A stable coordinate that is not a coordinate

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

A polynomial \(f\in k[x_1,\ldots,x_n]\) is a one-stable coordinate if it becomes a coordinate after adjoining one variable: it is a coordinate of \(k[x_1,\ldots,x_n,w]\). The stable coordinate conjecture asked whether every such polynomial is already a coordinate of \(k[x_1,\ldots,x_n]\). This lesson proves that the answer is no in four variables, for the explicit polynomial

\[ \begin{gathered} f=x_1-2Q\bigl(Q(x_2+x_4)+x_1x_4\bigr),\\ Q=x_2^2-x_4^2+x_1x_3, \end{gathered} \]

of degree five. Every fibre \(f=\lambda\) of this polynomial is isomorphic to affine three-space, and still none of these hypersurfaces can be moved onto a coordinate hyperplane by an automorphism of \(\mathbb A^4\). The proof follows the pattern of Cancellation fails in dimension four, with one simplification at the end.

We use Coordinates and the Abhyankar–Sathaye problem, Locally nilpotent derivations, Additive actions on line bundles and the line bundle of Section 4 of Cancellation fails in dimension four. The field \(k\) has characteristic zero.

Basic references are [OpenAI-stable] and [Shpilrain–Yu].

1. Stable coordinates

Coordinates were defined in Coordinates and the Abhyankar–Sathaye problem. If \(f\) is a coordinate of \(k[x_1,\ldots,x_n]\), it is a coordinate of \(k[x_1,\ldots,x_n,w]\), with \(w\) added to its complement. The stable coordinate conjecture [Shpilrain–Yu, Conjecture 3] asserts the converse. Over the complex numbers it holds in at most three variables [Shpilrain–Yu, Proposition 1.1]: in two variables by cancellation for curves and the theorem of Abhyankar and Moh, and in three variables by cancellation for affine surfaces together with Kaliman's theorem that a polynomial on \(\mathbb C^3\) whose general fibre is \(\mathbb C^2\) is a coordinate. The counterexample of this lesson was found in 2026 [OpenAI-stable]. A counterexample in five variables, the polynomial \(H\) of the cancellation lessons, is proved in A stable coordinate in five variables; the present example has four variables.

2. A presentation by a hypersurface

Let \(P=k[p,s,u,F,J]\) and make the definitions (2.1):

\[ \begin{aligned} x&=s^2-u^2+pF,\\ H&=x^2F-(1+2sx)J-pJ^2-u,\\ A&=P/(H). \end{aligned} \]

Proposition 2.1. There are polynomial coordinates \(p',s',F',J\) of \(A\), so \(A\cong k^{[4]}\), in which

\[ \begin{gathered} x=s'^2-J^2+p'F',\\ p=p'-2x\bigl(x(s'+J)+p'J\bigr). \end{gathered} \]

Under the identification \((p',s',F',J)=(x_1,x_2,x_3,x_4)\), the element \(p\in A\) becomes the polynomial \(f\).

Proof. Consider

\[ \begin{gathered} M=\begin{pmatrix}F&s-u\\ s+u&-p\end{pmatrix},\\ e=\begin{pmatrix}x\\-J\end{pmatrix},\qquad n=\begin{pmatrix}J\\x\end{pmatrix}. \end{gathered} \]

Then \(x=-\det M\), \(e^{\mathsf T}n=0\), and expanding gives \(H=e^{\mathsf T}Me-J-u\). Define \(F',s',u',p'\), keeping \(J\), by

\[ \begin{aligned} M'&=(I_2-2ne^{\mathsf T})M\\ &=\begin{pmatrix}F'&s'-u'\\ s'+u'&-p'\end{pmatrix}. \end{aligned} \]

