Curves containing all points of a finite projective Galois plane
aa r X i v : . [ m a t h . AG ] J u l Curves containing all points of a finite projective Galois plane
Gregory Duran Cunha
Abstract
In the projective plane
P G (2 , q ) over a finite field of order q , a Tallini curve is a plane irreducible (al-gebraic) curve of (minimum) degree q +2 containing all points of P G (2 , q ). Such curves were investigatedby G. Tallini [8, 9] in 1961, and by Homma and Kim [5] in 2013. Our results concern the automorphismgroups, the Weierstrass semigroups, the Hasse-Witt invariants, and quotient curves of the Tallini curves. Keywords: algebraic curve, finite field, automorphism, Weierstrass semigroup
A fundamental result on plane irreducible (algebraic) curves defined over a finite field F q is the Hasse-Weilbound S q ≤ q + 1 + ( n − n − √ q where n is the degree of the curve and S q is the number of its points lying in the projective plane P G (2 , q )of order q ; see [4, Section 9.6]. Any plane (possibly reducible) curve containing all points of P G (2 , q ) hasdegree at least q + 1, and if equality holds then the curve splits into the q + 1 lines of a pencil in P G (2 , q ). Acomplete classification of plane irreducible curves of degree q + 2 containing all points of P G (2 , q ) was givenby G. Tallini [8, 9]; see also [1], and [5]. Up to projective transformations in P G (2 , q ), each such curve X has homogeneous equation of type( aX + bX + cX ) ϕ − X ϕ + X ϕ = 0 , (1)where ϕ ij = X qi X j − X i X qj and a, b, c are elements in F q such that the cubic equation X − cX − aX − b = 0 (2)is irreducible over F q . In this paper, the above irreducible curve X of degree n = q + 2 is named Tallinicurve .G. Tallini proved that X has no singular points in P G (2 , q ). Homma and Kim [5, Section 3] extended hisresult to any point in P G (2 , K ) where K is the algebraic closure of F q . Therefore, X is a plane nonsingularcurve of genus g = ( n − n −
2) = q ( q + 1).G. Tallini showed that the automorphism group G q of X over F q contains a Singer cycle, that is, a cyclicsubgroup S of P GL (3 , q ) of order q + q + 1 acting on P G (2 , q ) as a regular permutation group. He alsoclaimed that G q may be a bit larger but only for some special curves, named harmonic and equianharmoniccurves in [8, 9]. More precisely, Homma and Kim proved [5, Theorem 5.4] that if G q with q > S then G q is the normalizer of S in P GL (3 , q ), that is, G q = S ⋊ C , the semidirect product of S by agroup C of order 3. 1n this paper we go on with the study of the Tallini curves, also from the function field point of view.We look at the Tallini curves in the projective plane P G (2 , K ) defined over the algebraic closure K of F q .Our Theorem 2.2 shows that up to projective equivalence in P G (2 , K ), the Tallini curve X is projectivelyequivalent to the curve X q +11 X + X q +12 X + X q +10 X = 0 . (3)For q = 2, X q is the famous plane Klein quartic whose automorphism group is isomorphic to P SL (2 , X q was first investigated in [7], and we refer to it as the Pellikaan curve . From theproof of Theorem 2.2, the smallest projective plane
P G (2 , q i ) where this equivalency occurs is in generalmuch larger than P G (2 , q ) as F q i turns out to be the smallest overfield of F q containing the roots of theequation X q + q +1 = ( α q − α ) q + q − where α is a root of (2). Here i divides q + q + 1 and the automorphismgroup of X in P G (2 , K ) is isomorphic to S ⋊ C where S is defined over F q but C is in general defined over F q i .For every divisor d of q + q + 1, the curve X q has a quotient curve X q /C d with respect to a cyclicgroup C d of order d . In case where q is a square, that is, q = p i with p prime and i ≥
1, the factorization p i + p i + 1 = ( p i + p i + 1)( p i − p i + 1) raises the question whether the quotient curve X q /C d with d = p i − p i + 1 is isomorphic to X p i . The answer is affirmative, see Theorem 5.3.We also show that X p is an ordinary curve, that is, its genus g = p ( p + 1) coincides with its Hasse-Witt invariant. For this purpose, we prove that no exact differential of K ( X p ) is regular, and then use theproperties of the Cartier operator to show that X p is ordinary. It should be noticed that this result does nothold true for q > p ; see [6]. P G (2 , q ) Let X be an irreducible plane curve of degree q + 2 defined over F q containing all points of P G (2 , q ). In hispaper [8], Tallini proved that such a curve exists and it has equation (1).Now, we recall some facts from [8]. The polar net of X is a net of conics and it is easy to see that this netis homaloidal with the three base points, namely A = ( α : 1 : α ), A = ( α : 1 : α ) and A = ( α : 1 : α )where α , α , α are the three solutions of (2) in F q . In particular, the points A i are conjugate over F q .Furthermore, they belong to X and they are simple points for it. We call each of these three points a basepoint of X .Also in [8], Tallini showed that the automorphism group of X contains a Singer cycle, that is, a cyclicsubgroup S of P GL (3 , q ) of order q + q + 1 acting on P G (2 , q ) as a regular permutation group. Moreover, S , viewed as a subgroup of P GL (3 , q ), fixes A , A , A .The following result was proved in [5]. Theorem 2.1.
