{
  "index": "1962-A-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. Let \\( S \\) be a set of rational numbers such that whenever \\( a \\) and \\( b \\) are members of \\( S \\), so are \\( a+b \\) and \\( a b \\), and having the property that for every rational number \\( r \\) exactly one of the following three statements is true:\n\\[\nr \\in S, \\quad-r \\in S, \\quad r=0\n\\]\n\nProve that \\( S \\) is the set of all positive rational numbers.",
  "solution": "Solution. If \\( r \\neq 0 \\), then either \\( r \\in S \\) or \\( -r \\in S \\). Since \\( r^{2}=(-r)^{2} \\), in either case we have \\( r^{2} \\in S \\).\n\nIn particular, \\( 1 \\in S \\). Then from the sum property it follows that every positive integer is in \\( S \\).\nIf \\( p \\) and \\( q \\) are positive integers, then \\( 1 / q^{2} \\in S \\) by our first result, and\n\\[\n\\frac{p}{q}=p q\\left(\\frac{1}{q^{2}}\\right) \\in S\n\\]\nby the product property. Thus every positive rational is in \\( S \\). Now the hypothesis implies that no negative rational is in \\( S \\) and \\( 0 \\notin S \\), so \\( S \\) is just the set of positive rational numbers.\n\nRemarks. The given properties of the set \\( S \\) characterize the set of positive elements in an abstract ordered field. Hence the result can be interpreted as meaning that there is only one way to make the field of rational numbers into an ordered field. See, for example, \"Birkhoff and MacLane, \"A Survey of Modern Algebra, Marmillan, New York, 1941, page 48.\n\nIf the product property is dropped from the hypothesis, then we can prove that \\( S \\) is either the set of positive rationals or the set of negative rationals. For each positive integer \\( q \\), choose \\( \\epsilon_{q}= \\pm 1 \\) so that \\( \\epsilon_{q} / q \\in S \\). The sum rule shows that \\( \\epsilon_{q}(p / q) \\in S \\) for all positive integers \\( p \\). Then the choice \\( p=q \\) gives \\( \\epsilon_{\\varphi}=\\epsilon_{1} \\) and therefore \\( \\epsilon_{1} r \\in S \\) for all positive rationals \\( r \\).",
  "vars": [
    "r",
    "a",
    "b",
    "p",
    "q"
  ],
  "params": [
    "S",
    "\\\\epsilon_q",
    "\\\\epsilon_1",
    "\\\\epsilon_\\\\varphi"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "ratnum",
        "a": "elemfirst",
        "b": "elemsecond",
        "p": "varpint",
        "q": "varqint",
        "S": "rationalset",
        "\\epsilon_q": "epsilonsubq",
        "\\epsilon_1": "epsilonsubone",
        "\\epsilon_\\varphi": "epsilonsubphi"
      },
      "question": "6. Let \\( rationalset \\) be a set of rational numbers such that whenever \\( elemfirst \\) and \\( elemsecond \\) are members of \\( rationalset \\), so are \\( elemfirst+elemsecond \\) and \\( elemfirst elemsecond \\), and having the property that for every rational number \\( ratnum \\) exactly one of the following three statements is true:\n\\[\nratnum \\in rationalset, \\quad-ratnum \\in rationalset, \\quad ratnum=0\n\\]\n\nProve that \\( rationalset \\) is the set of all positive rational numbers.",
      "solution": "Solution. If \\( ratnum \\neq 0 \\), then either \\( ratnum \\in rationalset \\) or \\( -ratnum \\in rationalset \\). Since \\( ratnum^{2}=(-ratnum)^{2} \\), in either case we have \\( ratnum^{2} \\in rationalset \\).\n\nIn particular, \\( 1 \\in rationalset \\). Then from the sum property it follows that every positive integer is in \\( rationalset \\).\nIf \\( varpint \\) and \\( varqint \\) are positive integers, then \\( 1 / varqint^{2} \\in rationalset \\) by our first result, and\n\\[\n\\frac{varpint}{varqint}=varpint varqint\\left(\\frac{1}{varqint^{2}}\\right) \\in rationalset\n\\]\nby the product property. Thus every positive rational is in \\( rationalset \\). Now the hypothesis implies that no negative rational is in \\( rationalset \\) and \\( 0 \\notin rationalset \\), so \\( rationalset \\) is just the set of positive rational numbers.\n\nRemarks. The given properties of the set \\( rationalset \\) characterize the set of positive elements in an abstract ordered field. Hence the result can be interpreted as meaning that there is only one way to make the field of rational numbers into an ordered field. See, for example, \\\"Birkhoff and MacLane, \\\"A Survey of Modern Algebra, Marmillan, New York, 1941, page 48.\n\nIf the product property is dropped from the hypothesis, then we can prove that \\( rationalset \\) is either the set of positive rationals or the set of negative rationals. For each positive integer \\( varqint \\), choose \\( epsilonsubq= \\pm 1 \\) so that \\( epsilonsubq / varqint \\in rationalset \\). The sum rule shows that \\( epsilonsubq(varpint / varqint) \\in rationalset \\) for all positive integers \\( varpint \\). Then the choice \\( varpint=varqint \\) gives \\( epsilonsubphi=epsilonsubone \\) and therefore \\( epsilonsubone ratnum \\in rationalset \\) for all positive rationals \\( ratnum \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "mapleleaf",
