{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-polynomial-ring-over-finite-commutative-ring-01",
  "mko_id": "mko-polynomial-ring-over-finite-commutative-ring",
  "title": {
    "zh-Hant": "有限交換環上的多項式環：在 (ℤ/4ℤ)[x] 中做加法、卷積與次數核對",
    "en": "Polynomial ring over a finite commutative ring: addition, convolution and degree checks in (ℤ/4ℤ)[x]"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-polynomial-ring-over-finite-commutative-ring-01",
    "source": {
      "mko_id": "mko-polynomial-ring-over-finite-commutative-ring",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "738ac2a85874f0a45327552d44abd7bfe8f789812c2f6c29c8f80d592ebb8a67",
      "english_sha256": "6e6e1f5cd46d4615a07facebc94f250a9638d9c5f3145355998233dac87d03bc"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "finite_sum_rules",
      "finite_counting_basics",
      "mul_zero",
      "power_laws",
      "function_def"
    ],
    "goal": {
      "zh-Hant": "在 R=ℤ/4ℤ 上取 f=1+2x、g=2x。依係數序列定義計算 f+g、f·g、f·f、g·g 並正規化；證明首項係數公式 (fg)_{m+n}=f_m g_n，說明為何此處 deg(fg)=deg f+deg g 失敗；最後證明 R[x] 不是有限環。",
      "en": "Over R=ℤ/4ℤ take f=1+2x and g=2x. Using the coefficient-sequence definition, compute and normalize f+g, f·g, f·f and g·g; prove the top-coefficient formula (fg)_{m+n}=f_m g_n and explain why deg(fg)=deg f+deg g fails here; finally prove that R[x] is not a finite ring."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "R=ℤ/4ℤ={0,1,2,3}，加法與乘法為整數運算後取除以 4 的餘數；完整的兩張表由此確定，R 是有限交換含一環，但不是整環（2·2=0）。",
          "en": "R=ℤ/4ℤ={0,1,2,3}, with addition and multiplication given by the integer operation followed by the remainder on division by 4; this fixes both complete tables, and R is a finite commutative unital ring but not an integral domain (2·2=0)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "R[x] 的元素是 R 中只有有限多項非零的係數序列 (a_k)_{k≥0}；寫成列表時去掉尾端的零（正規化）。兩多項式相等當且僅當每個係數相等。",
          "en": "An element of R[x] is a coefficient sequence (a_k)_{k≥0} in R with only finitely many nonzero terms; as a list it is written with trailing zeros removed (normalization). Two polynomials are equal exactly when every coefficient agrees."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "f=1+2x，即 f_0=1、f_1=2、其餘為 0；g=2x，即 g_0=0、g_1=2、其餘為 0。",
          "en": "f=1+2x, i.e. f_0=1, f_1=2 and all other coefficients 0; g=2x, i.e. g_0=0, g_1=2 and all others 0."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mod_reduce"
        ],
        "before_tex": "\\mathbb{Z}/4\\mathbb{Z}\\ \\text{tables}",
        "after_tex": "2+2=0,\\ 2\\cdot2=0,\\ 1\\cdot2=2,\\ 1\\cdot1=1,\\ 1\\neq0",
        "relation": "implies",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "從完整表讀出後面要用的項：2+2=4=1·4+0 得 0；2·2=4 得 0；1·2=2；1·1=1。又 1−0=1 不是 4 的倍數，所以 1≠0。2·2=0 而 2≠0，說明 R 有零因子，所以 MKO 假設 a2 禁止在此套用次數相加公式。",
          "en": "Read off the entries needed later from the complete tables: 2+2=4=1·4+0 gives 0; 2·2=4 gives 0; 1·2=2; 1·1=1. Also 1−0=1 is not a multiple of 4, so 1≠0. Since 2·2=0 with 2≠0, R has a zero divisor, which is why MKO assumption a2 forbids using degree additivity here."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "每項都可由取模重算（N1、N3）；本步只讀表，不推出表外的事實。",
            "en": "Each entry can be recomputed by reduction (N1, N3); the step only reads the tables and derives nothing beyond them."
