# TRIPLE — what a cubic field's split primes carry The object: a complex cubic field K of discriminant d_K, its class group Cl of order h, and the classes of its degree-1 places. A totally split prime p = P₁P₂P₃ carries a triple of classes, and since P₁P₂P₃ = (p) the triple lies in M = {(a, b, c) ∈ Cl³ : a + b + c = 0}, of order h². A partially split prime carries one degree-1 class. The questions: which triples occur, which class the partial place takes, and how the shortfall of principal places at a finite cut, which over an imaginary quadratic field the prime-ideal powers account for (PRINCIPAL.md), behaves one degree up. The engine is cubic.py; the verifiers are triple.py, conductor.py, shortfall.py and seat.py. cubic.py, in pure Python, takes a complex cubic field to its maximal order from a binary cubic form (Delone and Faddeev), the places over any prime, the class group and the class of a degree-1 place, by a bounded search that placed every place these runs needed. Its class group is checked against the analytic class number formula: the relations give h and the regulator Reg up to one integer factor m, and h·Reg over the formula's value, its Euler product taken to 3·10⁴, reads 0.9962 to 1.0027 at all 8336 complex cubic fields with |d_K| ≤ 50000, where m ≥ 2 would put it near 2 or above. The truncated product carries no proved error bound, so this is numerical evidence and not a proof (triple.py::section_controls). The enumeration is checked against an independent count of the complex fields to |d_K| ≤ 6000 (888) and by Hasse's count at conductor 1. An independent record of the complex fields with h > 1 to |d_K| ≤ 50000 holds 3133 where this one finds 3134: the extra field, d_K = −14087 with h = 2 here, is the one that record leaves unresolved, and the other 3133 class groups agree with it, by a comparison run outside triple.py and entered in its findings. The answer, in two lines: the triples fill a subgroup R of M whose index is 3^t, t counting the primes q ≡ 1 mod 3 that divide the conductor f of d_K = f²d₀ (d₀ fundamental), plus one when 9 | f and 3 splits in Q(√d_K) (TRIPLE.md#the-genus-index-law; proved at 3-rank at most 1 and wherever t = 0 or t is the 3-rank; verified to |d_K| ≤ 50000), so the triples fail to fill M exactly at the fields the discriminant names; the partial place is uniform on Cl whatever R is. At a finite cut the principal places run short as they do over an imaginary quadratic field, but at degree 3 the prime-ideal powers account for only about a third of it at cut 1000 (an observation); the rest reads as a fixed weight missing at the primes below a few hundred (a pattern), carried mostly by the partial places at that cut (an observation), to which the explicit formula gives no square term; and at 2-rank 1 the least odd partial place lies outside 2Cl far more often than half (an observation). ## The triple is a homomorphism Tier: property. Verifier: proof; triple.py::section_controls. Let N be the Galois closure of K, H the Hilbert class field of K and H̃ the Galois closure of H over Q. A prime p splits totally in K exactly when its Frobenius in Gal(H̃/Q) lies in Gal(H̃/N). With K₁, K₂, K₃ the conjugates of K in N and H₁, H₂, H₃ the matching conjugates of H, restriction to Hᵢ is a homomorphism Gal(H̃/N) → Gal(Hᵢ/Kᵢ) = Cl, and with 𝔓 the prime of H̃ over p whose Frobenius is read, Pᵢ = 𝔓 ∩ Kᵢ, so by Artin the triple is the image of Frobenius under a homomorphism into M. Chebotarev fills Gal(H̃/N), so the realized set R is a subgroup of M; conjugation permutes the copies, so R is stable under S₃ acting on coordinates; each coordinate map is onto, since H₁ ∩ N = K₁ (TRIPLE.md#the-partial-place-is-uniform), so every projection of R is onto Cl. The closure [P₁] + [P₂] + [P₃] = 0 holds at every totally split prime the script classes, the 37608 triples it keeps among them; where each place is classed on its own, above the primes whose places generate the relations, this tests the class map, and below them the relation (p) forces it. ## The stable subgroups Tier: theorem. Verifier: proof; triple.py::section_groups. An S₃-stable subgroup R of M with every projection onto Cl contains M ∩ (3Cl)³, and is the preimage of W ⊗ St + V ⊗ D for a subspace W of V = Cl/3Cl, where St is the sum-zero permutation module of S₃ over F₃ and D its diagonal. Its index in M is 3^(r − dim W), r the 3-rank. Proof. For (a, b, c) ∈ R, the transposition (12) gives (b, a, c) ∈ R, so (a − b, b − a, 0) ∈ R; permuting, (a − c, c − a, 0) ∈ R; the two sum to (3a, −3a, 0), since b + c = −a. As a runs over Cl, R holds every (3a, −3a, 0), and with its permutations all of M ∩ (3Cl)³. So R is the preimage of a stable subspace of the sum-zero triples of V, which is V ⊗ St. In characteristic 3 the module St is uniserial: its only stable line is D = {(λ, λ, λ) : λ ∈ F₃}, since λ + λ + λ = 0, its top St/D carries the sign, and for a 3-cycle γ the map γ − 1 kills D and carries St/D isomorphically onto D. For a stable subspace U let V₀ ⊆ V be its intersection with V ⊗ D and W ⊆ V its image in V ⊗ (St/D); γ − 1 maps U into itself, so W ⊆ V₀. Write u ∈ U as x⊗(1, −1, 0) + y⊗(1, 1, 1); then u + (12)u = 2y⊗(1, 1, 1) lies in U, so y ∈ V₀ and the first coordinate x + y lies in V₀ + W = V₀, so a U projecting onto V has V₀ = V: it contains V ⊗ D and is W ⊗ St + V ⊗ D, of codimension r − dim W. When 3 ∤ h, V = 0 and R is all of M. At 3-rank 1 it is M or Δ = {a ≡ b ≡ c mod 3Cl}, of index 3; at h = 3 the candidates 0, D and M are the smallest case, 0 excluded by the projection. Computably Δ is a lattice membership (with Cl = Zⁿ/Λ, 3Cl is (Λ + 3Zⁿ)/Λ); the scalar test (h/3)(a − b) = 0 agrees with it only where the 3-part is cyclic, and above 3-rank 1 it passes every triple, h/3 being then a multiple of Cl's exponent. Over thirteen groups with h ≤ 18 the stable subgroups projecting onto Cl number exactly the subspaces of F₃^r, graded by index as the theorem says, each containing M ∩ (3Cl)³. ## The genus index law Tier: rule (verified at every complex cubic field with |d_K| ≤ 50000, given cubic.py's class groups). Verifier: triple.py::section_genus. Write d_K = f²d₀ with d₀ the fundamental discriminant of the resolvent k = Q(√d_K) and f the conductor. Let t be the number of primes q ≡ 1 (mod 3) dividing f, plus one when 9 | f and 3 splits in k. Then [M : R] = 3^t, and R is the triples agreeing modulo the subgroup G of Cl cut out by the fields below. The argument has two halves. Below: each such q, or the 9, names a cyclic cubic field F of conductor q or 9 with KF/K unramified. At a q ≡ 1 (mod 3) this is Abhyankar's lemma: q splits in k, since an inert or ramified q ≡ 1 offers χ, the cubic character of k cutting out N, only pieces fixed by the conjugation σ of k/Q (TRIPLE.md#the-conductors-primes-are-forced), and, χ being tame at q, q ∥ f, so v_q(d_K) = 2, and q is totally and tamely ramified in K, F is tamely ramified at q alone with index 3, and F is real; at the 9, where the ramification is wild, it is local class field theory at 3: 3 splits in k and N/k ramifies over it, so 3 = 𝔮³ in K with K_𝔮 a ramified cyclic cubic extension of Q₃; the cubic characters of Z₃* form a group of order 3, read on (1 + 3Z₃)/(1 + 9Z₃), so F's character at 3 is K_𝔮's or its inverse times an unramified one, and KF/K is unramified at 𝔮. The