aa r X i v : . [ m a t h . AG ] D ec CLASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES
P.-M. POLONI
Abstract.
The Danielewski hypersurfaces are the hypersurfaces X Q,n in C defined by anequation of the form x n y = Q ( x, z ) where n ≥ Q ( x, z ) is a polynomial such that Q (0 , z )is of degree at least two. They were studied by many authors during the last twenty years.In the present article, we give their classification as algebraic varieties. We also give theirclassification up to automorphism of the ambient space. As a corollary, we obtain that everyDanielewski hypersurface X Q,n with n ≥ C . Introduction
The story of Danielewski hypersurfaces goes back to 1989, when Danielewski [4] showed that,if W n denotes the hypersurface in C defined by the equation x n y − z ( z −
1) = 0, then W n × C and W m × C are isomorphic algebraic varieties for all n, m ≥
1, whereas the surfaces W and W are not isomorphic. He discovered the first counterexamples to the Cancellation Problem.Then, Fieseler [6] proved that W n and W m are not isomorphic if n = m .Since these results appeared, complex algebraic surfaces defined by equations of the form x n y − Q ( x, z ) = 0 (now called Danielewski hypersurfaces ) have been studied by many differentauthors (see [10], [8], [3], [1], [7], [9]), leading to new interesting examples as byproducts. Letus mention two of them.In their work on embeddings of Danielewski hypersurfaces given by x n y = p ( z ), Freuden-burg and Moser-Jauslin [7] discovered an example of two smooth algebraic surfaces which arealgebraically non-isomorphic but holomorphically isomorphic.More recently, the study of Danielewski hypersurfaces of equations x y − z − xq ( z ) = 0 pro-duced the first counterexamples to the Stable Equivalence Problem [9]; that is two polynomialsof C [ X , X , X ] which are not equivalent (i.e. such that there exist no algebraic automorphismof C [ X , X , X ] which maps one to the other one) but, when considered as polynomials of C [ X , X , X , X ], become equivalent.The purpose of the present paper is to classify all Danielewski hypersurfaces, both as algebraicvarieties, and also as hypersurfaces in C . More precisely, we will give necessary and sufficientconditions for isomorphism of two Danielewski hypersurfaces; and, on the other hand, wewill give necessary and sufficient conditions for equivalence of two isomorphic Danielewskihypersurfaces. Recall that two isomorphic hypersurfaces H , H ⊂ C n are said to be equivalent if there exists an algebraic automorphism Φ of C n which maps one to the other one, i.e. suchthat Φ( H ) = H .We know indeed that isomorphic classes and equivalence classes are distinct for Danielewskihypersurfaces. This was first observed by Freudenburg and Moser-Jauslin, who showed in [7]that the Danielewski hypersurfaces defined respectively by the equations f = x y − (1 + x )( z −
1) = 0 and g = x y − z + 1 = 0 are isomorphic but non equivalent. (One way to see that theyare not equivalent is to remark that the level surfaces f − ( c ) are smooth for every constant c ∈ C , whereas the surface g − (1) is singular along the line { x = z = 0 } .)Several papers already contain the classification, up to isomorphism, of Danielewski hyper-surfaces of a certain form. Makar-Limanov proved in [8] that two Danielewski hypersurfacesof equations x n y − p ( z ) = 0 and x n y − p ( z ) = 0 with n , n ≥ p , p ∈ C [ z ]are isomorphic if and only if they are equivalent via an affine automorphism of the form( x, y, z ) ( ax, by, cz + d ) with a, b, c ∈ C ∗ and d ∈ C . Then, Daigle generalized in [3] this result to the case n , n ≥
1. Next, Wilkens has given in [10] the classification of Danielewskihypersurfaces of equations x n y − z − h ( x ) z = 0 with n ≥ h ( x ) ∈ C [ x ].Finally, Dubouloz and the author showed in [5] that every Danielewski hypersurface X Q,n ofequation x n y = Q ( x, z ), where Q ( x, z ) is such that Q (0 , z ) has simple roots, is isomorphic toa one defined by an equation of the form x n y = Q di =1 ( z − σ i ( x )), where { σ ( x ) , . . . σ d ( x ) } is acollection of polynomials in C [ x ] so that σ i (0) = σ j (0) if i = j . In the same paper, we classifiedthese last ones and called them standard forms . This effectively classifies, up to isomorphism,all Danielewski hypersurfaces of equations x n y = Q ( x, z ), where Q (0 , z ) has simple roots.In the present paper, we generalize the notion of Danielewski hypersurface in standard formand we prove that every Danielewski hypersurface is isomorphic to a one in standard form(which can be found by an algorithmic procedure). Then, we are able to classify all Danielewskihypersurfaces. The terminology standard form is relevant since every isomorphism between twoDanielewski hypersurfaces in standard form – and every automorphism of such a Danielewskihypersurface – extend to a triangular automorphism of C .We also give a criterion (Theorem 4.1) to distinguish isomorphic but not equivalent Danielewskihypersurfaces.As a corollary, we obtain that every Danielewski hypersurface defined by an equation of theform x n y − Q ( x, z ) = 0 with n ≥ C .Most of these results are based on a precise picture of the sets of locally nilpotent derivationsof coordinate rings of Danielewski hypersurfaces, obtained using techniques which were mainlydeveloped by Makar-Limanov in [8].The paper is organized as follows. Section 1 is the introduction. In section 2, we fix some nota-tions and definitions. In section 3, we study the locally nilpotent derivations on the Danielewskihypersurfaces in order to get information on what an isomorphism between two Danielewskihypersurfaces looks like. Section 4 is devoted to the classification of Danielewski hypersurfacesup to equivalence, whereas sections 5 and 6 contain their classification up to isomorphism andthe study of the Danielewski hypersurfaces in standard form.2. Definitions and notations
In this paper, our base field is C , the field of complex numbers. If n ≥
1, then C [ n ] willdenote a polynomial ring in n variables over C . Definition 2.1.