The curve X is non-singular.Proof. If P is a singular point of X then its orbit under S = h φ iO = { φ i ( P ) | ≤ i ≤ q + q } consists of singular points of X . Note that the size of O is 1 or q + q + 1. Since the number of singularpoints of X is at most q ( q + 1) /
2, then O = { P } . This yields that every singular point of X is fixed by S .Since S fixes only A , A and A , which are simple points, it follows that X has no singular points.2 heorem 2.2. Any Tallini curve is projectively equivalent to the Pellikaan curve over F q q q +1) .Proof. Let α , α , α ∈ F q be the distinct solutions of (2), and let Y be the image of X under the linearmap associated to the non-singular matrix M = α α α α α α . That is, Y is the curve given by G ( X , X , X ) = 0, where G = F (cid:16) X i =0 α i X i , X i =0 X i , X i =0 α i X i (cid:17) , (4)and X = v ( F ) is the Tallini curve. A straightforward computation gives G = c X q +10 X + c X q +10 X + c X q +11 X + c X q +11 X + c X q +12 X + c X q +12 X , where c ij = ( α i − α j ) ( α qi − α i )( α j − α qi ) , for 0 ≤ i, j ≤ . Note that c ij = 0 whenever α j = α qi . Since the Frobenius map acts transitively on { α , α , α } , it followsthat either ( α , α , α ) = ( α q , α q , α q ) or ( α , α , α ) = ( α q , α q , α q ). In the former case, we have c = c = c = 0 and then G = c X q +10 X + c X q +11 X + c X q +12 X , (5)whereas the latter case gives c = c = c = 0 and G = c X q +10 X + c X q +11 X + c X q +12 X . (6)We prove the result in the case G is given by (5), and then case (6) will follow analogously. Note that from( α , α , α ) = ( α q , α q , α q ), equation (5) can be written as( α q − α ) X q +10 X + ( α q − α ) q X q +11 X + ( α − α q ) X q +12 X = 0 . (7)Finally, one can easily check that the curve given by (7) is the image of X q under the transformation( X , X , X ) ( µX , λX , X )where µ, λ ∈ F q q q +1) satisfy λ q + q +1 = ( α q − α ) ( α q − α ) q + q +1 and µ = λ q +1 ( α q − α q ) α − α q . This finishes the proof. 3
Weierstrass semigroup at a base point
Let Σ = K ( x, y ), with xy q +1 + x q +1 + y = 0, be the function field of the Pellikaan curve X q and consider thefundamental triangle O = (0 : 0 : 1) , X ∞ = (1 : 0 : 0) , Y ∞ = (0 : 1 : 0) . The tangent lines to X q at O , X ∞ and Y ∞ are l Y = v ( Y ), l Z = v ( Z ) and l X = v ( X ), respectively. Notethat the points in l Y ∩ X q are O and X ∞ with I ( O, l Y ∩ X q ) = q + 1 and I ( X ∞ , l Y ∩ X q ) = 1 , the points in the intersection l Z ∩ X q are X ∞ and Y ∞ with I ( X ∞ , l Z ∩ X q ) = q + 1 and I ( Y ∞ , l Z ∩ X q ) = 1 , and the points in the intersection l X ∩ X q are Y ∞ and O with I ( Y ∞ , l X ∩ X q ) = q + 1 and I ( O, l X ∩ X q ) = 1 . From [4, Theorem 6.42] the principal divisor of x is given by( x ) = l X · X q − l Z · X q = I ( O, l X ∩ X q ) O + I ( Y ∞ , l X ∩ X q ) Y ∞ − I ( X ∞ , l Z ∩ X q ) X ∞ − I ( Y ∞ , l Z ∩ X q ) Y ∞ = O + ( q + 1) Y ∞ − ( q + 1) X ∞ − Y ∞ = O + qY ∞ − ( q + 1) X ∞ and the principal divisor of y is given by( y ) = l Y · X q − l Z · X q = I ( O, l Y ∩ X q ) O + I ( X ∞ , l Y ∩ X q ) X ∞ − I ( X ∞ , l Z ∩ X q ) X ∞ − I ( Y ∞ , l Z ∩ X q ) Y ∞ = ( q + 1) O + X ∞ − ( q + 1) X ∞ − Y ∞ = ( q + 1) O − qX ∞ − Y ∞ . For 1 ≤ n ≤ q + 1, the above relations gives (cid:16) yx n (cid:17) = ( q + 1 − n ) O + ( n ( q + 1) − q ) X ∞ − ( nq + 1) Y ∞ , hence the divisor of poles of y/x n is (cid:16) yx n (cid:17) ∞ = ( nq + 1) Y ∞ , for 1 ≤ n ≤ q + 1 . Theorem 3.1.