        "a": "pineapple",
        "b": "watermelon",
        "p": "buttercup",
        "q": "hummingbird",
        "S": "lighthouse",
        "\\\\epsilon_q": "windflower",
        "\\\\epsilon_1": "dragonfly",
        "\\\\epsilon_\\\\varphi": "rainshadow"
      },
      "question": "6. Let \\( lighthouse \\) be a set of rational numbers such that whenever \\( pineapple \\) and \\( watermelon \\) are members of \\( lighthouse \\), so are \\( pineapple+watermelon \\) and \\( pineapple watermelon \\), and having the property that for every rational number \\( mapleleaf \\) exactly one of the following three statements is true:\n\\[\nmapleleaf \\in lighthouse, \\quad-mapleleaf \\in lighthouse, \\quad mapleleaf=0\n\\]\n\nProve that \\( lighthouse \\) is the set of all positive rational numbers.",
      "solution": "Solution. If \\( mapleleaf \\neq 0 \\), then either \\( mapleleaf \\in lighthouse \\) or \\( -mapleleaf \\in lighthouse \\). Since \\( mapleleaf^{2}=(-mapleleaf)^{2} \\), in either case we have \\( mapleleaf^{2} \\in lighthouse \\).\n\nIn particular, \\( 1 \\in lighthouse \\). Then from the sum property it follows that every positive integer is in lighthouse.\nIf \\( buttercup \\) and \\( hummingbird \\) are positive integers, then \\( 1 / hummingbird^{2} \\in lighthouse \\) by our first result, and\n\\[\n\\frac{buttercup}{hummingbird}=buttercup hummingbird\\left(\\frac{1}{hummingbird^{2}}\\right) \\in lighthouse\n\\]\nby the product property. Thus every positive rational is in lighthouse. Now the hypothesis implies that no negative rational is in lighthouse and \\( 0 \\notin lighthouse \\), so \\( lighthouse \\) is just the set of positive rational numbers.\n\nRemarks. The given properties of the set \\( lighthouse \\) characterize the set of positive elements in an abstract ordered field. Hence the result can be interpreted as meaning that there is only one way to make the field of rational numbers into an ordered field. See, for example, \"Birkhoff and MacLane, \"A Survey of Modern Algebra, Marmillan, New York, 1941, page 48.\n\nIf the product property is dropped from the hypothesis, then we can prove that \\( lighthouse \\) is either the set of positive rationals or the set of negative rationals. For each positive integer \\( hummingbird \\), choose \\( windflower= \\pm 1 \\) so that \\( windflower / hummingbird \\in lighthouse \\). The sum rule shows that \\( windflower(buttercup / hummingbird) \\in lighthouse \\) for all positive integers \\( buttercup \\). Then the choice \\( buttercup=hummingbird \\) gives \\( rainshadow=dragonfly \\) and therefore \\( dragonfly mapleleaf \\in lighthouse \\) for all positive rationals \\( mapleleaf \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "irrationalnum",
        "a": "constantvalue",
        "b": "excludedelem",
        "p": "negativenum",
        "q": "numeratornum",
        "S": "negativepool",
        "\\\\epsilon_q": "granddelta",
        "\\\\epsilon_1": "hugeomega",
        "\\\\epsilon_\\\\varphi": "vasttheta"
      },
      "question": "6. Let \\( negativepool \\) be a set of rational numbers such that whenever \\( constantvalue \\) and \\( excludedelem \\) are members of \\( negativepool \\), so are \\( constantvalue+excludedelem \\) and \\( constantvalue excludedelem \\), and having the property that for every rational number \\( irrationalnum \\) exactly one of the following three statements is true:\n\\[\nirrationalnum \\in negativepool, \\quad-irrationalnum \\in negativepool, \\quad irrationalnum=0\n\\]\n\nProve that \\( negativepool \\) is the set of all positive rational numbers.",
