The Jacobian Conjecture

Q: Why does the hypothesis say Δ F = 1 rather than Δ F ≠ 0?

Because Δ F ≠ 0 as a polynomial is far too weak: F(x,y) = (x, xy) satisfies it and is not invertible. The intended hypothesis is that Δ F is a unit of K[X] , that is, a non-zero constant, and rescaling then normalises it to 1 . Over C this is the same as Δ F having no zero in C^n ; over R it is not.

Q: If the conjecture is true, how large can the inverse be?

Bounded by (4.13): F^-1 ≤ ( F)^n-1 . This bound is a theorem about maps that are already known to be invertible, so it does not presuppose the conjecture, and it is what makes verification a finite search.

Q: Does the conjecture have any content for linear maps?

None. For F(p) = Ap the hypothesis is A ≠ 0 and the conclusion is that A^-1 exists, which is linear algebra. All the difficulty is in how the non-linear part can conspire with itself.

Q: What goes wrong in characteristic p?

Differentiation loses information: p -th powers have zero derivative, so the Jacobian matrix cannot see them. The map x_1 ↦ x_1 + x_1^,p has derivative 1 and is not invertible. The same phenomenon is what makes the Weyl algebra behave differently in characteristic p ; see the positive characteristic page .

Q: How does the conjecture connect to the Weyl algebra?

From a Keller map one builds n commuting derivations D_1, … ,D_n of K[x_1, … ,x_n] dual to F_1, … ,F_n , in the sense that D_i(F_j) = δ_ij . Sending x_i ↦ F_i and ∂_i ↦ D_i defines an endomorphism of A_n . If that endomorphism is an automorphism, the D_i turn out to be locally nilpotent , and a theorem on locally nilpotent derivations then gives K[F_1, … ,F_n] = K[x_1, … ,x_n] .

Overview

The chain rule shows that a polynomial map $F : K^n → K^n$ with a polynomial inverse must have $Δ F = J(F)$ equal to a non-zero constant; that argument is on the polynomial maps page and takes three lines. The Jacobian conjecture asserts the converse. It is one of the most conspicuously simple unsolved problems in algebra.

What makes it hard is that the hypothesis is pointwise and the conclusion is global. Non-vanishing of $Δ F$ gives an inverse near every point, formally or analytically. Producing one inverse valid everywhere, and polynomial, is a different order of statement. The conjecture is false over fields of positive characteristic and false over $ℝ$ if the hypothesis is weakened from "constant" to "nowhere zero", so any proof must use characteristic zero and algebraic closedness in an essential way.

Half of the conjecture is elementary and worth isolating: if $Δ F ≠ 0$ then the coordinate functions $F_1, … ,F_n$ are algebraically independent, so the comorphism $F^#$ is injective. The whole difficulty is surjectivity. That is why the conjecture is usually restated as the assertion that $F_1, … ,F_n$ generate the polynomial ring.

The reason the conjecture appears in a book on D-modules is that this last form is exactly what one gets by feeding $F$ into the Weyl algebra. Chapter 4 constructs from a Keller map an endomorphism of $A_n$ , and shows that if that endomorphism is an automorphism then $F_1, … ,F_n$ generate. So the Dixmier conjecture on endomorphisms of the Weyl algebra implies the Jacobian conjecture; they are now known to be equivalent.

Definition

Let $K$ be a field of characteristic zero, $X = K^n$ , $K[X] = K[x_1, … ,x_n]$ , and let $F : X \to X$ be a polynomial map with coordinate functions $F_1, … ,F_n$ .

Keller map

$F$ is a Keller map if $Δ F = J(F)$ is a non-zero element of $K$ . Replacing $F_1$ by $F_1 / Δ F$ changes nothing essential, so one may always normalise to $Δ F = 1$ .

The Jacobian conjecture (open)Coutinho (4.2.1); Keller 1939

Let $F : K^n → K^n$ be a polynomial map over a field of characteristic zero with

Δ F = 1 identically on K^n.

(4.7)

Then $F$ has an inverse which is a polynomial map defined on the whole of $K^n$ .

This is a conjecture, not a theorem. It is proved for $n = 1$ and open for every $n ≥ 2$ , including $n = 2$ .

Equivalent formCoutinho (4.2.3)

The conjecture is equivalent to the following statement. If $F : K^n → K^n$ is a polynomial map with $Δ F = 1$ , then

K[F_1, … , F_n] = K[x_1, … , x_n].