The Weierstrass semigroup at a base point of X q is the semigroup generated by q + 1 , q +1 , . . . , ( q + 1) q + 1 .Proof. We may assume that the base point is Y ∞ . From the above discussion, q + 1 , q + 1 , . . . , ( q + 1) q + 1belong to the Weierstrass semigroup H ( Y ∞ ). Let H = h q + 1 , q + 1 , . . . , ( q + 1) q + 1 i be the semigroupgenerated by q + 1 , q + 1 , . . . , ( q + 1) q + 1. Since H ⊂ H ( Y ∞ ) and the number of gaps in H is q ( q + 1) / H ( Y ∞ ), the assertion follows.4 heorem 3.2. If the function field of X q is given by K ( x, y ) , with xy q +1 + x q +1 + y = 0 , then the divisorof the differential dx is ( dx ) = ( q + 2 q ) Y ∞ − ( q + 2) X ∞ . Proof.
The curve X q is constituted by two points in the infinity X ∞ , Y ∞ and the affine points P = ( a : b : 1),with ab q +1 + a q +1 + b = 0. The tangent line to X q at an affine point P = ( a, b ) is not vertical, in fact, supposeby contradiction that X = 0 is the tangent to X q = v ( F ) at P . Then ∂F∂Y = 0 in the point P = ( a, b ). Itmeans that ab q + 1 = 0 and hence a = 0. On the other hand, ab q +1 + a q +1 + b = 0 becomes − b + a q +1 + b = 0,and therefore a = 0, a contradiction. Thus, a primitive parametrization of X q at P is given by x ( t ) = a + t,y ( t ) = b + b t + · · · , b = b. Therefore, ord P dx = 0, for all affine point P in X q . So it follows that ( dx ) = nX ∞ + mY ∞ with n + m = 2 g − g = q ( q + 1) / X q . Since ( x ) = qY ∞ + O − ( q + 1) X ∞ and q + 1 is not divisibleby p , ord X ∞ dx = − ( q + 2), that is, n = − ( q + 2). From n + m = 2 g − m = q + 2 q . Again, let K ( X q ) = K ( x, y ), with xy q +1 + x q +1 + y = 0, be the function field of the Pellikaan curve X q andlet λ ∈ K be a primitive ( q + q + 1)-root of unity and define the linear collineations σ : ( X , X , X ) ( X , λX , λ q +1 X ) . and τ : ( X , X , X ) ( X , X , X ) . It is straightforward to see that the automorphism group of X q in P G (2 , K ) contains S ⋊ C as a subgroup,where S = h σ i and C = h τ i .Observe that S has order q + q + 1. Therefore, going back to the original equation (1) of X , S becomesa Singer cycle of P G (2 , q ).To show that S ⋊ C is actually the whole automorphism group of X q over K , we need some lemmas. Lemma 4.1. If Q is a point in X q and t Q is its tangent line, then I ( Q, X q ∩ t Q ) = (cid:26) q + 1 , if Q ∈ { X ∞ , Y ∞ , O } , if Q
6∈ { X ∞ , Y ∞ , O } . Proof.
