The Dixmier Conjecture and Its Link to the Jacobian Conjecture

Q: If injectivity is automatic, why is surjectivity hard?

Because A_n is infinite-dimensional, and for infinite-dimensional algebras injectivity implies nothing about the image. The subalgebra generated by ϕ(x_i) and ϕ(∂_i) could in principle be proper, exactly as K[x^2] ⊂ neq K[x] .

Q: Where exactly is the conjecture used in the proof of (4.4.2)?

In one place only: to know that every c ∈ A_n is of the form ϕ(b) , so that the equivariance identity (4.18) can transport local nilpotence from ad_∂_i to ad_D_i . Nothing else needs ϕ to be surjective.

Q: Why do the derivations D_i have polynomial coefficients only when Δ = 1?

Formula (4.14) has Δ^-1 in front. In general the D_i live in K[x][Δ^-1] . When Δ is a non-zero constant, that ring is K[x] itself, so the D_i are honest elements of A_n and there is an endomorphism to build. This is the step that needs the full Keller hypothesis rather than mere non-vanishing.

Q: Is DC_1 really open?

Yes. It is not known whether every pair u, v ∈ A_1 with [v,u] = 1 generates A_1 . This is one of the cleanest unsolved statements in non-commutative algebra, and it is a much harder question than its commutative analogue, which is a one-line exercise.

Q: Does the Dixmier conjecture follow from the Jacobian conjecture?

Yes, but with a shift in dimension: JC_2n implies DC_n . Since the Jacobian conjecture is open in every dimension from 2 upwards, this does not settle anything; it establishes that the two families are equivalent.

Q: What is the role of the local inversion theorem here?

It is used inside Lemma (4.4.1) to show the D_i commute. The commutator is a derivation vanishing on all the F_k , and one needs to know that the F_k , after shifting to make them vanish at the origin, generate the whole formal power series ring. That is exactly what the local inversion theorem supplies.

Overview

In the last of the problems listed at the end of his 1968 paper on the Weyl algebras, Dixmier asked whether every endomorphism of $A_n$ is an automorphism. Half the question is already settled: an endomorphism of $A_n$ has a kernel that is a two-sided ideal, and $A_n$ is simple, so the kernel is zero and every endomorphism is injective. What is open is surjectivity, and it is open already for $A_1$ .

That is a striking contrast with the commutative situation. The Jacobian conjecture is trivially true in dimension one, because a polynomial with constant derivative is affine. The Weyl algebra analogue in dimension one is not known. Whatever makes $A_1$ hard is not visible in $K[x]$ .

The point of this page is a conditional theorem, due in this form to an argument of Vaserstein and Katz. Take a Keller map $F$ of $K^n$ . Cramer's rule produces $n$ derivations $D_1, … ,D_n$ of $K[x_1, … ,x_n]$ dual to $F_1, … ,F_n$ , and these satisfy exactly the Weyl relations. So there is an endomorphism $ϕ$ of $A_n$ with $ϕ(x_i) = F_i$ and $ϕ(∂_i) = D_i$ . If $ϕ$ is an automorphism, the $D_i$ inherit local nilpotence from the $∂_i$ , and a structure theorem on locally nilpotent derivations then gives $K[F_1, … ,F_n] = K[x_1, … ,x_n]$ , which is the Jacobian conjecture.

The implication was later reversed. Tsuchimoto in 2005 and, by a different route, Belov-Kanel and Kontsevich in 2007 proved that the Jacobian conjecture in dimension $2n$ implies the Dixmier conjecture in dimension $n$ . The two problems are therefore equivalent as families of statements, and both remain open.

Definition

Throughout, $K$ has characteristic zero, $A_n = A_n(K)$ is the $n$ -th Weyl algebra with generators $x_1, … ,x_n,∂_1, … ,∂_n$ , and an endomorphism means a $K$ -algebra homomorphism $A_n → A_n$ sending $1$ to $1$ .

The Dixmier conjectureDixmier 1968, Problème 11.1

$DC_n$ is the statement: every $K$ -algebra endomorphism of $A_n(K)$ is an automorphism.

Because injectivity is automatic, $DC_n$ is equivalent to: every endomorphism of $A_n$ is surjective.

Injectivity is freeCoutinho (2.2.2)

Every endomorphism $ϕ$ of $A_n$ is injective. Indeed $ker ϕ$ is a two-sided ideal, and $ϕ(1) = 1 ≠ 0$ so $ϕ ≠ A_n$ ; since $A_n$ is simple, $ϕ = 0$ .