(4.8)

In words: the coordinate functions of a Keller map generate the polynomial ring.

Note

The equivalence is immediate from the dictionary of the previous page. Condition (4.8) says $F^#$ is surjective; injectivity is automatic by the lemma below; and a bijective $K$ -algebra homomorphism $K[X] → K[X]$ corresponds to a polynomial isomorphism by Coutinho (4.1.3).

Core Concepts

The case $n = 1$ , in full

Here $F$ is a single polynomial $F(x)$ and $J(F)$ is the $1 × 1$ matrix $(dF/dx)$ . The hypothesis says $dF/dx = 1$ , so $F = x + b$ for some $b ∈ K$ , an affine map with the obvious polynomial inverse. The conjecture is therefore true, and trivially so.

It is worth noticing exactly what the one-variable argument used: a polynomial whose derivative is constant has degree at most one. There is no analogue for $n ≥ 2$ . Elementary automorphisms such as $(x,y) ↦ (x + y^100, y)$ have $Δ F = 1$ and coordinate functions of arbitrarily large degree, so no degree bound can be read off the hypothesis. The one-dimensional proof gives no information whatever about the general case.

Local to global

Over $ℝ$ or $ℂ$ , the inverse function theorem provides an analytic local inverse near every point where $Δ F ≠ 0$ ; algebraically, the local inversion theorem provides a formal power series inverse near any point where the Jacobian determinant does not vanish. Both are cheap. Neither says the local inverses glue, and neither says the result is polynomial. The gap between them and the conjecture is the entire problem.

Why surjectivity is the whole content

The comorphism $F^#$ sends $y_i ↦ F_i$ , so its image is the subalgebra $K[F_1, … ,F_n]$ . Lemma (4.2.2) below shows the Jacobian hypothesis already forces injectivity: the $F_i$ satisfy no polynomial relation. What remains is whether the subalgebra they generate is everything. This is a genuinely different kind of question, one about generation rather than about non-degeneracy, and no invariant that only sees $Δ F$ can decide it.

Construction and Proof

The half that is a theorem

Algebraic independence of the coordinate functionsCoutinho (4.2.2)

Let $K$ have characteristic zero and let $F : X \to X$ be a polynomial map with $Δ F ≠ 0$ as an element of $K[X]$ . Then $F^#$ is injective; equivalently, $F_1, … ,F_n$ are algebraically independent over $K$ .

Proof

Suppose not. Among the non-constant polynomials $g ∈ K[y_1, … ,y_n]$ with $F^#(g) = g(F_1, … ,F_n) = 0$ , choose one of smallest total degree. Write $g_i = ∂ g / ∂ y_i$ .

Differentiate the identity $g(F_1, … ,F_n) = 0$ with respect to $x_j$ and use the chain rule (4.10). The result, for $j = 1, … ,n$ simultaneously, is the matrix equation (4.11): the row vector $v = (g_1(F), … ,g_n(F) )$ satisfies $v ⋅ J(F) = 0$ over $K[X]$ .

Since $Δ F ≠ 0$ , the matrix $J(F)$ is invertible over the field of rational functions $K(X)$ . Multiplying (4.11) on the right by $J(F)^-1$ gives $v = 0$ , that is, $g_i(F_1, … ,F_n) = 0$ for every $i$ .

Now examine each $g_i$ . It has total degree strictly less than $deg g$ . If $g_i$ were a non-zero constant $c$ , then $g_i(F_1, … ,F_n) = c ≠ 0$ , contradicting what we just proved. If $g_i$ were non-constant, it would be a non-constant element of $F^#$ of degree smaller than $deg g$ , contradicting minimality. So $g_i = 0$ for every $i$ .

In characteristic zero a polynomial all of whose partial derivatives vanish is constant. So $g$ is constant, contradicting the choice of $g$ . Hence $F^# = 0$ .

Two small strengthenings

The source states the hypothesis as " $Δ F(p) ≠ 0$ for every $p ∈ X$ " and then argues pointwise, concluding $v(p) = 0$ for all $p$ and hence $v = 0$ because $K$ is infinite. The version proved above assumes only that $Δ F$ is a non-zero polynomial, which is weaker, and works over the rational function field, so no appeal to infiniteness of $K$ is needed. Both routes are correct; the second is the one to remember, because in the geometric language the hypothesis " $Δ F ≠ 0$ " is just the statement that $F$ is dominant and separable.

