive inverse; associativity and commutativity descend from integer addition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "加法逆元必使和為0；1+2=0是模3結論。",
            "en": "Each inverse must sum to zero; 1+2=0 is specifically modulo 3."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "a,b\\in F_3",
        "after_tex": "\\begin{array}{c|ccc}*&0&1&2\\\\\\hline 0&0&0&0\\\\1&0&1&2\\\\2&0&2&1\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "九個積全部列出。整數乘法的結合律、分配律在取餘後仍成立，表沿對角對稱。",
          "en": "All nine products are listed. Integer associativity and distributivity survive reduction, and the table is symmetric."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "若換整數代表相差3的倍數，加乘結果仍相差3的倍數，故法則能下降。",
            "en": "Changing representatives by multiples of 3 changes sums and products by multiples of 3, so the laws descend."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "\\begin{array}{c|ccc}*&0&1&2\\\\\\hline 0&0&0&0\\\\1&0&1&2\\\\2&0&2&1\\\\\\end{array}",
        "after_tex": "1\\ne0,\\quad 1^{-1}=1,\\quad2^{-1}=2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1列保持元素；全部非零元素1與2均有表中雙側逆元。",
          "en": "The row of 1 fixes every element; both nonzero elements, 1 and 2, have two-sided inverses in the table."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "需核2·2=1及交換方向；不能把0加進逆元清單。",
            "en": "Check 2*2=1 in both orders; zero must not enter the inverse list."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S04"
        ],
        "before_tex": "2/2:=2\\cdot2^{-1}",
        "after_tex": "2/2=2\\cdot2=1\\quad(\\bmod 3)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [
          "denominator 2 is nonzero in F3"
        ],
        "explanation": {
          "zh-Hant": "除數2非零，因此除法以乘逆元定義，結果為1。",
          "en": "The denominator 2 is nonzero, so division is multiplication by its inverse and gives 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "反乘2得到2；此反查只因2可逆，並不允許0/0。",
            "en": "Multiplying back by 2 yields 2; this audit relies on invertibility and does not permit 0/0."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1"
        ],
        "before_tex": "a\\in D\\setminus\\{0\\},\\quad m_a(x)=ax",
        "after_tex": "m_a(x)=m_a(y)\\Rightarrow a(x-y)=0\\Rightarrow x=y",
        "relation": "implies",
        "rule": "calc",
        "conditions": [
          "a is nonzero",
          "D has no zero divisors"
        ],
        "explanation": {
          "zh-Hant": "在任意上述有限整環，無零因子與a非零證乘法自映射單射。",
          "en": "In the stated finite domain, absence of zero divisors and a nonzero a make its multiplication self-map injective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "最後消去不是預設逆元，而是從無零因子推出x-y=0，避免循環。",
            "en": "The cancellation uses no assumed inverse; the zero-divisor condition forces x-y=0, avoiding circularity."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "m_a:D\\to D\\text{ injective},\\quad |D|<\\infty",
        "after_tex": "m_a\\text{ surjective}\\Rightarrow\\exists b\\in D:\\ ab=1",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有限自映射的單射即滿射，所以1有前像b。交換性再給ba=1。",
          "en": "An injective finite self-map is surjective, so 1 has a preimage b; commutativity also gives ba=1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "滿射必對同一有限集合D使用；無限整環的單射未必滿射。",
            "en": "Surjectivity concerns the same finite set D; an infinite-domain injection need not be onto."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1"
        ],
        "before_tex": "t^2=t+1\\text{ over }F_2",
        "after_tex": "\\begin{array}{c|cccc}+&0&1&t&t+1\\\\\\hline 0&0&1&t&t+1\\\\1&1&0&t+1&t\\\\t&t&t+1&0&1\\\\t+1&t+1&t&1&0\\\\\\end{array}\\qquad\\begin{array}{c|cccc}*&0&1&t&t+1\\\\\\hline 0&0&0&0&0\\\\1&0&1&t&t+1\\\\t&0&t&t+1&1\\\\t+1&0&t+1&1&t\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個餘式0、1、t、t+1以t²=t+1約化，列出完整雙表。若改代表差為t²+t+1的倍數，加乘仍保持這個性質。",
          "en": "The four remainders 0,1,t,t+1 reduce by t^2=t+1, giving both complete tables. Sums and products preserve differences divisible by t^2+t+1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "因此多項式環法則下降至這四個餘式；表不是任意未認證的運算。",
            "en": "Thus polynomial ring laws descend to these four remainders; the tables are not arbitrary unqualified operations."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "S08"
        ],
        "before_tex": "\\begin{array}{c|cccc}*&0&1&t&t+1\\\\\\hline 0&0&0&0&0\\\\1&0&1&t&t+1\\\\t&0&t&t+1&1\\\\t+1&0&t+1&1&t\\\\\\end{array}",
        "after_tex": "1^{-1}=1,\\quad t^{-1}=t+1,\\quad(t+1)^{-1}=t,\\quad |F_4|=4",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "F₄ 全部三個非零元素有雙側逆元，因此4這個非質數也可以是域階。",
          "en": "All three nonzero elements of F4 have two-sided inverses, so the nonprime number 4 can be a field order."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "t(t+1)=t²+t=1；4=2²是質數冪，並未否證質數冪限制。",
            "en": "t(t+1)=t^2+t=1; 4=2^2 is a prime power and does not refute the prime-power restriction."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "S02",
          "S04",
          "S07"
        ],
        "before_tex": "F_3\\text{ ring laws and inverse list}",
        "after_tex": "F_3\\text{ is a field};\\quad D\\text{ finite domain}\\Rightarrow D\\text{ field}",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整F₃證書滿足域定義；有限整環引理由單射到滿射補足所有非零逆元。",
          "en": "The complete F3 certificate satisfies the field definition; the finite-domain lemma supplies every nonzero inverse through injective-to-surjective reasoning."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S04",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "域結論依全部非零元素，不是僅算2/2；未分類任意有限交換環。",
            "en": "The conclusion covers every nonzero element, not merely 2/2, and does not classify arbitrary finite commutative rings."