      "solution": "Solution. If \\( irrationalnum \\neq 0 \\), then either \\( irrationalnum \\in negativepool \\) or \\( -irrationalnum \\in negativepool \\). Since \\( irrationalnum^{2}=(-irrationalnum)^{2} \\), in either case we have \\( irrationalnum^{2} \\in negativepool \\).\n\nIn particular, \\( 1 \\in negativepool \\). Then from the sum property it follows that every positive integer is in \\( negativepool \\).\nIf \\( negativenum \\) and \\( numeratornum \\) are positive integers, then \\( 1 / numeratornum^{2} \\in negativepool \\) by our first result, and\n\\[\n\\frac{negativenum}{numeratornum}=negativenum\\,numeratornum\\left(\\frac{1}{numeratornum^{2}}\\right) \\in negativepool\n\\]\nby the product property. Thus every positive rational is in \\( negativepool \\). Now the hypothesis implies that no negative rational is in \\( negativepool \\) and \\( 0 \\notin negativepool \\), so \\( negativepool \\) is just the set of positive rational numbers.\n\nRemarks. The given properties of the set \\( negativepool \\) characterize the set of positive elements in an abstract ordered field. Hence the result can be interpreted as meaning that there is only one way to make the field of rational numbers into an ordered field. See, for example, \"Birkhoff and MacLane, \"A Survey of Modern Algebra, Marmillan, New York, 1941, page 48.\n\nIf the product property is dropped from the hypothesis, then we can prove that \\( negativepool \\) is either the set of positive rationals or the set of negative rationals. For each positive integer \\( numeratornum \\), choose \\( granddelta= \\pm 1 \\) so that \\( granddelta / numeratornum \\in negativepool \\). The sum rule shows that \\( granddelta(negativenum / numeratornum) \\in negativepool \\) for all positive integers \\( negativenum \\). Then the choice \\( negativenum=numeratornum \\) gives \\( vasttheta=hugeomega \\) and therefore \\( hugeomega irrationalnum \\in negativepool \\) for all positive rationals \\( irrationalnum \\)."
    },
    "garbled_string": {
      "map": {
        "r": "ykdlzgfa",
        "a": "jgvsychk",
        "b": "qmnplxoe",
        "p": "lwrtaudc",
        "q": "fznhbcji",
        "S": "qzxwvtnp",
        "\\epsilon_q": "sgfktvra",
        "\\epsilon_1": "kmbdwoze",
        "\\epsilon_\\varphi": "vjxqslme"
      },
      "question": "6. Let \\( qzxwvtnp \\) be a set of rational numbers such that whenever \\( jgvsychk \\) and \\( qmnplxoe \\) are members of \\( qzxwvtnp \\), so are \\( jgvsychk+qmnplxoe \\) and \\( jgvsychk qmnplxoe \\), and having the property that for every rational number \\( ykdlzgfa \\) exactly one of the following three statements is true:\n\\[\nykdlzgfa \\in qzxwvtnp, \\quad-ykdlzgfa \\in qzxwvtnp, \\quad ykdlzgfa=0\n\\]\n\nProve that \\( qzxwvtnp \\) is the set of all positive rational numbers.",
      "solution": "Solution. If \\( ykdlzgfa \\neq 0 \\), then either \\( ykdlzgfa \\in qzxwvtnp \\) or \\( -ykdlzgfa \\in qzxwvtnp \\). Since \\( ykdlzgfa^{2}=(-ykdlzgfa)^{2} \\), in either case we have \\( ykdlzgfa^{2} \\in qzxwvtnp \\).\n\nIn particular, \\( 1 \\in qzxwvtnp \\). Then from the sum property it follows that every positive integer is in \\( qzxwvtnp \\).\nIf \\( lwrtaudc \\) and \\( fznhbcji \\) are positive integers, then \\( 1 / fznhbcji^{2} \\in qzxwvtnp \\) by our first result, and\n\\[\n\\frac{lwrtaudc}{fznhbcji}=lwrtaudc\\, fznhbcji\\left(\\frac{1}{fznhbcji^{2}}\\right) \\in qzxwvtnp\n\\]\nby the product property. Thus every positive rational is in \\( qzxwvtnp \\). Now the hypothesis implies that no negative rational is in \\( qzxwvtnp \\) and \\( 0 \\notin qzxwvtnp \\), so \\( qzxwvtnp \\) is just the set of positive rational numbers.\n\nRemarks. The given properties of the set \\( qzxwvtnp \\) characterize the set of positive elements in an abstract ordered field. Hence the result can be interpreted as meaning that there is only one way to make the field of rational numbers into an ordered field. See, for example, \"Birkhoff and MacLane, \"A Survey of Modern Algebra, Marmillan, New York, 1941, page 48.\n\nIf the product property is dropped from the hypothesis, then we can prove that \\( qzxwvtnp \\) is either the set of positive rationals or the set of negative rationals. For each positive integer \\( fznhbcji \\), choose \\( sgfktvra= \\pm 1 \\) so that \\( sgfktvra / fznhbcji \\in qzxwvtnp \\). The sum rule shows that \\( sgfktvra(lwrtaudc / fznhbcji) \\in qzxwvtnp \\) for all positive integers \\( lwrtaudc \\). Then the choice \\( lwrtaudc=fznhbcji \\) gives \\( vjxqslme=kmbdwoze \\) and therefore \\( kmbdwoze\\, ykdlzgfa \\in qzxwvtnp \\) for all positive rationals \\( ykdlzgfa \\)."