          }
        },
        "leaf_basis": "mod_reduce + numeric_evaluation"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "G3",
          "S01"
        ],
        "before_tex": "f=1+2x,\\ g=2x",
        "after_tex": "f=(1,2),\\ g=(0,2),\\ \\deg f=\\deg g=1",
        "relation": "equivalent",
        "rule": "deg_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "寫成正規化的係數列表：f=(1,2)、g=(0,2)，索引 2 的係數起全為 0。兩者最後一項 2≠0，所以已正規化，且依次數定義 deg f=deg g=1。",
          "en": "Write the normalized coefficient lists f=(1,2) and g=(0,2), with all coefficients from index 2 on equal to 0. In both the last entry is 2≠0, so the lists are normalized, and by the definition of degree deg f=deg g=1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G2",
            "deg_def"
          ],
          "explanation": {
            "zh-Hant": "列表與多項式互相決定：由 (1,2) 讀回 1+2x，由 (0,2) 讀回 2x。",
            "en": "The lists and the polynomials determine each other: (1,2) reads back as 1+2x and (0,2) as 2x."
          }
        },
        "leaf_basis": "deg_def + def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "S01",
          "poly_ops"
        ],
        "before_tex": "f+g",
        "after_tex": "(1+0,\\ 2+2)=(1,0)",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "加法逐係數在 R 中進行：索引 0 的係數 1+0=1，索引 1 的係數 2+2=0（S01，N1），索引 2 的係數起 0+0=0。得到未正規化的列表 (1,0)。",
          "en": "Addition is coefficientwise in R: index 0 gives 1+0=1, index 1 gives 2+2=0 (S01, N1), and from index 2 on 0+0=0. The result is the unnormalized list (1,0)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "反向：(1,0)−g=(1−0, 0−2)=(1,2)=f（在 ℤ/4ℤ 中 −2=2），還原了 f+g 的左邊。",
            "en": "Backward: (1,0)−g=(1−0, 0−2)=(1,2)=f (in ℤ/4ℤ, −2=2), which recovers the left side f+g."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "f+g=(1,0)",
        "after_tex": "f+g=(1)=1,\\ \\deg(f+g)=0",
        "relation": "equivalent",
        "rule": "deg_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(1,0) 有尾端零。列表 (1,0) 與 (1) 都以零補足成同一個無限係數序列，逐項相同，所以去掉尾端零只是換一種寫法：f+g=1。正規化後最後一項是 1≠0，deg(f+g)=0。若不正規化而把 (1,0) 讀成次數 1，就會把同一多項式算出兩個次數。",
          "en": "(1,0) has a trailing zero. As sequences, (1,0) and (1) both continue with zeros and agree term by term, so dropping the trailing zero only rewrites the same polynomial: f+g=1. After normalization the last entry is 1≠0, so deg(f+g)=0. Reading (1,0) unnormalized as degree 1 would give one polynomial two degrees."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G2",
            "deg_def"
          ],
          "explanation": {
            "zh-Hant": "補回尾端零得到同一序列，雙向成立；次數只對正規化的列表讀取。",
            "en": "Putting the trailing zero back gives the same sequence, so the step is two-way; the degree is read only from the normalized list."
          }
        },
        "leaf_basis": "deg_def + def_unfold"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "poly_ops"
        ],
        "before_tex": "(fg)_k=\\sum_{i+j=k}f_ig_j",
        "after_tex": "fg=(0,2,0)",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "乘法是有限卷積。因 f_i=0（i≥2）且 g_j=0（j≥2），k≥3 時每個 i+j=k 的項都有一個因子為 0，所以只需算 k=0、1、2（S05.1 到 S05.3）。",
          "en": "Multiplication is finite convolution. Since f_i=0 for i≥2 and g_j=0 for j≥2, for k≥3 every term with i+j=k has a zero factor, so only k=0, 1, 2 need computing (S05.1 to S05.3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mul_zero",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "每個係數都是同一卷積公式的求值，可逐項重算；左右兩邊是同一個多項式。",
            "en": "Each coefficient is an evaluation of the same convolution formula and can be recomputed term by term; both sides are the same polynomial."
          }
        },
        "leaf_basis": "poly_ops + finite_sum_rules + mul_zero"
      },
      {
        "id": "S05.1",
        "uses": [
          "S02",
          "poly_ops"
        ],
        "before_tex": "(fg)_0=f_0g_0",
        "after_tex": "1\\cdot0=0",
        "relation": "equivalent",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "i+j=0 只有 (0,0)：f_0 g_0=1·0=0。",
          "en": "i+j=0 only for (0,0): f_0 g_0=1·0=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "恆等式 u·0=0，雙向成立。",
            "en": "The identity u·0=0 holds both ways."