The final step of the source's argument is compressed: it says only that $g_i$ has smaller degree than $g$ , which by itself is not yet a contradiction when $g_i$ is a non-zero constant. The two cases are separated explicitly above.

From the lemma to the reformulation

Proof that (4.7) and (4.8) are equivalent

Assume (4.8). Then $F^#$ is surjective, and it is injective by the lemma since $Δ F = 1 ≠ 0$ . So $F^#$ is an isomorphism of $K$ -algebras, and Coutinho (4.1.3) turns that into a polynomial inverse for $F$ , which is (4.7).

Conversely, if $F$ has a polynomial inverse $G$ , then $F^#$ is an isomorphism with inverse $G^#$ , so in particular surjective, and its image $K[F_1, … ,F_n]$ is all of $K[X]$ . Concretely, $x_j = G_j(F_1, … ,F_n)$ exhibits each variable as a polynomial in the coordinate functions.

Key Equations

Δ F = ( (∂ F_i)/(∂ x_j) )_1 ≤ i,j ≤ n ∈ K^ × .

(4.9)

The Keller hypothesis. Necessary for invertibility; conjecturally sufficient.

(∂)/(∂ x_j) ( g(F_1, … ,F_n) ) = ∑ _i=1^n (∂ g)/(∂ y_i)(F_1, … ,F_n) ⋅ (∂ F_i)/(∂ x_j).

(4.10)

The chain rule in the form used to prove algebraic independence.

v ⋅ J(F) = 0, v = ( ∂ g∂ y_1(F), ; … , ; ∂ g∂ y_n(F) ).

(4.11)

Equation (4.10) for all $j$ at once, with $g$ in the kernel of the comorphism. Invertibility of $J(F)$ over the field of rational functions forces $v = 0$ .

F = id + H, H homogeneous of degree 3, J(H) nilpotent.

(4.12)

The Bass-Connell-Wright and Yagzhev reduction: the conjecture in all dimensions follows from maps of this shape.

F^-1 ≤ ( F)^,n-1.

(4.13)

The degree bound on the inverse, valid whenever the inverse exists. It turns the conjecture into a statement about a bounded, finite-dimensional system of equations for each fixed $n$ and $deg F$ .

Variable Definitions

$K$: The ground field, of characteristic zero; usually $ℂ$ , where the conjecture is most often stated.
$F_1, … , F_n$: The coordinate functions of the map $F$ , polynomials in $x_1, … , x_n$ .
$J(F)$: The Jacobian matrix, with $(i,j)$ entry the partial derivative of $F_i$ with respect to $x_j$ .
$Δ F$: The Jacobian determinant $J(F)$ , an element of $K[x_1, … ,x_n]$ .
$F^#$: The comorphism, the $K$ -algebra homomorphism of $K[x_1, … ,x_n]$ sending $y_i$ to $F_i$ .
$K[F_1, … ,F_n]$: The subalgebra generated by the coordinate functions; the image of the comorphism.
$K(X)$: The field of rational functions in $x_1, … ,x_n$ , over which $J(F)$ becomes invertible as soon as $Δ F ≠ 0$ .
$H$: The non-linear part of a map written in the reduced form identity plus $H$ .

Properties and Behaviour

What is known

True for $n = 1$ , by the argument above, and vacuously for $n = 0$ .
Open for every $n ≥ 2$ . The two-variable case has been studied hardest and is still open. Moh verified it in 1983 for maps of $C^2$ of degree at most $100$ .
True for maps of degree at most two in any dimension, over any field of characteristic zero (Wang, 1980).
Reduction to cubic homogeneous maps. If the conjecture holds for all $n$ and all maps of the form (4.12), it holds in general (Bass-Connell-Wright 1982, following Yagzhev). The reduction increases the dimension, so it does not settle any fixed $n$ .
Injective implies bijective with polynomial inverse over $ℂ$ . This is a consequence of the Ax-Grothendieck theorem, recorded as Theorem 2.1 of the Bass-Connell-Wright survey. So over $ℂ$ it suffices to prove that a Keller map is injective.
Degree bound. When the inverse exists it satisfies (4.13). This is a theorem, proved without assuming the conjecture, and it is what makes verification for fixed $n$ and fixed degree a finite computation.
Equivalent to the Dixmier conjecture. Coutinho's Theorem (4.4.2) shows that the Dixmier conjecture in dimension $n$ implies the Jacobian conjecture in dimension $n$ ; Tsuchimoto and, independently, Belov-Kanel and Kontsevich proved the converse in the stable form that the Jacobian conjecture in dimension $2n$ implies the Dixmier conjecture in dimension $n$ . Both remain open. See the Dixmier conjecture page.