Since \((ne^{\mathsf T})^2=n(e^{\mathsf T}n)e^{\mathsf T}=0\), the matrix \(I_2-2ne^{\mathsf T}\) has determinant one and inverse \(I_2+2ne^{\mathsf T}\). So \(\det M'=\det M=-x\): the element \(x\), hence also \(e\) and \(n\), is the same polynomial when written through \(M'\) and \(J\). The substitution is therefore invertible, with inverse \(M=(I_2+2ne^{\mathsf T})M'\), and \(p',s',u',F',J\) are coordinates of \(P\). Half the difference of the off-diagonal entries of \(M'\) is \(u'=u-e^{\mathsf T}Me\), so \(H=-J-u'\). Eliminating \(u'=-J\) gives \(A=k[p',s',F',J]\). In this quotient, \(x=-\det M'=s'^2-u'^2+p'F'\), which is \(s'^2-J^2+p'F'\), and the lower right entry of \(M=(I_2+2ne^{\mathsf T})M'\) gives \(p=p'-2x\bigl(x(s'-u')+p'J\bigr)\), which is \(p'-2x\bigl(x(s'+J)+p'J\bigr)\). \(\square\)

The polynomial \(f\) takes the values \(0\) and \(1\) at \((0,0,0,0)\) and \((1,0,0,0)\), so \(p\) is a nonzero nonunit of \(A\).

3. The polynomial becomes a coordinate after one more variable

Put

\[ y=s+x(x+u^2),\qquad z=sx+pJ . \]

Expanding, as in Lemma 2.1 of A fourfold whose cylinder is affine space,

\[ \begin{aligned} &xy-z(z+1)\\ &\quad =p\,(H+u). \end{aligned}\tag{3.1} \]

Over \(k[p,p^{-1}]\), the elements \(x,y,z,u\) are coordinates of \(P[p^{-1}]\), with inverse formulas \(s=y-x(x+u^2)\), \(F=p^{-1}(x-s^2+u^2)\), \(J=p^{-1}(z-sx)\). Let \(\Delta=-p\,\partial/\partial u\) on \(P[p^{-1}]=k[p^{\pm1}][x,y,z,u]\). Differentiating the inverse formulas gives

\[ \begin{aligned} &\Delta(p)=0,\quad\Delta(u)\\ &\quad =-p,\quad\Delta(s)\\ &\quad =2pxu,\quad\Delta(F)\\ &\quad =-4sxu-2u,\quad\Delta(J)\\ &\quad =-2x^2u, \end{aligned}\tag{3.2} \]

so \(\Delta\) is a locally nilpotent derivation of \(P\). By (3.1), \(H=p^{-1}(xy-z(z+1))-u\), so \(\Delta(H)=p\) and \(\Delta^2(H)=0\).

Proposition 3.1. \(A[w]\) is a polynomial ring in five variables in which \(p\) is a coordinate. Consequently \(f\) is a coordinate of \(k[x_1,x_2,x_3,x_4,w]\).

Proof. Extend \(\Delta\) by \(\Delta(w)=0\). The automorphism \(\exp(w\Delta)\) of \(P[w]\) fixes \(p\) and sends \(H\) to \(H+pw\), so \(A[w]\cong P[w]/(H+pw)\) by an isomorphism fixing \(p\). Put \(x_0=s^2-u^2\) and

\[ \begin{gathered} L=x_0^2F-(1+2sx_0)J,\\ N=(1-2sx_0)F+4s^2J, \end{gathered} \]

a linear change of \(F,J\) over \(k[p,s,u]\) of determinant \(4s^2x_0^2+(1+2sx_0)(1-2sx_0)=1\), with inverse \(F=4s^2L+(1+2sx_0)N\), \(J=-(1-2sx_0)L+x_0^2N\). Expanding \(x=x_0+pF\) in \(H\) gives

\[ \begin{gathered} H=L-u+pQ_0,\\ Q_0=2x_0F^2+pF^3-2sFJ-J^2\in P . \end{gathered} \]

So \(H+pw=L-u+pw'\) with \(w'=w+Q_0\), and \(p,s,u,L,N,w'\) are coordinates of \(P[w]\). The relation lets us eliminate \(L=u-pw'\), and

\[ P[w]/(H+pw)\cong k[p,s,u,N,w'], \]

a polynomial ring with \(p\) among its coordinates. By Proposition 2.1, \(A=k[x_1,\ldots,x_4]\) with \(p=f\), so \(f\) is a coordinate of \(k[x_1,\ldots,x_4,w]\). \(\square\)

Proposition 3.2. For every \(\lambda\in k\), \(A/(p-\lambda)\cong k^{[3]}\). So every fibre \(f=\lambda\) is isomorphic to affine three-space.