          }
        },
        "leaf_basis": "mul_zero"
      },
      {
        "id": "S05.2",
        "uses": [
          "S02",
          "S01",
          "poly_ops"
        ],
        "before_tex": "(fg)_1=f_0g_1+f_1g_0",
        "after_tex": "1\\cdot2+2\\cdot0=2",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "i+j=1 有 (0,1) 與 (1,0) 兩項：1·2+2·0=2+0=2（N2）。",
          "en": "i+j=1 has the two terms (0,1) and (1,0): 1·2+2·0=2+0=2 (N2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "兩項都列出，沒有遺漏或重複；結果可重算。",
            "en": "Both terms are listed with none missing or repeated; the result can be recomputed."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce + numeric_evaluation"
      },
      {
        "id": "S05.3",
        "uses": [
          "S02",
          "S01",
          "poly_ops"
        ],
        "before_tex": "(fg)_2=f_0g_2+f_1g_1+f_2g_0",
        "after_tex": "0+2\\cdot2+0=0",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "i+j=2 有三項；f_0 g_2=1·0=0、f_2 g_0=0·0=0，中間 f_1 g_1=2·2=4≡0（S01，N3）。所以 (fg)_2=0：首項係數的乘積被零因子消去。",
          "en": "i+j=2 has three terms: f_0 g_2=1·0=0 and f_2 g_0=0·0=0, while the middle term f_1 g_1=2·2=4≡0 (S01, N3). So (fg)_2=0: the product of the leading coefficients is killed by a zero divisor."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mod_reduce",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "三項逐一可重算；在 ℤ 中 2·2=4≠0，但在 ℤ/4ℤ 中為 0。",
            "en": "All three terms can be recomputed; in ℤ, 2·2=4≠0, but in ℤ/4ℤ it is 0."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce + mul_zero"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "S05.1",
          "S05.2",
          "S05.3",
          "G2"
        ],
        "before_tex": "fg=(0,2,0)",
        "after_tex": "fg=(0,2)=2x,\\ \\deg(fg)=1\\neq 1+1",
        "relation": "equivalent",
        "rule": "deg_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "去掉尾端零得 fg=(0,2)=2x，最後一項 2≠0，所以 deg(fg)=1。但 deg f+deg g=2：次數相加公式在此失敗，原因正是 S05.3 中 2·2=0。",
          "en": "Removing the trailing zero gives fg=(0,2)=2x; its last entry is 2≠0, so deg(fg)=1. But deg f+deg g=2: degree additivity fails here, precisely because 2·2=0 in S05.3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G2",
            "deg_def"
          ],
          "explanation": {
            "zh-Hant": "正規化可逆（補回尾端零）；次數讀自正規化列表。",
            "en": "Normalization is reversible (put the trailing zero back); the degree is read from the normalized list."
          }
        },
        "leaf_basis": "deg_def + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S01",
          "poly_ops"
        ],
        "before_tex": "f\\cdot f",
        "after_tex": "(1\\cdot1,\\ 1\\cdot2+2\\cdot1,\\ 2\\cdot2)=(1,0,0)=1",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同樣的卷積：索引 0 的係數 1·1=1；索引 1 的係數 1·2+2·1=4≡0（N4）；索引 2 的係數 2·2=0；k≥3 為 0。正規化得 f·f=1。所以 f=1+2x 雖是 1 次多項式，卻是 R[x] 的可逆元，f^{-1}=f；在整環上次數 1 的多項式不可能有這種性質。",
          "en": "The same convolution: index 0 gives 1·1=1; index 1 gives 1·2+2·1=4≡0 (N4); index 2 gives 2·2=0; indices from 3 on give 0. Normalizing, f·f=1. So f=1+2x, though of degree 1, is a unit of R[x] with f^{-1}=f; over an integral domain a polynomial of degree 1 could not behave like this."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "三個係數逐一可重算；f·f 與 1 逐項相同。",
            "en": "The three coefficients can be recomputed one by one; f·f and 1 agree term by term."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S02",
          "S01",
          "poly_ops"
        ],
        "before_tex": "g\\cdot g",
        "after_tex": "(0\\cdot0,\\ 0\\cdot2+2\\cdot0,\\ 2\\cdot2)=(0,0,0)=0",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "索引 0 的係數 0·0=0；索引 1 的係數 0·2+2·0=0；索引 2 的係數 2·2=0；k≥3 為 0。所以 g·g 是零多項式，沒有次數：兩個非零多項式之積為 0，R[x] 本身有零因子。",
