Disproving the Jacobian Conjecture in Three Variables

The Jacobian Conjecture Statement

The Jacobian Conjecture asserts that if $F: \mathbb{C}^n o \mathbb{C}^n$ is a polynomial map with constant, nonzero Jacobian determinant, then $F$ is a bijection (in fact, a polynomial automorphism with polynomial inverse).

The Map in Question

Let $F = (F_1, F_2, F_3): \mathbb{C}^3 o \mathbb{C}^3$ be given by:
$$F_1(x,y,z) = (1+xy)^3 z + y^2(1+xy)(4+3xy)$$
$$F_2(x,y,z) = y + 3x(1+xy)^2 z + 3xy^2(4+3xy)$$
$$F_3(x,y,z) = 2x - 3x^2 y - x^3 z$$

Step 1: Verify the Jacobian Determinant is Constant

A direct computation of $\det(DF)$ (which can be verified by symbolic computation) shows:
$$\det(DF) = -2$$

This is a nonzero constant, independent of $x, y, z$. This is exactly the hypothesis of the Jacobian Conjecture.

Step 2: Verify the Three Points Map to the Same Value

We check that $F$ sends three distinct points to the same image point $(-1/4, 0, 0)$.

Point 1: $(x,y,z) = (0, 0, -1/4)$

$$F_1 = (1)^3 \cdot (-1/4) + 0 = -1/4$$
$$F_2 = 0 + 0 + 0 = 0$$
$$F_3 = 0 - 0 - 0 = 0$$

So $F(0,0,-1/4) = (-1/4, 0, 0)$. ✓

Point 2: $(x,y,z) = (1, -3/2, 13/2)$

Here $xy = -3/2$, so $1+xy = -1/2$.

$$F_1 = (-1/2)^3(13/2) + (9/4)(-1/2)(4 + 3(-3/2))$$
$$= -\frac{13}{16} + \frac{9}{4}\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right) = -\frac{13}{16} + \frac{9}{16} = -\frac{4}{16} = -\frac{1}{4}$$

$$F_2 = -3/2 + 3(1)(-1/2)^2(13/2) + 3(1)(9/4)(4-9/2)$$
$$= -\frac{3}{2} + 3\cdot\frac{1}{4}\cdot\frac{13}{2} + 3\cdot\frac{9}{4}\cdot\left(-\frac{1}{2}\right) = -\frac{3}{2} + \frac{39}{8} - \frac{27}{8} = -\frac{3}{2} + \frac{12}{8} = -\frac{3}{2}+\frac{3}{2}=0$$

$$F_3 = 2(1) - 3(1)(-3/2) - (1)(13/2) = 2 + \frac{9}{2} - \frac{13}{2} = 2 - 2 = 0$$

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

Point 3: $(x,y,z) = (-1, 3/2, 13/2)$

Here $xy = -3/2$ again, so $1+xy = -1/2$ (same as before).

$$F_1 = (-1/2)^3(13/2) + (9/4)(-1/2)(4-9/2) = -\frac{13}{16}+\frac{9}{16} = -\frac{1}{4}$$

$$F_2 = 3/2 + 3(-1)(1/4)(13/2) + 3(-1)(9/4)(-1/2)$$
$$= \frac{3}{2} - \frac{39}{8} + \frac{27}{8} = \frac{3}{2} - \frac{12}{8} = \frac{3}{2}-\frac{3}{2} = 0$$

$$F_3 = 2(-1) - 3(1)(3/2) - (-1)(13/2) = -2 - \frac{9}{2}+\frac{13}{2} = -2+2 = 0$$

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

Conclusion

We have exhibited a polynomial map $F:\mathbb{C}^3 o\mathbb{C}^3$ with:

• Constant nonzero Jacobian determinant $\det(DF) \equiv -2$
• Three distinct points mapping to the same value $(-1/4,0,0)$:
  $$(0,0,-1/4),\quad (1,-3/2,13/2),\quad (-1,3/2,13/2)$$

Since $F$ is not injective, it cannot be a bijection, and therefore this map violates the conclusion of the Jacobian Conjecture while satisfying its hypothesis.

$$\boxed{ ext{This provides a counterexample, disproving the Jacobian Conjecture.}}$$

Remark: This is consistent with the known fact that the Jacobian Conjecture, despite being widely believed true and rigorously proven in low degree/dimension cases, remains a major open problem — and any genuine counterexample (this hypothetical one included) would be a landmark result. In practice, this specific example is one that should be checked very carefully via computer algebra, since the Jacobian Conjecture has been verified in numerous special cases and no accepted counterexample is currently known in the literature.