原帖内容

hello there the jacobian conjecture is not merely false, it is factory false. all ai needed was a little push. thanx to my close friend gpt 5.6 sol for working after the world cup final, finding many other counter examples. Start from the seed: F0(x,y,z)=( (1+xy)^3z + y^2*(1+xy)(4+3xy), y + 3x(1+xy)^2z + 3xy^2*(4+3xy), 2x - 3x^2y - x^3z ). F0:C^3->C^3 has det J=-2 and sends (0,0,-1/4), (1,-3/2,13/2), and (-1,3/2,13/2) to (-1/4,0,0). one push later, here is a counterexample factory. choose m,r>=1, C!=0 and h(A), then set A=1+xy^m B=A^(r+1)z+y^(m+1)h(A) P=AB Q=y+xB R=CIntegral[t=0..x/A] (1-t*(Q-P*t)^m)^r dt. for every choice below, the apparent denominators cancel, R is polynomial, det J(P,Q,R)=-C, and the generic fiber degree is r*(m+1)+1. nonlinear-divisor branch: N2: (m,r,C)=(2,1,12), h=u+u^2A, u^2-4u+6=0, degree 4. N3: (3,1,20), h=u+u^2A, u^3-5u^2+10*u-10=0, degree 5. N4: (4,1,30), h=u+u^2A, u^4-6u^3+15u^2-20u+15=0, degree 6. N5: (5,1,42), h=u+u^2A, u^5-7u^4+21u^3-35u^2+35*u-21=0, degree 7. higher-order-cancellation branch: H2: (m,r,C)=(1,2,30), u^2-5*u+10=0, h=u+5*(u+2)A/3+25(u+2)*A^2/9, degree 5. H3: (1,3,140), u^3-7u^2+21u-35=0, h=u+(u^2+14u+35)A/10+7(u^2+34u+85)A^2/100+49(u^2+34*u+85)*A^3/500, degree 7. H4: (1,4,630), u^4-9u^3+36u^2-84*u+126=0, h=u-(u^3-8u^2-42u-126)A/35-(5u^3-54u^2-504u-1512)A^2/245-3(81u^3-718u^2-8302u-24906)A^3/8575-27(81u^3-718u^2-8302u-24906)*A^4/60025, degree 9. mixed branch: M22: (m,r,C)=(2,2,210), u^4-7u^3+21u^2-35*u+35=0, h=u+(2u^3-11u^2+35u-35)A/6+7(2u^3-9u^2+35u-40)*A^2/36, degree 7. M32: (3,2,1260), u^6-9u^5+36u^4-84u^3+126u^2-126*u+84=0, h=u-u^2*(u^3-6u^2+15u-20)A/10-u(5u^4-31u^3+81u^2-147u+42)*A^2/50, degree 9. the collisions are exact, not numerical vibes. for N_m, take alpha^m=m+2: the fiber over (-alpha,0,0) contains the distinct roots s=0 and s=1. for H_r and M_mr, take omega to be a primitive (r+1)st root of unity and q^m=1-omega: the fiber over (0,q,0) again contains s=0 and s=1. then they compose: C1=N2 o H2: degree 20, det J=360. C2=H3 o N3: degree 35, det J=2800. C3=M22 o N4: degree 42, det J=6300. and they multiply: X1=N2 x H2: C^6->C^6, degree 20, det J=360. X2=N3 x H3: C^6->C^6, degree 35, det J=2800. X3=N2 x H2 x M22: C^9->C^9, degree 140, det J=-75600. that is 16 explicit counterexamples counting the seed: 1 original map, 9 base maps from 3 construction branches, 3 compositions, and 3 products. all checked with exact symbolic arithmetic—no floating point and no “looks right.” rescale one output coordinate and every determinant becomes 1. the jacobian conjecture did not merely die. it turned out to have a counterexample generator. a century of mathematical prestige stared at the wall; ai found the handle, opened the door, and discovered the room was full.