          "en": "Index 0 gives 0·0=0; index 1 gives 0·2+2·0=0; index 2 gives 2·2=0; from index 3 on 0. So g·g is the zero polynomial, which has no degree: two nonzero polynomials multiply to 0, and R[x] itself has zero divisors."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mod_reduce",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "每個係數都可重算為 0；零多項式與 (0,0,0) 是同一序列。",
            "en": "Each coefficient recomputes to 0; the zero polynomial and (0,0,0) are the same sequence."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce + mul_zero"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S06",
          "S07",
          "S08"
        ],
        "before_tex": "\\deg f=m,\\ \\deg g=n",
        "after_tex": "(fg)_k=0\\ (k>m+n),\\ (fg)_{m+n}=f_mg_n",
        "relation": "theorem_application",
        "rule": "L_top_coeff",
        "conditions": [
          "f\\neq0,\\ g\\neq0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_top_coeff（在 S09.1、S09.2 證明），索引大於 m+n 的積係數全為零，索引 m+n 的係數恰為首項係數之積；只有積非零時才把前者寫成次數至多 m+n，零多項式依本稿約定沒有次數。在本例 m=n=1，f_1=g_1=2，f_1 g_1=0，所以 f·g、f·f、g·g 的索引 2 的係數都是 0，與 S06 到 S08 一致：前兩者次數降到 1 與 0，後者成為零多項式。",
          "en": "By Lemma L_top_coeff (proved in S09.1 and S09.2), all product coefficients above m+n vanish and the coefficient at m+n is the product of the leading coefficients. Only when the product is nonzero is this expressed as degree at most m+n; under this guide's convention the zero polynomial has no degree. Here m=n=1 and f_1=g_1=2, so f_1 g_1=0 and the coefficient at index 2 of f·g, f·f and g·g is 0, matching S06 to S08: the first two drop to degrees 1 and 0, and the last becomes the zero polynomial."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "L_top_coeff"
          ],
          "explanation": {
            "zh-Hant": "引理只需要 R 交換，不需要整環；它說的是「何時」次數相加，而不是總是相加。",
            "en": "The lemma needs only a commutative ring, not an integral domain; it says when degrees add, not that they always do."
          }
        },
        "leaf_basis": "poly_ops + finite_sum_rules + mul_zero + deg_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "poly_ops",
          "deg_def",
          "mul_zero"
        ],
        "before_tex": "k>m+n,\\ i+j=k",
        "after_tex": "f_ig_j=0,\\ \\text{so}\\ (fg)_k=0",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 i≤m 且 j≤n，則 i+j≤m+n<k，矛盾；所以 i>m 或 j>n。依次數定義 f_i=0 或 g_j=0，該項為 0（u·0=0、0·u=0）。卷積和的每項都是 0，所以 (fg)_k=0。",
          "en": "If i≤m and j≤n, then i+j≤m+n<k, a contradiction; so i>m or j>n. By the definition of degree, f_i=0 or g_j=0, and the term is 0 (u·0=0 and 0·u=0). Every term of the convolution sum is 0, so (fg)_k=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "deg_def",
            "mul_zero",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "只用到次數定義與乘零；不宣稱 k≤m+n 的係數非零。",
            "en": "Only the definition of degree and multiplication by zero are used; it is not claimed that coefficients with k≤m+n are nonzero."
          }
        },
        "leaf_basis": "mul_zero + deg_def + de_morgan_logic + finite_sum_rules"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "poly_ops",
          "deg_def"
        ],
        "before_tex": "(fg)_{m+n}=\\sum_{i+j=m+n}f_ig_j",
        "after_tex": "(fg)_{m+n}=f_mg_n",
        "relation": "equivalent",
        "rule": "poly_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 i+j=m+n 的各項中：若 i<m 則 j>n，g_j=0；若 i>m 則 f_i=0。只剩 (i,j)=(m,n) 一項，所以 (fg)_{m+n}=f_m g_n。因此 deg(fg)=m+n 當且僅當 f_m g_n≠0。S09.1 與 S09.2 合起來證得引理 L_top_coeff。",
          "en": "Among the terms with i+j=m+n: if i<m then j>n and g_j=0; if i>m then f_i=0. Only the term (i,j)=(m,n) remains, so (fg)_{m+n}=f_m g_n. Hence deg(fg)=m+n exactly when f_m g_n≠0. S09.1 and S09.2 together prove Lemma L_top_coeff."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_ops",
            "mul_zero",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "去掉的項全為 0，加上或去掉 0 不改變和，所以兩式相等，雙向成立。",
            "en": "The removed terms are all 0, and adding or removing zeros does not change the sum, so the two expressions are equal both ways."