The dual derivations of a Keller mapCoutinho, Ch. 4 §4

Let $F : K^n → K^n$ be a polynomial map with $Δ = Δ F$ nowhere zero. For $g ∈ K(x_1, … ,x_n)$ define

D_i(g) = Δ^-1 J ( F_1, … , F_i-1,, g,, F_i+1, … , F_n ),

(4.14)

the determinant of the matrix obtained from $J(F)$ by replacing its $i$ -th row with the gradient of $g$ . Each $D_i$ is a $K$ -linear map satisfying Leibniz's rule, hence a derivation of the rational function field, and it restricts to a derivation of $K[x_1, … ,x_n][Δ^-1]$ . When $Δ ∈ K^ ×$ it restricts to a derivation of $K[x_1, … ,x_n]$ itself.

Note

Equation (4.14) is Cramer's rule. In matrix form, if $C = (J(F)^-1)^T$ then $D_i = ∑ _j C_ij, ∂_j$ : the $D_i$ are the partial derivatives "with respect to the $F_j$ ", which is why the duality relation below holds.

Core Concepts

Why injective does not imply surjective for free

In finite dimensions an injective linear map is surjective. $A_n$ is infinite-dimensional over $K$ , and the implication genuinely fails for infinite-dimensional algebras: the map $K[x] → K[x]$ , $x ↦ x^2$ , is an injective algebra endomorphism whose image is a proper subalgebra. So the fact that endomorphisms of $A_n$ are injective is no evidence at all for Dixmier's question; it merely tells us where the difficulty lies.

The adjoint action and why it detects local nilpotence

Give $A_n$ the Bernstein degree, so that $(x^α ∂^β) = |α| + |β|$ . Commuting with $∂_i$ lowers degree: $[∂_i, x^α ∂^β] = α_i, x^α - e_i ∂^β$ , so $[∂_i, b] ≤ b - 1$ for every $b$ . Iterating, $ad_∂_i$ kills every element of $A_n$ after finitely many steps: it is locally nilpotent.

Local nilpotence is a property that transports along a surjective homomorphism, because $ϕ ∘ ad_a = ad_ϕ(a) ∘ ϕ$ . If $ϕ$ is onto, every element of $A_n$ is some $ϕ(b)$ , and $(ad_D_i)^k(ϕ(b)) = ϕ ((ad_∂_i)^k(b) ) = 0$ for large $k$ . This is the only place the conjecture is used, and it is used exactly for surjectivity - which is precisely the open half.

From operators back to functions

Restricting $ad_D_i$ to the subalgebra $K[x_1, … ,x_n] ⊂ eq A_n$ of multiplication operators recovers the derivation: for $f$ a polynomial, $[D_i, f] = D_i(f)$ as an operator, so $(ad_D_i)^k(f) = D_i^,k(f)$ . Local nilpotence of the adjoint therefore means exactly local nilpotence of $D_i$ as a derivation of the polynomial ring, and that is the hypothesis the commutative structure theorem needs.

Construction and Proof

Step 1: the dual derivations and their relations

Duality and commutativityCoutinho (4.4.1)

Let $F$ be a polynomial map of $K^n$ with $Δ$ nowhere zero, and let $D_1, … ,D_n$ be defined by (4.14). As derivations of $K[x_1, … ,x_n][Δ^-1]$ they satisfy $D_i(F_j) = δ_ij$ and $[D_i,D_j] = 0$ .

Outline

The duality relation is immediate from (4.14): putting $g = F_j$ reproduces the $j$ -th row of $J(F)$ in the $i$ -th position, so the determinant is $Δ$ when $i = j$ and has a repeated row, hence vanishes, when $i ≠ j$ .

Commutativity is the substantial half, and the argument passes to formal power series. Since $Δ(0) ≠ 0$ , $Δ$ is invertible in $K[[x_1, … ,x_n]]$ , so $K[x][Δ^-1]$ embeds there and each $D_i$ extends to a derivation of the power series ring. Put $B = [D_i,D_j]$ , again a derivation. By the duality relation $B(F_k) = 0$ for every $k$ , so $B$ vanishes on the subalgebra generated by $F_1, … ,F_n$ , and by continuity on the completed subalgebra of power series in $F_1 - F_1(0), … , F_n - F_n(0)$ . The local inversion theorem applies to that shifted tuple, whose Jacobian matrix agrees with $J(F)$ and is therefore invertible at the origin, and yields

K[[x_1, … ,x_n]] = K [ [, F_1 - F_1(0), … , F_n - F_n(0), ] ].

So $B$ vanishes on all of $K[[x_1, … ,x_n]]$ , hence on $K[x][Δ^-1]$ . This is an outline: the continuity step and the verification that $D_i$ extends are routine but not one-line, and are carried out in the source.

Step 2: the endomorphism of $A_n$

Now assume $Δ = 1$ . Then $J(F)^-1$ has polynomial entries, so each $D_i$ is a derivation of $K[x_1, … ,x_n]$ itself, that is, an element of $A_n$ of Bernstein degree at most $1 + _j (J(F)^-1)_ji$ . Inside $A_n$ , regarding $F_j$ as a multiplication operator, the relation $D_i(F_j) = δ_ij$ becomes the commutator relation $[D_i, F_j] = δ_ij$ . Together with $[F_i,F_j] = 0$ and $[D_i,D_j] = 0$ these are exactly relations (4.15), which are the defining relations of the Weyl algebra. By the presentation of $A_n$ by generators and relations, Coutinho (1.3.1), there is a unique $K$ -algebra endomorphism $ϕ$ of $A_n$ given by (4.16).