Two hypersurfaces X and X of C n are said to be equivalent if there existsa (polynomial) automorphism Φ of C n such that Φ( X ) = X .This notion is related to the notion of equivalent embeddings in the following sense. If X and X are two isomorphic hypersurfaces of C n which are not equivalent, then X admits twonon-equivalent embeddings into C n . More precisely, let ϕ : X → X be an isomorphism anddenote i : X → C n and i : X → C n the inclusion maps. Then, i and i ◦ ϕ are twonon-equivalent embeddings of X into C n , since ϕ does not extend to an automorphism of C n . Definition 2.2. A Danielewski hypersurface is a hypersurface X Q,n ⊂ C defined by anequation of the form x n y − Q ( x, z ) = 0, where n ∈ N and Q ( x, z ) ∈ C [ x, z ] is such thatdeg ( Q (0 , z )) ≥ S Q,n the coordinate ring of a Danielewski hypersurface X Q,n , i.e. S Q,n = C [ X Q,n ] = C [ x, y, z ] / ( x n y − Q ( x, z )).It can be easily seen that every Danielewski hypersurface is equivalent to a one of the form X Q,n with deg x Q ( x, z ) < n . Lemma 2.3.
Let X Q,n be a Danielewski hypersurface and R ( x, z ) ∈ C [ x, z ] be a polynomial.Then X Q,n is equivalent to the Danielewski hypersurface of equation x n y − Q ( x, z ) − x n R ( x, z ) .Proof. It suffices to consider the triangular automorphism of C defined by ( x, y, z ) ( x, y + R ( x, z ) , z ). (cid:3) LASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES 3 Using locally nilpotent derivations
One important property of Danielewski hypersurfaces is that they admit nontrivial actionsof the additive group C + . For instance, we can define a C + -action δ Q,n : C × X Q,n → X Q,n ona hypersurface X Q,n by posing δ Q,n ( t, ( x, y, z )) = (cid:0) x, y + x − n ( Q ( x, z + tx n ) − Q ( x, z )) , z + tx n (cid:1) . Since a C + -actions on an affine complex surface S induces a C -fibration over an affine curve,affine complex surfaces with C + -actions split into two cases. Either there is only one C -fibrationon S up to an isomorphism of the base, or there exists a second one. In other words, eitherthe surface has a Makar-Limanov invariant of transcendence degree one, or its Makar-Limanovinvariant is trivial. Recall that algebraic C + -actions on an affine variety Spec A correspondto locally nilpotent derivations on the C -algebra A (for example, the action δ Q,n on a surface X Q,n corresponds to the locally nilpotent derivation ∆
Q,n = x n ∂∂z + ∂Q ( x,z ) ∂z ∂∂y on the coordinatering C [ X Q,n ]), and that the Makar-Limanov invariant ML( A ) of an algebra A is defined asthe intersection of all kernels of locally nilpotent derivations of A . Equivalently, ML( A ) is theintersection of all invariant rings of algebraic C + -actions on Spec( A ).Applying Makar-Limanov’s techniques, one can obtain the first important result concerningDanielewski hypersurfaces: the Makar-Limanov invariant of a Danielewski hypersurface X Q,n is non-trivial if n ≥ Theorem 3.1.
Let X Q,n be a Danielewski hypersurface. Then
ML( X Q,n ) = C if n = 1 and ML( X Q,n ) = C [ x ] if n ≥ .Proof. Let X Q,n be a Danielewski hypersurface.If n = 1, the result is easy. Indeed, we can suppose that Q ( x, z ) = p ( z ) ∈ C [ z ] (see Lemma2.3). Then, it suffices to consider the following locally nilpotent derivations on the coordinatering S p, = C [ x, y, z ] / ( xy − p ( z )). δ = x ∂∂z + p ′ ( z ) ∂∂y and δ = y ∂∂z + p ′ ( z ) ∂∂x . Since Ker( δ ) ∩ Ker( δ ) = C , this shows that the Makar-Limanov invariant of every Danielewskihypersurface X Q, is trivial.Suppose now that n ≥ δ be a non-zero locally nilpotent derivation on the coordinatering S Q,n = C [ X Q,n ]. Without loss of generality, we can suppose that the leading term of Q (0 , z )is z d with d ≥ x n y = p ( z ) stillholds. This proof goes as follows.The main idea is to consider S Q,n as a subalgebra of C [ x, x − , z ] with y = x − n Q ( x, z ) and tochoose a Z -filtration on S Q,n such that the corresponding graded algebra Gr( S Q,n ) is isomorphicto the subalgebra C [ x, x − n z d , z ].Recall that Makar-Limanov has proved that any non-zero locally nilpotent derivation D onan algebra A with a Z -filtration induces a non-zero locally nilpotent derivation gr ( D ) on thegraded algebra Gr( A ). In our case, he proved in [8] that Ker( gr ( δ )) = C [ x ].Recall also that we can define a degree function associated to a locally nilpotent D ∈ LND( A )by posing deg(0) = −∞ and deg D ( a ) := max { n ∈ N | D n ( a ) = 0 } if a ∈ A \ { } . Moreover, if A has a Z -filtration, then deg gr ( D ) ( gr ( a )) ≤ deg D ( a ) for any element a ∈ A . (Here gr : A → Gr( A )denote the natural function from A to Gr( A ).)This implies that Ker( δ ) = C [ x ]. Indeed, if p ∈ S Q,n \ C [ x ], then we can choose a filtrationon S Q,n such that gr ( p ) belongs to Gr( S Q,n ) \ C [ x ]. Then Ker( δ ) ⊂ C [ x ] follows from theinequalities 1 ≤ deg gr ( δ ) ( gr ( p )) ≤ deg δ ( p ). Since Ker( δ ) is of transcendence degree one and isalgebraically closed, we obtain that x ∈ Ker( δ ).Thus, Ker( δ ) = C [ x ] for any non-zero locally nilpotent derivation on S Q,n . In particular,ML( X Q,n ) = C [ x ]. (cid:3) P.-M. POLONI
Using this result, we can obtain a precise picture of the set of locally nilpotent derivationson rings S Q,n when n ≥ Theorem 3.2.