What is false

Positive characteristic. The map $F(x_1, … ,x_n) = (x_1 + x_1^,p,, x_2, … , x_n)$ over a field of characteristic $p$ has $J(F)$ equal to the identity matrix, since the derivative of $x_1^,p$ is zero, yet $F$ is not invertible.
The real conjecture with the weakened hypothesis. Pinchuk (1994) constructed a polynomial map $R^2 → R^2$ whose Jacobian determinant is strictly positive at every real point but which is not injective. This does not contradict the conjecture, because that Jacobian determinant is not constant, but it kills the tempting real version.

Examples and Special Cases

Elementary maps of any degree

$F(x,y) = (x + f(y),, y )$ has $J(F) = pmatrix 1 & f'(y) 0 & 1 pmatrix$ and $Δ F = 1$ , with inverse $(u - f(v),, v )$ . Since $f$ is arbitrary, the conjecture's hypothesis places no bound on the degree of a Keller map.

A map with non-constant Jacobian

$F(x,y) = (x,, xy)$ has $J(F) = pmatrix 1 & 0 y & x pmatrix$ and $Δ F = x$ . The hypothesis fails on the line $x = 0$ , and indeed $F$ collapses that line to a point and misses every point $(0, c)$ with $c ≠ 0$ . Note that $F^#$ is still injective, consistent with Lemma (4.2.2), which needs only $Δ F ≠ 0$ as a polynomial.

Characteristic $p$ Coutinho, Ch. 4 Ex. 5.4

Let $k$ have characteristic $p > 0$ and let $F(x_1, … ,x_n) = (x_1 + x_1^,p,, x_2, … , x_n)$ . Then $∂(x_1 + x_1^,p)/∂ x_1 = 1 + p,x_1^,p-1 = 1$ , so $J(F)$ is the identity and $Δ F = 1$ .

But $F$ is not invertible. The subalgebra $k[x_1 + x_1^,p]$ is generated by a single element of degree $p$ , so every non-constant element of it has degree divisible by $p$ ; in particular it does not contain $x_1$ . Hence $k[F_1, … ,F_n] ≠ k[x_1, … ,x_n]$ and (4.8) fails. Any proof of the conjecture must therefore use characteristic zero somewhere, and the natural place is the step "all partial derivatives vanish implies constant".

Pinchuk's real map

In 1994 Pinchuk produced explicit polynomials $P, Q ∈ R[x,y]$ , of degrees $10$ and $25$ , such that the map $(x,y) ↦ (P,Q)$ has Jacobian determinant strictly positive at every point of $R^2$ , yet is not injective. The determinant is a non-constant polynomial that happens to have no real zero, so the map is not a Keller map. The example shows that "the Jacobian never vanishes on real points" is genuinely weaker than "the Jacobian is constant", and that the difference is not a technicality.

Worked Example

A cubic Keller map of $K^2$ with nilpotent Jacobian, inverted by hand

Step 1 - the map
Write $s = x + y$ and set

$F(x,y) = ( x + s^3, ; y - s^3 ), s = x + y.$

This has the reduced shape (4.12): $F = id + H$ with $H = (s^3, -s^3)$ homogeneous of degree $3$ .
Step 2 - the non-linear part has nilpotent Jacobian
Since $∂ s / ∂ x = ∂ s / ∂ y = 1$ , every partial derivative of $s^3$ equals $3s^2$ . Hence

$J(H) = 3s^2 pmatrix 1 & 1 -1 & -1 pmatrix, J(H)^2 = 9s^4 pmatrix 1 & 1 -1 & -1 pmatrix^2 = 0,$

because the $2 × 2$ matrix shown squares to zero. So $J(H)$ is nilpotent of index $2$ .
Step 3 - the Jacobian determinant is 1
For any $2 × 2$ matrix $N$ one has $(I + N) = 1 + tr N + N$ . Here $tr J(H) = 3s^2 - 3s^2 = 0$ and $J(H) = -9s^4 + 9s^4 = 0$ , so

