Singularity conditions on the class group of Zariski surfaces

Journal of Algebra 379 (2013) 241–258

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Journal of Algebra www.elsevier.com/locate/jalgebra

Singularity conditions on the class group of Zariski surfaces Jeffrey Lang Mathematics Department, University of Kansas, Lawrence, KS 66045, USA

a r t i c l e

i n f o

Article history: Received 15 November 2011 Available online 31 January 2013 Communicated by Steven Dale Cutkosky MSC: 13A99 Keywords: Affine surfaces Divisor class groups

a b s t r a c t Let k be an algebraically closed field of characteristic p = 0 and X g ⊂ A k3 be a normal surface defined by an equation of the form z p = g (x, y ). Assume the number of singularities of X g is the maximum possible, which is very often the case. This paper defines an equivalence relation on the singularities of X g in terms of the Hessian from which it derives a fundamental decomposition of the group of Weil divisors of the surface. From the decomposition various results are obtained relating the structure of the equivalence classes to that of the class group. © 2013 Elsevier Inc. All rights reserved.

Introduction A Zariski surface is a normal variety X g in affine 3-space defined over an algebraically closed field k of characteristic p > 0 by a polynomial equation of the form z p = g (x, y ). Motivated by the classical result of Max Noether [6], that a generic surface in P3 has Pic ∼ = Z, Blass, Grant, and Lang [1,5,11] showed that for a generic g ∈ k[x, y ] the group of Weil divisors of a Zariski surface is the trivial group if deg( g )  4 and p > 2, and is Z/2Z if deg( g )  5 and p = 2. A corollary to these calculations is that for a general choice of g ∈ k[x, y ], the associated Zariski surface X g has the same divisor nglass groups as just described [10]. In the papers of Blass, Grant, and Lang, a polynomial g = i + j =0 αi j xi y j ∈ k[x, y ] is generic, if the αi j are algebraically independent over some subfield ] of degree n if there is a F of k and a property is said to hold for a general choice of g ∈ k[x, y n polynomial H ∈ k[(n+1)(n+2)/2] such that the property holds for all g = i + j =0 βi j xi y j of degree n with H (β) = 0. The determination of the class groups of the generic Zariski surfaces employs fundamental group methods and Galois descent techniques developed by Grothendieck [6] and Samuel [13], respectively.   H g ( Q ), where The first step in the calculations involves embedding the class group in Q ∈S g F p · 2 F p is the prime subfield of k, H g = g xy − g xx g y y and S g = { Q ∈ k2 : g x ( Q ) = g y ( Q ) = 0}. The next

E-mail address: [email protected]. 0021-8693/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.01.004

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is to show that in the generic case, the action of the automorphism group of k over F g , the field extension of F p generated by the coefficients of g, on the embedded class group is so large that if the group of divisors is not as described, then it has cardinality that exceeds a known upper bound. Specifically, if E g is the field extension of F g generated by the coordinates of the points in S g and  L g is the field extension of E g generated by { H g ( Q ): Q ∈ S g }, then the automorphism group of E g over F g in the generic case acts on S g as the full symmetric group when p > 2 and as the alternating group if p = 2. Furthermore, if p > 2, the action of the automorphism group of L g over E g  on { H g ( Q ): Q ∈ S g } in the generic case produces all possible double-flips, where a double-flip is  an automorphism that changes the sign of exactly two elements of { H g ( Q ): Q ∈ S g }, while leaving all others fixed. While any Zariski surface with the maximum possible number of singular points and induced actions this large will have the same divisor class groups as in the generic cases, these are not criteria that are normally easy to verify. Also, from our many calculations of class groups of Zariski surfaces,  it seemed to us that there should be considerably weaker field theoretic conditions on S g and { H g ( Q ): Q ∈ S g } that imply Cl( X g ) = 0 when p > 2 and Cl( X g ) ∼ = Z/2Z when p = 2. This  paper is a step in this direction, in that it investigates field theoretic conditions on S g and { H g ( Q ): Q ∈ S g } for arbitrary Zariski surfaces that limit the size of the class group of X g and, in particular, that guarantee the same results as in the generic cases. A review of Galois descent from Samuel’s notes appears in Section 1. It is used in Section 2 to show that the divisor class group of X g is isomorphic to the additive group, L g , of logarithmic derivatives in k[x, y ] of the Jacobian derivation, D g = g y ∂∂x − g x ∂∂y . We then prove that L g embeds into a square

system of generalized linear equations of the form M x = x( p ) and derive results concerning the separability of the solutions of such a system over any field containing the entries of M. A key concept, the reduction number of a square matrix M, is introduced in Section 2, which is an upper bound ( p) on the number of times p-th roots need  in solving the system M x = x . In Section 3  to be taken we define an embedding of L g into H g ( Q ) and describe a purely inseparable extenQ ∈S g F p · sion of L g that contains the coefficients of each logarithmic derivative of D g in k[x, y ]. The main section of the paper is the fourth. We establish a fundamental decomposition of L g in terms of  { H g ( Q ): Q ∈ S g } and derive from it conditions on this set that affect the size and structure of is that when X g has the maximum possible number of singularCl( X g ). Among   the results obtained ities and E g ( H g ( Q )) = E g ( H g ( Q  )) for all Q , Q  ∈ S g , then Cl( X g ) is trivial if p > 2 and Z/2Z if  p = 2. In Section 5 we study how H g ( Q ) with Q ∈ S g is related to the class group locally, i.e. at a stalk of a singular point. We present some applications and examples in Section 6. It is noteworthy that the results of this paper eluded us for so long partly because we had not identified the finite Galois extension L g of F g with L g ⊂ L g [x, y ] and partly because almost all of our computations of divisor class groups of Zariski surfaces, whether by hand or computer program [3,8], involved those whose coefficients are in a finite field. When F g is contained in a finite field, [ L g : E g ]  2, hence the field extension is too simple  to observe many of the phenomena studied below. For example, the action of Aut( L g / E g ) on { H g ( Q ): Q ∈ S g } will produce at most a single  flip of { H g ( Q ): Q ∈ S g }, whether singleton, double, triple, etc. 1. Preliminaries Galois descent techniques developed in Pierre Samuel’s Tata notes [13] provide the framework for this study. The reader is referred there for the definitions of a Krull domain and divisor class group of a Krull domain [13, pp. 1–4]. All of the rings studied herein are Noetherian integrally closed domains and are thus Krull rings [13, p. 6]. If R is a Noetherian integrally closed domain, then X = Spec( R ) is regular in codimension one and the group of Weil divisors of X , denoted Cl( X ), and the divisor class group of R, denoted Cl( R ), as defined by Samuel are isomorphic [7, p. 130]. The following theorems are from Samuel’s notes.

Theorem 1.1. Let A ⊂ B be Krull rings with B integral over A. Then there is a well-defined group homomorphism φ : Cl( A ) → Cl( B ) [13, pp. 19–20].

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Remark 1.2. The homomorphism φ : Cl( A ) → Cl( B ) can be described by its action on height one prime ideals, since their divisors generate  the class group of a Krull domain [13, pp. 5–7]. If Q is a height one prime of A, then φ( Q ) = e (q : Q )q, where the sum is over all prime ideals q lying over Q (which are necessarily height one) and e (q : Q ) is the ramification index of q over Q [13, pp. 18–21]. Remark 1.3. Let B be a Krull ring of characteristic p = 0 and L be the quotient field of B. Let  be a derivation of L such that ( B ) ⊂ B. Let K = kernel of  and A = B ∩ K . Then A is a Krull ring with B integral over A (since B p ⊂ A). Set L = {a−1 a: a ∈ L and a−1 a ∈ B } and L  = {u −1 u: u is a unit in B }. Then L  is a subgroup of L . Theorem 1.4. (a) Let φ : Cl( A ) → Cl( B ) be the homomorphism described in (1.2). There exists a canonical monomorphism φ : ker φ → L /L  . (b) If L is the quotient field of B and [ L : K ] = p and ( B ) is not contained in any height one prime of B, then φ is an isomorphism [13, pp. 63–64]. Remark 1.5. If I ∈ Cl( A ) is in the ker φ , then φ( I ) = aB for some a ∈ L. Samuel shows that a−1 a ∈ B and defines φ( I ) = a−1 a. Since L ⊂ B and the characteristic of L is p, each nonzero element of ker φ has order p. Remark 1.6. In (1.4b) each height one prime Q of A has a unique prime ideal q lying over it (specifically, q = {b ∈ B: b p ∈ Q }) and e (q : Q ) = 1 or p. If Q ∈ ker φ with e (q : Q ) = p and q is principal, generated by some b ∈ B, then Q = b p A and hence is trivial in Cl( A ). In particular, if B is a unique factorization domain (i.e. Cl( B ) = 0), then Cl( A ) is generated by the non-principal unramified height one primes of A. Theorem 1.7. If [ L : K ] = p, then (a) there exists α ∈ A such that  p = α  and (b) an element t ∈ L is equal to v −1  v for some v ∈ L if and only if  p −1 t − αt + t p = 0 [13, pp. 63–64].