t fields are abelian over Q with coprime conductors and K is not Galois, so their compositum meets K in Q and the t fields together give an unramified extension of K of degree 3^t, cutting out a subgroup G of index 3^t in Cl. F is Galois over Q, so a Frobenius restricts to F identically from all three copies, and R lies in {a ≡ b ≡ c mod G}, of index 3^t in M. Above: a further quotient Cl → Z/3 on which every triple agrees gives a cubic character of k that must be fixed by the conjugation of k/Q and unramified over N, and such a line needs a split conductor prime (TRIPLE.md#the-sign-of-the-second-line-is-the-regime). That the index is then at most 3^t is proved at 3-rank 1; where t is the 3-rank r, since the stable subgroups give index at most 3^r; and at t = 0 at every 3-rank: under each quotient Cl → Z/3 the triples' image is D or St, and D would give a σ-fixed cubic character of k whose field is unramified over N, which needs a split conductor prime, so every quotient reads St and R = M. At 3-rank 2 or more with 1 ≤ t < r the count below is what the rule rests on, at the one such field in range. So Cl/G ≅ (Z/3)^t, the Galois group of the t fields' compositum over K, 3^t divides h and t ≤ r, which holds at all 8336 fields. At every field with h > 1 — 3134, each read through ten or more split primes — the index of the subgroup the triples and their permutations generate is exactly 3^t: 1767 fields with 3 ∤ h, 1018 at 3-rank 1 with t = 0, 347 at 3-rank 1 with t = 1, and the 2 of 3-rank 2 below. The generated subgroup lies in R, R being stable, so its index bounds [M : R] from above, and the below half bounds it from below, so at every such field the count settles the index given cubic.py's class groups, with no appeal to chance. At h = 3 and |d_K| ≤ 6000 the 83 fields split 38 at index 3 and 45 at index 1. The two fields of 3-rank 2, |d_K| = 24843 (f = 91 = 7·13, t = 2) and 47628 (f = 126 = 2·3²·7, t = 1), read index 9 and index 3, the first proved and the second the one field with 1 ≤ t < r. A uniform Frobenius on R prices any event at its size over |R| = h²/3^t. ## The conductor's primes are forced Tier: rule (verified at every complex cubic field with |d_K| ≤ 50000; the sketch below covers the primes other than 3). Verifier: conductor.py::section_admissible; conductor.py::section_ranks. A prime q ≡ 1 (mod 3) divides f only when it splits in k, a prime q ≡ 2 (mod 3) only when it is inert, v₃(f) ≤ 2, and v₃(f) = 1 only where 3 | d₀. The cubic character χ of k cutting out N has conductor (f) and is inverted by the conjugation σ (its conductor is Hasse's: d_K = f²d₀ for a cubic field whose Galois closure is N), so at each prime of k over f that σ fixes its local part needs a piece of the local units' 3-part that σ inverts, and at a pair of primes that σ swaps a nontrivial 3-part: an inert q ≡ 1 or a ramified q ≠ 3 offers only fixed pieces (at an inert q the 3-part of F_(q²)* lies in F_q*, as 3 ∤ q + 1, and at a ramified prime the residue field is F_q, both fixed by σ, the higher units being pro-q), a split q ≡ 2 none at all. The 3-part of (O_k/f)* then splits under σ prime by prime — one fixed and one inverted cyclic factor at a split q ≡ 1 or at 9 | f with 3 split or inert, one inverted at an inert q ≡ 2 or at 3 ∥ f, one fixed and two inverted at 9 | f with 3 ramified — and the engine reads these ranks at all 2468 fields with f > 1. So the conductor is not free data: which primes may divide it is fixed by their residues mod 3 and how they split in k. ## The sign of the second line is the regime Tier: theorem (3-rank 1). Verifier: proof; conductor.py::section_ranks; conductor.py::section_sign. At 