We have this result for the points in the fundamental triangle, see section 3. Suppose Q = ( a, b ) with ab q +1 + a q +1 + b = 0 and ab = 0. The tangent line t Q to X q at Q is given by T ( X, Y ) = ba X + a q +1 b Y + ab q +1 = 0 . A primitive parametrization of X q at Q is given by x ( t ) = a + ty ( t ) = b − b a q +2 t − b q +3 a q +3 t − b q +4 a q +4 t − · · · . Therefore T ( x ( t ) , y ( t )) = − ( b/a ) q +2 t + · · · has order 2, that is, I ( Q, X q ∩ t Q ) = 2.5 emma 4.2. Every automorphism in
Aut( X q ) preserves the triangle { X ∞ , Y ∞ , O } .Proof. Let α ∈ Aut( X q ) and suppose α : P Q with P ∈ { X ∞ , Y ∞ , O } and Q / ∈ { X ∞ , Y ∞ , O } . Considerthe lines t Q and l Q given by t Q : T ( X, Y ) = 0 l Q : L ( X, Y ) = 0such that t Q is the tangent line to X q at Q and l Q is a secant line through Q . Consider the curve C of degree q − T ( X, Y ) L ( X, Y ) q − = 0 . By the Lemma 4.1, I ( Q, X q ∩ C ) = I ( Q, X q ∩ t Q ) + ( q − I ( Q, X q ∩ l Q ) = 2 + ( q −
2) = q. Observe that W := X q · C is a canonical divisor such that L ( W − qQ ) = L ( W − ( q + 1) Q ). Thus, byRiemann-Roch Theorem, ℓ (( q + 1) Q ) = ℓ ( qQ ). Hence q + 1 is a gap number at Q , but this is a contradictionas q + 1 is a non-gap at P . Theorem 4.3.
Aut( X q ) = S ⋊ C .Proof. Let α ∈ Aut( X q ). Since X q is non-singular, α can be represented as a matrix A in P GL (3 , K ). By theLemma 4.2, α preserves the fundamental triangle. First, suppose that α fixes all vertices of the fundamentaltriangle, then A = ξ η
00 0 1 . Since α preserves X q , ξη q +1 xy q +1 + ξ q +1 x q +1 + ηy = 0in K ( X q ) = K ( x, y ). Hence η = ξ q +1 and ξ q + q +1 = 1. Therefore α ∈ S . Now, suppose that α fixes novertices of the fundamental triangle. In that case, α = τ or α = τ . To complete the proof we only need toshow that the case when α fixes only one point in the fundamental triangle does not happen. Suppose that α only fixes the origin, thus α interchanges X ∞ and Y ∞ . Hence, A = ξ η . where ξ, η ∈ K . Since α preserves X q , ξη q +1 yx q +1 + ξ q +1 y q +1 + ηx = 0in K ( x, y ). It means that, there exists c = 0 in K such that ξη q +1 Y X q +1 + ξ q +1 Y q +1 + ηX = c ( XY q +1 + X q +1 + Y ) , which is a contradiction. The cases when α fixes only X ∞ or Y ∞ are analogues.Theorems 4.3 and 2.2 have the following corollary.6 orollary 4.4. The automorphism group of X is S ⋊ C , where S is defined over F q but C is defined over F q i with i = 3( q + q + 1) . Remark 4.5.
By a result of Cossidente and Siciliano, see [2], and [4, Theorem 11.110], if a plane nonsingularcurve C of P G (2 , q ) of degree q + 2 has an automorphism group S ⋊ C where S is a Singer cycle of P G (2 , q )then C is projectively equivalent to the Pellikaan curve X q where equivalency is meant in P G (3 , q ). It shouldbe noted that the authors in [2] claimed their result was valid under a weaker condition, namely when theautomorphism group of C contains S . But this turns out incorrect by Theorem 2.2 . Suppose that q is a square, say q = p i , i ≥
1. Thus q + q + 1 = ( p i + p i + 1)( p i − p i + 1). Let λ be aprimitive ( q + q + 1)-root of unity in K . A straightforward computation shows that α defined by α ( x ) = λx, α ( y ) = λ q +1 y, is a K -automorphism of K ( X q ) of order q + q + 1, where K ( X q ) = K ( x, y ), with xy q +1 + x q +1 + y = 0, isthe function field of X q .Let h = α p i + p i +1 . The group H = h h i has order p i − p i + 1. The next results provide equations for thequotient curve X q /H and one of those equations turns out to be XY p i +1 + X p i +1 + Y = 0 , which is indeed the equation of the Tallini curve X p i . Proposition 5.1.
The quotient curve X q /H is isomorphic to the curve given by the equation X p i + p i +1 Y q +1 + Y + 1 = 0 . Proof.