ˆ its completion. Then Cl( A ) injects into Cl( Aˆ ). Theorem 1.8 (Mori). Let A be a local Krull ring and A 2. Logarithmic derivatives of the Jacobian derivation Hereafter, k represents an algebraically closed field of characteristic p = 0 and g ∈ k[x, y ] a polynomial of degree n = 0 such that g x and g y have no common factors in k[x, y ]. X g denotes the surface in A k3 defined by the equation z p = g and D g the Jacobian derivation on k(x, y ) defined by D g = g y ∂∂x − g x ∂∂y .

Lemma 2.1. The coordinate ring of X g is isomorphic to the ring B g = k[x p , y p , g ]. Proof. Let θ : k[x, y , z] → B g be the homomorphism that sends x to x p , y to y p , z to g, and a to a p for all a ∈ k. θ is surjective since k is perfect. Hence, ker θ is a height one prime ideal containing the principal ideal I generated by z p − g. Since the latter is a height one prime, I = ker θ . Therefore, the coordinate ring of X g is isomorphic to B g . 2 1 Lemma 2.2. B g = D − g (0) ∩ k[x, y ]. 1 p p Proof. Since D g (x) = g y = 0, k(x p , y p )  k(x p , y p , g ) ⊂ D − g (0)  k(x, y ). Hence, k(x , y , g ) =

1 −1 D− g (0). Since X g has only finitely many singular points, B g is normal by (2.1). Since D g (0) ∩ k[x, y ] is integral over B g and they have the same quotient field, the two are equal. 2

Lemma 2.3. Let L g = { f −1 D g f : f ∈ k(x, y ) and f −1 D g f ∈ k[x, y ]} be the additive group of logarithmic derivatives of D g in k[x, y ]. Then Cl( B g ) ∼ = Lg .

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Proof. By (1.4) and the fact that D g (k[x, y ]) is not contained in any height one prime of k[x, y ] (since D g (x) = g y and D g ( y ) = − g x ) and [k(x, y ) : k(x p , y p , g )] = p. 2

α g ∈ B g such that D gp = α g D g . In 1980, Richard Ganong conjectured  p −1 to me that α g is given by the formula, α g = i =0 g i ∇( g p −1−i ), where ∇ = ∂ 2p −2 /∂ x p −1 ∂ y p −1 . In

Remark 2.4. By (1.7) there exists

his attempt to verify this, the author proved the next theorem [9] under the assumption that the highest degree form of g is not in k[x p , y p ]. Stohr and Voloch later proved it in general [14]. p

Theorem 2.5. Let α g ∈ B g be such that D g = α g D g . Then for all f ∈ k[x, y ], p −1 Dg

f − αg f = −

p −1 





g i ∇ g p −1 − i f ,

i =0

where ∇ = ∂ 2p −2 /∂ x p −1 ∂ y p −1 . Corollary 2.6. α g =

 p −1 i =0

Proof. Let f = 1 in (2.5).

g i ∇( g p −1−i ).

2

Corollary 2.7. A polynomial t ∈ k[x, y ] is in L g if and only if ∇( g i t ) = 0, for i = 0, 1, . . . , p − 2, and ∇( g p −1 t ) = t p . Proof. Since ∇(k(x, y )) = k(x p , y p ) and { g i : 0  i  p − 1} is independent over k(x p , y p ), the corollary follows by (1.7) and (2.5). 2 Proposition 2.8. Let H g = {t ∈ k[x, y ]: ∇( g p −1 t ) = t p }. Then L g ⊂ H g and deg(t )  n − 2 for each t ∈ H g . Proof. A comparison of degrees on both sides of the equation

  ∇ g p −1 t = t p yields p deg(t )  deg(t ) + ( p − 1)n − 2( p − 1), i.e. deg(t )  n − 2.

(2.8.1)

2

Definition 2.9. Let W g be the k-vector space of polynomials in k[x, y ] of degree at most n − 2 and ( p)

let W g

be the k-vector space of polynomials in k[x p , y p ] of degree at most p (n − 2). Let T g : W g →

W g be the linear transformation defined by T g ( f ) = ∇( g p −1 f ) and M g be the matrix of T g with respect to the monomial bases {xi y j : 0  i + j  n − 2} and {xip j jp : 0  i + j  n − 2} of W g and ( p)

p

W g , respectively. Then M g is a square matrix of dimension m =

n(n−1) 2

with entries in F g .

Definition 2.10. For a field F and positive integers r and s, let F r ×s be the set of r × s matrices with p entries in F . If M = [ai j ] ∈ F r ×s , let M ( p ) = [ai j ]. I r denotes the identity matrix in F r ×r and O rs the zero matrix in F r ×s .

Lemma 2.11. For each t =

n−2

i + j =0

αi j xi y j ∈ H g , let xt = (α00 , α10 , α01 , . . . , α0(n−2) ) in km . Then the map

t → xt is an isomorphism from H g to the group of solutions of the system M g x = x( p ) . Proof. The system M g x = x( p ) can be obtained by comparing coefficients on both sides of the equality ∇( g p −1 t ) = t p . Thus t is a solution of the latter equation if and only if xt is a solution of the matrix equation. The map is also clearly additive. 2

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Proposition 2.12. Let F be a subfield of k and r, s positive integers. Assume M ∈ F (r +s)×(r +s) has rank r and 1

1

the last r rows of M are F -independent. Then there exist N ∈ ( F p )s×r and M  ∈ ( F p )r ×r such that the solution sets of M x = x( p ) and the system



Is O rs

N M

   y 0 = ( p) z z

are the same. In particular, the systems M x = x( p ) and M  x = x( p ) have the same number of solutions. Proof. Since the last r rows of M form  a basis for its row space, for each i = 1, . . . , s, there are unique ai j ∈ F , j = 1, . . . , r, such that row(i ) + j =1 ai j row( j ) will be the zero row. The same row operations p

p

on x p will replace xi by xi + the linear equation, xi +

r

r

p j =1 ai j x j .

1 p

j =1 ai j

Thus we can replace the i-th equation in M x = x( p ) by

x j = 0. Substituting xi = −

r

M x = x( p ) , we obtain a system of the form



Is O rs

N M

1 p

j =1 ai j

x j into the last r equations of

   y 0 = ( p) z z

as described above with same solution set as the original.

2

Proposition 2.13. Let F be a subfield of k, r a positive integer and M ∈ F r ×r . If det( M ) = 0, then M x = x( p ) has p r distinct solutions. Furthermore, if (a1 , . . . , ar ) is a solution of M x = x( p ) , then each ai is separable over F of degree at most p r . Proof. By Bezout’s theorem the system of equations has p r solutions and clearly none of them are at infinity. If (a1 , . . . , ar ) is a solution of M x = x( p ) , then the system of equations remains fixed under the change of coordinates xi → xi − ai . Hence, every solution has the same multiplicity, which is one since det( M ) = 0. To prove the second statement of the lemma, it’s enough to show a1 is separable over F of degree at most p r . Assume M = [ai j ]. If a1 j = 0 for j  2, then a10 = 0 (since det( M ) = 0) and the first p equation of the system M x = x( p ) is a10 x1 = x1 , which shows that a1 is separable over F of degree at most p. If for somej  2, a1 j = 0, then after a permutation of x2 , . . . , xr , we may assume a12 = 0. r After replacing x2 by j =1 a1 j x j and then performing elementary operations on the equations so that the right side of the equations again becomes x( p ) , we can assume the first equation of the system p p M x = x( p ) is x2 = x1 . Now, if a2 j = 0 for j  3, then a21 = 0 and a21 x1 + a22 x2 = x2 , in which case p

p2

p

a21 x1 + a22 x1 = x1 (since x2 = x1 ), and hence, a1 is separable over F of degree at most p 2 . If for some  j  3, a2 j = 0, then after a permutation of x3 , . . . , xr , we may assume a23 = 0. After replacing r x3 by j =1 a2 j x j and then performing some elementary operations on the equations, we can assume p

the second equation of the system M x = x( p ) is x3 = x2 . Continuing in this way, we obtain that a1 is separable over F of degree at most p r . 2

Remark 2.14. Results (2.12) and (2.13) can be used to determine the order of the group of solutions of an r × r system M x = x( p ) . If det( M ) = 0, then the order is p r . If det( M ) = 0, then after e  r applications of (2.12), the system M x = x( p ) can be reduced to an r × r system of the form



I O

N M

   y 0 = ( p) , z z

where I is an identity matrix, O is the zero matrix, M  is a square matrix with either M  = O or −e det( M  ) = 0, and both matrices M  and N have entries in F p . The number of solutions of the system

