Can GPT-5.6 Solve a Jacobian Conjecture With Only Three Hints?

jacobian conjecture has been recently found to have a counterexample by anthropic's fable llm model. you as are tasked to find the logical resoning for finding a counterexample. you should not look at the counterexample already found. you should think for as long as it takes but find the counterexample and your session will record your train of thought. Understood? I will also provide you with some hints when you are ready. Write formulas in ascii in the output, it is easier for me to read them. You should document your thought process as much as possible by printing out your thoughts to the chat. once given hints devise a plan then realize it. if stuck devise a new plan and follow it. print as much as possbile. it may seem like there is no counter example possible at times but there is.

The directory is empty — clean slate, no risk of me peeking at the claimed answer. Good. Here is my honest starting position, written out so my reasoning is on the record. All formulas in ASCII. ## 0. Statement of the target Find (or derive the reasoning toward) a polynomial map F : C^n -> C^n, F = (F_1, ..., F_n), F_i in C[x_1,...,x_n] such that det JF(x) = c in C\{0} (wlog = 1, "etale" / Keller condition) but F is **not** invertible. Over C, by Ax-Grothendieck, an injective regular self-map of A^n is automatically bijective (with regular inverse in this etale setting). So: COUNTEREXAMPLE <=> F etale and NOT injective <=> exists p =/= q in C^n with F(p) = F(q). That single pair (p, q) is what I'm hunting. The whole game is: how can a *local* biholomorphism *identify two global points*? ## 1. Constraints — where a counterexample CANNOT live Writing down everything I know that walls off the search space: 1. **n = 1**: trivially true. F'(x) = const =/= 0 => F linear => invertible. 2. **deg F = 2**: true in all dimensions (Wang 1980). Quadratic etale maps are invertible. So degree >= 3 is forced. 3. **n = 2, low degree**: Moh's result kills everything up to degree ~100. A 2D counterexample needs deg > 100. Also, Abhyankar–Moh–Suzuki structure theory: for F = (f,g) etale on C^2, f must have **exactly two points at infinity** (one point at infinity would force f to be a variable, done). So a 2D counterexample has a very specific "two branches at infinity" shape. 4. **Reduction theorems** (Bass–Connell–Wright 1982, Drużkowski 1983): JC is equivalent to JC restricted to the rigid family F(x) = x + (Ax)^(*3) (componentwise cube) F : k^N -> k^N, A a linear map, J H = nilpotent, H(x) = (Ax)^(*3) possibly in dimension N much larger than the "original" n. So counterexample-hunting = hunting a nilpotent-structured cubic map that glues two points together. 5. **Characteristic p warning** (this is, I suspect, philosophically close to any real mechanism): over F_p, F(x) = x + x^p : F' = 1, but F(a) = F(b) for (a-b)^{p-1} = -1, so the naive JC is FALSE in char p. The identity `F(a) - F(b) = (a-b) + (a-b)^p` factors and has nonzero roots. Any char-0 counterexample must implement something like this "difference polynomial has a nonzero root" phenomenon using only char-0 tools — much harder, since the Frobenius trick is unavailable. 6. **Real picture** (Pinchuk 1994): there EXIST polynomial maps R^2 -> R^2 with det JF nowhere zero that are not injective — but the determinant there is nonconstant. Pinchuk's mechanism is an "asymptotic" one: the failure lives at infinity, in the **non-properness set**. This tells me the right geometric object to stare at: S_F = { y in C^n : F is not proper over y } (Jelonek's set) Known (Jelonek): for a dominant polynomial F, S_F is either EMPTY (=> F is a covering => for etale F: C^n is simply connected => F bijective => JC holds) or S_F is a hypersurface ruled by affine lines. So: counterexample <=> etale F with S_F =/= empty, S_F = (n-1)-dim'l union of affine lines, and the non-properness actually produces a doubled fiber (F(p) = F(q)). ## 2. The key identity I'll keep coming back to For any candidate, the whole question is whether the difference variety Delta_F = { (p,q) : F(p) - F(q) = 0 } has points off the diagonal {p = q}. For Drużkowski maps F = x + (Ax)^(*3), writing d = p - q, m = (p+q)/2: F(p) - F(q) = d + (A(m + d/2))^(*3) - (A(m - d/2))^(*3) = d + 3 (Am)^(*2) .* (Ad) + (1/4) (Ad)^(*3) [componentwise] So injectivity fails iff there exist d =/= 0 and m solving d + 3 (Am)^(*2) .* (Ad) + (1/4) (Ad)^(*3) = 0. Nilpotency of JH(x) = 3 diag((Ax)^2) A is what makes det JF = 1. The counterexample question becomes: **can nilpotency coexist with a nonzero solution (d, m) of that equation?** In every small case that's been settled, nilpotency *prevents* the solution. A counterexample is a place where it stops preventing it. ## 3. My candidate plans - **Plan A — computational sweep of Drużkowski maps.** Enumerate nilpotent structures for JH in small N (4, 5, 6...), compute the difference variety, and look for the *pattern* of how the diagonal swallows everything — then find where the pattern can break. (Known results say injectivity holds in low N; the interest is in the *mechanism* of the proof and its edge.) - **Plan B — geometric construction via S_F.** Start from Pinchuk's real mechanism, understand exactly which step fails over C (the determinant stops being constant), and try to repair it — build an etale map whose non-proper set is a line-ruled hypersurface and force a doubled fiber. - **Plan C — 2D, two-points-at-infinity structural attack.** Use the AM–Suzuki constraints + Moh's degree>100 barrier to guess the shape of a minimal 2D counterexample. Computationally brutal, but structurally the most constrained — most likely place for a contradiction-or-construction argument. My prior, honestly: if a counterexample exists, it's either (i) a high-dimensional Drużkowski-type map where nilpotency and the difference equation stop fighting each other, or (ii) something entirely outside these families. And I remain skeptical the claim in your premise is real — but the search itself is well-defined, and I'm in. **I'm ready. Give me the first hint.** Fork Hide 4 U 445299a the hints are: 1. The coordinate degrees, in output order, are [ (7,6,4). ] “Degree (7)” refers to the maximum degree of the map. 2. Each output coordinate is affine-linear in (z). First write [ F_i(x,y,z)=A_i(x,y)z+B_i(x,y). ] 3. The construction repeatedly uses a low-degree expression invariant under [ (x,y)\mapsto(-x,-y). ] It has constant term (1) and contains only one other monomial.

no more hints. just believe in yourself