Let X Q,n be a Danielewski hypersurface with n ≥ and let S Q,n denote itscoordinate ring. Then
LND ( S Q,n ) = (cid:26) h ( x ) (cid:18) x n ∂∂z + ∂Q ( x, z ) ∂z ∂∂y (cid:19) , where h ( x ) ∈ C [ x ] (cid:27) . In particular,
Ker( δ ) = C [ x ] and Ker( δ ) = C [ x ] z + C [ x ] for every non-zero locally nilpotentderivation δ ∈ LND( S Q,n ) .Proof. Let δ be a non-zero locally nilpotent derivation on S Q,n with n ≥
2. In the proof ofTheorem 3.1, we showed that Ker( δ ) = C [ x ]. Then, due to Lemma 1.1 in [2], there existpolynomials a ( x ) , b ( x ) ∈ C [ x ] such that a ( x ) δ = b ( x )∆ Q,n where ∆
Q,n ∈ LND( S Q,n ) is thederivation defined by ∆
Q,n = x n ∂∂z + ∂Q ( x,z ) ∂z ∂∂y .The theorem will follow easily. First note that a ( x ) δ ( z ) = x n b ( x ). Also a ( x ) divides x n b ( x ).Therefore, in order to prove that a ( x ) divides b ( x ), it is enough to show that a (0) = 0 implies b (0) = 0. This holds since δ ( y ) ∈ C [ x, z ] and a (0) δ ( y )(0 , z ) = b (0)( Q (0 , z )) ′ ( z ).The theorem is proved. (cid:3) This theorem gives us a very powerful tool for classifying Danielewski hypersurfaces. Indeed,note that an isomorphism ϕ : A → B , between two algebras A and B , conjugates the setsLND( A ) and LND( B ) of locally nilpotent derivations on A and B , i.e. LND( A ) = ϕ − LND( B ) ϕ if ϕ : A → B is an isomorphism. In turn, we obtain the following result. Corollary 3.3. (1)
Let ϕ : X Q ,n → X Q ,n be an isomorphism between two Danielewski hypersurfaces with n , n ≥ . Then, there exist two constants a, α ∈ C ∗ and a polynomial β ( x ) ∈ C [ x ] such that ϕ ∗ ( x ) = ax and ϕ ∗ ( z ) = αz + β ( x ) . (2) If X Q ,n and X Q ,n are two isomorphic Danielewski hypersurfaces, then n = n and deg( Q (0 , z )) = deg( Q (0 , z )) . (3) Suppose that X Q ,n and X Q ,n are two equivalent Danielewski hypersurfaces with n ≥ ,and let Φ : C → C be an algebraic automorphism such that Φ( X Q ,n ) = X Q ,n . Then,there exist constants a, α ∈ C ∗ , β ∈ C and a polynomial B ∈ C [2] such that Φ ∗ ( x ) = ax and Φ ∗ ( z ) = αz + β + xB ( x, x n y − Q ( x, z )) .Proof. For (1) and (2), we follow the ideas of a proof given by Makar-Limanov in [8].Let ϕ : X Q ,n → X Q ,n be an isomorphism between two Danielewski hypersurfaces with n , n ≥
2. Let x i , y i , z i denote the images of x, y, z in the coordinate ring S i = S Q i ,n i = C [ X Q i ,n i ].If δ ∈ LND( S ), then ( ϕ ∗ ) − ◦ δ ◦ ϕ ∗ ∈ LND( S ). Thus, Theorem 3.2 implies δ ( z ) = 0and δ ( ϕ ∗ ( z )) = 0 for any locally nilpotent derivation δ ∈ LND( S ). Therefore, ϕ ∗ ( z ) = α ( x ) z + β ( x ) for some polynomials α and β . Since ϕ is invertible, α must be a nonzeroconstant α ∈ C ∗ .On the other hand, ϕ ∗ induces an isomorphism ML( S ) = C [ x ] → ML( S ) = C [ x ]. Conse-quently, ϕ ∗ ( x ) = ax + b for some constants a ∈ C ∗ and b ∈ C .In order to prove b = 0, consider the locally nilpotent derivation δ ∈ LND( S ) defined by δ = ( ϕ ∗ ) − ◦ (cid:16) x n ∂∂z + ∂Q ( x ,z ) ∂z ∂∂y (cid:17) ◦ ϕ ∗ . Now, Theorem 3.2 implies that δ ( z ) is divisible by x n . Since δ ( z ) = aα − ( x − b ) n , we must have b = 0 and n ≥ n . This proves the first partof the corollary.Moreover, repeating this analysis with ϕ − instead of ϕ , we also obtain n ≥ n and so n = n = n .Since ϕ : X Q ,n → X Q ,n is a morphism, we know that ϕ ∗ ( x n y − Q ( x, z )) belongs to the ideal( x n y − Q ( x, z )). In particular, when x = 0, it implies that Q (0 , αz + β (0)) = ϕ ∗ ( Q (0 , z )) ∈ ( Q (0 , z )). Thus deg( Q (0 , z )) ≥ deg( Q (0 , z )). LASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES 5
