Introduction and statement

The Abhyankar–Sathaye conjecture asks whether a polynomial \(f\in\mathbb C[z_1,\ldots,z_n]\) satisfying \[\mathbb C[z_1,\ldots,z_n]/(f)\cong\mathbb C^{[n-1]}\] must be a coordinate, that is, the first component of a polynomial automorphism of \(\mathbb A^n_{\mathbb C}\). Here \(\mathbb C^{[m]}\) denotes a polynomial ring in \(m\) independent variables, and the displayed isomorphism is required to fix \(\mathbb C\). Geometrically, the question asks whether every hypersurface embedding \(\mathbb A^{n-1}_{\mathbb C}\hookrightarrow\mathbb A^n_{\mathbb C}\) can be carried to a coordinate hyperplane by an ambient polynomial automorphism.

For \(n=2\), the question is answered affirmatively by the Abhyankar–Moh–Suzuki Theorem [1, 13]. The higher-dimensional problem belongs to a family of questions about polynomial maps with affine-space fibres. Sathaye [11] distinguished the coordinate, or epimorphism, question from the weaker question whether an affine-space zero fibre forces all fibres to be affine space. A coordinate polynomial defines a projection after an automorphism, so its fibres are all isomorphic to \(\mathbb A^{n-1}_{\mathbb C}\); an isomorphism of the zero fibre alone supplies no such global trivialization. Kraft [7] surveys the embedding problem and its relation to algebraic group actions. Among the early affirmative results, Sathaye [11] proved rectifiability for planes defined by polynomials linear in one of three variables in characteristic zero, and Russell [10] extended this result to arbitrary fields.

In four variables, Kaliman, Vénéreau and Zaidenberg [6] and Maubach [8] established affirmative results for equations \(a(x)y+b(x,z,t)\) with \(a\ne0\): an affine-three-space zero fibre can be straightened by an ambient automorphism. Here \(x,y,z,t\) are independent variables, and the coefficient of \(y\) depends on only one of them. Further algebraic treatments appear in work of El Kahoui, Essamaoui and Ouali [2] and Ghosh, Gupta and Pal [5]. Ghosh, Gupta and Pal also study linear hypersurfaces with multivariable coefficients under additional hypotheses on the coefficients and their divisors [4]. Ghosh [3] proves an affirmative embedding theorem in characteristic zero for \(a(x_m)b(x_1,\ldots,x_{m-1})y+f(z,t)+x_m\), where the displayed variables are independent, \(b\ne0\), and \(a\) has at least two distinct roots over the algebraic closure. These results illustrate the role of restrictions on the defining equation; an abstract polynomial presentation of the quotient does not itself provide an automorphism of the ambient polynomial ring.

We give a counterexample in ambient dimension four, and hence in every ambient dimension at least four. The three-variable case is not addressed.

The construction starts from the cusp parametrization \((u^3,-u^2)\) and perturbs its two entries by independent multiples of \(h\). In \(R=\mathbb C[h,u,v,w]\), put \[ \begin{aligned} x&=u^3+hv,\qquad y=-u^2+hw,\\ s&=S(h,u,v,w) :=2u^3v+3u^4w+h(v^2-3u^2w^2)+h^2w^3. \end{aligned} \tag{1}\] Since \(x^2+y^3\) vanishes modulo \(h\), it is divisible by \(h\) in \(R\). The displayed polynomial \(s\) is its quotient, so \(x^2+y^3=hs\). Define \(p\) and \(F\) by \[ p(x,y,s)=-2s^2x+3sy^2-3s^3y, \qquad F=h-p(x,y,s)-1. \tag{2}\] Thus \(F\) is a polynomial in the four independent variables \(h,u,v,w\).

The critical point \(P=(2,0,-1/2,1/2)\) lies on the fibre \(F=-1\). The zero fibre is smooth, since it is isomorphic to affine three-space.