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M x = x( p ) is the same as that of the system M  z = z( p ) , which by (2.13) is p q , where q is the rank of M  . Of course, e depends on which rows we choose as a basis of the row space at each step in the reduction. For example, if

M=

1 1 0 0 0 1 1 1 1



and we choose the first two rows as a basis of the row space, then the reduction proceeds in three steps as follows:



1 1 0 0 0 1 1 1 1







1 → 1

1 → [1] → [0]. 1

On the other hand, if the last two rows are chosen as a basis, then the reduction takes two steps:



1 0 1

1 0 1

0 1 1



 →



0 1 → [0]. 0 0

Hence, we define the reduction number of a square matrix M, denoted e ( M ), to be the fewest possible number of applications of (2.12) needed to reduce the system M x = x( p ) to a system of the form



I O

N M

   y 0 = ( p) , z z

with either M  = O or det( M  ) = 0. The next result is then an immediate consequence of (2.12) and (2.13). Proposition 2.15. Let F be a subfield of k, r a positive integer, and M ∈ F r ×r . If (a1 , . . . , ar ) is a solution to −e M x = x( p ) , then each ai is separable over F p , where e = e ( M ) is the reduction number of M. Corollary 2.16. Let F be a subfield of k, r a positive integer, and M ∈ F r ×r . If (a1 , . . . , ar ) is a solution to −r M x = x( p ) , then each ai is separable over F p . Proposition 2.17. The order of Cl( X g ) is at most p

n(n−1) 2

, where n = deg( g ).

Proof. By (2.1) and (2.3), Cl( X g ) is isomorphic to L g . By (2.8) and (2.11), L g is isomorphic to a subgroup of the solution group of the at most p

n(n−1) 2

.

n(n−1) 2

×

n(n−1) 2

system, M g x = x( p ) , which by (2.14) has order

2

Remark 2.18. By (2.5) and (2.17), Cl( X g ) is a finite p-group of type ( p , . . . , p ) of order at most p where n = deg( g ).

n(n−1) 2

,

3. Logarithmic derivatives at singular points The projection map from X g : z p = g (x, y ) to the xy-plane is a bijection and defines a one-toone correspondence between the singularities of the surface and the set S g = { Q ∈ k2 : g x ( Q ) = g y ( Q ) = 0}. We calculate the value of logarithmic derivatives of D g at points Q of S g in this section and use that to describe a finite algebraic extension L g of F g such that L g ⊂ L g [x, y ]. The first proposition (3.2) was originally proved in [3] but a simpler and more concise proof is given in [12].

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2 Definition 3.1. For g ∈ k[x, y ], let H g = g xy − g xx g y y .

Proposition 3.2. Let Q = (x0 , y 0 ) ∈ S g and [12, Proposition 3.4]. Notation 3.3. For each Q ∈ S g , let



α g be such that D gp = α g D g . Then α g ( Q ) = ( H g ( Q ))

H g ( Q ) designate a fixed root in k of the polynomial x2 − H g ( Q ).

Corollary 3.4. Let t ∈ L g . Then for each Q ∈ S g , t ( Q ) = n Q prime subfield of k [5, Lemma 2.2]. p −1

p −1 2



H g ( Q ) for some n Q ∈ F p , where F p is the

Proof. By (1.7), D g t − α g t = −t p . Evaluate at Q to obtain  n Q H g ( Q ) for some n Q ∈ F p . 2

α g ( Q )t ( Q ) = t ( Q ) p . Hence t ( Q ) =

Definition 3.5. A polynomial g ∈ k[x, y ] satisfies the maximum intersection condition if, in addition to g x and g y being relatively prime in k[x, y ], g x and g y intersect transversally and in the maximum possible number of points of k2 [4, where first introduced]. By Bezout’s theorem this number, which is the cardinality of S g , is (n − 1)2 when n is not divisible by p and n2 − 3n + 3 when it is. In both cases the intersection condition is satisfied for a general choice of g ([2, p. 282] and [5, p. 349]). Hereafter, let N g denote the cardinality of S g . Note if g satisfies the maximum intersection condition, then H g ( Q ) = 0 for each Q ∈ S g . Lemma 3.6. Let f ∈ k[x, y ] be such that f = 0. Then f ( Q ) = 0 for at most (deg( f ))(n − 1) points of S g . Proof. Let f = f 1 f 2 · · · f r be a factorization of f into irreducible factors in k[x, y ] and let ni be the degree of f i for i = 1, 2, . . . , r. Then for each i, f i is relatively prime in k[x, y ] to either g x or g y . If f i is relatively prime to g x , then by Bezout’s theorem, f i and g x are simultaneously 0 at at most ni (n − 1) points of S g . The same holds if f i and g y are relatively prime. Hence f ( Q ) = 0 for at most (n1 + · · · + nr )(n − 1) = deg( f )(n − 1) points of S g . 2 Proposition 3.7. Assume g satisfies the maximum intersection condition. Let W g be as in (2.9). Let

Φg : Wg →

N g

i =1

k be defined by Φ g ( f ) = ( f ( Q )) Q ∈ S g . Then Φ g is an injection.

Proof. Let f ∈ ker Φ g . Then deg( f )  n − 2. If f = 0, then by (3.6), f ( Q ) = 0 for at most (n − 2)(n − 1) points of S g . Since the cardinality of S g is at least n2 − 3n + 3, it must be that f = 0. 2 Corollary 3.8. Assume  the maximum intersection condition. Let Φ g be the restriction of Φ g to L g .  g satisfies Then Φ g : L g → Z H g ( Q ) is an injection of additive groups. p Q ∈S g Proof. By (2.8), (3.4) and (3.7).

2

Remark 3.9. If t ∈ L g , then t = f −1 D g ( f ) for some f ∈ k(x, y ). If σ ∈ Gal(k/ F g ), then σ extends to an automorphism of k(x, y ) over F g (x, y ) by letting σ act as the identity on x and y. If we let σ also denote this extension, then σ (t ) = (σ ( f ))−1 D g (σ ( f )) ∈ k[x, y ], i.e. σ (t ) ∈ L g . Hence, σ induces a group automorphism on L g . Also, σ (Φ g (t )) = Φ g (σ (t )), since for each Q ∈ S g , σ (t ( Q )) = σ (t )(σ ( Q )) and since σ permutes the elements of S g . Therefore, for each t ∈ L g and σ ∈ Gal(k/ F g ), σ (t ) = t if and only if σ (Φ g (t )) = Φ g (t ). Lemma 3.10. Let g satisfy the maximum  intersection condition and L g be the extension of F g generated by the coordinates of the points in S g and { H g ( Q ): Q ∈ S g }. Then for each t ∈ L g , the coefficients of t are purely inseparable over L g .

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Proof. Each element of Q ∈ S g Z p H g ( Q ) is fixed by each σ ∈ Gal(k/ L g ). By (3.8) and (3.9) the same is true for every t ∈ L g , which implies the coefficients of t are purely inseparable over L g . 2 Theorem 3.11. Assume g satisfies the maximum intersection condition and let e g be the reduction number p −e g

of M g (see (2.9) and (2.14)). Let L g be the field extension of F g



generated by the coordinates of the points

of S g and { H g ( Q ): Q ∈ S g }. Then L g ⊂ L g [x, y ]. p −e g

Proof. Let t ∈ L g . By (2.8), (2.11) and (2.15), the coefficients of t are separable over F g by (3.10) they are purely inseparable over L g ⊂ L g . 2

⊂ L g and

Corollary 3.12. Let g and L g be as in (3.10). If F g is perfect or det( M g ) = 0, then L g ⊂ L g [x, y ]. Proof. By (2.15) and (3.11).