          }
        },
        "leaf_basis": "poly_ops + mul_zero + add_zero + finite_sum_rules"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "S01",
          "poly_ops"
        ],
        "before_tex": "h=1+x=(1,1),\\ g=(0,2)",
        "after_tex": "hg=(0,2,2)=2x+2x^2,\\ \\deg(hg)=2",
        "relation": "implies",
        "rule": "L_top_coeff",
        "conditions": [
          "h_1g_1=2\\neq0"
        ],
        "explanation": {
          "zh-Hant": "對照：h=1+x 的首項係數是 1。卷積：索引 0 的係數 1·0=0；索引 1 的係數 1·2+1·0=2（N5）；索引 2 的係數 1·2=2≠0（N6）。所以 deg(hg)=2=1+1。依 L_top_coeff，次數相加恰在首項係數之積非零時成立；只要首項係數有一個是 1，就不會被零因子消去。",
          "en": "Contrast: h=1+x has leading coefficient 1. Convolution: index 0 gives 1·0=0; index 1 gives 1·2+1·0=2 (N5); index 2 gives 1·2=2≠0 (N6). So deg(hg)=2=1+1. By L_top_coeff, degrees add exactly when the product of the leading coefficients is nonzero; if one leading coefficient is 1, the zero divisor cannot kill it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "L_top_coeff"
          ],
          "explanation": {
            "zh-Hant": "結論依賴 h_1 g_1=2≠0 這個具體核對，不是由「通常次數相加」推得。",
            "en": "The conclusion rests on the explicit check h_1 g_1=2≠0, not on 'degrees usually add'."
          }
        },
        "leaf_basis": "poly_ops + mod_reduce + deg_def"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "x^n=e_n,\\ m\\neq n",
        "after_tex": "e_m\\neq e_n",
        "relation": "implies",
        "rule": "mko_polyring_def",
        "conditions": [
          "1\\neq0\\ \\text{in}\\ R"
        ],
        "explanation": {
          "zh-Hant": "x^n 的係數序列 e_n 在索引 n 的係數為 1、其餘為 0。若 m≠n，比較索引 m 的係數：(e_m)_m=1，(e_n)_m=0；由 S01，1≠0，所以兩序列不同，e_m≠e_n。因此 n↦x^n 是 ℕ 到 R[x] 的單射。",
          "en": "The coefficient sequence e_n of x^n is 1 at index n and 0 elsewhere. If m≠n, compare index m: (e_m)_m=1 and (e_n)_m=0; by S01, 1≠0, so the sequences differ and e_m≠e_n. Hence n↦x^n is an injective map from ℕ into R[x]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "G2",
            "mko_polyring_def"
          ],
          "explanation": {
            "zh-Hant": "關鍵前提是 1≠0：在零環中 1=0，所有 x^n 都等於 0，R[x] 只有一個元素。",
            "en": "The key premise is 1≠0: in the zero ring 1=0, every x^n equals 0, and R[x] has a single element."
          }
        },
        "leaf_basis": "function_def + injective_surjective_def + def_unfold"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "finite_counting_basics"
        ],
        "before_tex": "n\\mapsto x^n\\ \\text{injective}",
        "after_tex": "R[x]\\ \\text{is not finite}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "假設 R[x] 有限，恰有 N 個元素。由 S11，x^0、x^1 直到 x^N 是 R[x] 中 N+1 個互異元素，即 {0,1,直到 N} 到一個 N 元集合的單射；它的像是 N+1 個元素的子集，但有限集的子集元素不多於該集合，矛盾。所以 R[x] 不是有限環。每個固定次數上界的部分是有限的，例如次數小於 3 的多項式恰有 4^3=64 個（N7），但次數沒有上界。",
          "en": "Suppose R[x] were finite with exactly N elements. By S11, x^0, x^1, up to x^N are N+1 distinct elements of R[x], an injection of {0, 1, up to N} into a set of N elements; its image is a subset with N+1 elements, but a subset of a finite set has no more elements than the set, a contradiction. So R[x] is not a finite ring. Each part with a fixed bound on the degree is finite, for example there are exactly 4^3=64 polynomials of degree less than 3 (N7), but the degree is unbounded."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "finite_counting_basics",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "需要對每個 N 都有 N+1 個互異元素，這由 S11 的一般論證提供，不是由檢查有限多個 N 得到。",
            "en": "It needs N+1 distinct elements for every N, which the general argument of S11 provides; it does not come from checking finitely many values of N."
