Galois points for a normal hypersurface
aa r X i v : . [ m a t h . AG ] J u l GALOIS POINTS FOR A NORMAL HYPERSURFACE
SATORU FUKASAWA & TAKESHI TAKAHASHI
Abstract.
We study Galois points for a hypersurface X with dim Sing( X ) ≤ dim X −
2. Thepurpose of this article is to determine the set ∆( X ) of Galois points in characteristic zero:Indeed, we give a sharp upper bound of the number of Galois points in terms of dim X anddim Sing( X ) if ∆( X ) is a finite set, and prove that X is a cone if ∆( X ) is infinite. To achieveour purpose, we need a certain hyperplane section theorem on Galois point. We prove thistheorem in arbitrary characteristic. On the other hand, the hyperplane section theorem hasother important applications: For example, we can classify the Galois group induced from aGalois point in arbitrary characteristic and determine the distribution of Galois points for aFermat hypersurface of degree p e + 1 in characteristic p > Introduction
Let the base field K be an algebraically closed field of characteristic p ≥ X ⊂ P n +1 be anirreducible and reduced hypersurface of dimension n and of degree d ≥ s ( X ) := dim Sing( X ).H. Yoshihara introduced the notion of Galois point (see [3, 7, 10, 11, 12]). If the function fieldextension K ( X ) /K ( P n ), induced from the projection π P : X P n from a point P ∈ P n +1 , isGalois, then the point P is said to be Galois. In this paper a Galois point P means Galois pointwhich is contained in the smooth locus X sm or P n +1 \ X , except for Subsection 3.1. Let ∆( X )(resp. ∆ ′ ( X )) be the set of all Galois points contained in X sm (resp. P n +1 \ X ). We are interestedin the sets ∆( X ) and ∆ ′ ( X ). If p = 0, n = 1 and X is smooth, then Yoshihara determined thesets ∆( X ) and ∆ ′ ( X ) ([10]). If p > n = 1 and X is smooth, then the first author determinedin most cases ([2, 3]). If p = 0 and X is a smooth quartic surface, then Yoshihara determinedthe set ∆( X ) completely ([11]). If K = C , n ≥ X is smooth, then Yoshihara gave sharpupper bounds on the cardinalities of ∆( X ) and ∆ ′ ( X ) ([12]). If p = 0 and X is a normal quarticsurface, then the second author determined the set ∆( X ) ([9]). However, for a higher dimensionalhypersurface with large singularities, there are few results.The purpose of this article is to determine the sets ∆( X ) and ∆ ′ ( X ) for a hypersurface ofdimension n with s ( X ) = dim Sing( X ) ≤ n − p = 0. We define the dimension of the emptyset as −
1. Our results are:
Theorem 1.
Let X ⊂ P n +1 be a hypersurface of dimension n ≥ and degree d ≥ with s = s ( X ) ≤ n − in characteristic p = 0 and let m = m ( n, s ) := [( n + s + 1) / where [ ∗ ] means theinteger part of ∗ . Assume that ∆( X ) is a finite set. Mathematics Subject Classification.
Key words and phrases.
Galois point, hypersurface. (I-0) If d = 4 and s = − then ♯ ∆( X ) ≤ m ( n, s ) + 1) . The equality holds if and only if X isprojectively equivalent to the hypersurface defined by X m +1 X + · · · + X m +1 X m + X m +1 + · · · + X n +1 = 0 . (I-1) If d = 4 , s ≥ and n + s is odd, then ♯ ∆( X ) ≤ m ( n, s ) − s −
1) + ( s + 2) . The equalityholds if and only if X is projectively equivalent to the hypersurface defined by X m +1 X + · · · + X m − s − X m − s − + X m − s ( X m − s − + · · · + X m )+ X m +1 + · · · + X m − s − + aX m − s = 0 where a = 0 or . (I-2) If d = 4 and s = 0 and n is even then ♯ ∆( X ) ≤ m ( n, s ) + 1 . The equality holds if andonly if X is projectively equivalent to the hypersurface defined by X m +1 X + · · · + X m +1 X m + X m +1 + · · · + X m = 0 . (I-3) If d = 4 and s = 1 and n + s is even then ♯ ∆( X ) ≤ m ( n, s ) − . The equality holds ifand only if X is projectively equivalent to the hypersurface defined by X m − X + · · · + X m − X m − + X m − + · · · + X m − + G = 0 where G ∈ K [ X m − , X m − , X n +1 ] has a multiple component. (I-4) If d = 4 , s ≥ and n + s is even, then ♯ ∆( X ) ≤ m ( n, s ) − s −
1) + ( s + 2) . Furthermore,the equality holds if and only if X is projectively equivalent to the hypersurface defined byone of the followings: (i) X m − s X + · · · + X m − s − X m − s − + X m − s + · · · + X m − s − + G = 0 where s = 2 and G ∈ K [ X m − s , . . . , X n +1 ] has a multiple component; or (ii) X m +1 X + · · · + X m − s − X m − s − + X m − s X m − s − + A m − s X m − s · · · + A m X m + X m +1 + · · · + X m − s − + G = 0 where s ≥ and A m − s , . . . , A m , G ∈ K [ X m − s , X n +1 ] . (II) If d ≥ , then ♯ ∆( X ) ≤ m ( n, s ) + 1 . Furthermore, the equality holds if and only if X isprojectively equivalent to the hypersurface defined by X m +1 X d − + · · · + X m − s X d − m − s − + A m − s X d − m − s + · · · + A m X d − m + G = 0 where A m − s , . . . , A m , G ∈ K [ X m +1 , . . . , X n +1 ] are homogeneous polynomials with deg A m − s = · · · = deg A m = 1 and deg G = d . (Note that the polynomial A m − s X m − s + · · · + A m X d − m does not appear if s = − .) Corollary 1.
Let λ := [ n/ and let X ⊂ P n +1 be a hypersurface of dimension n ≥ and degree d = 4 with s ( X ) ≤ n − in characteristic p = 0 . Assume that ∆( X ) is a finite set. Then, ♯ ∆( X ) ≤ λ + 1) . Furthermore, the equality holds if and only if X is projectively equivalent tothe hypersurface defined by X λ +1 X + · · · + X λ +1 X λ + X λ +1 + · · · + X n +1 = 0 . Theorem 2.
Assume that p = 0 , s ( X ) ≤ n − and ∆ ′ ( X ) is finite. ALOIS POINTS FOR A NORMAL HYPERSURFACE 3 (1) If s = − , then ♯ ∆ ′ ( X ) ≤ n + 2 . Furthermore, the equality holds if and only if X isprojectively equivalent to the Fermat hypersurface. (2) If s ≥ , then ♯ ∆ ′ ( X ) ≤ n − s . Furthermore, the equality holds if and only if X isprojectively equivalent to the hypersurface defined by X d + · · · + X dn − s − + G = 0 where G ∈ K [ X n − s , . . . , X n +1 ] has a multiple component. Theorem 3.