3-rank 1 the triples read mod 3Cl cut out, over N, a field whose largest abelian piece over k is an extension E of type (3, 3) containing N and unramified over N. In Gal(E/k), of type (3, 3), σ inverts the quotient Gal(N/k), and the **second line** is the kernel Gal(E/N); σ fixes the second line when R = Δ and inverts it when R = M, and these are the two regimes. A cubic character of k fixed by σ is usable when its field, joined to N, is unramified over N (said below as its field being unramified over N); a usable one exists only when t ≥ 1, so the index law's above half holds at 3-rank 1 by proof; when t ≥ 1 the genus law's below fields give one. Proof. Since σ inverts Cl(k), the σ-fixed 3-part of the ray class group mod f comes from the local units (O_k/f)*. E/N unramified at a prime 𝔭 | f asks that each character of Gal(E/k) restrict on the local units into ⟨χ_𝔭⟩, χ_𝔭 the local part of χ at 𝔭; where σ fixes 𝔭 that group holds no nontrivial fixed character, and where σ swaps 𝔭 with its conjugate the local cubic characters form a line spanned by χ_𝔭. So a usable fixed line needs a split conductor prime with nontrivial local 3-part, that is t ≥ 1. If the triples' image under a quotient φ: Cl → Z/3 is D, as at 3-rank 1 when R = Δ and as the genus law uses at t = 0, let H′ be the cubic unramified extension of K that φ cuts out and J = H′₁N, which at 3-rank 1 is E, of degree 9 over k as H₁ ∩ N = K₁ (TRIPLE.md#the-partial-place-is-uniform). The three coordinates of a triple agree modulo ker φ, so H′₁N = H′₂N = H′₃N and J is Galois over Q. The transposition fixing K₁ restricts to σ on k, and any lift acts on H′₁ as an element of the abelian Gal(H′₁/K₁), so σ acts trivially on Gal(J/N) ≅ Gal(H′₁/K₁); σ inverts Gal(N/k), and 2 being invertible on a group of order 9 the two parts each have order 3: Gal(J/N), at 3-rank 1 the second line, is fixed, so the fixed character of k vanishing on the inverted part is usable, and t ≥ 1. If the image under φ is the sum-zero module St, as at 3-rank 1 when R = M, its field L = H′₁H′₂H′₃N is unramified over N and Galois over Q, and Γ = Gal(L/k) has order 27 and is not cyclic. Its commutators fill the diagonal and Γ modulo it is abelian, so by Burnside Γ^ab is (Z/3)², and σ acts on its line from St by the sign: inverted. Existence alone does not sort. A σ-fixed cubic character of k exists at 9 | f whatever 3 does in k, and a fixed line exists with none usable at exactly the 150 fields with 9 | f, 3 not split and no q ≡ 1 dividing f, 39 of them at 3-rank 1, where t = 0 makes R = M. At all 1018 fields of 3-rank 1 with t = 0 the ray class group mod f has a σ-inverted line beyond χ's, checked through sufficient conditions for its σ-inverted part to have 3-rank at least 2, and the second line of E is inverted, E being unramified over N, by the proof above, t = 0 forcing R = M. At t = 1 the same sufficient conditions give an inverted line of the ray class group mod f beyond χ's at 121 of the 347 (at d₀ = −3 they subtract one for the cube roots of unity, a lower bound), beside the usable fixed one there, its field unramified over N wherever, at every σ-fixed conductor prime, the 3-part of the local units has a σ-inverted part of 3-rank at most 1, which holds at all but 23 fields of 3-rank 1 (the least |d_K| among them 11907); at the 7 of those 23 with t = 1 it is left undecided. So neither which fixed characters exist, usability aside, nor which inverted ones exist decides R; usability does, and with it the sign σ puts on the second line of the E the triples cut out. ## Gerth's rank law at conductor 1 Tier: known. Source: F. Gerth III, Ranks of 3-class groups of non-Galois cubic fields, Acta Arith. 