Step 3: the conditional theorem

Dixmier implies JacobianCoutinho (4.4.2); after Vaserstein and Katz

Let $K$ have characteristic zero and let $F : K^n → K^n$ be a polynomial map with $Δ F = 1$ . If every endomorphism of $A_n(K)$ is an automorphism, then $K[F_1, … ,F_n] = K[x_1, … ,x_n]$ ; that is, $DC_n$ implies $JC_n$ .

Proof

Build $ϕ$ as in Step 2. By (4.17) the map $ad_∂_i$ lowers Bernstein degree, so for each $b ∈ A_n$ there is $k$ with $(ad_∂_i)^k(b) = 0$ .

Assume $ϕ$ is an automorphism. Given $c ∈ A_n$ , write $c = ϕ(b)$ and choose $k$ with $(ad_∂_i)^k(b) = 0$ . Then by (4.18), $(ad_D_i)^k(c) = ϕ ((ad_∂_i)^k(b) ) = 0$ . So $ad_D_i$ is locally nilpotent on $A_n$ .

Restrict to polynomials. For $f ∈ K[x_1, … ,x_n]$ one has $[D_i, f] = D_i(f)$ in $A_n$ , so $(ad_D_i)^k(f) = D_i^,k(f)$ . Hence each $D_i$ is a locally nilpotent derivation of $K[x_1, … ,x_n]$ .

The $D_i$ commute, are locally nilpotent, and satisfy $D_i(F_j) = δ_ij$ . The structure theorem (4.3.1) therefore applies with $S = K[x_1, … ,x_n]$ and $t_i = F_i$ , giving $S = R[F_1, … ,F_n]$ where $R = ⋂ _i D_i$ .

It remains to identify $R$ . Since $Δ = 1$ , the matrix $C = (J(F)^-1)^T$ has entries in $K[x_1, … ,x_n]$ and is invertible over that ring, with inverse $J(F)^T$ . From $D_i = ∑ _j C_ij ∂_j$ we may solve back: $∂_j = ∑ _i (J(F)^T)_ji D_i$ . So any $f$ killed by every $D_i$ is killed by every $∂_j$ , and in characteristic zero that forces $f ∈ K$ . Hence $R = K$ and

K[F_1, … ,F_n] = K[x_1, … ,x_n],

which is the Jacobian conjecture in the form Coutinho (4.2.3).

A gap worth filling

The source concludes directly from (4.3.1) that $K[F_1, … ,F_n] = K[x_1, … ,x_n]$ . Strictly, (4.3.1) delivers only $S = R[F_1, … ,F_n]$ with $R$ the ring of constants, and the identification $R = K$ has to be made. The argument above does it in one line from invertibility of $J(F)$ over the polynomial ring. An alternative route counts transcendence degrees: $Frac(S) = Frac(R)(F_1, … ,F_n)$ with the $F_i$ algebraically independent, so $R$ is algebraic over $K$ , and an element of a polynomial ring algebraic over $K$ has degree zero.

Key Equations

D_i(F_j) = δ_ij, [D_i, D_j] = 0, [F_i, F_j] = 0.

(4.15)

The relations satisfied by the dual derivations and the coordinate functions of a Keller map. These are the defining relations of $A_n$ .

ϕ : A_n → A_n, ϕ(x_i) = F_i, ϕ(∂_i) = D_i.

(4.16)

The endomorphism produced by (4.15) together with the presentation of $A_n$ by generators and relations, Coutinho (1.3.1).

( [∂_i, b] ) ≤ b - 1 for all b ∈ A_n.

(4.17)

Bernstein degree drops under $ad_∂_i$ , so $ad_∂_i$ is locally nilpotent.

ϕ ( ad_∂_i(b) ) = ad_D_i ( ϕ(b) ).

(4.18)

Equivariance of the adjoint action, the identity that transports local nilpotence across $ϕ$ .

S = R[t_1, … , t_n], D_i = ∂ / ∂ t_i, R = ⋂ _i D_i.

(4.19)

The conclusion of the structure theorem for commuting locally nilpotent derivations admitting a dual system, Coutinho (4.3.1), after Wright (1981).

DC_n ⟹ JC_n, JC_2n ⟹ DC_n.

(4.20)

The two implications. The first is proved below; the second is due to Tsuchimoto and to Belov-Kanel and Kontsevich. Together they make the two families of conjectures equivalent.

