# An explicit noncoordinate polynomial[2pt]with affine three-space zero fibre


hypersetup{
  pdftitle={An explicit noncoordinate polynomial with affine three-space zero fibre}}


maketitle

Abstract: 
We give an explicit polynomial F in four complex variables whose zero fibre is a
copy of CC^3 and whose gradient vanishes at (2,0,-tfrac12,tfrac12). A
polynomial whose gradient vanishes somewhere cannot be the first component of a
polynomial automorphism of CC^4. This answers the Abhyankar--Sathaye question
negatively in ambient dimension four; adjoining variables gives counterexamples
in every dimension at least four. The proof rests on one elementary lemma, and
every identity in this note can be checked by hand.


# The question

All polynomials in this note have complex coefficients. Write
R_n=CC[z_1,...,z_n] and CC^{[m]}=CC[t_1,...,t_m].

[coordinate]
A polynomial fin R_n is a coordinate if there are
f_2,...,f_nin R_n for which Phi=(f,f_2,...,f_n)colonCC^ntoCC^n is
a bijection with polynomial inverse. Such a Phi is a polynomial
automorphism of CC^n, and f is its first component.


For instance z_1 is a coordinate, and so is z_1+z_2^2, because
(z_1,z_2)|->(z_1+z_2^2,z_2) has the polynomial inverse
(w_1,w_2)|->(w_1-w_2^2,w_2).


If fin R_n is a coordinate, then R_n/(f)congCC^{[n-1]}.


Proof. 
In the coordinates in which f is the first coordinate,
R_n=CC[w_1,...,w_n] and R_n/(f)=CC[w_2,...,w_n].


Hence every fibre of a coordinate, in particular its zero fibre, is a copy of
CC^{n-1}. The question is whether the converse holds.

[Abhyankar--Sathaye]
If fin R_n satisfies R_n/(f)congCC^{[n-1]}, must f be a coordinate?


Geometrically: can every copy of CC^{n-1}, cut out inside CC^n by one
polynomial equation, be carried to a coordinate hyperplane by a polynomial
automorphism of CC^n? For n=2 the answer is yes, by the Abhyankar--Moh
theorem~[cite] and Suzuki~[cite]: an embedded line can be straightened.
In dimension four the answer is yes when the equation is linear in one variable
and the coefficient of that variable depends on only one variable, by Kaliman,
V'en'ereau and Zaidenberg~cite[Theorem~4.2(a)]{KVZ} and
Maubach~cite[Corollaries~4.2--4.3]{Maubach}; further linear cases appear in
[cite]. Below we answer the question negatively for
n=4 (Section~sec:fibre) and, by adjoining variables, for all nge4
(Section~sec:higher). The three-variable case is not addressed here.

# The polynomial

The starting point is the cusp. Since (u^3)^2+(-u^2)^3=0, the map
u|->(u^3,-u^2) parametrizes the curve x^2+y^3=0. We perturb both entries
by independent multiples of a new variable h. In R=CC[h,u,v,w] put

  x=u^3+hv, y=-u^2+hw,
  s=S(h,u,v,w):=2u^3v+3u^4w+h(v^2-3u^2w^2)+h^2w^3.

Expanding gives x^2=u^6+2hu^3v+h^2v^2 and
y^3=-u^6+3hu^4w-3h^2u^2w^2+h^3w^3; the two u^6 terms cancel, so

  x^2+y^3=hs.

The perturbation has turned the cusp equation into a divisible expression.

The second ingredient is an identity in independent indeterminates x,y,s. Put

  p(x,y,s)=-2s^2x+3sy^2-3s^3y, X=x+s^3, Y=y-s^2.

Since X^2=x^2+2s^3x+s^6 and Y^3=y^3-3s^2y^2+3s^4y-s^6, the two s^6 terms
cancel and

  X^2+Y^3=x^2+y^3-sp(x,y,s).

Now define

  F=h-p(x,y,s)-1in R,

with x,y,s the polynomials (eq:xyz); thus
F=h-1+2s^2x-3sy^2+3s^3y in the four variables h,u,v,w.

Combining (eq:cusp), (eq:shift) and (eq:F) gives the identity
in R

  X^2+Y^3=s(1+F),

because x^2+y^3-sp=s(h-p)=s(1+F). On the zero fibre F=0 it reads
X^2+Y^3=s, a surface that is a graph over the two free variables X and Y.
That is where the copy of CC^3 on the zero fibre comes from. The work below
produces a third coordinate on that fibre, together with its polynomial inverse.


The polynomial F of (eq:F) satisfies
[
  R/(F)congCC^{[3]}
   and 
  grad F(2,0,-tfrac12,tfrac12)=0.
]
Consequently F is not a coordinate of R.


