Jacobian conjecture

General results and reductions

The case of polynomials over a field ${\displaystyle \mathbb {K} }$ of characteristic zero can be reduced to ${\displaystyle \mathbb {K} =\mathbb {C} }$ using the Lefschetz principle.^[12] Further, if ${\displaystyle F}$ is injective, it can be shown to already be bijective with a regular inverse^[34] (cf. the Ax–Grothendieck theorem).

Many special cases and reductions of the Jacobian conjecture were established in the decades before it was disproved in 2026 for ${\displaystyle n>2}$. In light of the counterexample, the positive partial results now describe conditions that any counterexample must violate, while the reductions show that counterexamples of quite special forms must exist.

The existence of a polynomial inverse is obvious if ${\displaystyle F}$ is simply a set of functions linear in the variables, because then the inverse will also be a set of linear functions. However, unlike the 1-dimensional case, in two or more dimensions, there exist nonlinear polynomial maps with constant Jacobian determinant and a polynomial inverse. A simple quadratic example is given by

${\displaystyle u=x^{2}+y+x}$

${\displaystyle v=x^{2}+y}$

so that the Jacobian determinant is

${\displaystyle J_{F}=\left|{\begin{matrix}1+2x&1\\2x&1\end{matrix}}\right|=(1+2x)(1)-(1)2x=1.}$

In this case the inverse exists as the polynomials

${\displaystyle x=u-v}$

${\displaystyle y=v-(u-v)^{2}.}$

Stuart Sui-Sheng Wang proved the Jacobian conjecture for polynomials of degree 2,^[35] so any counterexample must have degree at least 3. Hyman Bass, Edwin Connell, and David Wright showed that the general case follows from the special case where the polynomials are of degree 3, or even more specifically, of cubic homogeneous type, meaning of the form ${\displaystyle F=(X_{1}+H_{1},\dots ,X_{n}+H_{n})}$, where each ${\displaystyle H_{i}}$ is either zero or a homogeneous cubic.^[36] Ludwik Drużkowski showed that one may further assume that the map is of cubic linear type, meaning that the nonzero ${\displaystyle H_{i}}$ are cubes of homogeneous linear polynomials.^[37] These reductions introduce additional variables and therefore do not preserve a fixed dimension.

Edwin Connell and Lou van den Dries proved that if the Jacobian conjecture is false, then it has a counterexample with integer coefficients and Jacobian determinant 1.^[38]

Let ${\displaystyle \mathbb {K} [X]}$ denote the polynomial ring ${\displaystyle \mathbb {K} [X_{1},\dots ,X_{n}]}$ and ${\displaystyle \mathbb {K} [F]}$ denote the ${\displaystyle \mathbb {K} }$-subalgebra generated by ${\displaystyle f_{1},\dots ,f_{n}}$. For a given ${\displaystyle F}$, the Jacobian condition implies invertibility if and only if ${\displaystyle \mathbb {K} [X]=\mathbb {K} [F]}$. Keller (1939) proved the birational case, that is, where the two fields ${\displaystyle \mathbb {K} (X)}$ and ${\displaystyle \mathbb {K} (F)}$ are equal. The case where ${\displaystyle \mathbb {K} (X)}$ is a Galois extension of ${\displaystyle \mathbb {K} (F)}$ was proved by Andrew Campbell for complex maps^[39] and in general by Michael Razar^[40] and, independently, by David Wright.^[25] No counterexample can therefore be birational or define a Galois extension; consistent with this, the 2026 counterexample is generically three-to-one.^[4]

Michiel de Bondt and Arno van den Essen^[41][42] and Ludwik Drużkowski^[43] independently showed that the general case of the conjecture is equivalent to the special case of complex maps of cubic homogeneous type with a symmetric Jacobian matrix, so counterexamples of this form must also exist. They further showed that the conjecture holds for maps of cubic linear type with a symmetric Jacobian matrix, over any field of characteristic ${\displaystyle 0}$; no counterexample of this more restricted form is therefore possible, so the cubic-linear and symmetric reductions cannot be combined.