$Δ F = ( I + J(H) ) = 1 + 0 + 0 = 1.$

Written out, $J(F) = pmatrix 1 + 3s^2 & 3s^2 -3s^2 & 1 - 3s^2 pmatrix$ , whose determinant is $(1+3s^2)(1-3s^2) + 9s^4 = 1 - 9s^4 + 9s^4 = 1$ . So $F$ is a Keller map.
Step 4 - invert it
The key observation is that the two corrections cancel: $F_1 + F_2 = x + s^3 + y - s^3 = x + y = s$ . So the quantity $s$ is preserved by $F$ . Writing $(u,v) = F(x,y)$ we get $u + v = s$ , hence $x = u - (u+v)^3$ and $y = v + (u+v)^3$ , that is

$G(u,v) = ( u - (u+v)^3, ; v + (u+v)^3 ),$

which is again a polynomial map, and $G = F^-1$ by construction. Note $F = 3$ , $G = 3$ , comfortably inside the bound (4.13), which for $n = 2$ reads $G ≤ 3$ .
Step 5 - numerical check and the algebraic form
Take $(x,y) = (1,1)$ , so $s = 2$ and $s^3 = 8$ . Then $F(1,1) = (1 + 8,, 1 - 8) = (9, -7)$ . Now $u + v = 9 - 7 = 2$ and $(u+v)^3 = 8$ , so $G(9,-7) = (9 - 8,, -7 + 8) = (1,1)$ . The point returns.

In the form (4.8): $K[F_1,F_2]$ contains $F_1 + F_2 = x + y$ , hence contains $(x+y)^3$ , hence contains $F_1 - (x+y)^3 = x$ and therefore $y$ as well. So $K[F_1,F_2] = K[x,y]$ , verifying the conjecture for this map directly.

Result

$F(x,y) = (x + (x+y)^3,; y - (x+y)^3 )$ satisfies $Δ F = 1$ and is invertible, with inverse $G(u,v) = (u - (u+v)^3,; v + (u+v)^3 )$ . It is a cubic homogeneous Keller map of the exact type to which the general conjecture reduces, and the mechanism that makes it work - a linear form annihilated by the non-linear part - is the only mechanism understood in general.

Applications and Industry Use

In a mathematics topic, this section covers downstream use inside mathematics, computing and engineering rather than a manufactured product.

Endomorphisms of the Weyl algebra. The route taken in Chapter 4 turns a Keller map into an endomorphism of $A_n$ and reduces the conjecture to a question about automorphisms of the Weyl algebra. This is the only known link between the conjecture and non-commutative algebra, and it is the reason the problem appears in a D-module text at all.
Affine algebraic geometry. The conjecture is a testing ground for questions about the automorphism group of affine space, tameness, and the structure of étale endomorphisms; progress on it usually produces progress on those.
Dynamical systems. Keller maps arise as time-one maps and as changes of coordinates for polynomial vector fields; the Markus-Yamabe problem on global asymptotic stability is a close relative, and its planar case was settled while the Jacobian conjecture was not.
Computer algebra benchmarks. Verifying a Keller map's invertibility is a standard stress test for elimination and Gröbner-basis implementations; the degree bound (4.13) is what makes the test terminate.
Cryptography. Multivariate schemes rest on the difficulty of inverting a polynomial map that the designer can invert; the conjecture's landscape - easy necessary conditions, hard sufficiency - is the same landscape those schemes live in, though over finite fields where the conjecture is false.

Standards and Codes

For mathematics, the relevant standards are notation, numeric and markup standards together with reference implementations.

ISO 80000-2 governs the notation used here for the number systems $ℝ$ and $ℂ$ , for partial derivatives, and for $det$ and $tr$ set in upright roman type.
ISO/IEC 40314 (MathML 3.0) is the encoding of every expression on this page.
There is no standardised name. "Jacobian conjecture", "Keller's problem" and "the Keller-Jacobian problem" all appear in the literature; "Keller map" for a map satisfying the hypothesis is now standard and is used here.
The problem appears as Problem 16 in Smale's 1998 list of mathematical problems for the next century, which is the usual citation when its status is being asserted rather than used.

Computational Notes

Read this as the manufacturing section of the template: how the object is actually built by machine, at what cost, and where the computation stops being decidable.