Assume that p = 0 and s ( X ) ≤ n − . If the set ∆( X ) ∪ ∆ ′ ( X ) is infinite, then X is a cone. Furthermore, there exist linear spaces M , M ⊂ P n +1 with M ∩ M = ∅ and anirreducible hypersurface Y ⊂ M satisfying the following conditions: ( i ) X = Y ♯M ; ( ii ) M hasthe maximal dimension as a vertex of X ; ( iii ) the set ∆( Y ) ∪ ∆ ′ ( Y ) is finite; and ( iv ) ∆( X ) = (∆( Y ) ♯M ) \ M and ∆ ′ ( X ) = (∆ ′ ( Y ) ♯M ) \ M , where ♯ means a linear join of algebraic sets. To prove our Theorems, we need a hyperplane section theorem below. We denote by G P ( X )the Galois group induced from P if P is a Galois point for a hypersurface X . Theorem 4.
Let X ⊂ P n +1 ( n ≥ ) be an irreducible hypersurface of degree d ≥ with ≤ s ( X ) ≤ n − (resp. s ( X ) = − ) in characteristic p ≥ . Let P ∈ P n +1 be a Galois point for X .Then, the followings hold: (i) A general hyperplane H passing through P satisfies the following condition: ( ⋆ ) the hyperplane section X H := X ∩ H is an irreducible hypersurface in H ∼ = P n ofdegree d with s ( X H ) = s ( X ) − (resp. s ( X H ) = − ), and a general tangent space of X H does not contain P . (ii) Let H be a hyperplane passing through P and satisfying the condition ( ⋆ ) . Then, the point P is Galois with respect to X H . (iii) The Galois groups are isomorphic: G P ( X ) ∼ = G P ( X H ) for any hyperplane H passingthrough P and satisfying ( ⋆ ) . In fact, this Theorem will be used for the proof of Propositions 1 and 2 in Subsection 3.1. Itis remarkable, as in these Propositions, that the birational transformation induced by σ ∈ G P ( X )can be extended to a projective transformation on P n +1 when p = 0.The hyperplane section theorem has other important applications. One application is to classifythe Galois group. Combining Theorem 4 with a result on the structure of G P ( C ) for a smoothplane curve C with a Galois point P obtained in [2], we have the following: Corollary 2.
Let X ⊂ P n +1 be a hypersurface of degree d ≥ with s ( X ) ≤ n − in characteristic p ≥ . Let P ∈ X (resp. P ∈ P n +1 \ X ) be a Galois point for X and d − p e l (resp. d = p e l ),where l is not divisible by p if p > .Then, l divides p e − and the Galois group G P is isomorphic to the semidirect product of ( Z /p Z ) ⊕ e (as a normal subgroup) and Z /l Z (as a quotient). ALOIS POINTS FOR A NORMAL HYPERSURFACE 4
Another application is to determine the distribution of Galois points for a Fermat hypersurface F n ( q + 1) ⊂ P n +1 of degree q + 1 ≥ p > q is a power of p . Homma [5] determined thedistribution of Galois points for a Hermitian curve in P , which is defined by X q Z + XZ q − Y q +1 = 0.Note that the Hermitian curve is projectively equivalent to F ( q + 1) over F q . We will generalizeHomma’s result. Theorem 5.
A point P ∈ P n +1 is Galois for F n ( q + 1) if and only if P ∈ P n +1 is F q -rational. Note that Theorem 5 implies that Theorems 1 and 2 do not hold in p >
0. In the final section,we give examples which are not cones and have infinitely many Galois points. Especially, Example2 implies that Theorem 3 does not hold in p >
Proof of Theorem 4
In this section, X ⊂ P n +1 is an irreducible hypersurface of degree d ≥ s ( X ) =dim Sing( X ) ≤ n −
2, otherwise specified. We denote by ˇ P n +1 the dual projective space, whichparameterizes hyperplanes in P n +1 . We denote by T x X ⊂ P n +1 the projective tangent space at x ∈ X sm . We have the Gauss map γ : X sm → ˇ P n +1 ; x T x X . We note the following elementaryfact: Note 1.
A hyperplane H coincides with the tangent space T x X ⊂ P n +1 at a smooth point x ∈ X if and only if the scheme X ∩ H is singular at x ∈ X ∩ H .Let V P ⊂ ˇ P n +1 be the set of all hyperplanes which pass through P and let S X ⊂ ˇ P n +1 be theset of all hyperplanes which contain some irreducible component Y of the singular locus of X withdim Y = s ( X ). We have the following assertion of Bertini type in arbitrary characteristic (cf. theproof of [4, II, 8.18]). Note 2.
Assume that 0 ≤ s ( X ) ≤ n − s ( X ) = − H ∈ V P \ ( γ ( X sm ) ∪ S X ), the hyperplane cut X H := X ∩ H is an irreducible hypersurface of degree d with s ( X H ) = s ( X ) − s ( X H ) = − P ∈ P n +1 is called a strange center if the tangent space T x X contain P for a generalpoint x ∈ X . Note 3.
The projection π P : X P n is generically finite and separable if and only if P ∈ P n +1 is not a strange center.Now we assume that π P is separable and generically finite onto its image. Then, the differentialmap d Q π P : T Q X → T π P ( Q ) P n is isomorphic at a general point Q ∈ X , where T Q X is the Zariskitangent space at Q . We denote by U P the maximal open set of X sm such that the differential map d Q π P : T Q X → T π P ( Q ) P n is isomorphic for any Q ∈ U P . We denote by Σ P ⊂ ˇ P n +1 the finite setconsisting of hyperplanes H such that X ∩ H is contained in X \ U P (as a set). On the other hand,we note the following: Note 4. If X ∩ H is an integral scheme, then T Q ( X H ) = T Q X ∩ H for any smooth point Q of X H . ALOIS POINTS FOR A NORMAL HYPERSURFACE 5
Combining above Notes, we have the following:
Lemma 1.
Let X ⊂ P n +1 be an irreducible hypersurface of degree d with ≤ s ( X ) ≤ n − (resp. s ( X ) = − ). Assume that P is not a strange center. Then, we have the followings: (i) V P \ ( γ ( X sm ) ∪ S X ∪ Σ P ) = ∅ . (ii) Let H ∈ V P \ ( γ ( X sm ) ∪ S X ∪ Σ P ) . Then, H satisfies the condition ( ⋆ ) . Now assume that P is a Galois point. Let G P be the Galois group. Then, we find easily that P is not a strange center. Therefore, V P \ ( γ ( X sm ) ∪ S X ∪ Σ P ) = ∅ as in Lemma 1. Then, we canconsider σ ∈ G P as a birational map from X to itself. Needless to say, σ is defined and isomorphicat a general point of X . We denote by the maximal open subset U σ such that σ is defined andisomorphic over U σ . It follows from a certain elementary lemma [4, V. Lemma 5.1] that X \ U σ isof codimension ≥
2. Therefore, we have:
Note 5.
For any hyperplane H ∈ V P \ ( γ ( X sm ) ∪ S X ∪ Σ P ), X H ∩ U σ is non-empty.We also define U G := T σ ∈ G P U σ . We have the following natural property in Galois extension(cf. [8, III. 7.1]): Lemma 2.
Let P be a Galois point. For any points Q ∈ U G and R ∈ U P with π P ( Q ) = π P ( R ) ,there exists an element σ ∈ G P such that σ ( Q ) = R .Proof. Assume that σ ( Q ) = R for any σ ∈ G P . We take a function f ∈ T σ O σ ( Q ) ∩ O R such that f ( σ ( Q )) = 0 for any σ ∈ G P and f ( R ) = 0. Then, we consider the function g = Q σ σ ∗ f . For any σ ∈ G P , we have σ ∗ g = g . Therefore, we find that g ∈ O π P ( R ) . Note that g ∈ m π P ( R ) because of π ∗ P m π ( R ) = m R by R ∈ U P , where m R is the maximal ideal of the local ring of O R . This impliesthat 0 = g ( π P ( Q )) = g ( π P ( R )) = 0. This is a contradiction. (cid:3) Proof of Theorem 4.