Proof. For \(\lambda=0\): modulo \(p\), \(H\equiv L-u\), and \(P/pP=k[s,u,L,N]\), so \(A/pA\cong k[s,u,N]\). For \(\lambda\ne0\): by (3.1) and the coordinates \(x,y,z,u\), \(A/(p-\lambda)\) is \(k[x,y,z,u]/\bigl(xy-z(z+1)-\lambda u\bigr)\), and eliminating \(u\) gives \(k[x,y,z]\). \(\square\)

4. A filtration and its graded ring

Give the generators of \(P\) the weights \(\omega(p,s,u,F,J)=(-1,0,0,1,1)\), and let \(\mathcal F_dA\) be the image in \(A\) of the span of the monomials of weight at most \(d\). Then \(x\) has weight zero, and (4.1) holds:

\[ \begin{gathered} H=H_*-u,\\ H_*=x^2F-(1+2sx)J-pJ^2, \end{gathered} \]

where \(H_*\) is homogeneous of weight one and \(u\) has weight zero.

Proposition 4.1. The filtration is exhaustive and separated, and its associated graded ring is the domain

\[ G=\operatorname{gr}A\cong P/(H_*), \]

graded by the weights; the classes of \(p,s,u,F,J\) are nonzero, of degrees \(-1,0,0,1,1\). For \(0\ne a\in A\) put \(\deg a=\min\{d:a\in\mathcal F_dA\}\). This is a degree function on \(A\), and the generators \(p,s,u,F,J\) are exact for it.

Proof. Every element has a polynomial representative, so the filtration is exhaustive. A monomial of weight at most \(-r\) contains \(p\) at least \(r\) times, so \(\mathcal F_{-r}A\subset p^rA\). In the factorial ring \(A\cong k^{[4]}\), a nonzero element is divisible by only finitely many powers of the nonzero nonunit \(p\), so \(\bigcap_rp^rA=0\) and the filtration is separated.

The natural map \(P\to\operatorname{gr}A\) sends a homogeneous polynomial of weight \(d\) to its class in \(\mathcal F_dA/\mathcal F_{d-1}A\); it is surjective. A homogeneous \(Q_d\) of weight \(d\) lies in its kernel exactly when \(Q_d-R=qH\) for some \(R\) of weight less than \(d\) and some \(q\in P\). If \(q\ne0\), the top-weight part of \(qH\) is \((\operatorname{top}q)H_*\), because \(P\) is a graded domain, and it must equal \(Q_d\). So the kernel is \((H_*)\); conversely \(H_*=H+u\equiv u\) has class zero in degree one.

To see that \(H_*\) is prime, use the coordinates \(x,y,z,u\) of \(P[p^{-1}]\): there \(pH_*=xy-z(z+1)\) by (3.1) and (4.1), an irreducible polynomial, and \(p\) is a unit. Modulo \(p\), \(H_*\equiv x_0^2F-(1+2sx_0)J\ne0\). As in Lemma 1.1 of Cancellation fails in dimension four, \(H_*\) is irreducible in \(P\), hence prime. It is not associated to any of the five variables, so their classes are nonzero.