Shpilrain and Yu [12] used the ambient critical locus to distinguish isomorphic hypersurfaces with inequivalent embeddings. They asked whether the gradient must be nowhere zero when the zero fibre is isomorphic to a coordinate hyperplane. Theorem 1 gives a negative answer in ambient dimension four, and Corollary 2 preserves this obstruction after adjoining variables. The distinction is between the intrinsic geometry of one fibre and the behaviour of its defining polynomial on the ambient space.

The main algebraic ingredient is Lemma 3. It turns a presentation over a commutative base ring \(B\) with \(x^2+y^3=hs\) and \((h,y)=B\) into a polynomial extension of that base. Two compatible linear equations, with coefficients generating the unit ideal, give explicit inverse maps even when the base has zero divisors. In our example, a translation identifies the auxiliary algebra cut out by \(x^2+y^3=hs\) and \(h=1+p(x,y,s)\) with a polynomial ring in two variables; the lemma supplies the third variable. This proves that the zero fibre is affine space. An identity in the original four-variable ring then exhibits a critical point on the fibre over \(-1\), which rules out a coordinate.

The polynomials in (1) also have an invariant-theoretic interpretation. The triangular derivation \(D=h\partial_u-3u^2\partial_v+2u\partial_w\) of \(R\) annihilates \(h,x,y,s\). They arise in the triangular monomial kernel calculation of Maubach [9]. The proof below uses the cusp relation directly through Lemma 3.

Section 2 proves the lifting lemma over an arbitrary commutative base ring. Section 3 identifies the auxiliary algebra, establishes the zero-fibre isomorphism, and proves the critical-point obstruction and its persistence after adjoining variables. Appendix 4 records the resulting polynomial coordinates and their inverse.

An elementary lifting lemma

The next lemma provides the polynomial parameter used to identify the zero fibre. Its unit-ideal hypothesis allows us to solve two compatible linear equations without dividing by either coefficient. This point matters because the coefficients need not be units, and the base ring may have zero divisors.

In the following statement, \(S(H,U,V,W)\) denotes the universal polynomial specified in (1), with the indicated arguments. All rings are commutative with identity.

The affine-space fibre and the critical point

We first identify the base algebra to which Lemma 3 applies. After obtaining the coordinates on the zero fibre, we use the same polynomial identities in the ambient ring to locate a critical point on another fibre.

Explicit coordinates on the zero fibre

The proof of Theorem 1 already supplies both directions of the isomorphism. We collect them here, including a polynomial certificate for the unit-ideal condition, so that the parametrization can be used directly.

To make the isomorphism (15) explicit, take in the algebra \(B\) of (12) \[\begin{align*} \alpha&=1+2s^2x+4s^5,\\ \beta&=(3s^3-3sy)(1+2s^2x)-4s^4y^2. \tag{17}\end{align*}\] The following identity in the free polynomial ring proves \(\alpha h+\beta y=1\) in \(B\): \[ \alpha h+\beta y-1 =(1+2s^2x)(h-1-p)-4s^4(x^2+y^3-sh). \tag{18}\]

The three polynomial coordinates on \(F=0\) are \[ X=x+s^3,\qquad Y=y-s^2,\qquad T=\alpha u+\beta(v+uw), \tag{19}\] with all expressions interpreted by (1) and (17). Together they define a map from the zero fibre to \(\mathbb A^3_{\mathbb C}\). Conversely, start from independent \(X,Y,T\), reconstruct \(s,x,y,h\) by (14), and compute \(\alpha,\beta\) by (17). Set \[\begin{align*} u&=hT-\beta x,\qquad g=yT+\alpha x,\\ w&=\alpha(1+\beta y)(y+u^2) +\beta^2(s-hg^2+2uyg),\\ v&=g-uw. \tag{20}\end{align*}\] These are exactly the inverse maps from Lemma 3. In particular, they are defined on every point of \(\mathbb A^3_{\mathbb C}\) and every point of the zero fibre; no open subset has been removed.