Assume that P is Galois. We have Theorem 4 (i) by Lemma 1. Let d be thedegree of the function field extension K ( X ) /π ∗ P K ( P n ). Let H be any hyperplane passing through P and satisfying the condition ( ⋆ ). Then, we consider a group morphism φ ( H, P ) : G P ( X ) → G ; σ σ | X H where G = { σ ∈ Bir( X H ) | σ ( X H ∩ l ) ⊂ X H ∩ l for a general line l such that P ∈ l ⊂ H } . Itfollows from Note 5 that σ is defined and isomorphic at a general point of X H . We also find that σ ( X H ) = X H . Therefore, φ ( H, P ) is well-defined. Now we prove that φ ( H, P ) is an injection.We assume that σ | X H = id X H and σ = id X . Then, σ ( Q ) = Q for any Q ∈ X H . It follows fromLemma 2 that the cardinality X H ∩ l \ P is at most d − l in H containing P .If dim X H = 1, then l is a tangent line of X H by B´ezout theorem. However, this is a contradictionwith ( ⋆ ). If dim X H >
1, then this is also a contradiction by taking some linear section of X H andusing Lemma 1. Since the order of G is at most d , φ is an isomorphism. Therefore, P is a Galoispoint for X H . We have (ii) and (iii). (cid:3) Remark 1.
It seems to be already known that the assertion in Theorem 4 holds when p = 0, s ( X ) ≤ H ∈ V P is assumed to be general in (ii) (cf. [9], [11, Proposition 2.5], [12]). Yoshihara[14] also informed the first author an idea of the proof in this case. ALOIS POINTS FOR A NORMAL HYPERSURFACE 6 Galois points for a normal hypersurface in characteristic zero
In this section, we assume that p = 0.3.1. Preliminaries.
Let X ⊂ P n +1 be an irreducible and reduced hypersurface of degree d and P a point in P n +1 . If P X , then we define the multiplicity of X at P as 0. Let ( X : X : · · · : X n +1 ) be a system of homogeneous coordinates, ( x , . . . , x n +1 ) = ( X /X , X /X , . . . , X n +1 /X )a system of affine coordinates, and we assume that P = (0 , , . . . , F ( X , X , . . . , X n +1 ) = 0be the defining homogeneous equation of X , and f ( x , . . . , x n +1 ) = F (1 , x , . . . , x n +1 ) = P di =0 f i its dehomogenized polynomial, where f i is a homogeneous part of f with degree i . The multiplicityis equal to m if and only if f m = 0 and f i = 0 for any i < m . Lemma 3.
Let ≤ m ≤ d − . The following three conditions are equivalent :(1) m The multiplicity at P is m , f m divides f m +1 , and f m + i = (cid:18) d − mi (cid:19) f m (cid:18) f m +1 ( d − m ) f m (cid:19) i ( i = 0 , , . . . , d − m − m By taking a suitable projective transformation fixing the point P , the defining equation canbe given by g m ( x , . . . , x n +1 ) + g d ( x , . . . , x n +1 ) = 0 , where g m and g d are homogeneous polynomials of degree m and degree d , respectively. (3) m The point P is of multiplicity m and Galois, and the birational map induced by σ is arestriction of a projective transformation of P n +1 for any σ ∈ G P ( X ) .Furthermore, P is Galois and the Galois group G P ( X ) is a cyclic group of order d − m , if one ofthe conditions holds.Proof. First let us prove the implication (1) m ⇒ (2) m . Since f m +1 / (( d − m ) f m ) is a homogeneouspolynomial of degree one, let us put h := f m +1 / (( d − m ) f m ) = m x + m x + · · · + m n +1 x n +1 ,where m , · · · , m n +1 ∈ K . Let ˆ X = X + h ( X , . . . , X n +1 ). Then, F ( ˆ X − h, X , · · · , X n +1 ) = f m ( ˆ X − h ) d − m + · · · + f m + i ( ˆ X − h ) d − m − i + · · · + f d . The coefficient F d − m − i of ˆ X d − m − i is i X k =0 (cid:18) d − m − kd − m − i (cid:19) f m + k ( − h ) i − k . By using (1) m , we have F d − m − i = f m h i i X k =0 ( − i − k (cid:18) d − m − kd − m − i (cid:19)(cid:18) d − mk (cid:19) = ( − i f m h i (cid:18) d − md − m − i (cid:19) i X k =0 ( − k (cid:18) ik (cid:19) . Since P ik =0 ( − k (cid:0) ik (cid:1) = 0 if i >
0, we have the assertion of (2) m .Now we prove (2) m ⇒ (1) m . Let φ be the projective transformation as in the condition (2) m and let ψ be the inverse, and let ˆ X i = ψ ∗ X i for 0 ≤ i ≤ n + 1. Then, by the assumption, g m ( ˆ X , . . . , ˆ X n +1 ) ˆ X d − m + g d ( ˆ X , . . . , ˆ X n +1 ) = 0. Let A ψ be a matrix representing ψ . Since ALOIS POINTS FOR A NORMAL HYPERSURFACE 7 ψ ( P ) = P , we may assume A ψ = · · · a , a , · · · a ,n +1 ... ... ... a n +1 , a n +1 , · · · a n +1 ,n +1 . Let ˆ g m ( X , . . . , X n +1 ) := g m ( ˆ X , . . . , ˆ X n +1 ), let ˆ g d ( X , . . . , X n +1 ) := g d ( ˆ X , . . . , ˆ X n +1 ) and h := a ,n +1 X + · · · + a n +1 ,n +1 X n +1 . Then, we have F = ˆ g m ( h + X ) d − m + ˆ g d and f m + i = ˆ g m (cid:18) d − mi (cid:19) h i . Therefore, f m +1 /f m = ( d − m ) h and (cid:18) d − mi (cid:19) f m (cid:18) f m +1 ( d − m ) f m (cid:19) i = (cid:18) d − mi (cid:19) f m h i = (cid:18) d − mi (cid:19) ˆ g m h i = f m + i for i = 0 , . . . , d − m − m ⇔ (3) m is the same as the proof for n = 1 (See [13, Proposition 1], and also[6, Proposition 4]). (cid:3) Below, we assume that s ( X ) ≤ n − X sm ∪ ( P n +1 \ X ). Wehave the following Proposition, by Lemma 3 and Theorem 4. Proposition 1.