The strong real Jacobian conjecture was the assertion that a real polynomial map with a nowhere vanishing Jacobian determinant has a smooth global inverse. That is equivalent to asking whether such a map is topologically a proper map, in which case it is a covering map of a simply connected manifold, hence invertible. Sergey Pinchuk constructed a counterexample to the strong real Jacobian conjecture of total degree 35.^[44] Because Pinchuk's maps have nonconstant Jacobian determinant, they did not disprove the Jacobian conjecture itself.

For ${\displaystyle n\geq 1}$, the Dixmier conjecture ${\displaystyle \mathrm {DC} _{n}}$ states that every endomorphism of the ${\displaystyle n}$-th Weyl algebra ${\displaystyle A_{n}}$ over a field of characteristic zero is an automorphism. For each ${\displaystyle n}$, the Dixmier conjecture for ${\displaystyle A_{n}}$ implies the ${\displaystyle n}$-dimensional Jacobian conjecture.^[36] Conversely, Yoshifumi Tsuchimoto^[45] and, independently, Alexei Belov-Kanel and Maxim Kontsevich^[46] showed that the Jacobian conjecture in ${\displaystyle 2n}$ variables implies the Dixmier conjecture for ${\displaystyle A_{n}}$.

Kossivi Adjamagbo and Arno van den Essen gave a self-contained algebraic proof and established the more precise chain of implications

${\displaystyle \mathrm {JC} _{2n}\Longrightarrow \mathrm {PC} _{n}\Longrightarrow \mathrm {DC} _{n}\Longrightarrow \mathrm {JC} _{n},}$

where ${\displaystyle \mathrm {PC} _{n}}$ denotes the Poisson conjecture for the canonical complex Poisson algebra in ${\displaystyle 2n}$ variables.^[47] Consequently, the all-dimensional forms of the Jacobian, Poisson, and Dixmier conjectures are equivalent.

Since the 2026 counterexample disproves ${\displaystyle \mathrm {JC} _{n}}$ for every ${\displaystyle n>2}$, the above implications show that ${\displaystyle \mathrm {DC} _{n}}$ and ${\displaystyle \mathrm {PC} _{n}}$ are also false for every ${\displaystyle n>2}$, while both conjectures remain open for ${\displaystyle n=1,2}$.

In two dimensions

The conjecture remains open in two dimensions as of 2026^[update]. Much of the study of the two-dimensional conjecture concerns the geometry of a hypothetical counterexample at infinity. In two variables, a pair of polynomials is called a Jacobian pair if its Jacobian determinant is a nonzero constant. Shreeram Abhyankar proved that each component of a Jacobian pair has at most two points at infinity.^[16] He also showed that the planar conjecture is equivalent to each of several assertions about all Jacobian pairs, including that each component has only one point at infinity, that the Newton polygon of each component is a triangle, and that one of the two component degrees divides the other.^[16] Using characteristic sequences and approximate roots, Abhyankar proved the conjecture in the case of two characteristic pairs and later sharpened this to what he called the "two plus epsilon characteristic pairs" case.^[48] His subsequent work developed the more general "three minus epsilon characteristic pairs" setting.^[49]

Another line of work has progressively restricted the possible degree pairs and Newton polygons of a counterexample. Raymond C. Heitmann showed that any counterexample ${\displaystyle (P,Q)}$ must satisfy ${\displaystyle \gcd(\deg P,\deg Q)\geq 16.}$^[50] Jorge A. Guccione, Juan J. Guccione, and Christian Valqui subsequently gave an elementary proof of Heitmann's result using refinements of Abhyankar's methods, proved that ${\displaystyle \gcd(\deg P,\deg Q)\neq 2p}$ for every prime ${\displaystyle p}$, and obtained further restrictions on the possible shape of a counterexample.^[51] Leonid Makar-Limanov later introduced a Newton polyhedron associated with a minimal counterexample, obtaining a sharper estimate for its geometric degree and a new proof of Abhyankar's result for two characteristic pairs.^[52]

More recent geometric approaches study hypothetical Jacobian pairs through compactifications of the affine plane. Alexander Borisov formulated the resulting constraints on the maps between the Picard groups of suitable compactifications as an essentially combinatorial problem and described several possible configurations, which he called "frameworks".^[15] A framework is not itself a counterexample: each gives a system of equations whose solution would produce a Jacobian pair.^[15]