          }
        },
        "leaf_basis": "contradiction + finite_counting_basics + injective_surjective_def"
      },
      {
        "id": "S13",
        "uses": [
          "S01",
          "G2",
          "function_def"
        ],
        "before_tex": "x^2,\\ x^4\\in R[x]",
        "after_tex": "x^2\\neq x^4,\\ \\text{but}\\ t^2=t^4\\ \\text{for every}\\ t\\in R",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "作為多項式，x^2 與 x^4 在索引 2 的係數係數為 1 與 0，所以不同（S11）。作為 R 上的函數，逐點代入全部 4 個元素：0^2=0^4=0，1^2=1^4=1，2^2=4≡0 與 2^4=16≡0，3^2=9≡1 與 3^4=81≡1，兩函數相同（N8 對 R 的全部 4 個元素逐一核對）。R 到 R 的函數只有 4^4=256 個（N9），是有限集；R[x] 不是函數集合，不能因為函數有限就說 R[x] 有限。",
          "en": "As polynomials, x^2 and x^4 have coefficients 1 and 0 at index 2, so they differ (S11). As functions on R, substitute all 4 elements: 0^2=0^4=0, 1^2=1^4=1, 2^2=4≡0 and 2^4=16≡0, 3^2=9≡1 and 3^4=81≡1, so the two functions agree (N8 checks all 4 elements of R one by one). There are only 4^4=256 functions from R to R (N9), a finite set; R[x] is not the set of functions, and the finiteness of the functions does not make R[x] finite."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "function_def",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "函數相同只靠對 R 的全部 4 點核對（對有限定義域這是完整核對）；多項式不同靠係數比較。兩個結論不衝突，因為相等的標準不同。",
            "en": "Equality of the functions rests on checking all 4 points of R (a complete check, since the domain is finite); inequality of the polynomials rests on comparing coefficients. The two conclusions do not conflict, because the notions of equality differ."
          }
        },
        "leaf_basis": "function_def + mod_reduce + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S04",
          "S06",
          "S07",
          "S08",
          "S09",
          "S12"
        ],
        "before_tex": "\\text{S04, S06 to S09, S12}",
        "after_tex": "f+g=1,\\ fg=2x,\\ f^2=1,\\ g^2=0;\\ R[x]\\ \\text{infinite}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：f+g=1（需正規化），fg=2x（次數 1，不是 2），f·f=1（1 次多項式可逆），g·g=0（R[x] 有零因子）；首項係數公式說明次數何時相加；1≠0 使單項式互異，R[x] 不是有限環。",
          "en": "Summary: f+g=1 (normalization needed), fg=2x (degree 1, not 2), f·f=1 (a degree-1 polynomial is a unit), g·g=0 (R[x] has zero divisors); the top-coefficient formula explains when degrees add; 1≠0 makes the monomials distinct, so R[x] is not a finite ring."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S06",
            "S07",
            "S08",
            "S09",
            "S12"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿目標中的計算、次數說明與無限性。足夠條件 H：S04、S06 到 S09、S12；H⇒G 由合取。H 由 G1 到 G3 正向證成，沒有把 G 當作前提，也沒有使用次數相加公式。",
            "en": "Goal G: the computations, the degree explanation and infiniteness stated in the goal. Sufficient condition H: S04, S06 to S09 and S12; H⇒G by conjunction. H is proved forwards from G1 to G3; G is never used as a premise, and degree additivity is never assumed."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 (ℤ/4ℤ)[x] 中，f+g=1（正規化後次數 0），fg=2x（次數 1≠2），f·f=1（f=1+2x 是可逆元），g·g=0（兩個非零多項式之積為 0）。對任何交換環，(fg)_{m+n}=f_m g_n，所以次數相加恰在首項係數之積非零時成立。因 1≠0，單項式 1、x、x² 等互異，R[x] 不是有限環，雖然 R 只有 4 個元素。",
      "en": "In (ℤ/4ℤ)[x], f+g=1 (degree 0 after normalization), fg=2x (degree 1≠2), f·f=1 (f=1+2x is a unit) and g·g=0 (two nonzero polynomials multiply to 0). Over any commutative ring (fg)_{m+n}=f_m g_n, so degrees add exactly when the product of the leading coefficients is nonzero. Since 1≠0, the monomials 1, x, x² and so on are distinct, and R[x] is not a finite ring, although R has only 4 elements."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "R=ℤ/4ℤ 只有 4 個元素，所以 R[x] 是有限環",
          "en": "R=ℤ/4ℤ has only 4 elements, so R[x] is a finite ring"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1≠0 使 1、x、x² 等互異（S11），R[x] 含任意多個元素（S12）。",
          "en": "1≠0 makes 1, x, x² and so on distinct (S11), so R[x] has arbitrarily many elements (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "deg((2x)(2x))=1+1=2",
          "en": "deg((2x)(2x))=1+1=2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(2x)(2x)=4x²=0 是零多項式，沒有次數（S08）。",