    },
    "kernel_variant": {
      "question": "Let $K/\\mathbb{Q}$ be a totally real Galois number field of degree $d\\ge 2$ and fix once and for all an embedding  \n\\[\n\\sigma_{1}\\colon K\\hookrightarrow\\mathbb{R}.\n\\]\nFor every $\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q})$ put  \n\\[\n\\sigma_{\\tau}:=\\sigma_{1}\\circ \\tau\\colon K\\longrightarrow\\mathbb{R}.\n\\]\nBecause $K/\\mathbb{Q}$ is Galois,  \n\\[\n\\Sigma:=\\{\\sigma_{\\tau}\\}_{\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q})}\n\\]\nis the complete set of $d$ real embeddings of $K$.\n\nFor a subset $S\\subseteq K$ write $-S:=\\{-\\alpha\\colon \\alpha\\in S\\}$.  \nA subset $P\\subset K$ is called a {\\em positive cone} (or {\\em ordering cone}) in $K$ if  \n\n(OC1) $\\alpha,\\beta\\in P\\Longrightarrow\\alpha+\\beta,\\ \\alpha\\beta\\in P;$  \n\n(OC2) (Trichotomy) for every $\\alpha\\in K$ exactly one of  \n\\[\n\\alpha=0,\\qquad \\alpha\\in P,\\qquad \\alpha\\in -P\n\\]\nholds.\n\nWrite  \n\\[\nK^{++}:=\\{\\alpha\\in K\\colon \\sigma(\\alpha)>0\\text{ for all }\\sigma\\in\\Sigma\\}\n\\]\nfor the set of totally positive elements of $K$.\n\n(a) Show that there exists no positive cone $P\\subset K$ that is invariant under $\\operatorname{Gal}(K/\\mathbb{Q})$, i.e.  \n\\[\n\\tau(P)=P\\quad\\forall\\,\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q}).\n\\]\nDeduce in particular that $K^{++}$ itself is not a positive cone.\n\n(b) For $\\sigma\\in\\Sigma$ define  \n\\[\nK^{+}_{\\sigma}:=\\{\\alpha\\in K\\colon \\sigma(\\alpha)>0\\}.\n\\]\nProve that every $K^{+}_{\\sigma}$ is a positive cone and that every ordering of $K$ is of the form $K^{+}_{\\sigma}$ for a unique $\\sigma\\in\\Sigma$.  Conclude that $K$ has exactly $d$ distinct orderings.\n\n(c) Prove that  \n\\[\nK^{++}=\\bigcap_{\\sigma\\in\\Sigma}K^{+}_{\\sigma}.\n\\]\nIn other words, the set of totally positive elements is precisely the intersection of all positive cones on $K$ even though it is not itself an ordering cone.",
      "solution": "Throughout the usual order on $\\mathbb{R}$ is denoted by “$>$”.\n\n(b)  Classification of the orderings (proved first and afterwards used in part (a))\n\nStep 1.  Each $K^{+}_{\\sigma}$ is a positive cone.  \nBecause $\\sigma\\colon K\\to\\mathbb{R}$ is a field homomorphism, (OC1) and (OC2) are inherited from $\\mathbb{R}$.  Concretely, if $\\alpha,\\beta\\in K^{+}_{\\sigma}$ then $\\sigma(\\alpha),\\sigma(\\beta)>0$, hence $\\sigma(\\alpha+\\beta),\\sigma(\\alpha\\beta)>0$ and therefore $\\alpha+\\beta,\\alpha\\beta\\in K^{+}_{\\sigma}$.  Trichotomy follows from the trichotomy law in $\\mathbb{R}$ applied to $\\sigma(\\alpha)$.\n\nStep 2.  Every ordering of $K$ is obtained in this way.  \n\nLet $\\le$ be an ordering of $K$ and let $P_{\\le}$ be its positive cone.  \nDenote by $K^{\\mathrm{rc}}$ the real closure of the ordered field $(K,\\le)$.  Because $K$ is a number field, the extension $K^{\\mathrm{rc}}/K$ is algebraic; hence $K^{\\mathrm{rc}}$ is algebraic over $\\mathbb{Q}$.