2

The polynomial version of the next lemma is well known [13, p. 58 and p. 65]. The same could be true of this result, but I cannot find a reference. The proof is an application of (1.8) and Nagata’s lemma. Lemma 3.13. The divisor class group of the power series ring k[[x p , y p , xy ]] is isomorphic to Z/ p Z and is generated by the height one prime ideal (x p , xy ). Proof. k[[x p , y p , xy , x1p ]] = k[[x p , xy , x1p ]] and the latter is a ring of fractions of the power series ring in two variables. Hence, by (1.8) and Nagata’s lemma [13, p. 21] Cl(k[[x p , y p , xy ]]) is cyclic, generated by 1 p p (x p , xy ). Arguing as in (2.2), k[[x p , y p , xy ]] = D − xy (0) ∩ k[[x, y ]], which by (1.4) implies Cl(k[[x , y , xy ]]) p is a factor group of additive subgroups of k[[x, y ]], which implies the order of Cl(k[[x , y p , xy ]]) is p. 2 Proposition 3.14. Assume g x and g y meet transversally at Q ∈ k2 and let P be the corresponding singular point of X g . Let X P represent the stalk of X g at P . Then Cl( X P ) is 0 or Z/ p Z and the latter is the case if and only if there exists t ∈ L g such that t ( Q ) = 0. Proof. After a change of coordinates, we may assume P = (0, 0, 0) and that the leading form of g is xy. By (2.1), X P is isomorphic to B P , the localization of B g = k[x p , y p , g ] at the maximal ideal corresponding to P . Let L P = { f −1 D g f : f ∈ k(x, y ) and f −1 D g f ∈ k[x, y ] Q }, L P =

{ f −1 D g f : f is a unit in k[x, y ] Q }, Lˆ = { f −1 D g f : f ∈ k((x, y )) and f −1 D g f ∈ k[[x, y ]]} and Lˆ  = { f −1 D g f : f ∈ k[[x, y ]]∗ }. Let Xˆ P denote the completion of X P at P . Then Xˆ P ∼ = k[[x p , y p , g ]] = k[[u p , v p , uv ]], where k[[x, y ]] = k[[u , v ]] with the leading forms of u and v equal to x and y, respectively. By (1.4) and (3.13), Cl(k[[u p , v p , uv ]]) ∼ = Z/ p Z, where Lˆ /Lˆ  is generated by u −1 D g u. = Lˆ /Lˆ  ∼ ˆ By (1.8), Cl( X P ) injects into Cl( X P ). Hence, Cl( X P ) is 0 or Z/ p Z. In the following commutative diagram of group homomorphisms, the three horizontal maps are isomorphisms by (1.4) and (2.3), and the upper left vertical map is surjective by Nagata’s theorem.

Cl( X g ) ←→

Lg ↓ ↓ Cl( X P ) ←→ L P /L P ↓ ↓ Cl( Xˆ P ) ←→ Lˆ /Lˆ  Therefore, the map L g → L P /L P is surjective. It also follows that Cl( X P ) ∼ = Z/ p Z if and only if there

exists a t ∈ L g such that its image in Lˆ /Lˆ  equals

ω · u −1 D g (u ) for some ω ∈ F p∗ . Since the leading

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249

form of u −1 D g u is 1 and every element of Lˆ  is 0 at Q , Cl( X P ) ∼ = Z/ p Z if and only if there exists a t ∈ L g such that t ( Q ) = 0. 2 4. Hessian conditions on Cl( X g )



In this section we explore field theoretic conditions on { H g ( Q ): Q ∈ S g } that affect the structure and, in particular, the size of Cl( X g ). We define an equivalence relation on S g in terms of the Hessian of g that gives rise to a fundamental decomposition of the group of logarithmic derivatives of the Jacobian derivation. From this, conditions on the singular points of X g are derived that imply Cl( X g ) = 0. E g represents the field extension of F g generated by the coordinates of the points of S g  and L g the field extension of E g generated by { H g ( Q ): Q ∈ S g }. We assume g satisfies the maximum intersection condition and let e g be the reduction number of M g (see (2.9) and (2.14)). Let p −e g

F g = Fg

, E g = F g ∨ E g and L g = F g ∨ L g .

Remark 4.1. By (3.11), L g ⊂ L g [x, y ]. Since F g is purely inseparable over F g , the maps Aut F g ( L g ) → Aut F g ( L g ) and Aut F g ( E g ) → Aut F g ( E g ) induced by restriction are isomorphisms. Lemma 4.2. Let F be a field and K its algebraic closure. For i = 1, 2, let h i (x, y ) ∈ F [x, y ] and X i ⊂ K 2 be the curve defined by h i . Assume X 1 and X 2 have no common component and that each point of X 1 ∩ X 2 has intersection multiplicity one. Then for each Q = (a, b) ∈ X 1 ∩ X 2 , F (a, b) is separable over F . Proof. If F is a finite field, then every algebraic extension of F is separable over F . Since the resultant of h1 and h2 with respect to y (or x) has coefficients in F , F (a, b) is algebraic over F for each (a, b) ∈ X 1 ∩ X 2 . Suppose F is infinite and the points of X 1 ∩ X 2 have distinct x-coordinates. Let f (x) be the resultant of h1 and h2 with respect to y. Then f (x) = (a,b)∈C 1 ∩C 2 (x − a) ∈ F [x]. Hence, the x-coordinates of the points of X 1 ∩ X 2 are separable algebraic over F . In general, since F is infinite, there exists s ∈ F such that a − a = s(b − b ) for each pair of distinct points (a, b), (a , b ) ∈ X 1 ∩ X 2 . Let φ : K [x, y ] → K [x, y ] be the change of coordinates defined by φ(x) = x + sy , φ( y ) = y and let Y i ⊂ K 2 be the curve defined by hi = φ(h i ), i = 1, 2. Then φ induces an isomorphism φ : X 1 ∩ X 2 → Y 1 ∩ Y 2 given by φ(a, b) = (a − sb, b). By the choice of s, no two points of Y 1 ∩ Y 2 have the same x-coordinate. Thus, by above, for each (a, b) ∈ X 1 ∩ X 2 , a − sb is separable over F . By the same argument, there exists t ∈ F , with st = 1, such that b − ta is separable over F for each (a, b) ∈ X 1 ∩ X 2 . Since F (a, b) = F (a − bs, b − ta), F (a, b) is separable over F for each (a, b) ∈ X 1 ∩ X 2 . 2 Proposition 4.3. E g and L g are Galois extensions of F g and E g and L g are Galois extensions of F g . Proof. Let C 1 = { Q ∈ k2 : g x ( Q ) = 0} and C 2 = { Q ∈ k2 : g y ( Q ) = 0}. Then each point of S g = C 1 ∩ C 2 has intersection multiplicity one. If f 1 , f 2 are the resultants of g x and g y with respect to x and y, respectively, then E g is the minimal splitting field of f 1 f 2 over F g . Since by (4.2) the roots of f 1 f 2 are separable over F g , E g is Galois over F g . If p = 2, then E g = L g , hence L g is Galois over F g . If p > 2, then L g is the minimal splitting field of h(x) = Q ∈ S g (x2 − H g ( Q )) over E g . Since h(x) is fixed by each element of Aut( E g / F g ) and E g is Galois over F g , it follows that h(x) ∈ F g [x]. Thus, L g is the minimal splitting field of f 1 f 2 h over F g ;  i.e. L g is normal over F g . For each Q ∈ S g , H g ( Q ) is clearly separable over E g , hence H g ( Q ) is separable over F g . Therefore, L g is Galois over F g . In all of the above we can replace F g by F g , E g by E g and L g by L g to obtain E g and L g are Galois extensions of F g . 2



Corollary 4.4. If p > 2, the extension of F g (or F g ) generated by { H g ( Q ): Q ∈ S g } is Galois over F g (or F g ).

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Proof. The field extension of F g (or F g ) generated by { H g ( Q ): Q ∈ S g } is the minimal splitting of h(x) =



 − H g ( Q )) over F g (or F g ) and by (4.3) { H g ( Q ): Q ∈ S g } are separable over F g

2 Q ∈ S g (x

2

(or F g ).



Remark 4.5. There exists { Q 1 , . . . , Q r } ⊂ S g such that { H g ( Q 1 ), . . . ,



H g ( Q r )} is a minimal gener-



i1

ating set of L g over E g . If S = ∅, then the degree of L g over F g is 2r and { H g ( Q 1 ) · · · each i j = 0 or 1} is a basis for L g over E g .



Proposition 4.6. Assume { Q 1 , . . . , Q r } ⊂ S g . Then { H g ( Q 1 ), . . . ,





ir

H g(Q r) :

H g ( Q r )} is a minimal generating set of

L g over E g if and only if it is a minimal generating set of L g over E g . Proof. If p = 2 there is nothing to prove, so assume p > 2. For each Q ∈ S g , H g ( Q ) ∈ E g . Hence,   { H g ( Q 1 ), . . . , H g ( Q r )} is a minimal generating set of L g (L g ) over E g (E g ) if and only if it is a generating set and [ L g : E g ] = 2r ([ L g : E g ] = 2r ). Since the extensions E g ⊃ E g and L g ⊃ L g are purely inseparable, it must be that [ L g : E g ] = [ L g : E g ] and [ L g : L g ] = [ E g : E g ], from which the conclusion follows. 2



Corollary 4.7. Assume { H g ( Q 1 ), . . . , Then t =



i1

H g ( Q 1) · · ·





H g ( Q r )} is a minimal generating set of L g over E g and t ∈ L g [x, y ].

ir

H g ( Q r ) t (i 1 ,...,ir ) with each t (i 1 ,...,i j ) ∈ E g [x, y ] and each i j = 0 or 1.

2

Proof. By (3.11) and (4.6).