          "en": "(2x)(2x)=4x²=0 is the zero polynomial, which has no degree (S08)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x² 與 x⁴ 在 ℤ/4ℤ 的每個元素取值相同，所以在 R[x] 中 x²=x⁴",
          "en": "x² and x⁴ take the same value at every element of ℤ/4ℤ, so x²=x⁴ in R[x]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "多項式相等是逐係數相等；索引 2 的係數係數 1≠0（S13）。",
          "en": "Polynomial equality is coefficientwise; the coefficients at index 2 are 1≠0 (S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "1+2x 是 1 次多項式，所以不可能是 R[x] 的可逆元",
          "en": "1+2x has degree 1, so it cannot be a unit of R[x]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 ℤ/4ℤ 中 (1+2x)²=1+4x+4x²=1（S07），所以 1+2x 確為 1 次可逆元。需要整環前提的是負例主張背後的「用次數相加排除正次數可逆元」論證，不是這個模 4 恆等式；此處係數環有零因子，不能套用那個次數論證。",
          "en": "In ℤ/4ℤ, (1+2x)²=1+4x+4x²=1 (S07), so 1+2x is indeed a degree-one unit. The integral-domain premise belongs to the rejected claim's degree-additivity argument against positive-degree units, not to this mod-4 identity. The coefficient ring here has zero divisors, so that degree argument cannot be applied."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_polyring_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-polynomial-ring-over-finite-commutative-ring",
          "sha256": "738ac2a85874f0a45327552d44abd7bfe8f789812c2f6c29c8f80d592ebb8a67",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "R 為有限交換含一環時，R[x] 的每個元素是有限支撐係數序列；加法逐係數，乘法為有限卷積。若 1 不等於 0，單項式 1、x、x 平方等依 degree 互異，因此普通 R[x] 不稱有限環。",
          "en": "For a finite commutative unital ring R, each element of R[x] is a finite-support coefficient sequence. Addition is coefficientwise and multiplication is finite convolution. If 1 differs from 0, the monomials 1, x, x² and so on are distinct by degree, so ordinary R[x] is not a finite ring."
        }
      },
      {
        "id": "mod_reduce",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在 ℤ/nℤ 中，整數 m 代表它除以 n 的餘數 r（0≤r<n）；兩個整數代表同一元素當且僅當 n 整除兩者之差；和與積是整數和、整數積的餘數。",
          "en": "In ℤ/nℤ an integer m stands for its remainder r on division by n (0≤r<n); two integers stand for the same element exactly when n divides their difference; sums and products are the remainders of the integer sum and product."
        }
      },
      {
        "id": "poly_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式運算：(f+g)_k=f_k+g_k；(fg)_k=Σ_{i+j=k} f_i g_j（有限和，在 R 中計算）。",
          "en": "Polynomial operations: (f+g)_k=f_k+g_k and (fg)_k=Σ_{i+j=k} f_i g_j (a finite sum computed in R)."
        }
      },
      {
        "id": "deg_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零多項式 f 的次數 deg f 是使 f_k≠0 的最大 k；零多項式沒有次數。",
          "en": "The degree deg f of a nonzero polynomial f is the largest k with f_k≠0; the zero polynomial has no degree."
        }
      },
      {
        "id": "L_top_coeff",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "在任何交換環上，若 deg f=m、deg g=n，則 k>m+n 時 (fg)_k=0，且 (fg)_{m+n}=f_m g_n。因此 deg(fg)=m+n 當且僅當 f_m g_n≠0；否則 fg=0 或 deg(fg)<m+n。",
          "en": "Over any commutative ring, if deg f=m and deg g=n, then (fg)_k=0 for k>m+n and (fg)_{m+n}=f_m g_n. Hence deg(fg)=m+n exactly when f_m g_n≠0; otherwise fg=0 or deg(fg)<m+n."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "(2+2)%4",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S05.2",
        "lhs": "(1*2+2*0)%4",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S05.3",
        "lhs": "(2*2)%4",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "(1*2+2*1)%4",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "(1*2+1*0)%4",
        "rhs": "2"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "(1*2)%4",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "4**3",
        "rhs": "64"
      },
      {
        "id": "N8",
        "step": "S13",
        "vars": [
          "t"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(t**2)%4",
        "rhs": "(t**4)%4"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "4**4",