\n\nLemma 1.  Let $F$ be a real closed field that is algebraic over $\\mathbb{Q}$.  \nThen there exists a {\\em unique} order–preserving field embedding  \n\\[\n\\iota\\colon F\\hookrightarrow\\mathbb{R}.\n\\]\n\nProof of Lemma 1 (existence and uniqueness).  \n\nExistence.  \nSince $F/\\mathbb{Q}$ is algebraic, $F$ is of countable transcendence degree $0$ over $\\mathbb{Q}$, hence in particular {\\em archimedean}.  (One way to see this is to embed $\\mathbb{Q}$ into $F$; because $F$ is real closed and algebraic, no element of $F$ can be infinitely large with respect to $\\mathbb{Q}$.)  \nBy Hölder’s theorem an archimedean ordered field admits a unique order–preserving embedding into the real numbers.  Thus an embedding $\\iota\\colon F\\hookrightarrow\\mathbb{R}$ exists.\n\nUniqueness.  \nIf $\\jmath\\colon F\\hookrightarrow\\mathbb{R}$ is any other order–preserving embedding, then $\\jmath(F)$, being algebraic over $\\mathbb{Q}$, lies in the real algebraic numbers  \n\\[\n\\mathbb{R}^{\\operatorname{alg}}:=\\mathbb{Q}^{\\operatorname{alg}}\\cap\\mathbb{R}\\subset\\mathbb{R}.\n\\] \nBoth $\\iota$ and $\\jmath$ therefore land in $\\mathbb{R}^{\\operatorname{alg}}$.  The restrictions  \n\\[\n\\iota,\\jmath\\colon F\\longrightarrow\\mathbb{R}^{\\operatorname{alg}}\n\\]\nare isomorphisms of real closed fields that are the identity on $\\mathbb{Q}$.  \nBecause a real closed field possesses a {\\em unique} ordering, any order–preserving automorphism of $\\mathbb{R}^{\\operatorname{alg}}$ fixing $\\mathbb{Q}$ must be the identity.  Hence $\\jmath=\\iota$, proving uniqueness. \\qed\n\nApplying Lemma 1 with $F=K^{\\mathrm{rc}}$ yields an order–preserving embedding  \n\\[\n\\iota\\colon K^{\\mathrm{rc}}\\hookrightarrow\\mathbb{R}.\n\\]\nRestricting $\\iota$ to $K$ gives a field embedding $\\sigma:=\\iota|_{K}\\colon K\\longrightarrow\\mathbb{R}$.  Since $K$ is totally real, $\\sigma$ must equal $\\sigma_{\\tau}$ for a {\\em unique} $\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q})$.  Finally, for $\\alpha\\in K$ we have  \n\\[\n0<_{\\le}\\alpha\\;\\Longleftrightarrow\\;0<\\iota(\\alpha)=\\sigma(\\alpha),\n\\]\nso $P_{\\le}=K^{+}_{\\sigma}$.\n\nStep 3.  Distinct embeddings yield distinct cones.  \nSuppose $K^{+}_{\\sigma_{\\tau_{1}}}=K^{+}_{\\sigma_{\\tau_{2}}}$ with $\\tau_{1}\\neq\\tau_{2}$.  Choose $\\theta\\in K$ with $\\sigma_{\\tau_{1}}(\\theta)\\neq\\sigma_{\\tau_{2}}(\\theta)$ and pick $r\\in\\mathbb{Q}$ strictly between the two conjugates.  Then  \n\\[\n\\sigma_{\\tau_{1}}(\\theta-r)>0,\\qquad \\sigma_{\\tau_{2}}(\\theta-r)<0,\n\\]\ncontradicting the equality of the cones.  Hence $\\tau_{1}=\\tau_{2}$.\n\nSteps 1–3 establish a bijection  \n\\[\n\\Sigma\\longrightarrow\\{\\text{orderings of }K\\},\\qquad\n\\sigma\\longmapsto K^{+}_{\\sigma},\n\\]\nso $K$ possesses exactly $d$ distinct orderings.\n\n(a)  Non–existence of a Galois–stable ordering  \n\nAssume, to the contrary, that a positive cone $P\\subset K$ satisfies $\\tau(P)=P$ for every $\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q})$.  \nBy part (b) there is a unique $\\sigma\\in\\Sigma$ such that $P=K^{+}_{\\sigma}$.  \nFor any $\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q})$ we compute\n\\[\nP=\\tau(P)=\\tau\\bigl(K^{+}_{\\sigma}\\bigr)=K^{+}_{\\sigma\\circ\\tau^{-1}}.\n\\]\nUniqueness of the representing embedding forces\n\\[\n\\sigma=\\sigma\\circ\\tau^{-1}\\qquad\\forall\\tau\\in\\operatorname{Gal}(K/\\mathbb{Q}).\n\\]\nPrecomposing by $\\sigma_{1}^{-1}$ gives $\\tau^{-1}=\\operatorname{id}_{K}$ and consequently $\\tau=\\operatorname{id}_{K}$.  Therefore the Galois group is trivial, contradicting $d\\ge 2$.  No such $P$ can exist.  \n\nSince $K^{++}$ is clearly Galois–stable (by definition it is fixed by every automorphism), the previous paragraph implies that $K^{++}$ is not a positive cone: it violates trichotomy (OC2).\n\n(c) Intersection of all positive cones  \n\nThe inclusion $K^{++}\\subseteq\\bigcap_{\\sigma\\in\\Sigma}K^{+}_{\\sigma}$ is immediate from the definitions.  \nConversely, if $\\alpha$ lies in the intersection, then $\\sigma(\\alpha)>0$ for every $\\sigma\\in\\Sigma$, i.e. $\\alpha\\in K^{++}$.  Hence  \n\\[\nK^{++}=\\bigcap_{\\sigma\\in\\Sigma}K^{+}_{\\sigma}.\n\\]\nAlthough this intersection is strictly smaller than any individual ordering cone, part (a) shows that it cannot itself be an ordering cone.  \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.538219",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional structure:  The problem moves from ℚ to an arbitrary totally real number field K of degree d, introducing d conjugates and algebraic tools (norms, traces, embeddings).  \n• Additional axioms:  Beyond additive and multiplicative closure and trichotomy, one must handle inverses, stability under all embeddings, norm compatibility, and an Archimedean density condition.  \n• Deeper theory invoked:  The solution uses concepts from algebraic number theory (embeddings, totally positive elements, norms), quadratic form theory (positive–definite trace form, Gram/Cholesky decomposition), and Dirichlet–style density arguments.  \n• Multi-step reasoning:  \n  1. Deduce basic consequences (1∈S, positive rationals in S).  \n  2. Show every element of S is totally positive using conjugate stability.  \n  3. Prove a non-trivial representation theorem (every totally positive α is a sum of squares) via linear-algebraic arguments over number fields.  \n  4. Leverage that representation together with closure properties to force K^{++}⊂S.  \n• Non-trivial lemmas:  The proof of Lemma 2 (sum-of-squares representation) requires familiarity with positive-definite quadratic forms and their matrix decompositions—substantially more sophisticated than the elementary squaring trick of the original problem.  \n• Broader conceptual payoff:  The result classifies the positive cone in any totally real number field, showing that the original uniqueness phenomenon for ℚ is but the first case of a much richer algebraic fact."