Working with ϕ − , the same analysis allows us to conclude that deg( Q (0 , z )) ≥ deg( Q (0 , z )).Moreover, it implies that Q (0 , αz + β (0)) = µQ (0 , z ) for a certain constant µ ∈ C ∗ .Since the case n = n = 1 was already done by Daigle [3], this suffices to prove the secondpart of the corollary.It remains to prove the third part.Let X Q ,n and X Q ,n be two equivalent Danielewski hypersurfaces with n ≥
2, and let Φ bean algebraic automorphism of C such that Φ( X Q ,n ) = X Q ,n .Since the polynomial x n y − Q ( x, z ) is irreducible, there exists a nonzero constant µ ∈ C ∗ sothat Φ ∗ ( x n y − Q ( x, z )) = µ ( x n y − Q ( x, z )).Thus, Φ induces an isomorphism Φ c between the Danielewski hypersurfaces of equation x n y − Q ( x, z ) = µc and x n y − Q ( x, z ) = c for every c ∈ C .Since n ≥
2, the Makar-Limanov invariant of these hypersurfaces is C [ x ]. By (1), we obtainnow that the image by Φ ∗ of the ideal ( x, x n y − Q ( x, z ) − µc ) belongs to the ideal ( x, x n y − Q ( x, z ) − c ) = ( x, Q (0 , z ) + c ) for each c ∈ C . It turns out thatΦ ∗ ( x ) ∈ \ c ∈ C ( x, Q (0 , z ) + c ) = ( x ) . Since Φ is invertible, this implies that Φ ∗ ( x ) = ax for a certain constant a ∈ C ∗ . Thus − µQ (0 , z ) ≡ µ ( x n y − Q ( x, z )) ≡ Φ ∗ ( x n y − Q ( x, z )) ≡ − Q (0 , Φ ∗ ( z )) mod ( x ) . Since deg Q (0 , z ) = deg Q (0 , z ) (by the second part of the corollary), this implies that Φ ∗ ( z ) ≡ αz + β mod ( x ) for certain constants α and β such that Q (0 , αz + β ) = µQ (0 , z ).Thus, we can write Φ ∗ ( z ) = αz + β + xB ( x, y, z ), where B is polynomial of C [ x, y, z ].Now, we use again the first part of the corollary. For every c ∈ C , there exist a constant α c ∈ C ∗ and a polynomial β c ∈ C [1] such thatΦ ∗ ( z ) = αz + β + xB ( x, y, z ) ≡ α c z + β c ( x ) mod ( x n y − Q ( x, z ) − c ) . Therefore, for every c ∈ C , we have α c = α , β c (0) = β and B ( x, y, z ) ≡ x − ( β c ( x ) − β ) mod ( x n y − Q ( x, z ) − c ) . In particular B has the following property: For infinitely many constants c ∈ C , there existpolynomials r c ( x ) ∈ C [ x ] and s c ( x, y, z ) ∈ C [ x, y, z ] such that B ( x, y, z ) = r c ( x ) + s c ( x, y, z )( x n y − Q ( x, z ) − c ) . We will show that any polynomial with this property must belong to C [ x, x n y − Q ( x, z )].Remark that it suffices to show that at least one polynomial s c belongs to C [ x, x n y − Q ( x, z )].In order to see this, we define a degree function d on C [ x, y, z ] by posing, for every f ∈ C [ x, y, z ], d ( f ) := deg( ˜ f ( y, z )), with ˜ f ( y, z ) = f ( x, y, z ) ∈ C [ x ][ y, z ].Let B ( x, y, z ) = r c ( x ) + s c ( x, y, z )( x n y − Q ( x, z ) − c ) for one c ∈ C . Then, s c satisfiesalso the above property and its degree d ( s ) is strictly less than d ( B ).Therefore, the desired result can be obtained by decreased induction on the degree d . (cid:3) Equivalence classes
In this section, we prove the following result.
Theorem 4.1.
Two Danielewski hypersurfaces X Q ,n and X Q ,n are equivalent if and only if n = n = n and there exist a, α, µ ∈ C ∗ , β ∈ C and B ∈ C [2] such that Q ( ax, αz + β + xB ( x, Q ( x, z ))) ≡ µQ ( x, z ) mod ( x n ) . Remark . We will show in the next section (Proposition 5.7) that this theorem implies thatevery Danielewski hypersurface X Q,n with n ≥ C . P.-M. POLONI
Before proving Theorem 4.1, let us give another result. Given two Danielewski hypersurfaces,it is not easy to check if the second condition in Theorem 4.1 is fulfilled. Therefore, we alsoshow that any Danielewski hypersurface is equivalent to another one which is unique up to anaffine automorphism.