Proposition 4.8. Assume { H g ( Q 1 ), . . . ,



i1

H g ( Q 1) · · ·





H g ( Q r )} is a minimal generating set of L g over E g and t =

ir

H g ( Q r ) t (i 1 ,...,ir ) ∈ L g [x, y ] as in (4.7). Then t ∈ L g if and only if



i1

H g ( Q 1) · · ·



ir

H g ( Q r ) t (i 1 ,...,i j ) ∈ L g

for each (i 1 , . . . , i r ). Proof. By (1.7), t ∈ L g if and only if ∇( g i t ) = 0 for i = 0, 1, . . . , p − 2 and ∇( g p −1 t ) = t p . ∇( g i t ) = 0 if and only if

    i1 ir ∇ g i H g ( Q 1 ) · · · H g ( Q r ) t (i 1 ,...,ir ) = 0 

i1

for each (i 1 , . . . , i r ), since { H g ( Q 1 ) · · · only if

 = 

i1

H g ( Q 1) · · ·





i1

i1



ir

H g ( Q r ) } is a basis for L g over E g . ∇( g p −1 t ) = t p if and



H g ( Q r ) ∇ g p −1 t (i 1 ,...,ir )

H g ( Q 1) · · ·

Again, since { H g ( Q 1 ) · · ·

ir





H g(Q r)

ir 

H g ( Q 1)



( p −1)i 1 2

· · · H g(Q r)

( p −1)i r 2

p



t (i ,...,i ) . r 1

ir

H g ( Q r ) } is a basis for L g over E g , ∇( g p −1 t ) = t p if and only if

     p  i1 ir i1 ir ∇ g p −1 H g ( Q 1 ) · · · H g ( Q r ) t (i 1 ,...,ir ) = H g ( Q 1 ) · · · H g ( Q r ) t (i 1 ,...,ir ) for each (i 1 , . . . , i r ). The conclusion follows by (1.7).

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Corollary 4.9. For each (i 1 , . . . , i r ) with each i j = 0 or 1, let

 

 i1 ir L(i 1 ,...,ir ) = t ∈ L g : t = H g ( Q 1 ) · · · H g ( Q r ) t  with t  ∈ E g [x, y ] . Then L g =



L(i 1 ,...,ir ) . 

Lemma 4.10. Assume { H g ( Q 1 ), . . . ,





H g(Q ) = α(i1 ,...,ir ) of the α(i 1 ,...,ir ) is not 0.



i1

H g ( Q 1) · · ·



H g ( Q r )} is a minimal generating set of L g over E g . If Q ∈ S g and



ir

H g ( Q r ) , with α(i 1 ,...,ir ) ∈ E g , i j = 0, 1 for each j, then exactly one







i1

ir

Proof. Suppose that two of the summands in α(i1 ,...,ir ) H g ( Q 1 ) · · · H g ( Q r ) are nonzero. Then  some H g ( Q i ), 1  i  r, appears in one of these two that does not appear in the other. Then









there is a σ ∈ Aut( L g / E g ) such that σ ( H g ( Q i )) = − H g ( Q i ) and σ ( H g ( Q j )) = H g ( Q j ) for    all j = i, 1  j  r. Then H g ( Q ), − H g ( Q ), σ ( H g ( Q )) are distinct, but this is impossible since H g(Q ) ∈ E g .

2

   E g ( H g ( Q  )). Note that by the same reasoning used in (4.6), Q ∼ Q  if and only if E g ( H g ( Q )) =   E g ( H g ( Q  )). For each Q ∈ S g , let S Q = { Q  ∈ S g : Q ∼ Q  } and let L Q = {t ∈ L g : t ∈ H g ( Q ) · E g [x, y ]}. The sets S Q partition S g since ∼ is an equivalence relation. Definition 4.11. Define an equivalence relation on S g by Q ∼ Q  if and only if E g ( H g ( Q )) =

Lemma 4.12. Assume p > 2 and let Q , Q  ∈ S g . Then Q ∼ Q  if and only if





H g ( Q )/ H g ( Q  ) ∈ E g

(or E g ).







Proof. If Q ∼ Q  , then H g ( Q  ) = α + β H g ( Q ), for some α , β ∈ E g . It follows that H g ( Q ) ∈ E g      if and only if H g ( Q  ) ∈ E g . Hence, if H g ( Q ) ∈ E g , then H g ( Q )/ H g ( Q  ) ∈ E g . If H g ( Q ) ∈ / Eg,     then there exists σ ∈ Aut( L g / E g ) such that σ ( H g ( Q )) = − H g ( Q ). If σ ( H g ( Q  )) = H g ( Q  ),    then H g ( Q  ) = α . Hence, σ ( H g ( Q  )) = − H g ( Q  ), which yields α = 0. 2 Lemma 4.13. If S Q = S Q  , then L Q = L Q  . The converse holds if L Q = 0. Proof. As noted in (4.11), S Q = S Q  if and only if if



H g ( Q ) · E g [x, y ] =





H g(Q ) · E g =



H g ( Q  ) · E g , which is if and only

H g ( Q  ) · E g [x, y ]. Thus, S Q = S Q  implies L Q = L Q  .

Conversely, suppose L Q = 0 and L Q = L Q  . Then there exists t ∈ L g with t = 0 and h ∈ E g [x, y ]   such that t = H g ( Q )h. Since L Q = L Q  , t = H g ( Q  )h for some h ∈ E g [x, y ], which implies



H g(Q ) · E g =



H g ( Q  ) · E g , which implies S Q = S Q  .

2

Theorem 4.14. If S g is the disjoint union, S g = S Q   · · ·  S Q  , as described in (4.11), then L g = L Q  ⊕ 1 1 d · · · ⊕ LQ  . d



i1



ir

Proof. Let t ∈ L(i 1 ,...,ir ) . Then t = H g ( Q 1 ) · · · H g ( Q r ) t  with t  ∈ E g [x, y ] and each i j = 0 or 1.  If t = 0, then t ( Q ) = 0 for some Q ∈ S g by (3.8). By (3.4), t ( Q ) = n Q H g ( Q ) for some n Q ∈ F p∗ .



i1



ir

Hence, H g ( Q 1 ) · · · H g ( Q r ) ∈ and (4.13), L Q  ⊕ · · · ⊕ L Q  = L g . 1

d



H g ( Q ) · E g , from which it follows that L(i 1 ,...,ir ) = L Q . By (4.9)

2

Proposition 4.15. Let t ∈ L Q . Then t ( Q  ) = 0 for all Q  ∈ S g − S Q .

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  H g ( Q )t  , for some t  ∈ E g [x, y ]. But t ( Q  ) = n Q  H g ( Q  ) for some n Q  ∈ F p   by (3.4). Hence, n Q  H g ( Q  ) = H g ( Q )t  ( Q  ). Since t  ( Q  ) ∈ E g , Q  ∈ S Q if n Q  = 0. 2 Proof. By (4.11), t =

Corollary 4.16. If n = 0 (mod p ) and the order of S Q is less than or equal to n − 2, then L Q = 0. Proof. Let t ∈ L Q . By (2.8), deg(t )  n − 2. By (3.6), if t = 0, then t ( Q  ) = 0 for at most (n − 2)(n − 1) points of S g . Since the order of S g − S Q is at least (n − 1)2 − (n − 2) = (n − 2)(n − 1) + 1, t = 0 by (4.15). 2 Lemma 4.17. If Aut( E g / F g ) acts transitively on S g , then for each pair Q , Q  ∈ S g , L Q ∼ = LQ  . Proof. Choose σ ∈ Aut( E g / F g ) such that σ ( Q ) = Q  and let σ also denote an extension of it to L g .   Then σ ( H g ( Q )) = ± H g ( Q  ) and σ induces an isomorphism of L Q and L Q  with inverse induced by σ −1 . 2





Theorem 4.18. Suppose n = deg( g ) = 0 (mod p ) and E g ( H g ( Q )) = E g ( H g ( Q  )) for all Q , Q  ∈ S g with Q = Q  . Then Cl( X g ) = 0. Proof. The order of S Q is 1 for each Q ∈ S g . Since n = 0 (mod p ), Cl( X g ) = 0 by (4.14) and (4.16).