          }
        },
        "leaf_basis": "mko_definition"
      }
    ],
    "result": {
      "zh-Hant": "F₃ 是域，2/2=1；有限整環的非零乘法映射給逆元。",
      "en": "F3 is a field and 2/2=1; nonzero multiplication maps supply inverses in a finite domain."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "Z/4Z 也是域。",
          "en": "Z/4Z is a field too."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2≠0但2·2=0，且2乘任何元素皆非1。",
          "en": "2 is nonzero yet 2*2=0, and no product 2*x is 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "0/0 在 F₃ 定義為1。",
          "en": "0/0 is defined as 1 in F3."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0乘任意元素均為0，沒有乘法逆元。",
          "en": "Zero times every element is zero, so it has no multiplicative inverse."
        }
      },
      {
        "claim": {
          "zh-Hant": "每個有限域的階都是質數。",
          "en": "Every finite field has prime order."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "已列出的 F₄ 雙表與逆元給出階4反例；4是質數冪而非質數。",
          "en": "The listed F4 tables and inverses give order 4, a prime power but not a prime."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設整數與 F₂ 多項式的環運算法則、同餘運算及有限集合單射必滿射；整環指無零因子。",
          "en": "Assume ring arithmetic of integers and F2 polynomials, congruence arithmetic, and injective self-maps of a finite set being surjective; an integral domain has no zero divisors."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "以完整有限環雙表核1不等於0、乘法交換及每個非零元素的雙側逆元。除法只對非零除數定義。有限整環成域是利用有限乘法映射單射推出滿射的定理步驟；域階為質數冪，不必是質數。",
          "en": "Use the complete finite ring tables to retain 1 distinct from 0, multiplicative commutativity and a two-sided inverse for every nonzero element. Division is defined only for nonzero denominators. A finite integral domain becomes a field through the finite injective-to-surjective multiplication-map theorem. A finite field has prime-power order, not necessarily prime order."
        },
        "source": {
          "mko_id": "mko-finite-field",
          "sha256": "a0f82af13b19a69bc7a3bab914cede8b8bae9db78543343f63b0b4086a684c5e",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "inverse2",
        "lhs": "(2*2)%3",
        "rhs": "1",
        "step": "S04"
      },
      {
        "id": "f4order",
        "lhs": "2**2",
        "rhs": "4",
        "step": "S09"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整 F₃ 雙表及有限整環引理；F₄ 僅作非質數階的明示邊界。",
      "en": "Complete F3 tables and a finite-domain lemma; explicit F4 only as the nonprime-order boundary."
    },
    "md_symbols": [
      [
        "F_3",
        "模3域",
        "field modulo 3"
      ],
      [
        "D",
        "任意有限整環",
        "arbitrary finite integral domain"
      ],
      [
        "m_a",
        "乘a自映射",
        "multiplication-by-a self-map"
      ]
    ],
    "md_error_controls": [
      {
        "title": "非零門檻 / Nonzero gate",
        "zh": "每次寫a/b前先核b≠0；F₃中的2是合法除數。",
        "en": "Before a/b check b is nonzero; 2 is a valid denominator in F3."
      },
      {
        "title": "有限性責任 / Finiteness obligation",
        "zh": "Z中的乘2映射單射但1無前像，故不可略去有限性。",
        "en": "Multiplication by 2 on Z is injective but misses 1, so finiteness cannot be omitted."
      }
    ],
    "md_open_items": []
  },
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    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
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}