{
  "packet_id": "noncoordinate-polynomial",
  "scope_summary": "Main claim: the explicit F in R=C[h,u,v,w] from (eq:xyz)--(eq:F) satisfies R/(F)\u2245C^[3] but is not a coordinate, since grad F(P)=0 at P=(2,0,-1/2,1/2) on F=-1; adjoining variables gives the same in every ambient dimension \u22654. Full source audited, including Lemma lem:lift, Theorem thm:main, Corollary cor:dimension, the cusp/shift/ambient identities, the unit-ideal argument, the C\u2245R/(F) elimination, and the appendix B\u00e9zout certificate. Contextual citations are not proof dependencies.",
  "audited_steps": [
    {
      "anchor": "eq:xyz and the sentence after it (x^2+y^3=hs)",
      "result": "checked",
      "evidence": "Expanding x=u^3+hv and y=-u^2+hw gives x^2+y^3=h(2u^3v+3u^4w+h(v^2-3u^2w^2)+h^2w^3). The bracket is exactly S, so the cusp identity holds in R."
    },
    {
      "anchor": "eq:shift--eq:shiftidentity",
      "result": "checked",
      "evidence": "With X=x+s^3 and Y=y-s^2, the s^6 terms cancel and X^2+Y^3-(x^2+y^3)=2xs^3-3y^2s^2+3ys^4=-s p for p=-2s^2x+3sy^2-3s^3y. Identity is universal in x,y,s, hence survives substitution of (eq:xyz)."
    },
    {
      "anchor": "Lemma lem:lift, reduction to (eq:Clinear)",
      "result": "checked",
      "evidence": "G=V+UW is invertible. Modulo hW=y+U^2, U^3+hV-x becomes hG-yU-x. The same congruence gives S(h,U,G-UW,W)=hG^2-2UyG+y^2W, because the non-hG^2 terms regroup as W(U^2-hW)^2-2UyG. Thus S-s becomes y^2W-b with b=s-hG^2+2UyG."
    },
    {
      "anchor": "Lemma lem:lift, compatibility and elimination of W",
      "result": "checked",
      "evidence": "y^2a-hb=(hG-yU)^2+y^3-hs=x^2+y^3-hs=0. From \u03b1h+\u03b2y=1, r=\u03b1(1+\u03b2y) and q=\u03b2^2 satisfy rh+qy^2=1. Then W_0=ra+qb solves both linear equations, and W-W_0=r(hW-a)+q(y^2W-b), so (hW-a,y^2W-b)=(W-W_0) without cancelling zero divisors. Hence C\u2245A."
    },
    {
      "anchor": "Lemma lem:lift, A\u2245B[T] via (eq:UG)",
      "result": "checked",
      "evidence": "U=hT-\u03b2x and G=yT+\u03b1x satisfy hG-yU=x. In A, h(\u03b1U+\u03b2G)=U+\u03b2x and y(\u03b1U+\u03b2G)=G-\u03b1x, so both composites with T\u21a6\u03b1U+\u03b2G are the identity. No division is used. Thus C\u2245B[T]."
    },
    {
      "anchor": "eq:B--eq:Bplane",
      "result": "checked",
      "evidence": "Eliminating h=1+p and applying the polynomial automorphism x=X-s^3, y=Y+s^2 converts x^2+y^3-s(1+p) into X^2+Y^3-s by (eq:shiftidentity). Then s=X^2+Y^3, so B\u2245C[X,Y] as C-algebras, with inverse (eq:Bparam)."
    },
    {
      "anchor": "Proof of thm:main, (h,y)=B",
      "result": "checked",
      "evidence": "In B/(h,y) the relations give x^2=0 and p\u2261-2s^2x, hence 1=2s^2x. Squaring yields 1=4s^4x^2=0, so the quotient is the zero ring and (h,y)=B."
    },
    {
      "anchor": "Proof of thm:main, C\u2245R/(F) and (eq:fiber)",
      "result": "checked",
      "evidence": "C is C[h,x,y,s,u,v,w] modulo (h-1-p,x^2+y^3-sh) and the three expressions setting x,y,s equal to (eq:xyz). Those three generators eliminate x,y,s. After substitution, x^2+y^3-sh is the zero polynomial by the cusp identity, and h-1-p becomes F. Thus C\u2245R/(F). Lemma lem:lift plus B\u2245C[X,Y] gives R/(F)\u2245C[X,Y,T]."
    },
    {
      "anchor": "eq:ambient and the values at P=(2,0,-1/2,1/2)",
      "result": "checked",
      "evidence": "In R, X^2+Y^3=hs-sp=s(h-p)=s(1+F). At P: x=-1, y=1, s=2(1/4)+4(1/8)=1, p=2, F=-1, X=Y=0. Direct evaluation matches the manuscript."
    },
    {
      "anchor": "Proof of thm:main, grad F(P)=0",
      "result": "checked",
      "evidence": "Differentiating the polynomial identity X^2+Y^3=s(1+F) and evaluating at P gives dF(P)=0. Independent chain rule: at P, dF=dh+2dx-3dy+2ds with dx=-(1/2)dh+2dv, dy=(1/2)dh+2dw, ds=(3/4)dh-2dv+3dw; all four coefficients cancel."
    },
    {
      "anchor": "Proof of thm:main, F is not a coordinate",
      "result": "checked",
      "evidence": "A polynomial automorphism and its polynomial inverse have Jacobian matrices inverse to each other, so the Jacobian determinant is a unit of C, hence a nonzero constant. If F were the first component, the first row at P would be grad F(P)=0, so the determinant would vanish. Contradiction."
    },
    {
      "anchor": "Corollary cor:dimension",
      "result": "checked",
      "evidence": "For n>4, F in R[z_1,...,z_{n-4}] has quotient (R/(F))[z]\u2245C^[n-1] by (eq:fiber). Partials in the new variables are identically zero and the original four partials vanish at P, so (P,0,...,0) is critical. The same Jacobian-row argument applies. The case n=4 is the theorem."
    },
    {
      "anchor": "Appendix eq:alphabeta--eq:certificate and eq:explicitinverse",
      "result": "checked",
      "evidence": "In the free ring, with c=1+2s^2x, expansion gives \u03b2y+c(1+p)=1-4s^4(x^2+y^3), which is exactly the condition that \u03b1h+\u03b2y-1 equals the displayed certificate. Thus \u03b1h+\u03b2y=1 in B. The formulas for u,g,w,v are the lemma's inverse maps with r=\u03b1(1+\u03b2y) and q=\u03b2^2; they are not needed for existence."
    }
  ],
  "objections": [],
  "unverified_obligations": [],
  "assessment": "no_demonstrated_error_found",
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