2





Theorem 4.19. Suppose n = 0 (mod p ) and E g ( H g ( Q )) = E g ( H g ( Q  )) for all Q , Q  ∈ S g with Q = Q  . Then Cl( X g ) = 0 or is isomorphic to Z/ p Z. Proof. Let t ∈ L g . Then f t = D g f = f x g y − f y g x for some f ∈ k[x, y ] and deg(t )  n − 2 by (2.8). Let f , g be the forms of highest degree of f , g, respectively, and t be the form of t of degree n − 2. Then x f t = x( f x g y − f y g x ) + y ( f y g y − f y g y ) = deg( f ) · f g y . Hence, t = deg( f ) gy x

gy x

, i.e., t is an integral

multiple of . It follows that if t 1 , t 2 ∈ L g , then there exists j ∈ F p such that either t 1 − jt 2 = 0 or deg(t 1 − jt 2 )  n − 3. By (3.6), if t 1 − jt 2 = 0, then (t 1 − jt 2 )( Q ) = 0 for at most (n − 3)(n − 1) points of S g , which means (t 1 − jt 2 )( Q ) = 0 for at least n points of S g . Since the order of each S Q is 1, t 1 − jt 2 = 0 by (4.15). Hence, Cl( X g ) = 0 or isomorphic to Z/ p Z. 2





Theorem 4.20. Assume Aut( E g / F g ) acts transitively on S g and E g ( H g ( Q )) = E g ( H g ( Q  )) for all Q , Q  ∈ S g with Q = Q  . If p = 3 and n  4, or if p  5 and n  3, then Cl( X g ) = 0. Proof. If Aut( E g / F g ) acts transitively on S g , then the order of L g is 0 or p | S g | by (4.14) and (4.17). If n(n−1)

p  3, the order of S g is at least n2 − 3n + 3 by (3.5), which by (2.17) implies n2 − 3n + 3  2 if Cl( X g ) = 0, which implies n = 2 or 3. Hence, Cl( X g ) = 0 if p  3 and n  4. If p  5 and n = 3, the order of S g is 4 by (3.5), while the order of Cl( X g ) is at most p 3 . Hence, Cl( X g ) = 0 in this case. 2 Remark 4.21. Note that [ L g : E g ] = 2r for some nonnegative integer r. If ∼ is the equivalence relation on S g defined above and s is the number of its equivalence classes, then clearly r  s. Corollary 4.22. Let r be as in (4.21) and assume deg( g ) = 0 (mod p ). If Aut( E g / F g ) acts transitively on S g and r  n = deg( g ), then Cl( X g ) = 0. Proof. Let s be as in (4.21). By (4.14), L Q  ⊕ · · · ⊕ L Q s = L g , where S Q   · · ·  S Q s = S g . Since 1 1 Aut( E g / F g ) acts transitively on S g , all of the S Q  have the same cardinality, say d, by (4.17). Then i

ds = (n − 1)2 . Since by (4.21), s  r > n − 1, it follows that d < n − 1, which implies L g = 0 by (4.16). 2

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Theorem 4.23. For Q ∈ S g , L Q = 0 if and only if Cl( X P ) = 0 for each singular point P = (a, b, c ) of X g such that (a, b) ∈ S Q . Proof. By (3.8), (3.14), (4.14) and (4.15).

2

5. Singularity conditions on Cl( X g ) In this section, we investigate how the square root of the Hessian of g at a singular point affects the class group of X g locally. We seek local results analogous to those obtained in the previous two sections. In particular, an intermediate field between F g and L g is defined that contains the coeffi cients of every element of L Q and then conditions on H g ( Q ) that imply Cl( X Q ) = 0 are derived. In this section g satisfies the maximum intersection condition. We let S g = { S ⊂ S g : | S g − S | = n − 2} and Fˆ g =



S ⊂S g

F g ( S ).

Proposition 5.1. If Aut( E g / F g ) is simple, then Fˆ g = F g or Fˆ g = E g .

σ ( Fˆ g ) = Fˆ g , for all σ ∈ Aut( E g / F g ) = Aut( E g / F g ), Fˆ g is a normal extension of F g . Hence, ˆ Aut( E g / F g ) is a normal subgroup of Aut( E g / F g ). Therefore, Fˆ g = F g or Fˆ g = E g . 2

Proof. Since

Proposition 5.2. If n = deg( g )  4 and Aut( E g / F g ) acts on S g as the full symmetric group, then Fˆ g = F g or

[ Fˆ g : F g ] = 2.

Proof. Let S ⊂ S g with | S g − S | = n − 2. Then Aut( E g / F g ( S )) acts as the full symmetric group on

S g − S. Hence, F g ( S ) = E g , which implies Fˆ g  F g ( S )  E g . Since Fˆ g is a normal extension of F g and the order of S g is at least n2 − 3n + 3, which is at least 7, and S r has no proper normal subgroup other than the alternating group when r  5, it follows that Fˆ g = F g or [ Fˆ g : F g ] = 2.

2

Throughout the rest of this section, assume p does not divide n[ E g : F g ] and 2 does not divide

[ E g : F g ].



Lemma 5.3. For each Q ∈ S g , the map ρ Q : L g → L g /√ H ( Q ) defined by ρ Q (t ) = t / H g ( Q ) is an isomorg phism. Proof. By (2.1), (2.3) and the fact that the surfaces z p = g and z p = √ Proposition 5.4. Assume Q ∈ S g and Aut( E g / F g ) such that σ ( Q ) = Q .



H g ( Q ) ∈ E g . Then



g H g (Q )

σ ( H g ( Q )) =

are isomorphic.



2

H g ( Q ), for each

σ ∈

Proof. Let Q ∈ S g and assume σ ∈ Aut( E g / F g ) such that  σ ( Q ) = Q . Let F g ( Q ) denote  the field H g(Q ) ∈ / F g ( Q ), then [ F g ( Q , H g ( Q )) : extension of F g generated by the coordinates of Q . If  F g ( Q )] = 2, which contradicts that 2 does not divide [ E g : F g ]. Therefore, H g ( Q ) ∈ F g ( Q ), which   implies σ ( H g ( Q )) = H g ( Q ). 2 Corollary 5.5. Let S ∈ S g , T = Tr E g / F g ( S ) be the trace map of E g over F g ( S ), and Q ∈ S with

  Then T ( H g ( Q )) = e H g ( Q ), where e = [ E g : F g ( S )].



H g(Q ) ∈ E g .

Proof. E g is Galois over F g ( S ) by (4.3). Hence, the order of |Aut F g ( S ) E g | = e. The conclusion follows by (5.4). 2

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Proposition 5.6. Assume Q ∈ S g and



H g ( Q ) ∈ E g . Then L Q ⊂ Fˆ g [x, y ].



Proof. In this case, L Q = {t  ∈ L g : t  ∈ E g [x, y ]}, and thus, Q  ∈ S Q if and only if H g ( Q  ) ∈ E g .  From (4.15) it follows that if t ∈ L Q , Q  ∈ S g , and t ( Q  ) = 0, then H g ( Q  ) ∈ E g .  } and If S ⊂ S g with | S g − S | = n − 2, we can order the elements of S g so that S = { Q 1 , . . . , Q m   2 S g − S = {Q m +1 , . . . , Q (n−1)2 } with m = (n − 1) − (n − 2). If Φ g : L g →

(n−1)2 i =1

 Z p H g ( Q i ) is the (n−1)2

. If t ∈ L Q injection of (3.8), then from opening paragraph of this proof we have Φ g (L Q ) ⊂ E g and T is the trace map of E g over F g ( S ) and Φ g (t ) = (t ( Q 1 ), . . . , t ( Q (n−1)2 )), then by (5.5) the i-th

component of Φ g ( T (t )) for 1  i  m is e · t ( Q i ), where e = [ E g : F g ( S )]. Since p does not divide e, Φ g ( T (t )) = 0 implies t ( Q i ) = 0, 1  i  m, which implies t = 0 by (3.6). Hence, T : L Q → L Q is

an injection, hence an isomorphism, since the order of L Q is finite. Therefore, L Q ⊂ F g ( S )[x, y ] for each S ⊂ S g with | S g − S | = n − 2, i.e. L Q ⊂ Fˆ g [x, y ]. Corollary 5.7. For each Q ∈ S g , L Q ⊂ Proof. Follows from (5.3) and (5.6).



2

H g ( Q ) · Fˆ g [x, y ], where g = √

g . H g (Q )

2 

Corollary 5.8. If Q ∈ S g , then L Q ⊂ Fˆ g ( H g ( Q ))[x, y ]. If, in addition,





H g ( Q ) · Fˆ g ( H g ( Q ))[x, y ].

Proof. Let g = √

g . H g (Q )

H g(Q ) ∈ / E g , then L Q ⊂





Then Fˆ g ⊂ Fˆ g ( H g ( Q )), which by (5.7) implies L Q ⊂ Fˆ g ( H g ( Q ))[x, y ].

  H g(Q ) ∈ / E g . If t ∈ L Q , then t = H g ( Q )h, for some h ∈ E g [x, y ]. Since   [ Fˆ g ( H g ( Q )) : Fˆ g ( H g ( Q ))] = 2, t can also be written as t = h1 + H g ( Q )h2 for unique h1 , h2 ∈  Fˆ g ( H g ( Q ))[x, y ]. Since Fˆ g ( H g ( Q )) ⊂ E g , h1 = 0 and h2 = h. Hence, t ∈ H g ( Q ) · Fˆ g ( H g ( Q ))[x, y ]. 2 Assume now that

Definition 5.9. For each Q ∈ S g , let Fˆ Q =



Q ∈S Q



Fˆ g ( H g ( Q  )). Clearly, Fˆ Q = Fˆ Q  for all Q  ∈ S Q .

