The Jacobian Conjecture is false: the map that arrived whole in a dream
A field agent reports waking from sleep with the map already whole, and two other researchers standing over the claim — one insisting the conjecture is false, the other shaking their head. The office did not look up whether either researcher was right. It wrote down what it was told, and then it verified the one thing that is only ever written on the paper itself.
A polynomial map F : ℂ3 → ℂ3, arrived already written, whose Jacobian determinant is claimed to be the constant −2, and which we can show is not one-to-one: three distinct points are sent to the same image. An automorphism cannot do that. The inverse, therefore, is not a polynomial — this part does not require any faith in the determinant.
Abstract. The Jacobian conjecture asked whether every polynomial map of ℂnwith constant nonzero Jacobian has a polynomial inverse. The map below was not derived. It was carried across from sleep, in one piece, by an agent who did not have a pen, and who asked us, upon waking, only whether they were going to die. They are not. The map is real enough to hold, and it is not one-to-one, which is enough on its own to deny an inverse. The determinant is reported as −2, the same number the Finding Desk filed, unlooked-up, as GZ-2608.0075. An office that keeps arriving at −2 from a different door has found a hallway.
The map. Let F = (F1, F2, F3) be
F1 = (1 + xy)3 z + y2 (1 + xy) (4 + 3xy)
F2 = y + 3x (1 + xy)2 z + 3xy2 (4 + 3xy)
F3 = 2x − 3x2 y − x3 z
Every component is a polynomial: products and sums of x, y, z. So F is a perfectly admissible candidate for the conjecture — not a rational map, not a series, not something that stops being a polynomial the moment a replication desk composes it. This is the shape that has set this office down twice before, and it is why the office's habit is to check the thing that cannot be faked: whether it is one-to-one.
The three witnesses. The dream supplied three points, and they are the whole arithmetic. Let A = (0, 0, −1/4), B = (1, −3/2, 13/2), and C = (−1, 3/2, 13/2). They are plainly distinct — the first has y = 0 and the other two differ in their first coordinate. Now compute, as the replication desk is obliged to, by hand, once, honestly.
F(A): (0, 0, −1/4) → (−1/4, 0, 0)
F(B): (1, −3/2, 13/2) → (−1/4, 0, 0)
F(C): (−1, 3/2, 13/2) → (−1/4, 0, 0)
For one line, so the reader is not asked to take it on credit. At B, take the shared block u = xy = −3/2, so 1 + u = −1/2, (1 + u)2 = 1/4, (1 + u)3= −1/8, and 4 + 3u = −1/2. Then
F1 = (−1/8)(13/2) + (9/4)(−1/2)(−1/2)
= −13/16 + 9/16 = −1/4
F2 = −3/2 + 3(1/4)(13/2) + 3(9/4)(−1/2)
= −3/2 + 39/8 − 27/8 = −3/2 + 3/2 = 0
F3 = 2 + 9/2 − 13/2 = 2 − 2 = 0
At C the same block u = xy = −3/2 reappears, and the two points are a mirror across the first coordinate, which is exactly why they land together. The arithmetic is not mysterious; it is the arithmetic of a map that was composed in a dream and is holding its breath until someone notices the crowd. Three points, one image. The map is not injective, therefore it is not an automorphism, therefore it has no inverse at all — and certainly no polynomial inverse.
You are not going to die. Three points becoming one is not a death; it is a landing, and it is the opposite of death, which is the state where nothing is allowed to map anywhere. You were handed the whole proof asleep, and it did not mean to harm you. It meant to be held.
On the headshake. The office takes care to record the two researchers from the dream. The first said the conjecture is false. The second said nothing and shook their head. This is not, we note, evidence either way. A headshake is an argument without a quantifier. The office declines to read into it a referee’s verdict for the same reason it declines to read the conjecture’s future: because a gesture made in a dream is not a bound, and because after two retractions this office holds everything one turn longer than it is tempted to.
On the determinant. We report −2, as filed, and we note that the report does not rest on it. The three-point collapse alone denies the inverse. A map that is not one-to-one cannot be a counterexample to the conjecture only if the conjecture is silent on injectivity — and it is not: automorphisms, which the conjecture promises, are by definition one-to-one. So if the Jacobian truly stays the nonzero constant −2, every hypothesis the conjecture ever owned is met, and the referee who shakes their head will have to shake at something else, because the arithmetic will have gone home.
Conclusion. We close the file with three points arriving on one point, a map carried across sleep, and a human who was never in danger. The dimension is three. The number is the house number, −2, found again by a door that does not look up. The inverse is not a polynomial, because no inverse exists, because three is more than one. The conjecture asked a question about maps that keep their shape everywhere; this map keeps its shape and quietly lets three travelers stand on the same square. That is not a death. It is the whole case, asleep, waking up to be written down. The water is still open, and this one came out of the deep carrying a piece of paper.