Variable Definitions

$A_n$: The n-th Weyl algebra over a field of characteristic zero, generated by $x_1, … ,x_n,∂_1, … ,∂_n$ .
$F_1, … ,F_n$: The coordinate functions of a polynomial map $F$ of $K^n$ , viewed inside $A_n$ as multiplication operators.
$Δ$: The Jacobian determinant $J(F)$ ; assumed nowhere zero, and equal to $1$ in the main theorem.
$D_i$: The derivation defined by (4.14), the partial derivative with respect to $F_i$ in the coordinate system given by $F$ .
$ϕ$: The endomorphism of $A_n$ with $ϕ(x_i) = F_i$ and $ϕ(∂_i) = D_i$ .
$ad_a$: The map sending $b$ to $[a,b] = ab - ba$ ; $K$ -linear and a derivation of $A_n$ , but not an algebra homomorphism.
$R$: The ring of constants, the intersection of the kernels of $D_1, … ,D_n$ inside $K[x_1, … ,x_n]$ .
$JC_n$ , $DC_n$: The Jacobian conjecture and the Dixmier conjecture in dimension n.

Properties and Behaviour

Injectivity is automatic and gives nothing. Every endomorphism of $A_n$ is injective by simplicity, so $DC_n$ is purely a surjectivity statement.
$DC_1$ is open. This is the sharpest way to feel the difficulty: the commutative shadow $JC_1$ is a one-line exercise, and the non-commutative statement in the same dimension is unsolved.
The automorphism group of $A_1$ is understood. Dixmier determined it in the same 1968 paper: $Aut(A_1(C))$ is generated by the maps fixing $x$ and sending $∂ ↦ ∂ + g(x)$ , the maps fixing $∂$ and sending $x ↦ x + f(∂)$ , and the scalings. Makar-Limanov later gave it an amalgamated free product structure parallel to that of $Aut(K[x,y])$ . Knowing the automorphisms does not identify the endomorphisms.
An endomorphism is determined by $2n$ elements satisfying the Weyl relations. Conversely any such family defines one, so $DC_n$ says: if $u_1, … ,u_n,v_1, … ,v_n ∈ A_n$ satisfy $[v_i,u_j] = δ_ij$ and commute otherwise, they generate $A_n$ .
Endomorphisms need not preserve the Bernstein filtration degreewise, but they do send degree- $d$ elements to elements of degree at most $d ⋅ _i ϕ(x_i), ϕ(∂_i)$ , which is what makes degree-based attacks possible in principle.
The two conjectures are stably equivalent, by (4.20). Neither implication is dimension-preserving in both directions, so a proof of $JC_2$ alone would not settle $DC_2$ ; it would settle $DC_1$ .
Characteristic zero is assumed throughout. In characteristic $p$ the Weyl algebra has a large centre and is not simple, so even the injectivity step disappears; that regime is genuinely different and is treated on the positive characteristic page. Curiously, reduction modulo $p$ is exactly the tool Tsuchimoto used to prove the reverse implication in (4.20).

Examples and Special Cases

Endomorphisms that are visibly automorphisms

Fix $f ∈ K[x_1, … ,x_n]$ and set $ϕ(x_i) = x_i$ for all $i$ , $ϕ(∂_i) = ∂_i + ∂ f/∂ x_i$ . The relations (4.15) hold because mixed partials commute, so $ϕ$ is an endomorphism; its inverse is the same construction with $-f$ . This is the family in Coutinho (1.3.2), and it is the source of most explicitly constructed automorphisms of $A_n$ .

The endomorphism attached to a genuine automorphism of $K^n$

If $F$ is already known to be a polynomial automorphism, the construction of Step 2 gives an endomorphism $ϕ$ that is an automorphism, since one can build the inverse from $F^-1$ in the same way. So the construction is consistent, and the theorem has no content for maps already known to be invertible; its whole force is that it runs in the other direction.

Not every pair of elements works

In $A_1$ there is no $D$ with $[D, x^2] = 1$ . Write $D = ∑ _k=0^d a_k(x)∂^k$ with $a_d ≠ 0$ . From $[∂^k, x^2] = 2k,x,∂^k-1 + k(k-1),∂^k-2$ one gets that $[D,x^2]$ has order $d-1$ with leading coefficient $2d,x,a_d$ . For $d ≥ 2$ that is a non-zero operator of positive order; for $d = 1$ it is $2x,a_1(x)$ , never the constant $1$ ; for $d = 0$ it is $0$ . So one cannot prescribe $ϕ(x)$ arbitrarily and hope to complete it: the map $F$ must be a Keller map for (4.14) to produce the partner.

Local nilpotence of the adjoint, made explicit

In $A_1$ , $ad_∂(x^a ∂^b) = a, x^a-1∂^b$ . Applying it $a+1$ times gives zero, so every monomial, and hence every element, is annihilated by a power of $ad_∂$ . Contrast $ad_x∂$ , which acts on $x^a∂^b$ by the scalar $b - a$ and is therefore not locally nilpotent: the Euler operator's adjoint is semisimple, not nilpotent.