We consider the following weaker condition than (3) : (3’) The point P ∈ X sm is Galois.Then, the three conditions (1) , (2) and (3 ′ ) are equivalent.Especially, if P ∈ X sm is Galois, then the birational transformation of X induced by σ ∈ G P ( X ) is a restriction of a projective transformation of P n +1 .Proof. The implications (1) ⇔ (2) and (2) ⇒ (3 ′ ) are clear from Lemma 3.Let us prove the implication (3 ′ ) ⇒ (1) by induction on the dimension n . If n = 1, thenit holds by [10, Proposition 5] and the implication of (2) ⇒ (1) in Lemma 3. We assume that n ≥ P is a smooth Galois point. Let H be a general hyperplane given by the equation x = a x + · · · + a n +1 x n +1 , where a i ∈ K . We put ˜ x = a x + · · · + a n +1 x n +1 . Then, fromTheorem 4, X H := X ∩ H satisfies the condition ( ⋆ ), and P is a smooth Galois point withrespect to X P . Hence, by the assumption of the induction, we have that f (˜ x, x , . . . , x n +1 ) = 0, f (˜ x, x , . . . , x n +1 ) /f (˜ x, x , . . . , x n +1 ) ∈ K [ x , . . . , x n − ] and f i +1 (˜ x, x , . . . , x n +1 ) = (cid:18) d − i (cid:19) f (˜ x, x , . . . , x n +1 ) (cid:18) f (˜ x, x , . . . , x n +1 )( d − f (˜ x, x , . . . , x n +1 ) (cid:19) i ( i = 0 , . . . , d − a , . . . , a n +1 ). This implies the assertion (1) . (cid:3) Corollary 3.
Assume that d = 4 . Let P = (1 : 0 : · · · : 0) be a smooth point of X . Then, P ∈ ∆( X ) if and only if f = 3 f f . From Lemma 1, we infer the following remark, which is useful to find smooth Galois points.
ALOIS POINTS FOR A NORMAL HYPERSURFACE 8
Remark 2. (1) If P ∈ X sm is a Galois point, then T P X ∩ X is a (possibly reducible) cone and P is itsvertex.(2) Let H = H ( F ) be the Hessian of F . If P ∈ X sm is a Galois point, then H ( F )( P ) = 0.Similar to Proposition 1, we have the following: Proposition 2.
We consider the following weaker condition than (3) : (3’) The point P ∈ P n +1 \ X is Galois.Then, the three conditions (1) , (2) and (3 ′ ) are equivalent.Especially, if P ∈ P n +1 \ X is Galois, then the birational transformation of X induced by σ ∈ G P ( X ) is a restriction of a projective transformation of P n +1 . Propositions 1 and 2 imply that some methods by Yoshihara [12] for smooth hypersurfaces areavailable also for normal hypersurfaces in our situation. We introduce the following:
Definition 1.
Let P be a Galois point. Then, the set F P := { Q ∈ P n +1 | σ ( Q ) = Q for any σ ∈ G P ( X ) } \ { P } is a hyperplane. We call F P the fixed hyperplane at P .Note that F P is defined by X = 0 if the defining polynomial has the form in Lemma 3 (2).Now we mention independent Galois points, which is a useful notion to count the number of Galoispoints ([12, Definition 4]). Definition 2.
A set of Galois points { P , . . . , P r } is said to be independent (or, simply, points P , . . . , P r are said to be independent) if for any two points P i and P j (0 ≤ i, j ≤ r ) all the Galoispoints for X lying on the line P i P j are exactly P i and P j .We have the following lemma also for normal hypersurfaces by copying the proof of [12, Lemma 3]. Lemma 4. If P , . . . , P r are independent Galois points, then we can choose coordinates ( X , . . . , X n +1 ) satisfying X j ( P i ) = δ ji ( ≤ i ≤ r , ≤ j ≤ n + 1 ) and a generator σ i of G P i ( ≤ i ≤ r ) hasa representation as diag[ ζ, . . . , ζ, , ζ, . . . , ζ ] , where is in i -th position and ζ = e d − (resp. e d ).Especially we have r ≤ n + 1 . Distribution of Galois points for a cone variety.
Firstly, we mention the distribution ofGalois points for a cone variety. We does not assume the normality of X in this subsection, thatis, s ( X ) may be n −
1. Assume that X ⊂ P n +1 is a cone. Then, there exists a linear space M and M in P n +1 with M ∩ M = ∅ and M ♯M = P n +1 such that Y := X ∩ M is an irreduciblehypersurface and M is the maximal vertex (i.e. M is the vertex with maximal dimension). Let n − a be the dimension of M (with 0 ≤ n − a ≤ n − M isdefined by X a +1 = · · · = X n +1 = 0, M is defined by X = · · · = X a = 0 and X is defined by F ( X , . . . , X a ) = 0. Lemma 5.
Let P ∈ M and Q ∈ ( P ♯M ) \ M . Then, P is Galois for X if and only if Q is Galoisfor X . ALOIS POINTS FOR A NORMAL HYPERSURFACE 9
Proof.
By direct computations, π P and π Q induce the same function field extension. (cid:3) Lemma 6.
Let P ∈ M . Then, P is Galois for Y if and only if P is Galois for X .Proof. By taking suitable coordinates, we may assume that P = (1 : 0 : · · · : 0), and π P : X P n is given by ( X : X : · · · : X n +1 ) ( X : · · · : X n +1 ). Let ( x , x , . . . , x n +1 ) =( X /X , X /X , . . . , X n +1 /X ) be a system of affine coordinates. We have K ( P a − ) = k ( x , . . . , x a ), K ( P n ) = K ( P a − )( x a +1 , . . . , x n +1 ), K ( X ) = K ( P n )( x | X ) and K ( Y ) = K ( P a − )( x | Y ). Let p X ( x ) and p Y ( x ) be minimal polynomials of x | X over K ( P n ) and of x | Y over K ( P a − ), respec-tively. Since X and Y are given by the same equation F ( X , . . . , X a ) = 0, we see that p X ( x ) = p Y ( x ). Let L X and L Y be splitting fields for p X ( x ) and p Y ( x ), respectively. Then, by [1, Theo-rem 29], we have that Gal( L X /K ( P n )) is isomorphic to the subgroup of Gal( L Y /K ( P a − )) having L Y ∩ K ( P n ) as its fixed field. Here, we note that L Y ∩ K ( P n ) = K ( P a − ), since x a +1 , . . . , x n +1 aretranscendental over K ( P a − ). Namely, we have that Gal( L X /K ( P n )) is isomorphic to Gal( L Y /K ( P a − )).Especially, we have [ L X : K ( P n )] = [ L Y : K ( P a − )]. Hence, note that [ K ( X ) : K ( P n )] = [ K ( Y ) : K ( P a − )], we conclude that K ( X ) /K ( P n ) is Galois if and only if K ( Y ) /K ( P a − ) is Galois. (cid:3) It follows from Lemmas 5 and 6 that we have the following:
Proposition 3.
Let X be a cone described as above. Then, ∆( X ) = (∆( Y ) ♯M ) \ M and ∆ ′ ( X ) = (∆ ′ ( Y ) ♯M ) \ M . In particular, if X is a cone, then ∆( X ) and ∆ ′ ( X ) are empty or infinite respectively. Using this proposition, Theorem 3 will be proved by Propositions 4, 5 and 6 which will be provedin the next subsection.3.3.
Inner Galois points.