Let P=(2,0,-frac12,frac12), in the variable order (h,u,v,w). There
x=u^3+hv=-1, y=-u^2+hw=1, s=1, and
p=-2s^2x+3sy^2-3s^3y=2, so F=h-p-1=-1. The critical point therefore lies on
the fibre over -1, not on the zero fibre; the zero fibre is smooth, being a copy
of CC^3. The isomorphism in the theorem is proved in
Section~sec:fibre, the vanishing of the gradient in
Section~sec:critical.


The polynomials (eq:xyz) are annihilated by the triangular derivation
D=h∂_u-3u^2∂_v+2u∂_w of R: a direct check gives
Dx=3hu^2-3hu^2=0, Dy=-2hu+2hu=0, Ds=0. They appear in Maubach's
calculation of kernels of triangular monomial derivations
cite[proof of Theorem~3.2, case~(3)]{MaubachMonomial}, but nothing below uses
that origin: the proof needs only (eq:cusp) and (eq:shift).


# A lifting lemma

One general statement is needed. Throughout, S denotes the polynomial of
(eq:xyz). In words: if a commutative ring B contains h,x,y,s with
x^2+y^3=hs, and if h and y generate the unit ideal in the sense that
alpha h+beta y=1 for some alpha,betain B, then the three equations
[
  u^3+hv=x, -u^2+hw=y, S(h,u,v,w)=s
]
in three new letters u,v,w describe a polynomial ring in one variable. The
hypothesis on h and y is exactly what lets us solve two linear equations
without dividing, which matters because h and y need not be invertible and
B need not be a domain.


Let B be a commutative ring and let h,x,y,sin B satisfy x^2+y^3=hs and
(h,y)=B. Then
[
  C=B[U,V,W]/(U^3+hV-x, -U^2+hW-y, S(h,U,V,W)-s)~= B[T].
]


Proof. 
Choose alpha,betain B with alpha h+beta y=1. Substitute G=V+UW (so
V=G-UW; this is an invertible change of generators of the polynomial ring).
Modulo the relation hW=U^2+y, a direct expansion of (eq:xyz) gives
[
  S(h,U,G-UW,W)=hG^2-2UyG+y^2W,
]
and the relation U^3+hV=x becomes hG-yU=x. Hence

  Ccong B[U,G,W]/(hG-yU-x, hW-U^2-y, y^2W-s+hG^2-2UyG).

Put A=B[U,G]/(hG-yU-x) and in A set a=U^2+y, b=s-hG^2+2UyG. The last
two relations of (eq:Clinear) read hW=a and y^2W=b. They are
compatible, since
[
  y^2a-hb=y^2(U^2+y)-h(s-hG^2+2UyG)
  =(hG-yU)^2+y^3-hs=x^2+y^3-hs=0.
]
With r=alpha(1+beta y) and q=beta^2 we have
rh+qy^2=(1-beta y)(1+beta y)+beta^2y^2=1, where the first step uses
alpha h=1-beta y. Hence W_0=ra+qb satisfies both equations: from
y^2a=hb one gets hW_0=(rh+qy^2)a=a and y^2W_0=(rh+qy^2)b=b. Moreover
[
  W-W_0=r(hW-a)+q(y^2W-b),
]
while hW-a=h(W-W_0) and y^2W-b=y^2(W-W_0). So in A[W] the ideal
(hW-a,,y^2W-b) equals (W-W_0), and evaluating at W_0 gives Ccong A.

It remains to identify A with B[T]. The B-algebra maps Ato B[T] and
B[T]to A given by
[
  U|-> hT-beta x, G|-> yT+alpha x,
  T|->alpha U+beta G
]
are inverse. Indeed, in A the relation hG-yU=x and alpha h+beta y=1 give
[
  h(alpha U+beta G)=U+beta x,
  y(alpha U+beta G)=G-alpha x,
]
so substituting U=hT-beta x and G=yT+alpha x into alpha U+beta G
returns T, and substituting T=alpha U+beta G into the first two returns U
and G. Thus Acong B[T], and Ccong A.

In the original letters, the parameter is T=alpha U+beta(V+UW); its inverse
recovers U and G from T, then W from W_0, then V=G-UW.