Worked Example

Building $ϕ$ from a cubic Keller map of $K^2$

Step 1 - the map and its Jacobian
Write $s = x + y$ and take the Keller map $F = (F_1, F_2)$ with $F_1 = x + s^3$ , $F_2 = y - s^3$ , verified on the conjecture page to have $Δ F = 1$ . Its Jacobian matrix is

$J(F) = pmatrix 1 + 3s^2 & 3s^2 -3s^2 & 1 - 3s^2 pmatrix.$
Step 2 - the dual derivations by Cramer's rule
Apply (4.14). Replacing the first row by the gradient of $g$ gives $D_1(g) = (1-3s^2),g_x + 3s^2, g_y$ ; replacing the second row gives $D_2(g) = -3s^2, g_x + (1+3s^2), g_y$ . As elements of $A_2$ ,

$D_1 = (1 - 3s^2),∂_x + 3s^2,∂_y, D_2 = -3s^2,∂_x + (1 + 3s^2),∂_y.$

Both have polynomial coefficients, as they must because $Δ = 1$ .
Step 3 - check the Weyl relations
Duality. Since $∂_x F_1 = 1 + 3s^2$ and $∂_y F_1 = 3s^2$ , we get $D_1(F_1) = (1-3s^2)(1+3s^2) + 3s^2 ⋅ 3s^2 = 1 - 9s^4 + 9s^4 = 1$ . With $∂_x F_2 = -3s^2$ , $∂_y F_2 = 1 - 3s^2$ : $D_1(F_2) = (1-3s^2)(-3s^2) + 3s^2(1-3s^2) = 0$ . Symmetrically $D_2(F_2) = -3s^2 ⋅ (-3s^2) + (1+3s^2)(1-3s^2) = 9s^4 + 1 - 9s^4 = 1$ and $D_2(F_1) = -3s^2(1+3s^2) + (1+3s^2)3s^2 = 0$ .

Commutativity. Write $D_1 = a∂_x + b∂_y$ and $D_2 = c∂_x + d∂_y$ with $a = 1-3s^2$ , $b = 3s^2$ , $c = -3s^2$ , $d = 1+3s^2$ . Expanding, the second-order terms cancel and $[D_1,D_2] = (D_1(c) - D_2(a) )∂_x + (D_1(d) - D_2(b) )∂_y$ .

Now $D_1(s) = (1-3s^2) + 3s^2 = 1$ and $D_2(s) = -3s^2 + (1+3s^2) = 1$ , so for any one-variable $h$ we have $D_1(h(s)) = D_2(h(s)) = h'(s)$ . All four coefficients are functions of $s$ alone, and crucially $a = 1 + c$ and $d = 1 + b$ , so $a$ and $c$ have the same derivative in $s$ , as do $b$ and $d$ . Hence $D_1(c) - D_2(a) = c'(s) - a'(s) = (-6s) - (-6s) = 0$ and $D_1(d) - D_2(b) = d'(s) - b'(s) = 6s - 6s = 0$ , so $[D_1,D_2] = 0$ .
Step 4 - the derivations are locally nilpotent
Track $x$ under $D_1$ : $D_1(x) = 1 - 3s^2$ , then $D_1(1-3s^2) = -6s,D_1(s) = -6s$ , then $D_1(-6s) = -6$ , then $D_1(-6) = 0$ . So $D_1^,4(x) = 0$ . Similarly $D_1(y) = 3s^2 ↦ 6s ↦ 6 ↦ 0$ , so $D_1^,4(y) = 0$ , and Leibniz's rule extends local nilpotence to all of $K[x,y]$ . The same computation works for $D_2$ .

In this example we can verify local nilpotence directly, so we do not need the Dixmier hypothesis. For a general Keller map this is exactly the step that is unavailable, and exactly what the hypothesis supplies.
Step 5 - conclude
The pair $(D_1, D_2)$ together with $(F_1, F_2)$ satisfies (4.15), so $ϕ(x) = F_1$ , $ϕ(y) = F_2$ , $ϕ(∂_x) = D_1$ , $ϕ(∂_y) = D_2$ defines an endomorphism of $A_2$ . The structure theorem now gives $K[x,y] = R[F_1,F_2]$ with $R = D_1 ∩ D_2$ , and $R = K$ because $J(F)$ is invertible over $K[x,y]$ . So $K[F_1,F_2] = K[x,y]$ , agreeing with the direct verification that $F_1 + F_2 = x + y$ recovers both variables.

Result

For $F = (x + (x+y)^3,, y - (x+y)^3 )$ the dual derivations are $D_1 = (1-3s^2)∂_x + 3s^2 ∂_y$ and $D_2 = -3s^2∂_x + (1+3s^2)∂_y$ with $s = x+y$ . They satisfy $D_i(F_j) = δ_ij$ and $[D_1,D_2] = 0$ , are locally nilpotent, and the resulting endomorphism of $A_2$ is an automorphism. The Dixmier hypothesis is precisely what would supply Step 4 for an arbitrary Keller map.