In this subsection, we consider only Galois points contained in X sm . Lemma 7. If l be a line lying on X , then the number of Galois points for X on l is zero, one,two or infinitely many. The last case occurs only if X is a cone.Proof. Suppose that there exist three Galois points P , P and P for X on l . Then, from Propo-sition 1, we may assume P = (1 : 0 : · · · : 0) and X is given by the equation X X d − + G ( X , . . . , X n +1 ) = 0 . Since l is contained in the tangent space at P , we may assume that l is given by the equation X = X = · · · = X n +1 = 0. The Gauss map γ is given by(( d − X X d − : X d − + ∂G/∂X : ∂G/∂X : · · · : ∂G/∂X n +1 ) . Here, we consider the restriction of γ to l , which is given by(0 : X d − + a X d − : da X d − : a X d − : · · · : a n +1 X d − )where a i ∈ K is the coefficient of X i X d − in G . Assume that the restriction γ | l is not a constantmap, i.e., there exists a non-zero coefficient a i for some i = 0 , . . . , n −
1. Then, the map γ | l : l → γ | l ( l ) ∼ = P is a finite morphism of degree d −
1. It is clear that the Galois point P is a ramification ALOIS POINTS FOR A NORMAL HYPERSURFACE 10 point of γ | l . So, other Galois point P and P must be also ramification points. However, thenumber of ramification points of γ | l must be two, this is a contradiction. So we have that γ | l is aconstant map. Especially, we have that T P X = T P X = T P X .By Remark 2, T P X ∩ X is a cone and P , P and P are its vertexes. We put that P = ( a :0 : 1 : 0 : · · · : 0), P = ( b : 0 : 1 : 0 : · · · : 0) ( a, b ∈ K ) and assume that a = 0. Then, calculatingthe local equation of T P X at P , we see that G (0 , , X , . . . , X n +1 ) is a homogeneous polynomialof degree d . Hence, we may assume that X is given by the equation X X d − + X ( G X d − + G X d − + · · · + G d − X ) + G d = 0 , where G i = G i ( X , X , X , . . . , X n +1 ) ( i = 0 , , · · · , d − , d ) is a homogeneous polynomial of degree i . Examining the condition (1) of Lemma 3 at P , we obtain that G = G = · · · = G d − = 0.Namely, X is given by the equation X X d − + G d ( X , X , X , . . . , X n +1 ) = 0 , and X is a cone with the vertex O = (0 : 0 : 1 : 0 : · · · : 0). (cid:3) Lemma 8.
Assume that d ≥ . If l is a line which does not lie on X , then the number of Galoispoints for X on l is at most one.Proof. Suppose that there exist two Galois points for X on the line l . Then, we denote them by P and P ′ . From Proposition 1, we may assume that P = (1 : 0 : · · · : 0) and X is given by theequation F = X X d − + G ( X , . . . , X n +1 ). Note that if P ′ ∈ X is on the hyperplane X = 0, then P ∈ T P ′ X . So, we infer from Remark 2 that P ′ is not on the hyperplane X = 0. Hence, we mayassume that P ′ = (1 : 1 : 0 · · · : 0) and l is given by the equations X = · · · = X n +1 = 0. Then,the local equation of X at P ′ is the following:( w + 1) d − + G (1 , u , . . . , u n +1 ) = 0 , where ( w, u , . . . , u n +1 ) = ( X /X − , X /X , . . . , X n +1 /X ) is a system of local coordinates.Putting G (1 , u , . . . , u n +1 ) = P di =0 g i , where g i = g i ( u , . . . , u n +1 ) ( i = 0 , , · · · , d ) is a homoge-neous polynomial of degree i . Examining Condition (1) in Lemma 3, we obtain that (cid:18) d − (cid:19) w + g = (cid:18) d − (cid:19) (( d − w + g ) (cid:0) d − (cid:1) w + g ( d −
1) (( d − w + g ) ! Comparing the coefficients of w of both sides, then we have a contradiction, if d ≥ (cid:3) Proposition 4. If d ≥ and ∆( X ) is infinite, then X is a cone.Proof. Suppose that ∆( X ) is an infinite set. Then, we infer from Lemma 4 that there exist threesmooth Galois points P , P , P which are collinear. By Lemma 8, the line l passing through P , P , P is contained in X . So, from Lemma 7, X is a cone. (cid:3) On the other hand, if d ≥
5, and if ∆( X ) is finite and non-empty, then the set ∆( X ) isindependent by Proposition 3 and Lemmas 7 and 8. Theorem 1 (II) is derived from the followinglemma which is a generalization of [12, Lemma 4]: ALOIS POINTS FOR A NORMAL HYPERSURFACE 11
Lemma 9.
Let m = [( n + s + 1) / . The cardinality of a set of independent inner Galois pointsis at most m + 1 . The equality holds if and only if X is projectively equivalent to the hypersurfacedefined by X m +1 X d − + · · · + X m − s X d − m − s − + A m − s X d − m − s + · · · + A m X d − m + G = 0 where A m − s , . . . , A m , G ∈ K [ X m +1 , . . . , X n +1 ] are homogeneous polynomials with deg A m − s = · · · = deg A m = 1 and deg G = d . (Note that the polynomial A m − s X d − m − s + · · · + A m X d − m does notappear if s = − .)Proof. Let P , . . . , P r form a set of independent inner Galois points. Suppose that r ≥ m +1. Thentake a system of coordinates ( X , . . . , X n +1 ) satisfying that X j ( P i ) = δ ji , where 0 ≤ i ≤ m + 1 and0 ≤ j ≤ n + 1. By Lemma 4, we can assume that σ i is a diagonal matrix diag[ ζ, . . . , ζ, , ζ, . . . , ζ ],where ζ = e d − . Since F σ = λ i F for λ i ∈ K \
0, we infer that F has the expression as F = A X d − + · · · + A m +1 X d − m +1 + G, where A i and G are forms in K [ X m +2 , . . . , X n +1 ], and deg A i = 1 and deg G = d . Let t :=dim h A , . . . , A m +1 i −
1. Then, 0 ≤ t ≤ n − m − ≤ m . We may assume that A , . . . , A t bea basis of the above vector space and A = X m +2 , . . . , A t = X m + t +2 . Then, A t +1 , . . . , A m +1 are represented by linear combinations of X m +2 , . . . , X m + t +2 . We consider the locus Γ which isdefined by X m +2 = · · · = X m + t +2 = ∂F∂X m +2 = · · · = ∂F∂X n +1 = 0. We find easily that the locus Γ iscontained in the singular locus of X . The dimension of Γ is at least ( n + 1) − ( t + 1 + n − m ) = m − t .Now, m − t ≥ m − n + 1 ≥ s + 1. This is a contradiction.Now assume that r = m . Similar to the above discussion, F has the expression as F = A X d − + · · · + A m X d − m + G, where A i and G are forms in K [ X m +1 , . . . , X n +1 ], and deg A i = 1 and deg G = d . Let t =dim h A , . . . , A m i −
1. Then, 0 ≤ t ≤ max { n − m, m } . We may assume that A , . . . , A t be a basis ofthe above vector space and A = X m +1 , . . . , A t = X m + t +1 . We consider the locus Γ which is definedby X m +1 = · · · = X m + t +1 = ∂F∂X m +1 = · · · = ∂F∂X n +1 = 0. We find easily that the locus Γ is containedin the singular locus of X . The dimension of Γ is at least ( n + 1) − ( t + 1 + n − m + 1) = m − t − s ≥ m − t −
1. We have t ≥ m − s −
1. Then, we have the assertion. (cid:3)
Now, we consider the case where d = 4. Lemma 10.