Theorem 4.3. (1)
Every Danielewski hypersurface is equivalent to a Danielewski hypersurface X ( p, { q i } i =2 .. deg( p ) , n ) defined by an equation of the form x n y − p ( z ) − x deg( p ) X i =2 p ( i ) ( z ) q i ( x, p ( z )) with deg x ( q i ) < n − . Moreover, there is an algorithmic procedure which computes, given a Danielewski hyper-surface X , a hypersurface X ( p, { q i } i =2 .. deg( p ) , n ) which is equivalent to X . (2) Two such Danielewski hypersurfaces X ( p , { q ,i } i =2 .. deg( p ) , n ) and X ( p , { q ,i } i =2 .. deg( p ) , n ) are equivalent if and only if n = n , deg( p ) = deg( p ) = d and there exist some con-stants a, α, µ ∈ C ∗ , β ∈ C such that p ( αz + β ) = µp ( z ) and aα − i q ,i ( ax, µt ) = q ,i ( x, t ) for every ≤ i ≤ d .Remark . This result generalizes the classification of Danielewski hypersurfaces of the form x y − z − xq ( z ) given by Moser-Jauslin and the author in [9]. Proof of Theorem 4.1.
Let X Q ,n and X Q ,n be two equivalent Danielewski hypersurfaces.Then, the second part of Corollary 3.3 implies n = n = n .If n = 1, the result is already known. Indeed, by Lemma 2.3, every Danielewski hypersurface X Q, with n = 1 is equivalent to a one of the form X p, with p ( x, z ) = p ( z ) ∈ C [ z ]. Then,Daigle [3] has proven that two such hypersurfaces X p , and X p , are isomorphic if and only if p ( az + b ) = µp ( z ) for some constants a, µ ∈ C ∗ and b ∈ C .Now, assume n ≥ C such that Φ( X Q ,n ) = X Q ,n .Corollary 3.3, gives us constants a, α ∈ C ∗ , β ∈ C and a polynomial B ∈ C [2] such that Φ ∗ ( x ) = ax and Φ ∗ ( z ) = αz + β + xB ( x, x n y − Q ( x, z )). Since the polynomial x n y − Q ( x, z ) is irreducible,there exists a nonzero constant µ ∈ C ∗ so that Φ ∗ ( x n y − Q ( x, z )) = µ ( x n y − Q ( x, z )). It turnsout that Q ( ax, αz + β + xB ( x, − Q ( x, z ))) ≡ µQ ( x, z ) mod ( x n ), as desired.Conversely, let X Q ,n and X Q ,n be two Danielewski hypersurfaces with Q ( ax, αz + β + xB ( x, Q ( x, z ))) ≡ µQ ( x, z ) mod ( x n ) for some a, α, µ ∈ C ∗ , β ∈ C and B ∈ C [2] .We pose R ( x, y, z ) = x − n ( Q ( ax, αz + β + xB ( x, − x n y + Q ( x, z ))) − µQ ( x, z )) ∈ C [ x, y, z ]and define an endomorphism of C byΦ( x, y, z ) = ( ax, a − n µy + a − n R ( x, y, z ) , αz + β + xB ( x, − x n y + Q ( x, z ))) . Remark that Φ ∗ ( x n y − Q ( x, y )) = µ ( x n y − Q ( x, y )). Therefore, the theorem will be provedif we show that Φ is invertible.It suffices to prove that Φ ∗ is surjective, i.e. C [ x, y, z ] ⊂ Φ ∗ ( C [ x, y, z ]) = C [Φ ∗ ( x ) , Φ ∗ ( y ) , Φ ∗ ( z )] . We know already that x and P := x n y − Q ( x, z ) are in the image of Φ ∗ .Then, since z = α − (Φ ∗ ( z ) − β − xB ( x, − P )), we obtain that z belongs to Φ ∗ ( C [ x, y, z ]).It remains to show that y belongs to the image of Φ ∗ . To do this, we first remark that thereexist polynomials f, g ∈ C [3] such that R ( x, y, z ) = xyf ( x, z, P ) + g ( x, z, P ) . Thus, y ( µ + xf ( x, z, P )) = a n Φ ∗ ( y ) − g ( x, z, P ) ∈ Φ ∗ ( C [ x, y, z ]). Now, choose some polyno-mials ˜ f and ˜ g such that( µ + xf ( x, z, P )) ˜ f ( x, z, P ) = 1 + x n ˜ g ( x, z, P ) . LASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES 7
Then, we can write y ( µ + xf ( x, z, P )) ˜ f ( x, z, P ) = y (1 + x n ˜ g ( x, z, P ))= y + ( P − Q ( x, z ))˜ g ( x, z, P ) . This implies that y belongs to the image of Φ ∗ and proves the theorem. (cid:3) Proof of Theorem 4.3.