Since \(G\) is a domain, the classes of nonzero elements multiply to nonzero classes, so \(\deg(ab)=\deg a+\deg b\); the other properties of a degree function are clear. Exactness of the generators is the definition of \(\mathcal F_dA\). \(\square\)

Write \(\tau\) for the class of \(p\) in \(G\). Let \(B=k[x,y,z,u]/(xy-z(z+1))\), the ring called \(R\) in Cancellation fails in dimension four. In \(G\), the elements \(x=s^2-u^2+\tau F\), \(y=s+x(x+u^2)\), \(z=sx+\tau J\) and \(u\) satisfy \(xy-z(z+1)=\tau H_*=0\), so they define a homomorphism \(B\to G\) into degree zero. After inverting \(\tau\), the inverse formulas of Section 3 give

\[ \begin{gathered} G[\tau^{-1}]=B[\tau,\tau^{-1}],\\ \deg B=0,\ \deg\tau=-1 . \end{gathered} \]

So \(B\to G\) is injective. In \(B\) put

\[ \begin{gathered} s_B=y-x(x+u^2),\\ f_0=x-s_B^2+u^2,\\ g_0=z-s_Bx,\qquad I=(f_0,g_0)\subset B . \end{gathered} \]

Inside \(B[\tau^{\pm1}]\), \(s=s_B\), \(F=f_0/\tau\) and \(J=g_0/\tau\), so

\[ G=B[\tau,f_0/\tau,g_0/\tau]=B[\tau,I/\tau]. \]

Lemma 4.2. For every \(\ell>0\), the degree-\(\ell\) component of \(G\) is \(G_\ell=\tau^{-\ell}I^\ell\).

Proof. \(G\) is spanned by the elements \(b\tau^r(f_0/\tau)^i(g_0/\tau)^j\) with \(b\in B\) and \(r,i,j\ge0\), of degree \(i+j-r\). If this degree is \(\ell\), the element is \(\tau^{-\ell}bf_0^ig_0^j\) with \(bf_0^ig_0^j\in I^{\ell+r}\subset I^\ell\). Conversely \(\tau^{-\ell}f_0^ig_0^j\in G\) for \(i+j=\ell\). \(\square\)

5. A coordinate would give a derivation of the quadric

Proposition 5.1. If \(p\) is a coordinate of \(A\), there are a nonzero locally nilpotent derivation \(E\) of \(B\) and an element \(0\ne h\in I\) with \(E(h)=0\).

Proof. Let \(A=k[p,q_1,q_2,q_3]\). Some \(q_i\) has positive degree: otherwise \(p\) and the \(q_i\) would all lie in the subalgebra \(\mathcal F_0A\), which does not contain \(F\). Renumber so that \(\deg q_1=\ell>0\), and let \(D=\partial/\partial q_2\), a nonzero locally nilpotent derivation of \(A\) with \(D(p)=D(q_1)=0\). By Propositions 6.1 and 6.2 of Locally nilpotent derivations, \(D\) induces a nonzero homogeneous locally nilpotent derivation \(D_0\) of \(G\), of some degree shift \(m\), with \(D_0(\tau)=0\) and \(D_0(\eta)=0\) for the class \(\eta\) of \(q_1\), which is homogeneous of degree \(\ell\).

Since \(D_0(\tau)=0\), \(D_0\) extends to a locally nilpotent derivation of \(G[\tau^{-1}]=B[\tau^{\pm1}]\). For \(b\in B\), \(D_0(b)\) has degree \(m\), and the degree-\(m\) component of \(B[\tau^{\pm1}]\) is \(\tau^{-m}B\); so \(D_0(b)=\tau^{-m}E(b)\) for a unique \(E(b)\in B\). Then \(E\) is a derivation of \(B\), with \(E^r(b)=\tau^{rm}D_0^r(b)\), so it is locally nilpotent. It is nonzero, since otherwise \(D_0\) would kill \(B\) and \(\tau\), hence \(G\). By Lemma 4.2, \(\eta=\tau^{-\ell}h\) with \(0\ne h\in I^\ell\subset I\), and \(0=D_0(\eta)=\tau^{-\ell-m}E(h)\). \(\square\)