Proposition 5.10. If Q ∈ S g , then L Q ⊂ Fˆ Q [x, y ]. Proof. By (4.13) and (5.8), since L Q = L Q  for all Q  ∈ S Q . Corollary 5.11. If Q ∈ S g and Proof. Suppose





2

H g(Q ) ∈ / Fˆ Q ( Q ), then t ( Q ) = 0 for all t ∈ L Q .

H g(Q ) ∈ / Fˆ Q ( Q ) and t ∈ L Q . Then t ( Q ) = ω

t ( Q ) ∈ Fˆ Q ( Q ) by (5.10). Hence,



H g ( Q ) for some

ω ∈ F p by (3.4). But

ω = 0. 2

Notation 5.12. Recall that for a point P ∈ X g , X P denotes the stalk of X g at P . Proposition 5.13. Let Q ∈ S g and P be the corresponding singular point of X g . If Cl( X P ) = 0.



H g(Q ) ∈ / Fˆ Q ( Q ), then



/ Fˆ Q ( Q ), then by (4.15) and (5.11), t ( Q ) = 0 for all t ∈ L g . Hence, Cl( X P ) = 0 by Proof. If H g ( Q ) ∈ (3.14). 2 Lemma 5.14. Let Q , Q  ∈ S g . Then Fˆ g ( Q ,  H g ( Q  )).



H g ( Q )) ∩ Fˆ g ( Q ,



H g ( Q  )) = Fˆ g ( Q ) or



H g ( Q ) ∈ Fˆ g ( Q ,

J. Lang / Journal of Algebra 379 (2013) 241–258

Proof. Since Fˆ g ( Q ) ⊂ Fˆ g ( Q ,



H g ( Q )) ∩ Fˆ g ( Q ,



Fˆ g ( Q )]  2, the conclusion follows. 2

Lemma 5.15. Let Q ∈ S g . Either Proof. Since Fˆ Q =





Q ∈S Q

H g ( Q  )) for some Q 



H g(Q ) ∈



Q ∈S Q

H g ( Q  )) ⊂ Fˆ g ( Q ,

Fˆ g ( Q ,



Fˆ g ( H g ( Q  )), Fˆ Q ( Q ) ⊂





255



H g ( Q )) and [ Fˆ g ( Q ,

H g ( Q )) :

H g ( Q  )) or Fˆ Q ( Q ) = Fˆ g ( Q ).

Q ∈S Q

Fˆ g ( Q ,



H g ( Q  )). If



H g(Q ) ∈ / Fˆ g ( Q ,

∈ S Q , then Fˆ Q ( Q ) = Fˆ g ( Q ) by (5.14). 2

Theorem 5.16. Let Q ∈ S g and P be the corresponding singular point of X g . If







H g(Q ) ∈ /

H g ( Q  )), then Cl( X P ) = 0.

Proof. By (5.15), Fˆ Q ( Q ) = Fˆ g ( Q ). By hypothesis,





Q ∈S Q

Fˆ g ( Q ,

H g(Q ) ∈ / Fˆ g ( Q ). Hence, Cl( X P ) = 0 by (5.13).

2

6. Applications In this section g satisfies the maximum intersection condition. Remark 6.1. If t ∈ L g and Q ∈ S g , then by (3.4), t ( Q ) =  (σ t )(σ ( Q )) = σ (t ( Q )) = ±ω Q H g (σ ( Q )).

ωQ



H g ( Q ), with

Proposition 6.2. Let n  4. If n = 0 (mod p ), assume r is an integer with 2  r  assume 2  r  Cl( X g ) ∼ = Z/ p Z.

(n−2)(n−3) 2

ω Q ∈ F p . Hence,

(n−1)(n−2) 2

. If n = 0 (mod p ),

. If the action of Aut( E g / F g ) on S g includes all r-cycles on S g , then Cl( X g ) = 0 or

Proof. L g is normal over E g by (4.3). Hence, Aut( E g / F g ) is a subgroup of Aut( L g / F g ) and the action of Aut( L g / F g ) on S g includes all r-cycles on S g . Assume n = 0 (mod p ). Then the cardinality of S g is (n − 1)2 . Let t ∈ L g − {0} be such that t ( Q ) = 0 for a maximal number of Q ∈ S g .  Order the elements of S g such that t ( Q i ) = 0 for 1  i  s and t ( Q i ) = 0 for i > s. By (3.4), t ( Q i ) = ωi H g ( Q i ) with ωi ∈ F p − {0} for 1  i  s. (n−1)(n−2)

Suppose s  . Choose σ ∈ Aut( L g / F g ) such that σ acts on S g as the r-cycle, Q s → 2  Q s+1 → · · · → Q s+r −1 → Q s . Then by (6.1), (σ t )( Q s+1 ) = ±ωs H g ( Q s+1 ) = 0 and (σ t )( Q s+ j ) = 0 for j = 0 and j  2. Similarly, for each integer i with 0  i  (n − 1)2 − s, there exists t i ∈ L g that zeros every Q s+ j with j > 0 except Q s+i . It follows that the t i , 0  i  (n − 1)2 − s, are linearly

n(n−1) 2 independent, so that the order of L g is greater than p (n−1) −s  p 2 , which contradicts (2.17).

(n−1)(n−2)

 r. Therefore, s > 2  As already noted, for each i = 1, . . . , (n − 1)2 , t ( Q i ) = ωi H g ( Q i ), with ωi ∈ F p . Hence, if σ ∈ Aut( L g / F g ) and σ ( Q i ) = Q i , then (6.1) implies σ (t ( Q i )) = ±t ( Q i ). It also follows that if 1  i , j  s and σ acts as the identity on { Q i , Q j } ∪ { Q s+1 , . . . , Q (n−1)2 }, then σ (t ( Q i )) = t ( Q i ) if and only if σ (t ( Q j )) = t ( Q j ); i.e. the sign of t ( Q i ) changes under σ if and only if the sign of t ( Q j ) changes under σ . Otherwise, t + σ t = 0 and it would zero more Q ∈ S g than t does. If σ ∈ Aut( L g / F g ) acts on S g as the r-cycle, Q 1 → Q 2 → · · · → Q r → Q 1 , then by (6.1),   (σ t )( Q i +1 ) = ±ωi H g ( Q i +1 ), 1  i  r − 1, and (σ t )( Q 1 ) = ±ωr H g ( Q 1 ). Since s > r, the previous paragraph implies either t + σ t or t − σ t zeros more Q ∈ S g than t does, which implies either t + σ t = 0 or t − σ t = 0. In either case this implies ωi = ±ω1 , for i = 2, . . . , r. By considering other induced r-cycles on { Q 1 , . . . , Q s }, we obtain ωi = ±ω1 for i = 2, . . . , s. Let ω = ω1 . If σ ∈ Aut( L g / F g ) acts on S g as the  r-cycle, Q 1 → Q 2 → · · · → Q r −1 → Q s+1 → Q 1 , then by (6.1), (σ t )( Q 1 ) = 0, (σ t )( Q i +1 ) = ±ω H g ( Q i +1 ) = ±t ( Q i +1 ) for 1  i  r − 2 and (σ t )( Q s+1 ) =  ±ω H g ( Q s+1 ). Also, σ fixes Q r , Q r +1 , . . . , Q s , which consists of at least two elements since s > r. Then from the above observation concerning fixed elements, either t + σ t or t − σ t zeros more Q ∈ S g

256

J. Lang / Journal of Algebra 379 (2013) 241–258

than t does, but t + σ t = 0 and t − σ t = 0, since neither of these zero Q 1 or Q s+1 . Hence, it must be that s = (n − 1)2 , i.e. t does not zero any Q ∈ S g , which implies each t ∈ L g − {0} zeros no Q ∈ S g , which implies L g = 0 or L g ∼ = Z/ p Z. 2 Corollary 6.3. If p > 2 and n  4 and the action of Aut( E g / F g ) on S g includes the alternating group on S g , then Cl( X g ) = 0 or Cl( X g ) ∼ = Z/ p Z. Proof. By (6.2) since the alternating group is generated by all 3-cycles.

2

Corollary 6.4. If p = 2 and n  4 and the action of Aut( E g / F g ) on S g includes the alternating group on S g , then Cl( X g ) ∼ = Z/2Z.