      }
    },
    "original_kernel_variant": {
      "question": "Let \\(K/\\mathbb{Q}\\) be a totally–real Galois number field of degree \\(d\\ge 2\\).  \nFix once and for all an embedding  \n\n  \\(\\sigma_{1}:K\\hookrightarrow\\mathbb{R}\\).\n\nFor every automorphism \\(\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q})\\) put  \n\n  \\(\\sigma_{\\tau}:=\\sigma_{1}\\circ \\tau :K\\longrightarrow\\mathbb{R}.\\)\n\nBecause \\(K/\\mathbb{Q}\\) is Galois, the map  \n\\[\n\\mathrm{Gal}(K/\\mathbb{Q})\\longrightarrow\\{\\text{real embeddings of }K\\},\\qquad \n\\tau\\longmapsto\\sigma_{\\tau}\n\\]\nis a bijection, so the family  \n\n  \\(\\Sigma=\\{\\sigma_{\\tau}\\}_{\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q})}\\)\n\nis precisely the set of all \\(d\\) real embeddings of \\(K\\).\n\nFor a subset \\(S\\subseteq K\\) write \\(-S:=\\{-\\alpha:\\alpha\\in S\\}\\).  \nA subset \\(S\\subset K\\) is called a \\(\\sigma_{1}\\)-positive cone if it satisfies\n\n(P1) (σ₁–positivity) \\(\\alpha\\in S\\ \\Longrightarrow\\ \\sigma_{1}(\\alpha)>0\\);\n\n(P2) (Trichotomy) For every \\(\\alpha\\in K\\) exactly one of  \n  \\(\\alpha=0,\\quad \\alpha\\in S,\\quad \\alpha\\in -S\\)  \n holds;\n\n(P3) (Galois stability) For every \\(\\alpha\\in S\\) and every \\(\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q})\\) one has \\(\\tau(\\alpha)\\in S\\).\n\n(i)  Prove that a \\(\\sigma_{1}\\)-positive cone is unique and equals  \n\n  \\(K^{++}:=\\{\\alpha\\in K:\\sigma(\\alpha)>0\\ \\text{for all }\\sigma\\in\\Sigma\\}.\\)\n\n(ii)  Show that \\(K\\) possesses exactly \\(d\\) orderings and that every ordering of \\(K\\) is obtained by pulling back the standard ordering of \\(\\mathbb{R}\\) through exactly one of the embeddings \\(\\sigma_{\\tau}\\ (\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q}))\\).\n\n—End of problem—",
      "solution": "Throughout “\\({>}\\)” denotes the usual order on \\(\\mathbb{R}\\).\n\n(i)  Uniqueness of the \\(\\sigma_{1}\\)-positive cone\n\nStep 1. \\(S\\subseteq K^{++}\\).  \nLet \\(\\alpha\\in S\\) and choose an arbitrary real embedding \\(\\sigma\\in\\Sigma\\); thus \\(\\sigma=\\sigma_{1}\\circ\\tau\\) for a unique \\(\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q})\\).  \nBy (P3) we have \\(\\tau(\\alpha)\\in S\\), whence (P1) yields  \n\\[\n\\sigma_{1}\\bigl(\\tau(\\alpha)\\bigr)>0.\n\\]\nBut \\(\\sigma_{1}\\circ\\tau=\\sigma\\), so \\(\\sigma(\\alpha)>0\\).  As \\(\\sigma\\) was arbitrary, every conjugate of \\(\\alpha\\) is positive, i.e. \\(\\alpha\\in K^{++}\\).  Therefore \\(S\\subseteq K^{++}\\).\n\nStep 2. \\(K^{++}\\subseteq S\\).  \nLet \\(\\alpha\\in K^{++}\\).  In particular \\(\\sigma_{1}(\\alpha)>0\\).  If \\(\\alpha\\notin S\\), then by (P2) we must have \\(-\\alpha\\in S\\); but (P1) would then give \\(\\sigma_{1}(-\\alpha)>0\\), contradicting \\(\\sigma_{1}(-\\alpha)=-\\sigma_{1}(\\alpha)<0\\).  Hence \\(\\alpha\\in S\\) and \\(K^{++}\\subseteq S\\).\n\nCombining Steps 1 and 2 gives \\(S=K^{++}\\), proving part (i).\n\n(ii)  Classification of all orderings of \\(K\\)\n\nLet \\(\\le\\) be an arbitrary ordering of \\(K\\) and denote its positive cone by  \n\\[\nP_{\\le}:=\\{\\alpha\\in K:0<_{\\le}\\alpha\\}.\n\\]\n\nStep 3. Embedding of the real closure into \\(\\mathbb{R}\\).  \nLet \\(R\\) be the real closure of \\((K,\\le)\\).  The extension \\(K\\hookrightarrow R\\) is algebraic, so every element of \\(R\\) is real algebraic.  By the Artin–Schreier theorem there exists an order–preserving embedding  \n\\[\n\\iota:R\\hookrightarrow\\mathbb{R}.\n\\]\n\nStep 4. Obtaining a real embedding of \\(K\\).  \nRestricting \\(\\iota\\) to \\(K\\) gives a field embedding  \n\\[\n\\sigma:=\\iota|_{K}:K\\longrightarrow\\mathbb{R}.