Testing the hypothesis is cheap: form $J(F)$ symbolically and take a determinant. If $Δ F$ is not a non-zero constant, stop.
Testing the conclusion is an elimination problem. Compute a Gröbner basis of the ideal generated by $u_1 - F_1, … , u_n - F_n$ in $K[x,u]$ under an order eliminating the $x$ variables; if $F$ is invertible the basis contains elements $x_j - G_j(u)$ exhibiting the inverse. The bound (4.13) caps $G_j$ , so a failure to find such elements within that degree is a proof of non-invertibility, not merely a timeout - provided the Gröbner computation itself completed.
Undetermined coefficients is a competitive alternative for small cases: posit $G_j$ with unknown coefficients up to degree $( F)^n-1$ , impose $G(F(x)) = x$ , and solve the resulting polynomial system. For $n = 2$ and $F = 3$ this is a small linear-in-the-unknowns problem after grading.
Scale. Verification is routine for $n = 2$ and low degree, and becomes hopeless quickly: the number of coefficients in a general map of degree $d$ in $n$ variables grows like $n n+dn$ , and the elimination cost is worse than exponential in that.
Software. Singular, Macaulay2 and SageMath all handle the elimination directly; van den Essen's monograph documents the algorithms used to search for counterexamples in the cubic homogeneous family. See Singular and SageMath.
A caution about numerics. Floating-point evaluation of $Δ F$ at sample points can suggest constancy that exact arithmetic refutes, and vice versa. Every claim about a Keller map should be checked symbolically.

Limits of Validity

Characteristic zero cannot be dropped. The counterexample in characteristic $p$ is elementary and definitive.
The hypothesis must be " $Δ F$ constant", not " $Δ F$ nowhere zero", unless $K$ is algebraically closed. Over $ℂ$ the Nullstellensatz makes the two agree; over $ℝ$ Pinchuk's map separates them.
Verified cases do not accumulate into a proof. Truth for degree at most two, and for degree at most $100$ in the plane, leaves the statement open, because there is no induction on degree available.
The cubic reduction changes the dimension. It proves "conjecture for all $n$ in cubic homogeneous form implies conjecture for all $n$ ", not the corresponding statement for a single fixed $n$ .
The equivalence with the Dixmier conjecture is not a solution. It relates two open problems; the transfer in one direction also shifts dimension, from $2n$ to $n$ .
Nothing here decides bijectivity of a general polynomial map. Deciding whether a given polynomial map of $C^n$ is bijective is algorithmically feasible for small inputs but has no known efficient method; the conjecture would give a one-line criterion, which is part of why it is attractive.

Failure Modes and Common Mistakes

Confusing injectivity of $F$ with injectivity of $F^#$

Lemma (4.2.2) concludes that $F^#$ is injective. That is a statement about polynomials, equivalent to algebraic independence of $F_1, … ,F_n$ , and equivalent geometrically to $F$ having dense image. It is not the statement that $F$ is injective on points. The map $F(x,y) = (x, xy)$ has injective comorphism and is not injective on points; the map $F(x) = x^2$ over $ℂ$ likewise. Over $ℂ$ injectivity of $F$ is what would finish the conjecture, and it is precisely what is not known.

Assuming a local inverse can be glued

The inverse function theorem is often invoked as though it almost proved the conjecture. It does not. Its output is a germ at each point, with no compatibility across points and no polynomiality. Over $ℝ$ , $F(x,y) = (x^3 + x,, y)$ has local inverses everywhere that do glue into a global inverse which is nevertheless not polynomial; Pinchuk's map has local inverses everywhere that do not glue into anything injective. Both failure modes occur.

The compressed final step of the standard proof

In the usual write-up of Lemma (4.2.2) the contradiction is stated as " $g_i$ has degree smaller than $g$ ". Taken literally that is not yet a contradiction, since $g_i$ could be a non-zero constant, which is not in the kernel by minimality of a non-constant choice. The constant case has to be dismissed separately, by observing that a non-zero constant is not sent to zero by an algebra homomorphism. The proof above splits the cases.

Treating published proofs as settled

The Jacobian conjecture has attracted an unusually large number of announced proofs, several by capable mathematicians, all of which have been withdrawn or refuted. Before relying on a claimed proof, check that it addresses the two known obstructions: it must fail in characteristic $p$ , and it must fail for Pinchuk's map. A proposed argument that does not visibly use characteristic zero and algebraic closedness is wrong.

Reading the reduction to degree three as a reduction of difficulty

It is easy to conclude from (4.12) that only cubic maps matter and therefore that a finite computation should settle the matter. It should not. The reduction is uniform in $n$ but requires all $n$ ; the dimension in the reduced problem is not the dimension you started in, and there is no bound on how far it grows.