Assume that d = 4 . Let P ∈ X be a Galois point, σ a generator of G P . (1) If a smooth point Q on X \ T P X is Galois then σ ( Q ) = Q . (2) Let P := P, P , P , P be distinct points in ∆( X ) . Further, we assume that these points arecollinear and the line l passing through these points is not contained in X . Then, by taking asuitable projective transformation, we can express X as X X + X + H ( X , . . . , X n +1 ) = 0 ,where H ( X , . . . , X n +1 ) is a homogeneous polynomial of degree four. (3) Assume the same assumptions in (2) and that X is not a cone. Then, we have that ∆( X ) \ T P X = { P , P , P } . ALOIS POINTS FOR A NORMAL HYPERSURFACE 12
Proof.
Let us prove the assertion (1). Suppose that σ ( Q ) = Q . By the assumtion, the line P Q X .Therefore, P and Q are independent Galois points. By Proposition 1, we may assume that P =(1 : 0 : · · · : 0) and Q = (0 : 1 : 0 : · · · : 0), and X is given by A X + A X + G ( X , . . . , X n +1 ) = 0where A , A ∈ K [ X , . . . , X n +1 ]. The tangent space at P is defined by A = 0 and A ( Q ) = 0.This contradicts the assumption.Let us prove the assertion (2). By Proposition 1, we may assume that P = (1 : 0 : · · · : 0), X is given by the equation X X + G ( X , . . . , X n +1 ) = 0. The point P is not contained in T P X nor F P . Therefore, we may assume that P = ( a : 1 : 0 : · · · : 0) and l is given by theequations X = · · · = X n +1 = 0. Let G ( X , . . . , X n +1 ) = G X + G X + G X + G X + G where G i = G i ( X , . . . , X n +1 ) is a homogeneous polynomial of degree i and G = − a . Let w := X /X − a , u := X /X , . . . , u n +1 := X n +1 /X . Then, F ( w + a, , u , . . . , u n +1 ) = ( w + a ) + ( − a ) + G + G + G + G = 0 . By applying Corollary 3 at P , we have(3 a + G ) = 3(3 a w + G )( w + G )as a polynomial. This implies that G = G = G = 0 and hence, the assertion.Let us prove the assertion (3). By the assertion in (2), we may assume that P = (1 : 0 : · · · : 0), X is given by X X + X + H ( X , . . . , X n +1 ) = 0, and l is given by the equations X = · · · = X n +1 = 0. Suppose that there exists another point Q in ∆( X ) \ T P X . By taking a suitableprojective transformation, we may assume that Q = ( a : 1 : 1 : 0 : · · · : 0). We put that H ( X , . . . , X n +1 ) = H X + H X + H X + H X + H , where H i = H i ( X , . . . , X n +1 ) is a homogeneous polynomial of degree i , and H = − a −
1. Let w := X /X − a , w := X /X − u = X /X , . . . , u n +1 := X n +1 /X . Then, F ( w + a, w +1 , , u , . . . , u n +1 ) = ( w +1)( w + a ) +( w +1) +( − a − H + H + H + H = 0 . By applying Corollary 3 at P , we have( a w + 3 a w + 4 w + H )(3 aw w + w + 4 w + H ) = 3(3 a w w + 3 aw + 6 w + H ) as a polynomial. This implies that H = H = H = 0 and a = −
1. Therefore, F = X X + X + H with H ∈ K [ X , . . . , X n +1 ] and X is a cone. (cid:3) Proposition 5. If d = 4 and ∆( X ) is infinite, then X is a cone.Proof. Assume that X is not a cone. It follows from Lemma 7 that if a line l contains three Galoispoints, then l X . Let P ∈ ∆( X ). Then, we have ∆( X ) \ F P ⊂ ∆( X ) \ T P X , because the line P Q contains four Galois points if Q ∈ ∆( X ) \ F P . It follows from Lemma 10 (3) that ∆( X ) \ F P is a finite set.Let P be a point in ∆( X ). Since ∆( X ) \ F P is finite, we can take a point P in ∆( X ) ∩ F P .Then, we infer that the set ∆( X ) \ ( F P ∩ F P ) = (∆( X ) \ F P ) ∪ (∆( X ) \ F P ) is finite and theset ∆( X ) ∩ ( F P ∩ F P ) is infinite. Hence, we can take a point P in ∆( X ) ∩ ( F P ∩ F P ). In thisway, we can take infinitely many points P , P , . . . such that P i ∈ ∆( X ) ∩ ( F P ∩ · · · ∩ F P i − ). ALOIS POINTS FOR A NORMAL HYPERSURFACE 13
Note that dim( F P ∩ · · · ∩ F P i ) = dim( F P ∩ · · · ∩ F P i − ) −
1, because P i F P i . We have that F P ∩ · · · ∩ F P n +2 = ∅ , thus we have a contradiction. (cid:3) Now we consider the case where ∆( X ) is finite and non-empty. Then, X is not a cone byProposition 3 in Subsection 3.2. Below in this subsection, we define r = r ( X ) := max ♯ { P , . . . , P b : independent Galois points on X } − P , . . . , P r form a set of independent Galois points. (Then, r ≤ m and r < m is possible.)We also define µ := ♯ { l : line | l contains four Galois points } − . We have that − ≤ µ ≤ r by the following Lemma: Lemma 11.
Let P , . . . , P r be independent Galois points and let Q ∈ X be a Galois point distinctfrom P , . . . , P r . Assume that X is not a cone. Then, (1) There exists i such that ♯ ( P i Q ∩ ∆( X )) = 4 . (2) If l , l are lines such that Q ∈ l i and ♯ ( l i ∩ ∆( X )) = 4 for i = 1 , , then l = l .Proof. Assume that the line P i Q consists exactly two Galois points for any i . Then, P , . . . , P r , Q are independent Galois points. This contradicts the assumption on r . We have (1). The assertionin (2) is derived from Lemma 10 (3). (cid:3) Let P , . . . , P µ be Galois points with lines l , . . . , l µ , where P i ∈ l i and l i contains four Galoispoints. Using Corollary 3, Lemmas 10 and 11, for a suitable coordinate, we find that X is definedby X µ +1 X + · · · + X µ +1 X µ + X µ +1 + · · · + X µ +1 + G = 0where G ∈ K [ X µ +2 , . . . , X n +1 ] (cf. [12, p. 532]). Then, by direct computations, the singularlocus is contained in a linear space defined by X = · · · = X µ +1 = 0. This implies that s ≥ ( n + 1) − (2 µ + 2) and hence µ ≤ ( n + s + 1) / − s −
1. Therefore, we have the following:
Lemma 12. µ ≤ m − s − .Proof of Theorem 1 (I-0)–(I-4). (0) Now, the number of inner Galois points is 4( µ + 1) + ( r − µ ).Note that µ ≤ r ≤ m and µ ≤ m − s −
1, by Lemmas 9 and 12. Therefore, we have an upperbound 4( m − s ) + ( s + 1).(1) We consider the case where n + s is odd. Then, 2 m − s = n + 1.(1-1) Assume that µ = m − s −