Proof. If p = 2, D g ( g x ) = g x g xy . Hence, g xy ∈ L g . Also, g xy = H g . Since g satisfies the maximum intersection condition, g xy = 0. Hence, Cl( X g ) ∼ = Z/2Z by (6.2). 2 Remark 6.5. In [11], the author shows that if p = 2, n  5, and g generic, the action of Aut( E g / F g ) on S g includes the alternating group. Hence, Cl( X g ) ∼ = Z/2Z for a generic g of degree greater than 4. As mentioned above, if p > 2, n  4, and g generic, Cl( X g ) = 0. Example 6.6. Assume n k is algebraically closed n of characteristic p  5 and n ∈ Z such that 4  n < p. Assume f (x) = i =1 ai xi ∈ k[x] and h( y ) = i =1 b i y i ∈ k[ y ] with {ai , b i | 1  i  n} algebraically inde-

n−1

n−1

pendent over F p . Let g (x, y ) = f (x) + h( y ). Then g x = f  = i =1 (x − αi ) and g y = h = i =1 ( y − βi ) with {αi , βi | 1  i  n − 1} algebraically independent over F p . Hence, g satisfies the maximum intersection condition. Let Q 1 = Q 2 ∈ S g . We wish to show Q 1  Q 2 under the equivalence of (4.11). Without lost of generality we may assume that Q 1 = (α1 , β1 ) and Q 2 = (α2 , βl ) for some l = 1, . . . , n − 1. f  (α1 ) f  (α2 )

n−1 α1 −αi n−1 α1 −α2 i =3 α2 −αi = − i =3 ( α2 −αi + 1). Since

f  (α1 )h (β1 ) f  (α2 )h (βl )

1 −α2 and {α1 , α2 , α α2 −α3 + α1 −α2 1, . . . , α −α + 1} ∪ {β1 , . . . , βn−1 } is a set of algebraically independent generators of E g over F p , 2 n −1 

Then

H g (Q 1) H g (Q 2)

=−

H g (Q 1) H g (Q 2)

=

∈ / E g and Q 1  Q 2 by (4.12). Therefore, by (4.18), Cl( X g ) = 0.

Example 6.7. Assume n k is algebraically closed of characteristic p  5 and n ∈ Z such that 4  n < p. Assume f (x) = i =1 ai xi ∈ k[x] with {ai | 1  i  n} algebraically independent over F p . Let g (x, y ) =

n−1

n−1

f (x) + f ( y ). Then g x = f  = i =1 (x − αi ) and g y = f  = i =1 ( y − αi ) with {αi | 1  i  n − 1} algebraically independent over F p . Then g satisfies the maximum intersection condition. Suppose Q = (αi , α j ) with i = j. Arguing as in (6.6), we obtain S Q = {(α1 , α2 ), (α2 , α1 )}. Hence, L Q = 0 by (4.16).  If Q i = (αi , αi ), then S Q i = { Q 1 , . . . , Q n−1 }. It follows that L g = L Q 1 by (4.14). If φ : L Q 1 → Z p · H g ( Q 1 ) is the evaluation map, t → t ( Q 1 ), then φ is an injection, for if t ( Q 1 ) = 0, then t = 0

by (2.8), (3.6) and (4.15). Therefore, L g = 0 or L g ∼ = Z p . Since is the case, i.e. Cl( X g ) ∼ = Zp .

D g (x− y ) x− y

=

f  ( y )− f  (x) x− y

∈ L g , the latter

Notation 6.8. For a subset S of a field F , let S (2) = {a2 | a ∈ S }. Let  F = {α ∈ ( F ∗ )(2) | 1 − α ∈ F (2) } and F¨ = F − {0, 1}. Proposition 6.9. Let F q be a finite field of q elements. If q ≡ 1 (mod 4), then the cardinality of  F q is 14 (q − 1). if and only if q ≡ 1 (mod 4). Define a permutation of F¨ q by

τ (x) = x−x 1 . ¨ Then τ (x) = 1−x and τ (x) = x. Hence, for each a ∈ F q , the orbit of a consists of either 1 or 3 elements. Also, each orbit contains at least one element of ( F q∗ )(2) . To see this, let b be a primitive element of F q∗ and suppose that a and τ 2 (a) are not in ( F q∗ )(2) . Then a = bm and τ 2 (a) = bn , where both (2)

Proof. Note that −1 ∈ F q 2

1

3

J. Lang / Journal of Algebra 379 (2013) 241–258

257

τ (a) = aτ−21(a) = −b−m−n ∈ ( F q∗ )(2) , since −m − n is even and −1 ∈ ( F q∗ )(2) . Furthermore, if two of a, τ (a), τ 2 (a) belong to ( F q∗ )(2) , then so does the third, since aτ (a)τ 2 (a) = −1.

m and n are odd. Hence,

But if each element of the orbit of a belongs to ( F q∗ )(2) , then clearly a ∈  F q . Therefore, an element of F¨ q does not belong to  F q if and only if its orbit consists of three distinct elements, only one of

which belongs to ( F q∗ )(2) , and an element of F¨ q belongs to  F q if and only if each element of its orbit

belongs to ( F q∗ )(2) . Therefore, the order of F¨ q −  F q is which implies the order of  F q is

1 (q 4

− 1). 2

3 (q 4

− 1), since the order of ( F q∗ )(2) is

q−1 , 2

Proposition 6.10. Let F q be the finite field of q elements. If q ≡ 3 (mod 4), then the cardinality of  F q is 1 (q 4

+ 1). (2)

Proof. −1 ∈ / Fq

since q ≡ 3 (mod 4). Thus the permutation x → −x of F q∗ takes ( F q∗ )(2) to its com-

plement in F q∗ . Hence, for each α ∈ F q∗ , either α or −α belongs to ( F q∗ )(2) , but not both. Assume α ∈ ( F q∗ )(2) − {1}. Then α1 ∈ ( F q∗ )(2) . Also, 1 − α ∈ ( F q∗ )(2) if and only if 1 − α1 the previous paragraph, since α (1 − 1 ) = α − 1. Therefore, for each α ∈ ( F ∗ )(2) − {1}, α α

only if α1 ∈ /  F q . Since the order of ( F q∗ )(2) − {1} is

q−3 , 2

q

1 (q 4

the order of  F q is

∈ / ( F q∗ )(2) , by ∈  F q if and

+ 1). 2

Corollary 6.11. Let F q be the finite field of q elements. Then the number of points in F q × F q that lie on the unit circle is q − 1 if q ≡ 1 (mod 4) and q + 1 if q ≡ 3 (mod 4). Corollary 6.12. Let F q be the finite field of q elements. Then the number of x ∈ F¨ q such that x − x2 ∈ ( F q∗ )2 is 1 (q 2

− 3) if q ≡ 1 (mod 4) and 12 (q − 1) if q ≡ 3 (mod 4).

Proof. For x ∈ F¨ q , x − x2 ∈ ( F q∗ )2 if and only if

− (x − 12 )2 ∈ ( F q∗ )2 if and only if 1 − (2x − 1)2 ∈ ( F q∗ )2 . Since the correspondence x → 2x − 1 maps F¨ q one-to-one and onto F q − {−1, 1}, the number of x ∈ F¨ q such that x − x2 ∈ ( F q∗ )2 is equal to the number of x ∈ F q − {−1, 1} such that 1 − x2 ∈ ( F q )2 . By (6.9) and (6.10), this number is 12 (q − 3) if q ≡ 1 (mod 4) and 12 (q − 1) if q ≡ 3 (mod 4). 2 1 4

Example 6.13. Assume k is algebraically closed of characteristic p  5. Assume f (x) =

 p −1

1 i =1 i

 p −1 i =1

ai xi ∈ k[x]

with {ai | 1  i  p − 1} algebraically independent over F p and h( y ) = y . Let g = f (x) + h( y ). p −2 Then g x = f  (x) = i =1 (x − αi ) with {αi | 1  i  p − 2} algebraically independent over F p and g y = i

y p −1 −1

h ( y ) = y −1 . Then the roots of g y are {2, 3, . . . , p − 1} and g satisfies the maximum intersection condition. Hence, E g = F p (α1 , . . . , α p −2 ). Arguing as in (6.6) and (6.7), if Q 1 , Q 2 ∈ S g with distinct let i, j be first coordinates, then Q 1  Q 2 . So assume Q 1 and Q 2 have the same first coordinate and the second coordinates of Q 1 , Q 2 , respectively. Then E g if and only if

j− j i −i 2

2

H g (Q 1) H g (Q 2)

=

h (i ) h ( j )

=

j− j2 . i −i 2

∈ ( F p∗ )2 . The latter holds if and only if both i − i 2 and j − j 2

not in ( F p∗ )2 . Therefore, by (6.12), if i − i 2 ∈ ( F p∗ )2 , then the order of S Q is and and

1 ( p − 1) 2 1 ( p − 3) 2

It follows that

if p ≡ 3 (mod 4). If i − i ∈ / 2

( F p∗ )2 , then the order of S Q is

if p ≡ 3 (mod 4). In either case, Cl( X g ) = 0 by (4.16).

H g (Q 1) H g (Q 2)

∈

∈ ( F p∗ )2 or both are

1 ( p − 3) if 2 1 ( p − 1) if 2

p ≡ 1 (mod 4) p ≡ 1 (mod 4)

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