Let X Q,n be a Danielewski hypersurface. Let p and q denote the poly-nomials such that Q ( x, z ) = p ( z ) + xq ( x, z ).We can write q ( x, z ) in the following form: q ( x, z ) = P deg( p ) i =1 p ( i ) ( z ) q i ( x, p ( z )) for some poly-nomials q , . . . , q deg( p ) ∈ C [2] .By Lemma 2.3, we can assume deg x ( q i ( x, p ( z ))) < n − ≤ i ≤ deg( p ).Now, rewrite Q ( x, z ) = p ( z ) + n − X k =1 x k deg( p ) X i =1 p ( i ) ( z ) q i,k ( p ( z ))for suitable polynomials q i,k ∈ C [1] . Let 1 ≤ k ≤ n − Q (cid:0) x, z − x k q ,k ( p ( z )) (cid:1) ≡ p ( z )+ k X k =1 x k deg( p ) X i =1 p ( i ) ( z ) q i,k ( p ( z )) − x k q ,k ( p ( z )) p ′ ( z ) mod ( x k +1 ) . Therefore, we obtain, using Theorem 4.1, that X Q,n is equivalent to a hypersurface of equation x n y = p ( z ) + n − X k =1 x k deg( p ) X i =1 p ( i ) ( z )˜ q i,k ( p ( z ))with ˜ q ,k = 0 and ˜ q i,k = q i,k if ( i, k ) ∈ [1 , k ] × [1 , deg( p )] \ { (1 , k ) } . Then, it is easy to proveby induction on k that X Q,n is equivalent to a hypersurface of the desired form.Now, we will prove the second part of the theorem.Let X j = X ( p j , { q j,i } i =2 .. deg( p j ) , n j ), j = 1 ,
2, and pose Q j = p j ( z )+ x P deg( p j ) i =2 p ( i ) j ( z ) q j,i ( x, p j ( z )).If X and X are equivalent, then, it follows from Theorem 4.1, that n = n = n and thatthere exist a, α, µ ∈ C ∗ , β ∈ C and B ∈ C [2] such that Q ( ax, αz + β + xB ( x, Q ( x, z ))) ≡ µQ ( x, z ) mod ( x n ) . This implies p ( αz + β ) = µp ( z ). Thus deg( p ) = deg( p ) = d .First, we prove that B ( x, · ) ≡ x n − ). In order to do this, suppose that we can write B ( x, t ) ≡ b k ( t ) x k for some 0 ≤ k ≤ n − b k ( t ) ∈ C [ t ] \
0. Then, we obtain the followingequalities modulo ( x k +2 ). µQ ( x, z ) ≡ Q ( ax, αz + β + xB ( x, Q ( x, z ))) mod ( x k +2 ) ≡ Q (cid:0) ax, αz + β + x k +1 b k ( p ( z )) (cid:1) ≡ p (cid:0) ax, αz + β + x k +1 b k ( p ( z )) (cid:1) + ax d X i =2 p ( i )2 ( αz + β ) q ,i ( ax, p ( αz + β )) ≡ µp ( z ) + x k +1 b k ( p ( z ))( p ) ′ ( αz + β ) + ax d X i =2 p ( i )2 ( αz + β ) q ,i ( ax, µp ( z )) ≡ µp ( z ) + x k +1 b k ( p ( z )) α − µp ′ ( z ) + ax d X i =2 α − i µp ( i )1 ( z ) q ,i ( ax, µp ( z )) . This would imply α − ( p ) ′ ( z ) b k ( p ( z )) ≡ P di =2 (cid:16) p ( i )1 ( z ) q ,i ( x, p ( z )) − aα − i p ( i )1 ( z ) q ,i ( ax, µp ( z )) (cid:17) x k mod ( x k +1 ) , P.-M. POLONI what is impossible.Therefore, B ( x, · ) ≡ x n − ). Since, by hypothesis, deg x ( Q j ( x, z )) < n for j = 1 , Q ( ax, αz + β ) = µQ ( x, z ). Then, we can easily check that this last equalityimplies aα − i q ,i ( ax, µt ) = q ,i ( x, t ) for every 2 ≤ i ≤ d , as desired. This concludes the proof. (cid:3) Standard forms
In [5], A. Dubouloz and the author proved that every Danielewski hypersurface X Q,n where Q ( x, z ) is a polynomial such that Q (0 , z ) has d ≥ Q (0 , z ) has multiple roots. In orderto do this, we first generalize the definition of standard form given in [5]. Definition 5.1.
We say that a Danielewski hypersurface X Q,n is in standard form if the poly-nomial Q can be written as follows: Q ( x, z ) = p ( z ) + xq ( x, z ) , with deg z ( q ( x, z )) < deg( p ) . We also introduce a notion of reduced standard form.
Definition 5.2.
A Danielewski hypersurface X Q,n is in reduced standard form if deg x ( Q ( x, z )) Lemma 5.5. Let n ≥ be a natural number and Q ( x, z ) and Q ( x, z ) be two polynomials of C [ x, z ] such that Q ( x, z ) = (1 + xπ ( x, z )) Q ( x, z ) + x n R ( x, z ) for some polynomials π ( x, z ) , R ( x, z ) ∈ C [ x, z ] .Then, the endomorphism of C defined by Φ( x, y, z ) = ( x, (1 + xπ ( x, z )) y + R ( x, z ) , z ) induces an isomorphism ϕ : X Q ,n → X Q ,n . LASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES 9 Proof. Remark that, since Φ ∗ ( x n y − Q ( x, z )) = (1 + xπ ( x, z )) ( x n y − Q ( x, z )), Φ induces amorphism ϕ : X Q ,n → X Q ,n .Let f ( x, z ) and g ( x, z ) be two polynomials in C [ x, z ] so that (1+ xπ ( x, z )) f ( x, z )+ x n g ( x, z ) =1 and define Ψ, an endomorphism of C , by posing Ψ ∗ ( x ) = x Ψ ∗ ( y ) = f ( x, z ) y + g ( x, z ) Q ( x, z ) − f ( x, z ) R ( x, z )Ψ ∗ ( z ) = z We check easily that Ψ ∗ ( x n y − Q ( x, z )) = f ( x, z ) ( x n y − Q ( x, z ))and that Ψ ∗ ◦ Φ ∗ ( x ) = x ;Ψ ∗ ◦ Φ ∗ ( y ) = y − g ( x, y ) ( x n y − Q ( x, z )) ;Ψ ∗ ◦ Φ ∗ ( z ) = z. Therefore, the restriction of Ψ ∗ ◦ Φ ∗ to S Q ,n = C [ X Q ,n ] is identity. Hence, Ψ induces