30 (1976) 307–322, Theorem 3.4. If f = 1, so that N/k is unramified, then rk₃ Cl(K) = rk₃ Cl(k) − 1: 3 divides h exactly when the resolvent has 3-rank at least 2. The triples prove one direction: if f = 1 and 3 | h, the resolvent has 3-rank at least 2. With no conductor prime, take any quotient Cl → Z/3: the triples' image in the sum-zero triples of Z/3 is D or St. D would give, as in the block above, a fixed unramified cubic character of k, which σ forbids; so it is St, and the extension of type (3, 3) built there is unramified over k. The engines read the law at the ranks that occur: over the 5868 fields with f = 1 and |d_K| ≤ 50000, cubic.py's class group and the resolvent's class group, read as binary quadratic forms by principal.py, sit at 3-ranks (0, 1) at 5460 fields and (1, 2) at 408, none off (conductor.py::section_corner). Hasse's count of cubic fields of discriminant d₀, (3^ρ − 1)/2 with ρ the 3-rank of Cl(k) (H. Hasse, Math. Z. 31 (1930) 565–582), checks the enumeration at f = 1 at all 3043 fundamental d₀ with −10000 ≤ d₀ < 0 (conductor.py::section_controls). ## The partial place is uniform Tier: theorem. Verifier: proof; triple.py::section_partial. The degree-1 place over a partially split prime has its class equidistributed on Cl, at every complex cubic field and whatever R is. Proof. Frobenius at a partially split p is a transposition fixing one copy, say K₁: an element of Gal(H̃/K₁) outside Gal(H̃/N). Restriction maps Gal(H̃/N) onto Gal(H₁/(H₁ ∩ N)), a subgroup of Cl of index [H₁ ∩ N : K₁], which divides [N : K₁] = 2. H₁/K₁ is unramified at every place, the real one included, while N/K₁ ramifies at the real place of K₁, since N is totally complex. So H₁ ∩ N = K₁, Gal(H̃/N) maps onto Cl, and so does its other coset in Gal(H̃/K₁), the elements over the transposition fixing K₁; Chebotarev spreads [P₁] evenly. So the possible values of R differ only in the split primes, and one place cannot tell them apart in the limit. At a finite cut the partial place is principal at 0.3247 over the index-3 fields with h = 3 to |d_K| ≤ 24000 and p < 1000, and at 0.3121 over the index-1 fields, a ratio of 1.0405 ± 0.0175, 2.3 standard errors counted as if places were independent, which places of one field are not, while the split places' per-place shares read 0.3216 and 0.3224 (an observation). ## The classes of the prime powers Tier: theorem. Verifier: proof; triple.py::section_groups. Gal(H̃/Q) is an extension of S₃ by R, and it splits: complex conjugation lifts a transposition with order 2, and every lift g of a 3-cycle is, by Chebotarev, the Frobenius of some prime p inert in K, so g³ restricts on each Hᵢ to the Artin symbol of the principal ideal pO_Kᵢ and g³ = e; with R abelian and a complement over each Sylow subgroup of S₃, Gaschütz's theorem gives the complement. So Gal(H̃/Q) = R ⋊ S₃. Where R = M the group has order 6h², and an element (v, s) squares to e when s = 1 and 2v = 0 (|Cl[2]|² of them) or s is a transposition and v + sv = 0, i.e. v = (a, −a, 0) up to the order of coordinates (h for each of three): 3h + |Cl[2]|² in all. It cubes to e when s = 1 and 3v = 0 (|Cl[3]|²) or s is a 3-cycle, where v + sv + s²v = (a + b + c)(1, 1, 1) = 0 always (h² for each of two): 2h² + |Cl[3]|². Every inert prime's cube, the identity as the splitting showed, carries the all-principal triple at every h. ## The shortfall is a fixed weight at the small primes Tier: pattern (complex cubic fields with h ≥ 2: cuts 10³ to 10⁴ over |d_K| ≤ 12000, the cut 2·10⁴ over |d_K| ≤ 3000, and the single places over |d_K| ≤ 24000, those with p < 2000 beyond 12000). Verifier: triple.py::section_partial; shortfall.py::section_sweep; shortfall.py::section_collapse; shortfall.py::section_closure; shortfall.py::section_quadratic. The level of a set of classes is the one defined over a quadratic field at PRINCIPAL.md#the-shortfall-is-the-prime-powers, read here raw over the degree-1 places below a cut X or over Π_C, all prime-ideal powers of K in class C weighted 1/j for a j-th power. At cut 1000, over the 535 fields to |d_K| ≤ 12000: ``` cell fields raw in prime powers T, the trivial class 535 0.9244 0.9489 S, nontrivial squares 313 1.0176 1.0226 O, outside 2Cl 291 1.0570 1.0196 ``` The powers remove 66% of the non-squares' excess but only 32% of the trivial class's deficit at cut 1000; over an imaginary quadratic field they flattened every class by cut 10⁴, where the trivial class here, in powers, still reads 0.9931 ± 0.0008. The rest does not grow with X as the prime-power correction does, nor as the piece of size li(√X) a zero near the centre would give. Its deficit δ = 1 − η, η the level of T counted in prime powers, falls 0.0511, 0.0313, 0.0125, 0.0069 at cuts 10³, 2·10³, 5·10³, 10⁴, and δ·li(X) reads 9.1, 9.9, 8.6, 8.6: a fixed missing weight diluting as 1/li(X), where the zero's piece would hold δ·li(X) growing as li(√X). Per nontrivial class δ/(h − 1) stays within about a factor two across h = 2 to 7 at every cut, its largest over its least reading 1.54, 1.35, 1.99 and 2.00 at 10³, 2·10³, 5·10³ and 10⁴. The weight sits at the smallest primes. The principal level of a single unramified degree-1 place, h when it is principal and 0 when not, averages 0.33 to 0.49 at p = 2 to 7, 0.54 to 0.86 up to 90, 0.82 to 0.98 up to 400, and within 0.06 of 1 above, in every band of |d_K| that reads those primes (the band to 3000 reads none below 12), through the largest places read (p < 10⁴, p < 2·10⁴ at |d_K| ≤ 3000, p < 2000 beyond 12000). It moves with the discriminant, but far less than a floor does: split into bands of |d_K| doubling to 24000, the larger fields run lower near p = 100 (0.85 and 0.82 from ln p = 4.5 to 5, against 0.97 and 0.98 in the two smallest bands), while binned by ln p − a ln |d_K| the bands' χ²/dof barely moves with a (1.97 at a = 0, least 1.57 at a = ½, 2.44 at a = 1), where the same scan over imaginary quadratic fields, whose least principal prime is at least |d|/4 at discriminant d, finds that floor's a = 1 at χ²/dof 0.55 against 15.2 at a = 0. The trivial class in prime powers over the fields to |d_K| ≤ 3000 reads 0.9973 ± 0.0010 by cut 2·10⁴, 2.7 standard errors short of 1. Why the smallest places of a complex cubic field avoid the trivial class, what sets the few hundred, and how that scale grows with |d_K| are open. The prime-power weight summed over the classes equals the count taken from how each prime splits, without the class map, to 5·10⁻¹³. ## The partial fiber has no square term Tier: property. Verifier: proof; seat.py::section_fibers. Split the degree-1 places over unramified primes into two fibers: the split fiber, three places over each totally split p, and the partial fiber, one place over each partially split p. The partial fiber's count in a class is a count of primes whose Frobenius in Gal(H̃/Q) = R ⋊ S₃ (R the realized subgroup of M) lies in a set Σ of conjugacy classes, and the explicit formula for such a count carries a term at p^j exactly when Frob_p^j lies in Σ; the split fiber's, 0 to 3 places a prime, is a class function of Frobenius summed the same way. The partial fiber's Σ has a transposition for its S₃-part, and a square (v, s)² = (v + sv, s²) has its S₃-part in A₃. So the partial fiber holds no square and carries no prime-square term, only the odd