A complementary line of work concerns the possible covering behavior of a counterexample. In 1971, Anatoli Vitushkin constructed a three-sheeted branched covering whose regular part is contractible and whose bifurcation set is homeomorphic to a plane.^[53] The example provided a topological model relevant to the three-sheeted case, but was not a polynomial map. S. Yu. Orevkov subsequently proved that the Jacobian of a three-sheeted polynomial map ${\displaystyle \mathbb {C} ^{2}\to \mathbb {C} ^{2}}$ cannot be a nonzero constant.^[54] Orevkov later constructed an analytic realization of the Vitushkin covering over a ball in ${\displaystyle \mathbb {C} ^{2}}$,^[55] while Nguyen Van Chau proved that the Vitushkin covering cannot occur in the polynomial case.^[56] Vitushkin also constructed noninvertible rational transformations of ${\displaystyle \mathbb {C} ^{2}}$ with nonzero constant Jacobian. One of his examples has a pole along a complex line, but on the complement of that line it defines a two-sheeted étale cover of the complement of a discriminant curve in ${\displaystyle \mathbb {C} ^{2}}$; in particular, it is generically two-to-one, but is not a polynomial counterexample.^[57]

A separate computational line has sought to exclude counterexamples of progressively higher degree. Tzuong-Tsieng Moh's 1983 computer-assisted argument, with its algorithm subsequently revised by Lih-Chung Wang in 2005, verified it for polynomials of degree at most 100.^[20][23] This bound was increased to 104 by Thuy Nguyen in 2025.^[58] In a 2022 preprint, Jorge Alberto Guccione, Juan José Guccione, Rodrigo Horruitiner, and Christian Valqui claimed that this bound can be increased to 124 except for the possible degree pair (72,108).^[59]

Counterexample for n > 2

On July 19, 2026, mathematician and Anthropic employee Levent Alpöge presented an explicit counterexample to the conjecture in three-dimensional space which he credited to Claude Fable 5 AI model.^[60][61] According to Abhishek Saha of the Queen Mary University of London, the counterexample is simple to verify in itself, but how Alpöge and Fable exactly arrived at it is unclear.^[62]

Given the polynomial map ${\displaystyle F:\mathbb {C} ^{3}\to \mathbb {C} ^{3},}$ where $${\displaystyle {\begin{aligned}F(x,y,z)={}{\bigl (}~\!&(1+xy)^{3}z+y^{2}(1+xy)(4+3xy),\\&y+3x(1+xy)^{2}z+3xy^{2}(4+3xy),\\&2x-3x^{2}y-x^{3}z~\!{\bigr )}.\end{aligned}}}$$

The Jacobian determinant of this function is the constant −2. However, the map is not invertible, as it maps multiple distinct points to the same image. For example, Alpöge gives $${\displaystyle {\begin{aligned}F{\bigl (}0,0,-{\tfrac {1}{4}}{\bigr )}=F{\bigl (}1,-{\tfrac {3}{2}},{\tfrac {13}{2}}{\bigr )}=F{\bigl (}{-1},{\tfrac {3}{2}},{\tfrac {13}{2}}{\bigr )}&={\bigl (}{-{\tfrac {1}{4}}},0,0{\bigr )}.\end{aligned}}}$$ It was not necessary to find such points explicitly. If a polynomial map ${\displaystyle F:\mathbb {C} ^{n}\to \mathbb {C} ^{n}}$ is injective, then it is a polynomial automorphism.^[34] Consequently each component of ${\displaystyle F}$ is a coordinate polynomial, and is therefore irreducible.^[63] Since both ${\displaystyle F_{1}}$ and ${\displaystyle F_{3}}$ have nontrivial factorizations, ${\displaystyle F}$ cannot be injective.