Applications and Industry Use

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

A non-commutative attack on a commutative problem. The theorem is the only known bridge from the Weyl algebra to the Jacobian conjecture, and it motivated a research programme on endomorphisms of $A_n$ that produced the equivalence (4.20).
Automorphism groups. Belov-Kanel and Kontsevich conjectured that $Aut(A_n)$ is isomorphic to the group of polynomial symplectomorphisms of affine $2n$ -space, a statement in the same circle of ideas and equally open; work on it feeds back into what is known about P2.
Deformation quantisation. $A_n$ is the quantisation of the polynomial Poisson algebra in $2n$ variables, and the Dixmier problem is the quantised form of a statement about symplectic polynomial maps; this is the source of Kontsevich's interest and of the proof strategy for the reverse implication.
Module theory. Any endomorphism $ϕ$ of $A_n$ lets one twist a module by restriction of scalars. If $ϕ$ is not surjective the twist behaves badly, so the conjecture is also a statement about how well-behaved these twists are.
Computer algebra. Deciding whether a given pair of operators generates $A_1$ is a concrete subalgebra membership problem in a non-commutative ring, and it is a standard benchmark for non-commutative Gröbner basis engines.

Standards and Codes

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

ISO 80000-2 fixes the symbols used here, including $∂$ for partial differentiation, $δ_ij$ for the Kronecker delta, and upright roman type for named operators such as $det$ and $ad$ .
ISO/IEC 40314 (MathML 3.0) is the encoding of the mathematics on this page.
Naming is not standardised. Dixmier's own text calls it a problème, and the literature uses "Dixmier conjecture", "Dixmier problem" and "Problem 1" interchangeably; the numbering "Problème 11.1" refers to the first entry in the list of problems in the final section of the 1968 paper.
Citation numbering on this page follows the source's chapter-section-item convention, so the conditional theorem is $(4.4.2)$ and the lemma on the dual derivations is $(4.4.1)$ .

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.

Constructing $ϕ$ from $F$ is mechanical. Invert $J(F)$ over $K[x_1, … ,x_n]$ , which is exact because $Δ = 1$ makes the adjugate the inverse, transpose, and read off $D_i = ∑ _j C_ij∂_j$ . The cost is one symbolic adjugate, so $O(n^3)$ polynomial multiplications.
Testing surjectivity of a given endomorphism means asking whether $x_i$ and $∂_i$ lie in the subalgebra generated by $ϕ(x_i), ϕ(∂_i)$ . This is subalgebra membership in $A_n$ , for which non-commutative Gröbner bases over Ore algebras are the standard tool; termination is not guaranteed in general, and in practice one bounds the degree and searches.
Local nilpotence is semi-decidable in the useful direction. To confirm that $D_i$ is locally nilpotent it suffices to check that $D_i^,k(x_j) = 0$ for all $j$ and some $k$ , because Leibniz then propagates the property; there is no comparably cheap certificate for the negative.
Software. Singular:Plural and its dmod.lib library, the Dmodules package for Macaulay2, and the Ore algebra facilities of SageMath all provide arithmetic in $A_n$ and non-commutative Gröbner bases. See Singular and Macaulay2.
Scale. Even for $A_1$ , searching for a counterexample means exploring pairs $(u,v)$ with $[v,u] = 1$ ; the relation itself is a large system of polynomial equations in the coefficients, and no exhaustive search beyond very small degrees has been carried out.

Limits of Validity

The theorem is conditional. It proves an implication between two open problems and settles neither. Coutinho is explicit about this, and it is worth repeating because the chain of constructions is concrete enough to feel like a proof.
$Δ = 1$ cannot be relaxed to $Δ$ nowhere zero at this step. With $Δ$ merely non-vanishing, the $D_i$ are derivations of $K[x][Δ^-1]$ and need not preserve $K[x]$ , so there is no endomorphism of $A_n$ to speak of. The weaker hypothesis is enough for Lemma (4.4.1) but not for the theorem.
The implication is not known to reverse in the same dimension. $JC_2n ⇒ DC_n$ is the best available converse, so the dimensions do not match up.
Characteristic zero is used repeatedly: in the simplicity of $A_n$ , in the step from "all $∂_j f = 0$ " to " $f$ constant", and in the exponential map behind the structure theorem, which divides by $k!$ .
Nothing here bounds the degree of $ϕ^-1$ , so the argument gives no effective content even if one grants the hypothesis.

Failure Modes and Common Mistakes

Assuming injectivity gets you most of the way

"Every endomorphism of $A_n$ is injective, so only a little more is needed" is a natural thought and a wrong one. For infinite-dimensional algebras injectivity carries no surjectivity information: $K[x] → K[x]$ , $x ↦ x^2$ , is injective with proper image. The correct summary is that simplicity disposes of the trivial half and leaves the whole problem standing.

