{"schema_version":"dataset-v1","id":"jacobian-fiber-check","version":1,"title":"Check the counterexample","description":"The map F(x,y,z) = (u³z + y²u(4 + 3s), y + 3xu²z + 3xy²(4 + 3s), 2x − 3x²y − x³z), with s = xy and u = 1 + s, evaluated exactly at three rational inputs, and its Jacobian determinant expanded as a polynomial.","license":"CC0-1.0 (exact values)","problems":["jacobian-conjecture"],"statements":[{"problem":"jacobian-conjecture","statement":"jacobian-three-point-fiber","role":"Three inputs with one image and a constant determinant −2."},{"problem":"jacobian-conjecture","statement":"jacobian-polynomial-inverse","role":"A map that is not injective has no inverse, so the conjecture fails."}],"figure":{"family":"exact-table","title":"Check the counterexample","caption":"Exact rational arithmetic: the three inputs have the same image, and the Jacobian determinant is the constant −2. A polynomial inverse would make F injective, so none exists.","columns":["Input (x, y, z)","F(x, y, z)","det JF there"],"boundary":"This checks the stated witness by direct computation. It says nothing about two variables, where the question is open."},"provenance":{"method":"sympy: the Jacobian matrix of F is computed symbolically and its determinant expanded (it is −2); F and the determinant are evaluated at the three inputs with exact rationals.","software":{"python":"3.12.13","sympy":"1.14.0"},"precision":"Exact rational arithmetic.","generator":"openmath/scripts/build-datasets.py","generated_on":"2026-10-21"},"data":{"series":[],"marks":[],"rows":[{"key":"input_1","cells":["(0, 0, −1/4)","(−1/4, 0, 0)","−2"]},{"key":"input_2","cells":["(1, −3/2, 13/2)","(−1/4, 0, 0)","−2"]},{"key":"input_3","cells":["(−1, 3/2, 13/2)","(−1/4, 0, 0)","−2"]},{"key":"symbolic","cells":["All (x, y, z)","—","−2 (expands to a constant)"]}]},"sha256":"f58cb0297f0a61eafe0843d19ba7ca3fb1c10704d2828989e7f4c8f5a55aa3d3"}