1. Then, X is projectively equivalent to the hypersurface definedby X m − s X + · · · + X m − s − X m − s − + X m − s + · · · + X m − s − + G = 0where G ∈ K [ X m − s , . . . , X n +1 ]. By direct computations, the singular locus of X is defined by X = · · · = X m − s − = ∂G∂X m − s = · · · = ∂G∂X n +1 = 0 . We have ∂G∂X m − s = · · · = ∂G∂X n +1 = 0 ALOIS POINTS FOR A NORMAL HYPERSURFACE 14 as a polynomial by counting the dimension of the singularity (and that X is not a cone). Then, G = 0 as a polynomial. This implies s = − s ≥ µ = m − s − r = m . Then, X is projectively equivalent to thehypersurface defined by X m +1 X + · · · + X m − s − X m − s − + A m − s − X m − s − + · · · + A m X m + X m +1 + · · · + X m − s − + G = 0where A m − s − , . . . , A m , G ∈ K [ X m − s ]. Therefore, we may assume that A m − s − = · · · = A m = X m − s and G = aX where a = 0 or 1. Then, ♯ ∆( X ) = 4( µ +1)+( m − µ ) = 4( m − s − s +2) =4( m − s ) + ( s − n + s is even.(2-1) Assume that µ = m − s −
1. Then X is projectively equivalent to the hypersurface definedby X m − s X + · · · + X m − s − X m − s − + X m − s + · · · + X m − s − + G = 0where G ∈ K [ X m − s , . . . , X n +1 ]. Assume that r ≥ µ + 1. Then, s ≥ G = X m − s +1 X m − s + G where G ∈ K [ X m − s +1 , . . . , X n +1 ]. The singular locus is defined by X = · · · = X m − s − = X m − s X m − s +1 = X m − s + ∂G ∂X m − s +1 = ∂G ∂X m − s +2 = · · · = ∂G ∂X m − s +1 = 0 . Then, by counting the dimension of the singularity of X , we should have G = 0 as a polynomialand hence, s = 0 and r = µ + 1 = m . Then, ♯ ∆( X ) = 4 m + 1. On the other hand, if s = 0, then r = µ and ♯ ∆( X ) = 4( m − s ). Then, by counting the dimension of the singularity of X , G shouldhave a multiple component.Using the result in (1-1), we find that the bound 4( m − s ) + ( s + 1) is sharp if and only if s = − s = 0 and n is even.(2-2) Assume that s ≥ µ = m − s − r = m . Then, X is projectively equivalent to thehypersurface defined by X m +1 X + · · · + X m − s − X m − s − + A m − s − X m − s − + · · · + A m X m + X m +1 + · · · + X m − s − + G = 0where A m − s − , . . . , A m , G ∈ K [ X m − s , X n +1 ]. Then, ♯ ∆( X ) = 4( µ + 1) + ( m − µ ) = 4( m − s −
1) +( s + 2) = 4( m − s ) + ( s − s ≥ µ = m − s − r = m , as in Example 1 below. (cid:3) We define t := n − dim T ≤ i ≤ r T P i X . We consider examples when n + s is even. Example 1.
Let s ≥ n + s is even.(i) A hypersurface in P n +1 defined by X m +1 X + · · · + X m − s X m − s − + X m − s +1 X m − s + X m − s ( X m − s +1 + · · · + X m )+ X m +1 + · · · + X m − s − = 0satisfies that dim Sing( X ) = s , r = m , t = m − s and µ = m − s − ALOIS POINTS FOR A NORMAL HYPERSURFACE 15 (ii) A hypersurface in P n +1 defined by X m +1 X + · · · + X m − s X m − s − + X m − s ( X m − s + · · · + X m )+ X m +1 + · · · + X m − s − + X m − s +1 = 0satisfies that dim Sing( X ) = s , r = m , t = m − s − µ = m − s − Outer Galois points.Lemma 13.
Let l ⊂ P n +1 be a line. Then, the cardinality of the set ∆ ′ ( X ) ∩ l is zero, one, twoor infinity. The last case occurs only if X is a cone.Proof. Let P , P , P ∈ ∆ ′ ( X ) ∩ l be distinct three outer Galois points and let σ ∈ G P be agenerator. From Proposition 2, we may assume that P = (1 : 0 : · · · : 0), X is defined by theequation X d + G ( X , . . . , X n +1 ) = 0 , where G ( X , . . . , X n +1 ) is a homogeneous polynomial of degree d , and σ = diag[ e d , . . . , e d , l is defined by X = · · · = X n +1 = 0. Note that l ∩ F P consists onlyone point. Therefore, P or P is not contained in F P at P . We may assume that P is so and P = (1 : 1 : 0 : · · · : 0), since F P is given by X = 0. Then, the local equation of X at P is thefollowing: ( w + 1) d + G (1 , u , . . . , u n +1 ) , where ( w, u , . . . , u n +1 ) = ( X /X − , X /X , . . . , X n +1 /X ) is a system of local coordinates. Weput that G (1 , u , . . . , u n +1 ) = P di =0 g i , where g i = g i ( u , . . . , u n +1 ) is a homogeneous polynomialof degree i . Examining the condition (1) in Lemma 3, we obtain that g i = (cid:18) di (cid:19) (cid:18) ( dw + g ) i d i (1 + g ) i − − w i (cid:19) , i = 1 , . . . , d − . Calculating the coefficients of w and w in the equation where i = 2, we see that g = 0 and g = 0.Hence, we see that g = · · · = g d − = 0. Thus, we conclude that X is defined by the following: X d + g d ( X , . . . , X n +1 ) = 0 , and X is a cone with a vertex Q = (0 : 1 : 0 : · · · : 0). (cid:3) Proposition 6. If ∆ ′ ( X ) is infinite, then X is a cone.Proof. Assume that ∆ ′ ( X ) is infinite. Then, by lemma 4, there exists three outer Galois pointswhich are collinear. By Lemma 13, we have that X is a cone. (cid:3) Lemma 14.
The cardinality of a set of independent outer Galois points is at most n − s + 1 . Theequality holds if and only if X is projectively equivalent to the hypersurface defined by X d + · · · + X dn − s = 0 . Therefore, X is smooth or a cone. ALOIS POINTS FOR A NORMAL HYPERSURFACE 16
Furthermore, if s ≥ and X is not a cone, then the cardinality of a set of independent outerGalois points is at most n − s . Then, the equality holds if and only if X is projective equivalent tothe hypersurface defined by X d + · · · + X dn − s − + G = 0 where G ∈ K [ X n − s , . . . , X n +1 ] has a multiple component.Proof. Let P , . . . , P r are independent outer Galois points. Then take a system of coordinates( X , . . . , X n +1 ) satisfying that X j ( P i ) = δ ji , where 0 ≤ i ≤ m + 1 and 0 ≤ j ≤ n + 1. By Lemma4, we can assume that σ i is a diagonal matrix diag[ ζ, . . . , ζ, , ζ, . . . , ζ ], where ζ = e d − . Since F σ = λ i F for λ i ∈ K \
0, we infer that F has the expression as F = X d + · · · + X dr + G, where G ∈ K [ X r +1 , . . . , X n +1 ]. Then, the singular locus is defined by X = · · · = X r = ∂G∂X r +1 = · · · = ∂G∂X n +1 = 0 . Therefore, s ≤ n + 1 − ( r + 1). The equality holds if and only if G = 0 as a polynomial, hence X is smooth or a cone.Assume that s ≥ X is not a cone. Then, r ≤ n , there exists j with r + 1 ≤ j ≤ n + 1 suchthat ∂G∂X j = 0 as a polynomial, and hence s ≤ n + 1 − ( r + 2). The equality holds if and only if G has a multiple component. (cid:3) Proof of Theorem 2.
