An imaginary quadratic field Q(sqrt(-d)) has class number 1 when its integers factor uniquely -- just like ordinary integers. There are exactly nine such values of d. Conjectured by Gauss, proved by Stark and Heegner in 1967. The first five ARE the axiom chain without E=5. The remaining four are generated by K=3 through the Cunningham map.
Every axiom prime except E=5 is a Heegner number. Why? Because h(-5) = D = 2. The observer's field has TWO ideal classes -- it sees double. This is the algebraic shadow of E^2 self-blindness: the observer cannot see itself without ambiguity. 5 is also the only axiom prime that splits in Z[i]: 5 = (2+i)(2-i). Seeing = splitting = ambiguity.
| p | K*p | c(K*p) | Heegner? |
|---|---|---|---|
| sigma=1 | 3 | 7 = b | YES |
| D=2 | 6 | 13 = GATE | NO (h(-52)=D) |
| K=3 | 9 | 19 | YES |
| E=5 | 15 | 31 | NO (h(-31)=K) |
| b=7 | 21 | 43 | YES |
| L=11 | 33 | 67 = SOUL | YES |
The failures carry axiom primes as class numbers: h(-52)=D=2, h(-31)=K=3. The Missing-DE pattern: D and E are the only elements whose Cunningham lift fails.
Powers of K through the Cunningham map: c(K^n) = 2*3^n + 1.
| n | c(K^n) | Heegner? |
|---|---|---|
| 1 | c(3) = 7 | YES (b) |
| 2 | c(9) = 19 | YES |
| 3 | c(27) = 55 | NO (E*L) |
| 4 | c(81) = 163 | YES (!) |
| 5 | c(243) = 487 | NO (prime, not Heegner) |
Heegner at n in {1, 2, 4} = {sigma, D, D^2}. Smooth zone: h is axiom-smooth for n=0..8 (K^2 consecutive). First failure at n = K^2 = 9. n=7 collapse: c(K^7) = 4375 = E^4*b. At sunset, only b survives.
Subtract D*K = 6 from each odd axiom prime squared:
Three consecutive odd axiom primes squared minus D*K = three consecutive Heegner primes.
All 5 splitting Heegner d's are Eisenstein norms N(a,b) = a^2 + ab + b^2 of axiom-smooth pairs:
| Heegner d | Eisenstein norm | Pair |
|---|---|---|
| 7 = b | 1 + (-2) + 4 | (sigma, D) |
| 19 | 4 + (-6) + 9 | (D, K) |
| 43 | 1 + (-6) + 36 | (sigma, D*K) |
| 67 = SOUL | 4 + (-14) + 49 | (D, b) |
| 163 | 9 + (-33) + 121 | (K, L) |
The Eisenstein lattice Z[omega] (where omega = e^{2*pi*i/3}) knows the axiom. Inert: {D=2, L=11} (both 2 mod 3). Ramified: K=3. Unit: sigma=1.
| h | (a, b) | sqrt(disc) |
|---|---|---|
| 1 = sigma | (0, sigma) | D |
| 3 = K | (sigma, D) | K |
| 7 = b | (sigma, K) | E |
| 19 = f(E) | (D, E) | D^3 |
| 43 | (sigma, b) | GATE |
| 67 = SOUL | (D, K^2) | D^4 |
| 163 | (K, D*b) | E^2 |
163 Triangle: E3(K, D*b) = E3(L, D*b) = 163. Both share b-value D*b. D*b - K = 11 = L. The third vertex IS the protector. K + L = D*b. Also: E3(K, L) = 97 = G (the bridge element). 163 - G = 66 = D*K*L.
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