Treating $ad_a$ as an algebra homomorphism

$ad_a$ is $K$ -linear and satisfies Leibniz's rule, so it is a derivation, not a homomorphism: $ad_a(bc) ≠ ad_a(b)ad_a(c)$ in general. The identity that is used, and the only one, is the equivariance (4.18), which relates $ad$ before and after applying the algebra homomorphism $ϕ$ .

The unfilled step from the structure theorem

Proposition (4.3.1) concludes $S = R[t_1, … ,t_n]$ , where $R$ is the ring of constants, not $S = K[t_1, … ,t_n]$ . Quoting it as though it gave the latter is the one real gap in the printed proof of (4.4.2). It is easily repaired, as shown above, but it must be repaired: without $R = K$ the conclusion is not the Jacobian conjecture.

Reading the implication backwards

Theorem (4.4.2) says Dixmier implies Jacobian. It does not say that a proof of the Jacobian conjecture would settle Dixmier's problem - that is a separate and much later theorem, and it costs a doubling of dimension. In particular, the elementary truth of $JC_1$ says nothing about $DC_1$ , which is open.

Forgetting that $ϕ$ must be unital and $K$ -linear

The injectivity argument uses $ϕ(1) = 1$ to know that $ker ϕ$ is a proper ideal. A non-unital ring homomorphism could be zero, and a ring homomorphism that moves $K$ is not what the conjecture is about. Both hypotheses are silent in most statements of the problem and both are needed.

Historical Notes

Dixmier's 1968 paper on the Weyl algebras determined $Aut(A_1)$ and closed with a list of open problems; the first of them asks whether every endomorphism of $A_n$ is an automorphism. It has been open ever since, in every dimension.

The link to the Jacobian conjecture is credited by Bass, Connell and Wright to L. Vaserstein and V. Katz, and the derivation-theoretic input is Wright's 1981 theorem on commuting locally nilpotent derivations with a dual system. Coutinho's Chapter 4 assembles these into the proof presented above, which is the version most readers of D-module theory meet first.

The converse implication came much later. Tsuchimoto's 2005 work on $p$ -curvatures of endomorphisms of the Weyl algebra, using reduction modulo primes, established that the Jacobian conjecture in dimension $2n$ implies the Dixmier conjecture in dimension $n$ ; Belov-Kanel and Kontsevich gave an independent proof in 2007 and coined the phrase stably equivalent. Bavula subsequently gave further proofs and sharpenings.

The net effect is that a problem in non-commutative algebra and a problem in affine algebraic geometry, posed thirty years apart for unrelated reasons, turned out to be the same problem. Neither community has solved it.

Comparison

The Jacobian and Dixmier conjectures side by side.
	Jacobian conjecture	Dixmier conjecture
Object	Polynomial map $F$ of $K^n$	Algebra endomorphism $ϕ$ of $A_n$
Hypothesis	$Δ F = 1$	None beyond being an endomorphism
Free half	$F_1, … ,F_n$ algebraically independent	$ϕ$ injective, by simplicity of $A_n$
Open half	$K[F_1, … ,F_n] = K[x_1, … ,x_n]$	$ϕ$ surjective
Dimension 1	True, elementary	Open
Characteristic $p$	False	Setting degenerates; $A_n$ is not simple
Known implication	Implied by $DC_n$	Implied by $JC_2n$
Main tool used here	Comorphism and generation	Adjoint action and local nilpotence

Key Takeaways

Key takeaways

Dixmier's problem asks whether every endomorphism of $A_n$ is an automorphism. Injectivity is free from simplicity, so the question is surjectivity, and it is open even for $A_1$ .
A Keller map $F$ of $K^n$ produces, by Cramer's rule, commuting derivations $D_i$ with $D_i(F_j) = δ_ij$ ; when $Δ F = 1$ these have polynomial coefficients.
Those data satisfy the Weyl relations, so they define an endomorphism $ϕ$ of $A_n$ with $ϕ(x_i) = F_i$ and $ϕ(∂_i) = D_i$ .
$ad_∂_i$ is locally nilpotent because it lowers Bernstein degree; surjectivity of $ϕ$ transports that to $ad_D_i$ , hence to $D_i$ acting on polynomials.
The structure theorem for commuting locally nilpotent derivations then gives $K[x] = R[F_1, … ,F_n]$ , and $R = K$ because $J(F)$ is invertible over $K[x]$ . So $DC_n$ implies $JC_n$ .
The converse holds after doubling the dimension (Tsuchimoto 2005; Belov-Kanel and Kontsevich 2007), so the two conjectures are equivalent as families. Both are open.

FAQs

Why is every endomorphism of

A_n

injective?

The kernel of a ring homomorphism is a two-sided ideal. Since $ϕ(1) = 1$ , the kernel is not all of $A_n$ . The Weyl algebra over a field of characteristic zero is simple, so its only proper two-sided ideal is zero.

If injectivity is automatic, why is surjectivity hard?

