Walk Through the Jacobian Conjecture Counterexample

If you came here looking for “jacobian conjecture explained for students,” the useful version is not a dramatic headline. It is a calculation: write down the polynomial map, form its Jacobian matrix, check that the determinant is the constant -2, and then watch three different input points land on the same output.

For broader context, the site’s AI-focused account covers the discovery story, and the basic overview explains the conjecture at definition level. Here we stay with the object on the page.

The counterexample map

The reported July 2026 counterexample is the polynomial map F from three variables to three variables given by F(x, y, z) = (F1, F2, F3), where:

F1 = (1 + x*y)^3*z + y^2*(1 + x*y)*(4 + 3*x*y)
F2 = y + 3*x*(1 + x*y)^2*z + 3*x*y^2*(4 + 3*x*y)
F3 = 2*x - 3*x^2*y - x^3*z

This is the map cited from Levent Alpöge’s announcement and reproduced by Kevin Buzzard on the Xena blog; its three coordinate polynomials have degrees 4, 6, and 7 among them.[1]

The Jacobian conjecture asks whether a polynomial map with constant nonzero Jacobian determinant must have a polynomial inverse. In student language: if the derivative matrix never loses volume anywhere, does the polynomial map have to be globally reversible?

A friendly two-variable example is G(x, y) = (x^2 + y + x, x^2 + y). Its Jacobian matrix has determinant 1, and this particular map really does have a polynomial inverse: from u = x^2 + y + x and v = x^2 + y, subtract to get x = u - v, then y = v - x^2. That is the behavior the conjecture hoped would always happen.[2]

The three-variable map above has the same kind of local signal: its Jacobian determinant is a nonzero constant. But it fails the global reversibility test in the bluntest possible way: several inputs give the same output.

Step 1: Form the Jacobian matrix

The Jacobian matrix of F is the 3 by 3 matrix of first partial derivatives:

J_F = [ dF1/dx   dF1/dy   dF1/dz ]
      [ dF2/dx   dF2/dy   dF2/dz ]
      [ dF3/dx   dF3/dy   dF3/dz ]

There is no prize for hand-expanding every entry. The meaningful check is whether det(J_F) simplifies to one constant polynomial. A computer algebra system is useful here because the unsimplified expression contains many canceling terms.

import sympy as sp

x, y, z = sp.symbols('x y z')

F1 = (1 + x*y)**3*z + y**2*(1 + x*y)*(4 + 3*x*y)
F2 = y + 3*x*(1 + x*y)**2*z + 3*x*y**2*(4 + 3*x*y)
F3 = 2*x - 3*x**2*y - x**3*z

J = sp.Matrix([F1, F2, F3]).jacobian([x, y, z])
sp.factor(J.det())

The expected output is:

-2

That single line is the first half of the verification. The determinant is not merely nonzero at a few sample points; it is the constant polynomial -2. So this map satisfies the hypothesis of the Jacobian conjecture as stated for polynomial maps in three variables.[1]

If you want a Sage-style version, the same calculation has the same shape: define the polynomial ring, put the three coordinate functions in a vector, build the derivative matrix, and ask for its determinant.

R.<x,y,z> = PolynomialRing(QQ)

F1 = (1 + x*y)^3*z + y^2*(1 + x*y)*(4 + 3*x*y)
F2 = y + 3*x*(1 + x*y)^2*z + 3*x*y^2*(4 + 3*x*y)
F3 = 2*x - 3*x^2*y - x^3*z

J = matrix([[diff(F1,x), diff(F1,y), diff(F1,z)],
            [diff(F2,x), diff(F2,y), diff(F2,z)],
            [diff(F3,x), diff(F3,y), diff(F3,z)]])
J.det().factor()

Again, the equality that matters is det(J_F) = -2. The determinant condition is passed.

Step 2: Check the collision

A polynomial inverse would make F one-to-one: two different inputs could not have the same output. The reported counterexample gives something even stronger than two colliding points. It gives three.[3][4]

The first point is almost too easy, which makes it a good place to start. Plug in x = 0, y = 0, z = -1/4:

F1 = (1 + 0)^3*(-1/4) + 0 = -1/4
F2 = 0 + 0 + 0 = 0
F3 = 0 - 0 - 0 = 0

So F(0, 0, -1/4) = (-1/4, 0, 0).