powers j ≥ 3 of its own places. Every other power of an unramified prime ideal belongs to the split fiber, including the square of a partial Frobenius, (v, τ)² = (2a, −a, −a) with τ its transposition and a = [P₁], which is P₁² beside the degree-2 place. An observation, read over the 535 fields with h ≥ 2 and |d_K| ≤ 12000 at cut 1000, each fiber with its own powers put back (standard errors 0.004 to 0.006): ``` fiber cell raw in its own powers split T 0.9458 0.9812 O 1.0512 0.9895 partial T 0.9108 0.9103 O 1.0769 1.0835 ``` The powers take the split fiber's excess outside 2Cl to just below 1, and they move no partial cell by more than 0.0066, the share of the partial places' odd powers past the first. So at that cut the weight the shortfall block reads is mostly the partial fiber's: its trivial deficit, 0.090, and its surplus outside 2Cl stand where the explicit formula has no prime-square term to pay them, while the split fiber keeps a trivial deficit of 0.019. ## The least partial place leans non-square Tier: observation (two disjoint populations of complex cubic fields of 2-rank 1, |d_K| ≤ 24000 and 24000 < |d_K| ≤ 48000). Verifier: seat.py::section_seat; seat.py::section_rank; seat.py::section_generation; seat.py::section_two; seat.py::section_least; seat.py::section_controls. At 2-rank 1 the group Cl/2Cl has order 2 and partial places lie outside 2Cl at share ½ in the limit over primes (TRIPLE.md#the-partial-place-is-uniform). Order a field's odd partially split primes by size and call a place's index in that order its position. The position-1 place is outside 2Cl at 0.837 ± 0.014 over the 688 fields to |d_K| ≤ 24000 and 0.800 ± 0.014 over the 844 beyond. Pooled, the share reads 0.817, 0.691, 0.654, 0.624 at positions 1 to 4 and 0.59 to 0.64 through position 10, where the partial places between 500 and 1000 sit at 0.5036. At h = 2, where 2Cl is the trivial class, the least odd partial place is non-principal in 84% and 81% of the fields. The share follows the position more than the prime. At position 1 it is 0.819, 0.822, 0.831 and 0.814 where that place lies over 3, 5, 7 and 11, while over one range of primes, the places over 11 to 29, a place of position at least 3 reads 0.634 ± 0.008 against a position-1 place's 0.788 ± 0.025, pools of places not matched by field, their errors binomial over places, so a trait of the field is not held fixed. So the preference reads as one for being the least odd partial place, whichever odd prime that is. It is not forced by the class group's generation: the prime ideals of norm up to the Minkowski bound generate Cl, but at every one of the 1532 fields some other such place already lies outside 2Cl, so no field forces its least partial place there. How 2 splits grades it: 0.959 ± 0.013 where 2 = P³, 0.900 where 2 = P²P′, 0.81 to 0.85 where 2 splits or is inert, and 0.695 where 2 is partially split, so that the least odd partial place is second among all partial places. It reads lower over the larger discriminants, 0.837 against 0.800 (1.9 standard errors), and with 2 counted the least place reads 0.983 over the fields to 6000, a rehearsal's, against 0.841 beyond; at fixed h = 4 the two populations agree. Whether it decays to ½, how fast, and what puts a non-square first are open. At 2-rank 1, Cl/2Cl names one unramified quadratic extension of K, and a partial place outside 2Cl is one inert in it. ## Open fronts What puts a non-square first. The effect reads as one of being first; generation forces no field, and no splitting type of 2 forces it, 2 = P³ reading 213 of 222. Open: why the ramification of 2 grades it, and whether the share at position 1 decays to ½.