6. The line bundle and the obstruction

Let \(C=k[a,d,b,c,u]/(ac-bd-1)\), with \(x=ab\), \(y=dc\), \(z=db\), and fibre weights \(\operatorname{wt}(a)=\operatorname{wt}(d)=1\), \(\operatorname{wt}(b)=\operatorname{wt}(c)=-1\), \(\operatorname{wt}(u)=0\). This is the ring \(\widetilde R\) of Cancellation fails in dimension four, Proposition 4.1, a Laurent bundle algebra over \(B\); and \(B\) is a smooth domain by Lemma 1.1 there, hence a regular Noetherian domain. Put

\[ v=a^2(x+u^2)-d^2=a^3b+a^2u^2-d^2 . \]

Lemma 6.1. \(IC=vC\).

Proof. In \(C[a^{-1}]\) put \(\xi=d/a\) and \(j=x+u^2-\xi^2\). The relation \(ac-bd=1\) gives \(z=\xi x\) and \(y=\xi+\xi^2x\), hence \(s_B=\xi-xj\), and then

\[ \begin{gathered} f_0=(1+2\xi x-x^2j)\,j,\\ g_0=x^2j,\qquad v=a^2j . \end{gathered} \]

Since \[ \begin{aligned} &(1-2\xi x)(1+2\xi x-x^2j)\\ &\quad+x^2(4\xi^2+j-2\xi xj)=1, \end{aligned} \] the two coefficients of \(j\) generate the unit ideal, so \(IC[a^{-1}]=jC[a^{-1}]=vC[a^{-1}]\). At a prime of \(C\) containing \(a\), we have \(bd=-1\), \(x=0\) and \(z=-1\), so \(g_0\equiv-1\) and \(v\equiv-d^2\) are units of the local ring, and both ideals are the unit ideal there. Two ideals that agree after localizing at every prime are equal. \(\square\)

Proposition 6.2. If \(E\) is a nonzero locally nilpotent derivation of \(B\) killing some \(0\ne h\in I\), then there is a nonzero locally nilpotent derivation \(\widetilde E\) of \(C\) that preserves the fibre weights and kills \(v\).

Proof. Theorem 3.2 of Additive actions on line bundles gives a weight-preserving locally nilpotent \(\widetilde E\) extending \(E\), nonzero because \(E\) is. By Lemma 6.1, \(h=vq\) in \(C\) with \(q\ne0\). Since \(\widetilde E(h)=E(h)=0\) and kernels are factorially closed (Proposition 3.2 of Locally nilpotent derivations), \(\widetilde E(v)=0\). \(\square\)

Proposition 6.3. No nonzero locally nilpotent derivation of \(C\) preserves the fibre weights and kills \(v\).

Proof. Suppose \(\widetilde E\) is one. Grade \(C\) by the auxiliary degrees \((a,d,b,c,u)\mapsto(0,1,-1,0,1)\); the relation is homogeneous of degree zero, and this grading commutes with the fibre weights. Let \(E'\) be the component of \(\widetilde E\) of largest auxiliary shift \(\lambda\), a nonzero locally nilpotent derivation preserving fibre weights (Proposition 5.1 of Locally nilpotent derivations). In \(v=a^3b+(a^2u^2-d^2)\), the first term has auxiliary degree \(-1\) and the second degree \(2\). So the component of degree \(2+\lambda\) of \(\widetilde E(v)=0\) is \(E'(a^2u^2-d^2)=0\).