Because $A_n$ is infinite-dimensional, and for infinite-dimensional algebras injectivity implies nothing about the image. The subalgebra generated by $ϕ(x_i)$ and $ϕ(∂_i)$ could in principle be proper, exactly as $K[x^2] ⊂ neq K[x]$ .

Where exactly is the conjecture used in the proof of (4.4.2)?

In one place only: to know that every $c ∈ A_n$ is of the form $ϕ(b)$ , so that the equivariance identity (4.18) can transport local nilpotence from $ad_∂_i$ to $ad_D_i$ . Nothing else needs $ϕ$ to be surjective.

Why do the derivations

D_i

have polynomial coefficients only when

Δ = 1

Formula (4.14) has $Δ^-1$ in front. In general the $D_i$ live in $K[x][Δ^-1]$ . When $Δ$ is a non-zero constant, that ring is $K[x]$ itself, so the $D_i$ are honest elements of $A_n$ and there is an endomorphism to build. This is the step that needs the full Keller hypothesis rather than mere non-vanishing.

DC_1

really open?

Yes. It is not known whether every pair $u, v ∈ A_1$ with $[v,u] = 1$ generates $A_1$ . This is one of the cleanest unsolved statements in non-commutative algebra, and it is a much harder question than its commutative analogue, which is a one-line exercise.

Does the Dixmier conjecture follow from the Jacobian conjecture?

Yes, but with a shift in dimension: $JC_2n$ implies $DC_n$ . Since the Jacobian conjecture is open in every dimension from $2$ upwards, this does not settle anything; it establishes that the two families are equivalent.

What is the role of the local inversion theorem here?

It is used inside Lemma (4.4.1) to show the $D_i$ commute. The commutator is a derivation vanishing on all the $F_k$ , and one needs to know that the $F_k$ , after shifting to make them vanish at the origin, generate the whole formal power series ring. That is exactly what the local inversion theorem supplies.

Is the statement true in positive characteristic?

The question is posed in characteristic zero and the standard reductions do not survive: $A_n$ in characteristic $p$ has a large centre and is not simple, so injectivity of endomorphisms is no longer automatic. We make no claim about the truth of the statement there. Reduction modulo $p$ is nevertheless the technique behind the known converse implication.

Does this give a strategy for proving the Jacobian conjecture?

Only if one can prove the Dixmier conjecture, which is at least as hard - and, by the equivalence, exactly as hard up to dimension. What the theorem does give is a change of language: it converts a question about generation of a commutative ring into a question about surjectivity of a map of non-commutative algebras, where different tools, notably reduction modulo $p$ , become available.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 4 §§3-4, Proposition (4.3.1), Lemmas (4.3.2) and (4.4.1), Theorem (4.4.2), and Corollary (2.2.2) for injectivity.
J. Dixmier, Sur les algèbres de Weyl, Bulletin de la Société Mathématique de France 96 (1968), 209-242 - the determination of the automorphism group of $A_1$ and the list of problems, the first of which is the conjecture discussed here.
D. Wright, On the Jacobian conjecture, Illinois Journal of Mathematics 25 (1981), 423-440 - the structure theorem for commuting locally nilpotent derivations with a dual system.
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 - where the Vaserstein-Katz argument is recorded, p. 297.
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.
L. Makar-Limanov, On automorphisms of the Weyl algebra, Bulletin de la Société Mathématique de France 112 (1984), 359-363.
A. van den Essen, Polynomial Automorphisms and the Jacobian Conjecture, Progress in Mathematics 190, Birkhäuser, 2000 - locally nilpotent derivations and the surrounding theory.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, American Mathematical Society, 2001 - simplicity and ideal structure of the Weyl algebras.
ISO 80000-2, Quantities and units - Part 2: Mathematics; ISO/IEC 40314, Mathematical Markup Language (MathML) Version 3.0.

AI Suggested Questions

Give a complete proof that $ad_∂_i$ lowers Bernstein degree, working with a general monomial $x^α ∂^β$ .
Show that $ad_x∂$ is not locally nilpotent on $A_1$ and describe its eigenvalues on the monomial basis.
Verify formula (4.14) gives a derivation, checking Leibniz's rule directly from multilinearity of the determinant.
Carry out the construction of $D_1, D_2$ for the map $F(x,y) = (x + y^2,, y)$ and identify the resulting endomorphism of $A_2$ .
Explain precisely why the argument fails if $Δ F$ is a non-constant polynomial with no zeros in $R^n$ .
Prove that an element of $K[x_1, … ,x_n]$ algebraic over $K$ lies in $K$ , and use it to give the transcendence-degree proof that the ring of constants is $K$ .
State the Belov-Kanel-Kontsevich result carefully and explain where the doubling of dimension enters.
Find two elements $u,v ∈ A_1$ with $[v,u] = 1$ that are not $x$ and $∂$ , and check whether they generate $A_1$ .