For the second point, x = 1 and y = -3/2, so xy = -3/2 and 1 + xy = -1/2. Also 4 + 3xy = 4 - 9/2 = -1/2. With z = 13/2:

F1 = (-1/2)^3*(13/2) + (9/4)*(-1/2)*(-1/2)
   = -13/16 + 9/16
   = -1/4

F2 = -3/2 + 3*(1)*(1/4)*(13/2) + 3*(1)*(9/4)*(-1/2)
   = -3/2 + 39/8 - 27/8
   = 0

F3 = 2*(1) - 3*(1)^2*(-3/2) - (1)^3*(13/2)
   = 2 + 9/2 - 13/2
   = 0

So F(1, -3/2, 13/2) = (-1/4, 0, 0) too.

The third point mirrors the second: x = -1 and y = 3/2, so xy is again -3/2, 1 + xy is again -1/2, and 4 + 3xy is again -1/2. With z = 13/2:

F1 = (-1/2)^3*(13/2) + (9/4)*(-1/2)*(-1/2)
   = -13/16 + 9/16
   = -1/4

F2 = 3/2 + 3*(-1)*(1/4)*(13/2) + 3*(-1)*(9/4)*(-1/2)
   = 3/2 - 39/8 + 27/8
   = 0

F3 = 2*(-1) - 3*(-1)^2*(3/2) - (-1)^3*(13/2)
   = -2 - 9/2 + 13/2
   = 0

So F(-1, 3/2, 13/2) = (-1/4, 0, 0). The three inputs are visibly distinct, but the output is identical.

Why this disproves the conjecture in dimension 3

Now the logic is short. The Jacobian determinant is the constant -2, so the map satisfies the conjecture’s nonzero-constant-determinant hypothesis. But three distinct points map to (-1/4, 0, 0), so the map is not injective.

A map with an inverse must be injective. If F had a polynomial inverse, it would certainly have an ordinary function inverse, and one output could not come from three different inputs. The collision alone rules out invertibility.

That is the whole student-checkable disproof in three variables: constant nonzero Jacobian determinant, but no inverse.

What changes in higher dimensions, and what does not

The significance is usually stated in terms of Smale’s Problem #16: the Jacobian conjecture was one of the problems Stephen Smale listed for the 21st century. A three-variable counterexample also gives counterexamples in every dimension n >= 3 by appending identity coordinates. For example, in four variables you can send (x, y, z, w) to (F1(x, y, z), F2(x, y, z), F3(x, y, z), w). The determinant stays nonzero constant, and the old collision remains.[5]

The two-variable case is different. This counterexample does not settle the plane case, and the plane Jacobian conjecture remains open.[5]

A note on the AI part

Reports say Alpöge used Anthropic’s Claude Fable 5 in the discovery process, and that the counterexample was checked quickly in systems including Sage, SymPy, Wolfram Alpha, and Lean.[6][7]

That is interesting, but it is not the same thing as a completed peer-review process. The announcement was fresh on July 19–20, 2026, and the exact model interaction has not been fully documented in the available reports.[6][7] The safe mathematical claim is narrower and stronger: the displayed polynomial map is short enough to verify directly, and the two checks above are the checks that matter.

A constant Jacobian determinant of -2 satisfies the hypothesis. Three distinct inputs sharing the output (-1/4, 0, 0) violate injectivity. So the Jacobian conjecture is disproved for dimensions at least 3, while the two-variable case remains open.

References

1. Human mathematicians are being outcounterexampled, Xena Blog, 2026-07-20
2. An explanation for undergraduated students about why the Jacobian Conjecture is hard?, Math StackExchange
3. Hacker News thread item 48973869, Hacker News
4. Anthropic Fable 5 disproves Jacobian conjecture, Mashable
5. Jacobian conjecture, Wikipedia
6. A Mathematician Used Claude Fable to Disprove the 87-Year-Old Jacobian Conjecture, Glitchwire
7. Claude Fable Jacobian Conjecture Counterexample, Kingy AI