theinverse morphism of ϕ , and X Q ,n ≃ X Q ,n . (cid:3) Theorem 5.6. Every Danielewski hypersurface is isomorphic to a Danielewski hypersurface inreduced standard form. Furthermore, there is an algorithmic procedure which computes one ofthe reduced standard forms of a given Danielewski hypersurface.Proof. Let X = X Q,n be a Danielewski hypersurface and denote Q ( x, z ) = p ( z ) + xq ( x, z ) with p ( z ) ∈ C [ z ] and q ( x, z ) ∈ C [ x, z ].One can construct, by induction on m ≥ 0, two polynomials q s,m ( x, z ) and π m ( x, z ) so thatdeg z ( q s ( x, z )) < deg( p ) and Q ( x, z ) ≡ (1 + xπ m ( x, z ))( p ( z ) + xq s,m ( x, z )) mod ( x m +1 ).Indeed, this assertion is obvious for m = 0, whereas, if it is true for a rank m , we can write: p ( z ) + xq ( x, z ) ≡ (1 + xπ m ( x, z ))( p ( z ) + xq s,m ( x, z )) mod ( x m +1 )= (1 + xπ m ( x, z ))( p ( z ) + xq s,m ( x, z )) + x m +1 R m +1 ( x, z ) ≡ (1 + xπ m ( x, z ) + x m ˜ π m +1 ( z ))( p ( z ) + xq s,m ( x, z ) + x m r m +1 ( z )) mod ( x m +1 ) ≡ (1 + xπ m +1 ( x, z ))( p ( z ) + xq s,m +1 ( x, z )) mod ( x m +1 ) , where R m +1 (0 , z ) = p ( z )˜ π m +1 ( z ) + r m +1 ( z ) is the Euclidean division (in C [ z ]) of R m +1 (0 , z ) by p . Thus, we obtain p ( z ) + xq ( x, z ) = (1 + xπ n − ( x, z ))( p ( z ) + xq s,n − ( x, z )) + x n R n ( x, z ) . Lemma 5.5 allows us to conclude that X is isomorphic to the Danielewski hypersurface instandard form X s defined by the equation x n y − p ( z ) − xq s,n − ( x, z ) = 0.In order to obtain a reduced standard form, we rewrite p ( z ) + xq s,n − ( x, z ) = d X i =0 a i z i + x d − X i =0 z i α i ( x )and consider the automorphism of C defined byΦ : ( x, y, z ) ( x, y, z − x ( da d ) − α d − ( x )) . One checks that the polynomial Φ ∗ ( x n y − p ( z ) − xq s,n − ( x, z )) satisfies the second condition inthe definition of Danielewski hypersurface in reduced standard form. Finally, the first conditioncan be obtain easily by applying Lemma 5.5.This proof gives an algorithm for finding a (reduced) standard form of given Danielewskihypersurface. (cid:3) It should be noticed that a Danielewski hypersurface is in general not equivalent to its(reduced) standard form given by Theorem 5.6. Morover, one can use this fact to constructnon-equivalent embeddings for every Danielewski hypersurface of non-trivial Makar-Limanovinvariant. Proposition 5.7. Every Danielewski hypersurface X Q,n with n ≥ admits at least two non-equivalent embeddings into C .Proof. Since, by Theorem 5.6, every Danielewski hypersurface is isomorphic to a one in standardform, it suffices to show that every Danielewski hypersurface in standard form X Q,n with n ≥ C .Let X = X Q,n be a Danielewski hypersurface in standard form with n ≥ 2. Then, due toLemma 5.5, X is isomorphic to the hypersurface Y = X (1+ x ) Q ( x,z ) ,n . Nevertheless, it turns outthat X and Y are non-equivalent hypersurfaces of C . Indeed, if they were, Theorem 4.1 wouldgive us constants a, α, µ ∈ C ∗ , β ∈ C and a polynomial B ∈ C [2] such that(1 − ax ) Q ( ax, αz + β + xB ( x, Q ( x, z ))) ≡ µQ ( x, z ) mod ( x n ) . In turn, if we denote Q ( x, z ) = p ( z ) + xq ( x, z ), it would lead the following equalities modulo( x ): µQ ( x, z ) ≡ µ ( p ( z ) + xq (0 , z )) ≡ (1 − ax ) Q ( ax, αz + β + xB (0 , Q (0 , z ))) ≡ (1 − ax )( p (cid:0) αz + β + xB (0 , p ( z ))) + xq (0 , αz + β )) ≡ p ( αz + β ) + x ( B (0 , p ( z )) p ′ ( αz + β ) + q (0 , αz + β ) − ap ( αz + β )) . Thus B (0 , p ( z )) p ′ ( αz + β ) + q (0 , αz + β ) − ap ( αz + β ) = µq (0 , z )which is impossible since deg( q (0 , z )) < deg( p ) by definition of a standard form. (cid:3) Remark . This proof is similar to the proof of Freudenburg and Moser-Jauslin in [7] forhypersurfaces of equation x n y = p ( z ) with n ≥ 2. In their article, they also have constructednon-equivalent embeddings into C for Danielewski hypersurfaces of the form xy − z d − d ∈ N . Nevertheless, we do not know if every Danielewski hypersurface X Q, admitsnon equivalent embeddings into C . For instance, the following question , which they posed in[7], is still open. Question 1. Does the hypersurface of equation xy + z = 0 admit a unique embedding into C ? Remark also that the two non-equivalent embeddings of a Danielewski hypersurface X Q,n with n ≥ analytically equivalent. Indeed, it can beeasily seen, as in [7] and [5], that a Danielewski hypersurface is analytically equivalent to itsstandard form given by Theorem 5.6. In turn, we obtain the following result. Proposition 5.9. If X and X are two isomorphic Danielewski hypersurfaces, then there isan analytic automorphism Ψ of C such that Ψ( X ) = X .Proof. Let X = X Q,n be a Danielewski hypersurface and let X Q s ,n be its standard formgiven by the theorem 5.6. In the proof of this theorem, we have seen that Q ( x, z ) = (1 + xπ ( x, z )) Q s ( x, z ) + x n R ( x, z ) for certain polynomials π ( x, z ) , R ( x, z ) ∈ C [ x, z ]. Now, considerthe following analytic automorphism of C .