Historical Notes

Ott-Heinrich Keller asked the question in 1939, in the setting of maps with integer coefficients, and the modern form over $ℂ$ emerged from that. Białynicki-Birula and Rosenlicht showed in 1962 that an injective polynomial self-map of complex affine space is surjective, which was later absorbed into the Ax-Grothendieck theorem and reduces the conjecture over $ℂ$ to injectivity.

Wang settled the case of degree at most two in 1980. Independently, Yagzhev in 1980 and Bass, Connell and Wright in 1982 obtained the reduction to cubic homogeneous maps, together with the degree bound on inverses; the 1982 survey remains the standard reference. Moh verified the planar case up to degree $100$ in 1983.

Pinchuk's 1994 example closed off the real version with the weakened hypothesis. Smale included the conjecture in his 1998 list of problems. In 2005 Tsuchimoto, and in 2007 Belov-Kanel and Kontsevich by a different route, proved that the Jacobian conjecture in dimension $2n$ implies the Dixmier conjecture in dimension $n$ ; combined with the implication in Coutinho's Chapter 4 this makes the two families of conjectures equivalent.

The conjecture is also notorious for the number of incorrect proofs it has attracted; van den Essen's monograph devotes space to the recurring errors, which is unusual for a research monograph and is itself a comment on the problem.

Comparison

Status of the conjecture and of its near neighbours.
Statement	Hypothesis	Status
Jacobian conjecture, $n = 1$	$Δ F ∈ K^ ×$ , char $0$	True, elementary
Jacobian conjecture, $n ≥ 2$	$Δ F ∈ K^ ×$ , char $0$	Open, including $n = 2$
Degree at most two, any $n$	$Δ F ∈ K^ ×$ , char $0$	True (Wang 1980)
Plane, degree at most 100	$Δ F ∈ C^ ×$	True (Moh 1983)
Cubic homogeneous form	$F = id + H$ , $J(H)$ nilpotent	Open; implies the general case
Positive characteristic	$Δ F = 1$ , char $p$	False
Real, non-vanishing Jacobian	$Δ F(p) ≠ 0$ on $R^n$	False (Pinchuk 1994)
Dixmier conjecture	Every endomorphism of $A_n$ is onto	Open; equivalent up to doubling the dimension
Injectivity over $ℂ$	$F$ injective	Implies bijectivity with polynomial inverse

Key Takeaways

Key takeaways

The conjecture: over a field of characteristic zero, $Δ F = 1$ should force $F$ to have a polynomial inverse. The converse implication is an easy theorem.
Equivalently, the coordinate functions of a Keller map should generate the polynomial ring: $K[F_1, … ,F_n] = K[x_1, … ,x_n]$ .
Algebraic independence of $F_1, … ,F_n$ follows from $Δ F ≠ 0$ and is proved by the chain rule plus a minimal-degree argument. Surjectivity of the comorphism is the entire open part.
True for $n = 1$ and for degree at most two; open for all $n ≥ 2$ ; false in characteristic $p$ ; false over $ℝ$ under the weakened hypothesis.
The general case reduces to maps identity-plus-homogeneous-cubic with nilpotent Jacobian, but the reduction raises the dimension.
It is equivalent, up to doubling the dimension, to the Dixmier conjecture on endomorphisms of the Weyl algebra - which is why it belongs in a D-module course.

FAQs

Is the conjecture known for

n = 2

No. The planar case is open. It has been verified for maps of degree at most $100$ (Moh, 1983) and many structural results are known, but there is no proof.

Why does the hypothesis say

Δ F = 1

rather than

Δ F ≠ 0

Because $Δ F ≠ 0$ as a polynomial is far too weak: $F(x,y) = (x, xy)$ satisfies it and is not invertible. The intended hypothesis is that $Δ F$ is a unit of $K[X]$ , that is, a non-zero constant, and rescaling then normalises it to $1$ . Over $ℂ$ this is the same as $Δ F$ having no zero in $C^n$ ; over $ℝ$ it is not.

If the conjecture is true, how large can the inverse be?

Bounded by (4.13): $F^-1 ≤ ( F)^n-1$ . This bound is a theorem about maps that are already known to be invertible, so it does not presuppose the conjecture, and it is what makes verification a finite search.

Does the conjecture have any content for linear maps?

None. For $F(p) = Ap$ the hypothesis is $A ≠ 0$ and the conclusion is that $A^-1$ exists, which is linear algebra. All the difficulty is in how the non-linear part can conspire with itself.