# The zero fibre is affine three-space

We prepare the base ring of Lemma~lem:lift. Let h,x,y,s now be four
independent indeterminates and set

  B=CC[h,x,y,s]/(h-1-p(x,y,s), x^2+y^3-sh),

writing h,x,y,s also for their classes in B. This is a new presentation: here
x,y,s are independent variables, not the polynomials (eq:xyz).
Eliminating h leaves x^2+y^3=s(1+p(x,y,s)), and the triangular change of
variables X=x+s^3, Y=y-s^2 of (eq:p) turns this into X^2+Y^3=s by
(eq:shift). Hence

  BcongCC[X,Y,s]/(X^2+Y^3-s)congCC^{[2]},

with inverse presentation
[
  s=X^2+Y^3, x=X-s^3, y=Y+s^2, h=1+p(x,y,s).
]

The elements h and y generate the unit ideal of B: modulo (h,y) the two
defining relations become x^2=0 and 1=2s^2x, and squaring the second gives
1=4s^4x^2=0. An explicit certificate is the pair

  alpha=1+2s^2x+4s^5,
  beta=(3s^3-3sy)(1+2s^2x)-4s^4y^2,

for which alpha h+beta y=1 in B, because the identity

  alpha h+beta y-1=(1+2s^2x)(h-1-p(x,y,s))-4s^4(x^2+y^3-sh)

holds in the free polynomial ring and both terms on the right vanish in B.

Apply Lemma~lem:lift over this B. Its algebra C is
[
  C=B[u,v,w]/(u^3+hv-x, -u^2+hw-y, S(h,u,v,w)-s).
]
In the polynomial ring CC[h,x,y,s,u,v,w] the last three relations are solved
for x,y,s by (eq:xyz): they read x=u^3+hv, y=-u^2+hw,
s=S(h,u,v,w). Under this substitution the relation x^2+y^3-sh becomes
identically zero by (eq:cusp), and the relation h-1-p(x,y,s) becomes the
equation F=0 of (eq:F). Therefore

  R/(F)~= Ccong B[T]congCC[X,Y,T].

This proves the first assertion of Theorem~thm:main: X, Y and T are
polynomial coordinates on the zero fibre. Appendix~app:coordinates records
both the forward and the inverse maps in terms of h,u,v,w.

# A point where the gradient vanishes

At P=(2,0,-frac12,frac12), in the variable order (h,u,v,w), the values of
the polynomials (eq:xyz) and (eq:p) are
[
  x=-1, y=1, s=1, p=2,
  X=x+s^3=0, Y=y-s^2=0, F=-1.
]
Differentiating the ambient identity (eq:ambient), X^2+Y^3=s(1+F),
gives
[
  2X,dX+3Y^2,dY=(1+F),ds+s,dF.
]
At P the left-hand side vanishes, 1+F=0 and s=1, hence dF(P)=0: all four
partial derivatives of F vanish at P.

To see why this excludes a coordinate, suppose F were the first component of a
polynomial automorphism Phi=(F,g_2,g_3,g_4) of CC^4 with polynomial
inverse Psi. The chain rule gives JPhi(P)· JPsi(Phi(P))=I, so
JPhi(P) is invertible. But its first row is grad F(P)=0, so the matrix is
not invertible. This contradiction proves that F is not a coordinate.

The obstruction uses a critical point of the defining polynomial, not of its zero
fibre; that is the ambient critical locus emphasized by Shpilrain and Yu
cite[S1]{SY}, who asked whether the gradient must be nowhere vanishing when the
zero fibre is a copy of a hyperplane. Theorem~thm:main answers that
question negatively in ambient dimension four.

# Higher ambient dimension


For every integer nge4 there is a polynomial fin R_n such that
R_n/(f)congCC^{[n-1]} and f is not a coordinate.


Proof. 
For n=4 take f=F. For n>4 take the same F, regarded as a polynomial in
z_1,...,z_n that does not involve z_5,...,z_n. By (eq:fiber),
[
  R_n/(F)=R[z_5,...,z_n]/(F)=CC[X,Y,T][z_5,...,z_n],
]
a polynomial ring in n-1 variables. All partial derivatives of F in the
added variables vanish, and grad F(P)=0, so the gradient vanishes at
(P,0,...,0) and the Jacobian argument of Section~sec:critical applies.


appendix
# Explicit formulas on the zero fibre

Both directions of the isomorphism (eq:fiber) are polynomial, and they are
spelled out here so that the zero fibre can be parametrized directly. All
expressions use (eq:xyz) and the pair (eq:alphabeta).

The forward map takes a point of the zero fibre F=0 to CC^3 by

  X=x+s^3, Y=y-s^2, T=alpha u+beta(v+uw).

The inverse map takes independent X,Y,T to a point of F=0 by

  s&=X^2+Y^3, & x&=X-s^3, notag
  y&=Y+s^2, & h&=1+p(x,y,s), notag
  u&=hT-beta x, & g&=yT+alpha x, notag
  v&=g-uw, & w&=alpha(1+beta y)(y+u^2)+beta^2(s-hg^2+2uyg).
  

These are exactly the maps of Lemma~lem:lift: the pair (u,g) is the pair
(U,G) of the lemma, the formula for w is W_0=ra+qb with
r=alpha(1+beta y) and q=beta^2, and v=g-uw is V=G-UW. No division
occurs, so both maps are polynomial on all of CC^3 and on all of the zero
fibre.

{99}
small


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