Given the definition of ${\displaystyle F}$ above, it follows that for any integer ${\displaystyle n\geq 4}$, the polynomial map $${\displaystyle \mathbb {C} ^{n}\to \mathbb {C} ^{n},\quad (x_{1},x_{2},x_{3},x_{4},\dots ,x_{n})\mapsto (F(x_{1},x_{2},x_{3}),x_{4},\dots ,x_{n})}$$ then gives a counterexample in ${\displaystyle n}$ variables. The same holds for any other counterexample in dimension 3.

The day after Alpöge found the counterexample, a geometric reformulation was announced by Andy Jiang, a doctoral student in mathematics at the University of Michigan, who credited it to "GPT".^[64]

Terence Tao, a mathematician from the University of California, Los Angeles (UCLA), discussed this geometric explanation of the counterexample using multiplication of binary forms. A generic binary cubic has three linear factors, and hence three ways to express it as the product of a distinguished linear factor and the remaining quadratic factor. After imposing a resultant normalization to remove the scaling ambiguity and restricting to a particular affine slice of the space of binary cubics, Tao obtained an étale, generically three-to-one map ${\displaystyle X\to \mathbb {C} ^{3}}$ from a three-fold ${\displaystyle X}$ inside ${\displaystyle \mathbb {C} ^{5}}$ (identified with ${\displaystyle \mathbb {C} ^{2}\times \operatorname {Sym} ^{2}\mathbb {C} ^{2}}$, pairs of linear and quadratic binary forms) which is explicitly polynomially isomorphic to ${\displaystyle \mathbb {C} ^{3}}$.^[4] This cubic factorization construction is thus similar to the earlier quadratic example by Vitushkin.^[57]

Separable Jacobian conjecture

Kossivi Adjamagbo [ht] suggested extending the Jacobian conjecture to characteristic ${\displaystyle p>0}$ by adding the hypothesis that ${\displaystyle p}$ does not divide the degree of the field extension ${\displaystyle \mathbb {K} (X)/\mathbb {K} (F)}$.^[8] This conjecture is called Adjamagbo's Jacobian conjecture after Kossivi Adjamagbo,^[7] or the separable Jacobian conjecture because Adjamagbo's condition guarantees that the field extension is separable.^[5][6] By design, the separable Jacobian conjecture coincides with the usual Jacobian conjecture for fields of characteristic zero, because given a field ${\displaystyle \mathbb {K} }$ of characteristic zero, zero never divides the degree of the field extension ${\displaystyle \mathbb {K} (X)/\mathbb {K} (F)}$ for any polynomial map ${\displaystyle F}$ over ${\displaystyle \mathbb {K} ^{n}}$.

For the separable Jacobian conjecture in characteristic two, in a 2026 preprint, Irit Huq-Kuruvilla claimed to have produced a counterexample in dimensions ${\displaystyle n\geq 3}$,^[5] and in another 2026 preprint, Romy Mondello claimed to have produced a counterexample in two dimensions.^[6] Finally, in a third 2026 preprint, Alexander Borisov, Ofer Gabber, and Adrian Vasiu claimed to have produced counterexamples to Adjamagbo's separable Jacobian conjecture for all positive characteristics in all dimensions ${\displaystyle n\geq 2}$.^[7] The case ${\displaystyle n=1}$ is true for the separable Jacobian conjecture for any positive characteristic ${\displaystyle p}$, since Adjamagbo's hypothesis that ${\displaystyle p}$ does not divide the degree of the field extension ${\displaystyle \mathbb {K} (X)/\mathbb {K} (F)}$ of the polynomial ${\displaystyle F}$ guarantees that the derivative of the polynomial is a nonzero constant only if the degree of the polynomial is ${\displaystyle 1,}$ and linear polynomial functions are invertible.

Maubach and Rauf's Jacobian Conjecture

Stefan Maubach and Abdul Rauf suggested an alternative to Adjamabgo's Jacobian conjecture, where they extended the Jacobian conjecture to characteristic ${\displaystyle p>0}$ by replacing the condition of a Keller map in the premise with what they call a strong Keller map.^[65] Maubach and Rauf later in the same article show that for characteristic zero fields, every Keller map is a strong Keller map, and thus that their version of the Jacobian conjecture coincides with the usual Jacobian conjecture for characteristic zero fields.^[65]