Now \(a^2u^2-d^2=(au-d)(au+d)\), and both factors are nonzero. By factorial closure, \(E'\) kills \(au-d\) and \(au+d\), hence \(d\) and \(au\), and then \(a\) and \(u\):

\[ E'(a)=E'(d)=E'(u)=0 . \]

As in Proposition 8.1 of Cancellation fails in dimension four, the element \(h=cE'(b)-bE'(c)\) satisfies \(ah=E'(b)\), \(dh=E'(c)\), lies in \(k[a,d,u]\), and is nonzero. Since \(E'\) preserves fibre weights, \(E'(b)\) has weight \(-1\) and \(h\) has weight \(-2\). But every nonzero homogeneous element of \(k[a,d,u]\) has nonnegative weight. \(\square\)

7. Conclusion

Theorem 7.1. The polynomial \(f=x_1-2Q(Q(x_2+x_4)+x_1x_4)\), with \(Q=x_2^2-x_4^2+x_1x_3\), is a coordinate of \(k[x_1,x_2,x_3,x_4,w]\) but not of \(k[x_1,x_2,x_3,x_4]\).

Proof. The first assertion is Proposition 3.1. If \(f\) were a coordinate, then by Proposition 2.1 so would be \(p\in A\); Proposition 5.1 would give a derivation \(E\) of \(B\) killing a nonzero element of \(I\), Proposition 6.2 would give a derivation of \(C\) that Proposition 6.3 excludes. \(\square\)

Reference: [OpenAI-stable, Theorem 1.1].

Corollary 7.2. For every \(\lambda\in k\), the hypersurface \(f=\lambda\) in \(\mathbb A^4\) is isomorphic to \(\mathbb A^3\), but no automorphism of \(\mathbb A^4\) carries it onto a coordinate hyperplane.

Proof. The first part is Proposition 3.2; in particular \((f-\lambda)\) is a prime ideal. If an automorphism carried the hypersurface onto a coordinate hyperplane, it would carry its prime ideal \((f-\lambda)\) onto the ideal of a coordinate \(g\); two generators of the same principal ideal differ by a nonzero constant, so \(f-\lambda\), and hence \(f\), would be a coordinate. \(\square\)

So this polynomial is a counterexample to the Abhyankar–Sathaye question of Coordinates and the Abhyankar–Sathaye problem of a different kind from the one constructed there: all of its fibres are affine spaces, and its gradient vanishes nowhere on them, since a critical point would make the fibre through it singular.

8. Exercises

Exercise 8.1 (easy). Check that \(f\) has degree five and compute \(f(1,0,0,0)\) and \(f(0,0,0,0)\).

Exercise 8.2 (medium). Verify the values (3.2) by differentiating the inverse formulas, and check that \(\Delta(H)=p\).

Exercise 8.3 (medium). Show that \(f\) has no critical point: \(\nabla f\) vanishes nowhere on \(k^4\), even over an algebraic closure of \(k\).

9. Solutions

8.1. \(Q\) has degree two and \(Q(x_2+x_4)+x_1x_4\) has degree three, so \(2Q(\cdots)\) has degree five, with leading part \(2(x_2^2-x_4^2+x_1x_3)^2(x_2+x_4)\neq0\). At \((1,0,0,0)\), \(Q=0\), so \(f=1\); at the origin \(f=0\).

8.2. From \(s=y-x(x+u^2)\): \(\Delta(s)=-2xu\,\Delta(u)=2pxu\). From \(pF=x-s^2+u^2\): \(p\Delta(F)=-2s\Delta(s)+2u\Delta(u)\), which is \(-4psxu-2pu\). From \(pJ=z-sx\): \(p\Delta(J)=-x\Delta(s)=-2px^2u\). Finally \(H=p^{-1}(xy-z(z+1))-u\) gives \(\Delta(H)=-\Delta(u)=p\).

8.3. By Proposition 3.2 over an algebraic closure \(\bar k\), every fibre \(f=\lambda\) with \(\lambda\in\bar k\) is isomorphic to \(\mathbb A^3\), hence smooth, with every local ring of embedding dimension three. If \(\nabla f(P_0)=0\) at a point with \(f(P_0)=\lambda\), then \(f-\lambda\) lies in the square of the maximal ideal of \(P_0\), and the local ring of the fibre at \(P_0\) would have embedding dimension four, as in Proposition 4.3 of Coordinates and the Abhyankar–Sathaye problem.

References