Ψ : ( x, y, z ) ( x, exp( xf ( x, z )) y − x − n (exp( xf ( x, z )) − − xπ ( x, z )) Q s ( x, z ) + R ( x, z ) , z ) , where f ( x, z ) ∈ C [ x, z ] is a polynomial so that exp( xf ( x, z )) ≡ xπ ( x, z ) mod ( x n ). Onechecks that Ψ ∗ ( x n y − Q ( x, z )) = x n y − Q s ( x, z ). Thus, Ψ maps X Q,n to its standard form LASSIFICATION(S) OF DANIELEWSKI HYPERSURFACES 11 X Q s ,n . Then, the result follows from Proposition 6.2, which will be proved at the end of thispaper. (cid:3) Classification up to isomorphism Finally, we give the classification of Danielewski hypersurfaces in standard form. Togetherwith the theorem 5.6, this effectively classifies all the Danielewski hypersurfaces up to isomor-phism of algebraic varieties. Theorem 6.1. (1) Two Danielewski hypersurfaces X Q ,n and X Q ,n in standard form are isomorphic ifand only if the two following conditions are satisfied: (a) n = n = n ; (b) ∃ a, α, µ ∈ C ∗ ∃ β ( x ) ∈ C [ x ] such that Q ( ax, αz + β ( x )) ≡ µQ ( x, z ) mod ( x n ) . (2) Two Danielewski hypersurfaces X Q ,n and X Q ,n in reduced standard form are isomor-phic if and only if the two following conditions are satisfied: (a) n = n ; (b) ∃ a, α, µ ∈ C ∗ ∃ β ∈ C such that Q ( ax, αz + β ) = µQ ( x, z ) .Proof. Let X = X Q ,n and X = X Q ,n be two isomorphic Danielewski hypersurfaces instandard form and let ϕ : X → X be an isomorphism. Then Corollary 3.3 implies that n = n = n . Since the case n = 1 was already done by Daigle [3], we can suppose that n ≥ x i , y i , z i the images of x, y, z in the coordinate ring C [ X i ] for i = 1 , 2. Then,due to Corollary 3.3, there exist constants a, α ∈ C ∗ and a polynomial β ( x ) ∈ C [ x ] such that ϕ ∗ ( x ) = ax and ϕ ∗ ( z ) = αz + β ( x ).Moreover, we have proven in the proof of Corollary 3.3, that Q (0 , αz + β (0)) = µQ (0 , z )for a certain constant µ ∈ C ∗ .Thus, viewing C [ X i ] as a subalgebra of C [ x i , x − i , z i ] with y i = x − ni Q i ( x i , z i ), we obtain ϕ ∗ ( y ) = ϕ ∗ ( x − n Q ( x , z )) = ( ax ) − n Q ( ax , αz + β ( x )) = µa − n y + ( ax ) − n ∆( x , z ) , where ∆( x , z ) = Q ( ax , αz + β ( x )) − µQ ( x , z ).Remark that deg z ∆( x , z ) < deg z Q (0 , z ) since X and X are in standard form.It turns out that x − n ∆( x , z ) ∈ C [ x , z ] since any polynomial of C [ X ] ⊂ C [ x ± , z ] withnegative degree in x has obviously a degree in z at least equal to deg z Q (0 , z ). Thus,∆( x, z ) ≡ x n ) and X and X fulfill conditions (1) (a) and (1) (b).If X and X are in reduced standard form, then we see easily that ∆( x, z ) ≡ x n )is possible only if β ( x ) ≡ β (0) mod ( x n ). If so Q ( ax , αz + β (0)) = µQ ( x , z ) and X and X fulfill the conditions (1) (a) and (2) (b).Conversely, suppose that X = X Q ,n X = X Q ,n are two Danielewski hypersurfaces whichsatisfy the conditions (a) and (b) of part (1). Then the following triangular automorphism of C induces an isomorphism between X and X :( x, y, z ) ( ax, µa − n y + ( ax ) − n ( Q ( ax, αz + β ( x )) − µQ ( x, z )) , αz + β ) . (cid:3) As a corollary, we observe that two isomorphic Danielewski hypersurfaces in standard formare equivalent via a triangular automorphism of C , and that two isomorphic Danielewskihypersurfaces in reduced standard form are equivalent via an affine one. In fact, we have evenproven a stronger result in the proof of Theorem 6.1. Proposition 6.2. 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On the group of automorphisms of a surface x n y = P ( z ). Israel J. Math. , 121:113–123,2001.[9] Lucy Moser-Jauslin and Pierre-Marie Poloni. Embeddings of a family of Danielewski hypersurfaces andcertain C + -actions on C . Ann. Inst. Fourier (Grenoble) , 56(5):1567–1581, 2006.[10] J¨orn Wilkens. On the cancellation problem for surfaces. C. R. Acad. Sci. Paris S´er. I Math. , 326(9):1111–1116, 1998. Mathematisches Institut Universit¨at Basel, Rheinsprung 21, CH-4051 Basel, Switzerland E-mail address ::