\n\\]\nBecause \\(K\\) is totally real, \\(\\sigma=\\sigma_{\\tau}\\) for a unique \\(\\tau\\in\\mathrm{Gal}(K/\\mathbb{Q})\\).\n\nStep 5. Identifying the positive cone.  \nIf \\(\\alpha\\in P_{\\le}\\) then \\(0<_{\\le}\\alpha\\), hence \\(0<\\iota(\\alpha)=\\sigma(\\alpha)\\); thus  \n\\(P_{\\le}\\subseteq K^{++}_{\\sigma_{\\tau}}\\), where  \n\\[\nK^{++}_{\\sigma_{\\tau}}:=\\{\\alpha\\in K:\\sigma_{\\tau}(\\alpha)>0\\}.\n\\]\nConversely, if \\(\\beta\\in K^{++}_{\\sigma_{\\tau}}\\) then \\(\\sigma_{\\tau}(\\beta)>0\\), so \\(0<_{\\le}\\beta\\).  Hence  \n\\(K^{++}_{\\sigma_{\\tau}}\\subseteq P_{\\le}\\), and therefore  \n\\[\nP_{\\le}=K^{++}_{\\sigma_{\\tau}}.\n\\]\n\nStep 6. Distinct embeddings give distinct orderings.  \nAssume \\(K^{++}_{\\sigma_{\\tau_{1}}}=K^{++}_{\\sigma_{\\tau_{2}}}\\) with \\(\\tau_{1}\\neq\\tau_{2}\\).  \nChoose \\(\\theta\\in K\\) such that \\(\\sigma_{\\tau_{1}}(\\theta)\\neq\\sigma_{\\tau_{2}}(\\theta)\\) (possible because the two embeddings are distinct).  Without loss of generality let  \n\n  \\(\\sigma_{\\tau_{2}}(\\theta)<\\sigma_{\\tau_{1}}(\\theta)\\).\n\nBy the density of \\(\\mathbb{Q}\\) in \\(\\mathbb{R}\\) there exists a rational number \\(r\\) satisfying  \n\n  \\(\\sigma_{\\tau_{2}}(\\theta)<r<\\sigma_{\\tau_{1}}(\\theta).\\)\n\nPut \\(\\beta:=\\theta-r\\in K\\).  Then  \n\\[\n\\sigma_{\\tau_{1}}(\\beta)=\\sigma_{\\tau_{1}}(\\theta)-r>0,\\qquad\n\\sigma_{\\tau_{2}}(\\beta)=\\sigma_{\\tau_{2}}(\\theta)-r<0.\n\\]\nConsequently \\(\\beta\\in K^{++}_{\\sigma_{\\tau_{1}}}\\setminus K^{++}_{\\sigma_{\\tau_{2}}}\\), contradicting the assumed equality of the two cones.  Hence \\(\\tau_{1}=\\tau_{2}\\).\n\nStep 7. Counting orderings.  \nThe map  \n\\[\n\\{\\text{orderings of }K\\}\\;\\longrightarrow\\;\\Sigma,\\qquad\n\\le\\;\\longmapsto\\;\\sigma_{\\tau}\\quad\\text{with}\\quad P_{\\le}=K^{++}_{\\sigma_{\\tau}}\n\\]\nis therefore bijective.  Hence \\(K\\) has exactly \\(d\\) orderings, each obtained by pulling back the standard order on \\(\\mathbb{R}\\) via a unique real embedding \\(\\sigma_{\\tau}\\).\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.447682",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional structure:  The problem moves from ℚ to an arbitrary totally real number field K of degree d, introducing d conjugates and algebraic tools (norms, traces, embeddings).  \n• Additional axioms:  Beyond additive and multiplicative closure and trichotomy, one must handle inverses, stability under all embeddings, norm compatibility, and an Archimedean density condition.  \n• Deeper theory invoked:  The solution uses concepts from algebraic number theory (embeddings, totally positive elements, norms), quadratic form theory (positive–definite trace form, Gram/Cholesky decomposition), and Dirichlet–style density arguments.  \n• Multi-step reasoning:  \n  1. Deduce basic consequences (1∈S, positive rationals in S).  \n  2. Show every element of S is totally positive using conjugate stability.  \n  3. Prove a non-trivial representation theorem (every totally positive α is a sum of squares) via linear-algebraic arguments over number fields.  \n  4. Leverage that representation together with closure properties to force K^{++}⊂S.  \n• Non-trivial lemmas:  The proof of Lemma 2 (sum-of-squares representation) requires familiarity with positive-definite quadratic forms and their matrix decompositions—substantially more sophisticated than the elementary squaring trick of the original problem.  \n• Broader conceptual payoff:  The result classifies the positive cone in any totally real number field, showing that the original uniqueness phenomenon for ℚ is but the first case of a much richer algebraic fact."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}