        "rhs": "256"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 範圍內工作：係數環 ℤ/4ℤ 以完整加法表與乘法表給出（取模 4），多項式為尾端零已正規化的有限支撐係數序列（假設 a1）；不假設係數環是域、整環、UFD 或 Euclidean domain，有零因子時不使用次數相加公式（假設 a2）。不處理多項式理想、商環或有限域；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the coefficient ring ℤ/4ℤ is given by its complete addition and multiplication tables (modulo 4), and polynomials are finite-support coefficient sequences normalized by removing trailing zeros (assumption a1); the coefficient ring is not assumed to be a field, integral domain, UFD or Euclidean domain, and the degree-additivity formula is not used in the presence of zero divisors (assumption a2). It does not treat polynomial ideals, quotient rings or finite fields, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "R=\\mathbb{Z}/4\\mathbb{Z}",
        "係數環，元素 0、1、2、3",
        "the coefficient ring, elements 0, 1, 2, 3"
      ],
      [
        "R[x]",
        "R 上的多項式環",
        "the polynomial ring over R"
      ],
      [
        "f_k",
        "f 的索引 k 的係數（x^k 的係數）",
        "the k-th coefficient of f (the coefficient of x^k)"
      ],
      [
        "\\deg f",
        "非零多項式中最大的 k 使 f_k≠0",
        "for nonzero f, the largest k with f_k≠0"
      ],
      [
        "e_n",
        "x^n 的係數序列：索引 n 的係數為 1，其餘為 0",
        "the coefficient sequence of x^n: 1 at index n, 0 elsewhere"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限係數環不給有限多項式環 / A finite coefficient ring does not give a finite polynomial ring",
        "zh": "MKO 指出的誤解：把 R[x] 也叫有限載體。係數只有 4 種選擇，但次數沒有上界；S11、S12 證明 R[x] 含任意多個互異的單項式。對非負整數 d，所有指數≥d 的係數均為 0 的部分（含零多項式）有 4^d 個元素。d=0 時只有零多項式；d=1 時這是對乘法封閉的常數子環，兩種情況都對乘法封閉；d≥2 時則不封閉，因 x^{d−1} 與 x 均在其中，而乘積 x^d 不在其中。",
        "en": "The misconception identified in the MKO is calling R[x] a finite carrier too. There are only 4 choices for each coefficient, but the degree is unbounded; S11 and S12 prove that R[x] contains arbitrarily many distinct monomials. For a nonnegative integer d, the subset whose coefficients of every exponent at least d vanish, including the zero polynomial, has 4^d elements. For d=0 it contains only the zero polynomial; for d=1 it is the constant subring. Both cases are multiplicatively closed. For d≥2 it is not closed: x^{d−1} and x belong to it, but their product x^d does not."
      },
      {
        "title": "有零因子時，次數相加未必成立 / Degree additivity need not hold with zero divisors",
        "zh": "MKO 的第二個誤解是硬套 deg(fg)=deg f+deg g。在 ℤ/4ℤ 上，(2x)(2x)=0 沒有次數，(1+2x)(2x)=2x 次數為 1，(1+2x)^2=1 次數為 0。正確的說法是 L_top_coeff：次數相加當且僅當首項係數之積非零（S09、S10）。",
        "en": "The second misconception identified in the MKO is to force deg(fg)=deg f+deg g. Over ℤ/4ℤ, (2x)(2x)=0 has no degree, (1+2x)(2x)=2x has degree 1, and (1+2x)^2=1 has degree 0. The correct statement is L_top_coeff: degrees add exactly when the product of the leading coefficients is nonzero (S09, S10)."
      },
      {
        "title": "先正規化再讀次數 / Normalize before reading the degree",
        "zh": "逐係數加法可能產生尾端零：f+g 先得 (1,0)。(1,0) 與 (1) 是同一多項式，次數是 0 而不是 1（S04）。比較或讀次數前都要先去掉尾端零。",
        "en": "Coefficientwise addition can produce trailing zeros: f+g first comes out as (1,0). (1,0) and (1) are the same polynomial, of degree 0, not 1 (S04). Remove trailing zeros before comparing or reading a degree."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "L_top_coeff 對任何交換環都已證明；S03 到 S08、S10 的計算只屬於 ℤ/4ℤ 與這幾個多項式。多項式理想、商環與有限域上的多項式（MKO 的下一步）不在本稿範圍。",
        "en": "L_top_coeff is proved for every commutative ring; the computations in S03 to S08 and S10 belong to ℤ/4ℤ and these particular polynomials. Polynomial ideals, quotient rings and polynomials over finite fields (the MKO's next step) are outside this guide."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "483fdc8d305e2bb342f7008783ce4641fe203322aebbac3fe3dbe5604f4c2e83",
    "original_source_sha256": "0bab65c7935d902ed18c55a3ac5fa3655cf979c044b1d2746c31e5cb7d687baf",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M089",
      "BR-M090"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}