What goes wrong in characteristic

p

Differentiation loses information: $p$ -th powers have zero derivative, so the Jacobian matrix cannot see them. The map $x_1 ↦ x_1 + x_1^,p$ has derivative $1$ and is not invertible. The same phenomenon is what makes the Weyl algebra behave differently in characteristic $p$ ; see the positive characteristic page.

How does the conjecture connect to the Weyl algebra?

From a Keller map one builds $n$ commuting derivations $D_1, … ,D_n$ of $K[x_1, … ,x_n]$ dual to $F_1, … ,F_n$ , in the sense that $D_i(F_j) = δ_ij$ . Sending $x_i ↦ F_i$ and $∂_i ↦ D_i$ defines an endomorphism of $A_n$ . If that endomorphism is an automorphism, the $D_i$ turn out to be locally nilpotent, and a theorem on locally nilpotent derivations then gives $K[F_1, … ,F_n] = K[x_1, … ,x_n]$ .

Is Pinchuk's map a counterexample to the conjecture?

No, and it is important not to say so. Its Jacobian determinant is a non-constant polynomial that happens to be positive at every real point, so it is not a Keller map. It refutes only the weakened real statement, and it shows why the distinction between "non-vanishing" and "constant" has to be maintained.

Would a proof for all

n

of the cubic homogeneous case settle everything?

Yes, that is exactly the content of the Bass-Connell-Wright and Yagzhev reduction. But it must be for all $n$ simultaneously; settling cubic maps in a fixed dimension settles only that dimension's cubic maps.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 4 §2, statements (4.2.1) and (4.2.3), Lemma (4.2.2), Exercises 5.3-5.5.
O.-H. Keller, Ganze Cremona-Transformationen, Monatshefte für Mathematik und Physik 47 (1939), 299-306 - the original question.
H. Bass, E. H. Connell and D. Wright, The Jacobian conjecture: reduction of degree and formal expansion of the inverse, Bulletin of the American Mathematical Society 7 (1982), 287-330 - the reduction to cubic homogeneous maps and the degree bound on the inverse.
S. S.-S. Wang, A Jacobian criterion for separability, Journal of Algebra 65 (1980), 453-494 - the conjecture for maps of degree at most two.
T. T. Moh, On the Jacobian conjecture and the configurations of roots, Journal für die reine und angewandte Mathematik 340 (1983), 140-212 - the planar case up to degree 100.
S. Pinchuk, A counterexample to the strong real Jacobian conjecture, Mathematische Zeitschrift 217 (1994), 1-4.
A. van den Essen, Polynomial Automorphisms and the Jacobian Conjecture, Progress in Mathematics 190, Birkhäuser, 2000 - the standard monograph, with algorithms and a discussion of failed proofs.
S. Smale, Mathematical problems for the next century, The Mathematical Intelligencer 20 (1998), 7-15 - Problem 16.
Y. Tsuchimoto, Endomorphisms of Weyl algebra and $p$ -curvatures, Osaka Journal of Mathematics 42 (2005), 435-452; A. Belov-Kanel and M. Kontsevich, The Jacobian conjecture is stably equivalent to the Dixmier conjecture, Moscow Mathematical Journal 7 (2007), 209-218.
ISO 80000-2, Quantities and units - Part 2: Mathematics; ISO/IEC 40314, Mathematical Markup Language (MathML) Version 3.0.

AI Suggested Questions

Prove that a polynomial map of $K^1$ with constant non-zero derivative is affine, and explain why no analogue holds for $n = 2$ .
Construct a cubic homogeneous map of $K^3$ with nilpotent Jacobian and verify by hand that it is invertible.
Show that $F(x,y) = (x, xy)$ has injective comorphism but is not injective on points, and identify the image.
Work out why the map $x_1 ↦ x_1 + x_1^,p$ fails to be invertible in characteristic $p$ , using degrees rather than the given hint.
Explain precisely which step of Lemma (4.2.2) uses characteristic zero, and what replaces it in characteristic $p$ .
Set up the undetermined-coefficient system that verifies the conjecture for all maps of $C^2$ of degree $3$ , and count the unknowns.
State the Ax-Grothendieck theorem and derive the reduction of the conjecture over $ℂ$ to injectivity.
Describe what Pinchuk's example does and does not show, and why its Jacobian determinant cannot be constant.