Assume that ∆ ′ ( X ) is finite and non-empty. Then, X is not a cone byProposition 3 in Subsection 3.2, and the set ∆ ′ ( X ) is independent by Lemma 13. Therefore, wehave the conclusion by Lemma 14. (cid:3) Galois points for a Fermat hypersurface of degree p e + 1 in p > p > q ≥ p . We consider the Fermat hypersurface F n ( q + 1) ⊂ P n +1 of degree q + 1:(1) X q +10 + X q +11 + · · · + X q +1 n +1 = 0 . Firstly, we prove the “if” part of Theorem 5. Homma gave an elegant proof if n = 1 by usingautomorphisms of Hermitian curve ([5, Claims 1 and 2]). We give an elementary proof. Proposition 7. If P ∈ P n +1 is F q -rational, then P is Galois for F n ( q + 1) .Proof. Let P = (1 : a : · · · : a n +1 ) and π P = ( X − a X : · · · : X n +1 − a n +1 X ) be the projectionfrom P . Then, we have a field extension K ( x , . . . , x n ) /K ( x , . . . , x n ) with f ( x , . . . , x n ) = (1 + a q +11 + · · · + a q +1 n + a q +1 n +1 ) x q +10 + ( a q x + · · · + a qn x n + a qn +1 ) x q + ( a x q + · · · + a n x qn + a n +1 ) x +( x q +11 + · · · + x q +1 n + 1) = 0.If P ∈ X and P is F q -rational, then f ( x , . . . , x n ) = ( a q x + · · · + a qn x n + a qn +1 ) x q + ( a q x + · · · + a qn x n + a qn +1 ) q x + ( x q +11 + · · · + x q +1 n + 1) = 0. It is not difficult to check that the extensionis Galois. ALOIS POINTS FOR A NORMAL HYPERSURFACE 17
We assume that P ∈ P n +1 \ X and P is F q -rational. Let β ( x , . . . , x n ) = ( a q x + · · · + a qn x n + a qn +1 ) / (1 + a q +11 + · · · + a q +1 n + a q +1 n +1 ) and ˆ x = x + β . Then, we have 0 = f ( x , x , . . . , x n ) = f ( ˆ x − β, x , . . . , x n ) = (1+ a q +11 + · · · + a q +1 n + a q +1 n +1 ) ˆ x q +1 + g ( x , . . . , x n ). Therefore, our extensionis cyclic. (cid:3) Next, we prove the “only-if” part. We will use the result of Homma [5, Claim 3] with n = 1,because we will use induction on the dimension n . Proposition 8. If P ∈ P n +1 is a Galois point for F n ( q + 1) , then P is F q -rational.Proof. Let P = (1 : a : · · · : a n +1 ). We may assume that i is an integer such that a i = 0 and1 + a q +1 i = 0 if and only if i ≤ i . Note that a i ∈ F q for any i > i because a i = 0 or 1 + a q +1 i = 0.Let 1 ≤ i ≤ i and let H i be a hyperplane defined by X i − a i X . Then, X ∩ H i is defined by(1 + a q +1 i ) X q +10 + · · · + X q +1 i − + X q +1 i +1 + · · · + X q +1 n +1 = 0 . We take a linear transformation φ i on H i ∼ = P n defined as follows: (1 : x : · · · : x i − : x i +1 : · · · : x n +1 ) (1 : x / q +1 q a q +1 i : · · · : x i − / q +1 q a q +1 i : x i +1 / q +1 q a q +1 i : · · · : x n +1 / q +1 q a q +1 i ). Then φ i ( X ∩ H i ) is the Fermat hypersurface in H i ∼ = P n . Note that H i satisfies the condition ( ⋆ ) in Theorem 4.We will prove the assertion by induction on the dimension n . If n = 1, then the assertion holdsby a result of Homma [5]. We assume that n ≥ P is a Galois point. It follows from Theorem4 (ii) that P is also Galois for X ∩ H i . By the assumption of the induction, φ i ( P ) is a F q -rationalpoint. Now we have φ i ( P ) = (1 : a / q +1 q a q +1 i : · · · : a n +1 / q +1 q a q +1 i ) . Therefore, ( a j / q +1 q a q +1 i ) q − = 1 for any i = j with 1 ≤ i, j ≤ i . By direct computations, wehave a i ∈ F q for any i . (cid:3) Examples
We give examples of hypersurfaces which have infinitely many Galois points and are not cones,in p > Example 2.
Let X ⊂ P be a hypersurface defined by F = ZW p − X p W − Y p +1 = 0 . Let P = (0 : 0 : 1 : 0), Q = (1 : 0 : 0 : 0) and let L be a line defined by Y = W = 0. Then, we havethe followings:(i) Sing( X ) = { P } .(ii) X is not a cone.(iii) L \ { P, Q } ⊂ ∆( X ). ALOIS POINTS FOR A NORMAL HYPERSURFACE 18
Proof.
Note that X is the closure of the image of a morphism φ : A → P ; ( x, y ) ( x : y : x p + y p +1 : 1) . The Gauss map γ is given by( ∂F/∂X : ∂F/∂Y : ∂F/∂Z : ∂F/∂W ) = (0 : − Y p : W p : − X p ) . We have (i) by the form of the Gauss map. We also have (ii) because γ is generically finite by that γ ◦ φ = (0 : − y p : 1 : − x p ). Now, we prove (iii). Let R = (1 : 0 : a : 0) and π R = ( Z − aX : Y : W )be the projection. Then, we have the field extension K ( x, y, z ) /K ( y, z ) with x p − ax − z + y p +1 = 0 . Therefore, this extension is Galois if a = 0. (cid:3) Example 3.
Let X ⊂ P be a hypersurface defined by F = ZW p − − X p W p − p − Y p = 0 . Let L be a line Y = W = 0, L be a line Z = W = 0 and H be a plane defined by W = 0. Then,we have the followings:(i) Sing( X ) = L .(ii) X is not a cone.(iii) H \ ( L ∪ L ) ⊂ ∆ ′ ( X ). Proof.
Note that X is the closure of the image of a morphism φ : A → P ; ( x, y ) ( x : y : x p + y p : 1) . The Gauss map γ is given by( ∂F/∂X : ∂F/∂Y : ∂F/∂Z : ∂F/∂W ) = (0 : 0 : W p − : − ZW p − ) . We have (i) by the form of the Gauss map. Now, we have γ ◦ φ ( x, y ) = (0 : 0 : 1 : − x p − y p ) . Therefore, the general fiber of the Gauss map is an irreducible curve defined by x + y p = c (for a suitable constant c ). Then, we have (ii). We prove (iii). Let R = (1 : a : b : 0) and π R = ( Z − aX : Y − bX : W ) be the projection. Then, we have the field extension K ( x, y, z ) /K ( y, z )with b p x p + x p − ax + z − y p +1 = 0 . Therefore, this extension is Galois if b = 0. (cid:3) Acknowledgements
The authors are grateful to Professor Hisao Yoshihara for helpful discussions. In particular, Yoshi-hara informed the first author an idea of the proof of Theorem 4 when p = 0, s ( X ) ≤ H is general. The first author was supported by Research Fellowships of the Japan Society for thePromotion of Science for Young Scientists. ALOIS POINTS FOR A NORMAL HYPERSURFACE 19
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E-mail address : [email protected] Division of General Education, Nagaoka National College of Technology, 888 Nishikatakai, Na-gaoka, Niigata 940-8532, Japan
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