Take ξ, η from K ( x, y ) = K ( X q ) given by ξ = x p i − p i +1 , η = x − ( q +1) y. Clearly h ( ξ ) = ξ and h ( η ) = η , then K ( ξ, η ) ⊂ K ( x, y ) H . Note that K ( x, y ) = K ( ξ, η )( x ) and T p i − p i +1 − ξ is a polynomial in K ( ξ, η )[ T ] which has x as a root. Thus [ K ( x, y ) : K ( ξ, η )] ≤ p i − p i + 1. Note that[ K ( x, y ) : K ( x, y ) H ] = ord( H ) = p i − p i + 1, hence [ K ( x, y ) H : K ( ξ, η )] = 1, therefore K ( x, y ) H = K ( ξ, η ).Finally, since xy q +1 + x q +1 + y = 0 and η = x − ( q +1) y we get x q +2 q +2 η q +1 + x q +1 + x q +1 η = 0 . Thus, x q + q +1 η q +1 + 1 + η = 0. Since ξ = x p i − p i +1 and q + q + 1 = ( p i + p i + 1)( p i − p i + 1) we get ξ p i + p i +1 η q +1 + η + 1 = 0 . Proposition 5.2.
The quotient curve X q /H is isomorphic to the curve given by the equation X p i + p i +1 + Y p i +1 + Y p i = 0 . roof. By the previous proposition, K ( X q /H ) = K ( x, y ), with x p i + p i +1 y q +1 + y + 1 = 0 , that is, x p i + p i +1 + 1 y q + 1 y q +1 = 0 . Putting ξ = x and η = 1 /y gives ξ p i + p i +1 + η q + η q +1 = 0 . Dividing by η p i + p i +1 and using q = p gives ξ p i + p i +1 η p i + p i +1 + 1 η p i +1 + 1 η p i = 0 . Replacing u = ξ/η and v = 1 /η gives u p i + p i +1 + v p i +1 + v p i = 0 . Theorem 5.3.
The quotient curve X q /H is isomorphic to the curve X p i given by the equation XY p i +1 + X p i +1 + Y = 0 . Proof.
Consider the function field K ( X p i ) = K ( x, y ), with xy p i +1 + x p i +1 + y = 0. We have that x ( y p i +1 + x p i ) + y = 0 . Raising to the p i -th power gives x p i ( y p i +1 + x p i ) p i + y p i = 0 . Multiplying by x gives x p i +1 ( y p i +1 + x p i ) p i + xy p i = 0 . This also can be written as, x p i + p i +1 + ( − xy p i − p i ( − xy p i ) = 0 . Putting u = x and v = − xy p i − u p i + p i +1 + v p i +1 + v p i = 0 . Note that K ( u, v ) = K ( x, y p i ) ⊂ K ( x, y ). Since xy p i +1 + x p i +1 + y = 0, y = − x p i +1 xy p i + 1 , that is, y belongs to K ( x, y p i ). Therefore, K ( u, v ) = K ( x, y ).8 The Hasse-Witt invariant
In this section q = p is a prime number. Let Σ = K ( x, y ), with xy p +1 + x p +1 + y = 0, be the function fieldof the Pellikaan curve X p , g its genus and γ its Hasse-Witt invariant. The partial derivative of F ( X, Y ) = XY p +1 + X p +1 + Y with respect to Y is F Y ( X, Y ) = XY p + 1.Consider ∆ Σ = { udx | u ∈ Σ } the differential module of Σ and C : ∆ (1)Σ → ∆ (1)Σ the Cartier operatordefined on the space of holomorphic differentials∆ (1)Σ = { w ∈ ∆ Σ | ( w ) ≥ } . Theorem 6.1.
The Hasse-Witt invariant of X p is equal to its genus.Proof. Let w be an exact differential in ∆ (1)Σ , that is, C ( w ) = 0. Then w can be written in the form w = ( u p + u p x + · · · + u pp − x p − ) dx. From [3], a basis for the K -vector space ∆ (1)Σ is given by B = (cid:26) x i y j F Y ( x, y ) dx | ≤ i + j ≤ p − (cid:27) . Thus u p + u p x + · · · + u pp − x p − = u ( x, y ) F Y ( x, y )where u ( X, Y ) is a polynomial in K [ X, Y ] of degree at most p −
1. Let u ( x, y ) = X i + j ≤ p − a ij x i y j . Since xy p +1 + x p +1 + y = 0 then 1 y = − xy p + 1 x p +1 . Thus we have u ( x, y ) F Y ( x, y ) = u ( x, y ) xy p + 1= 1 xy p + 1 X i + j ≤ p − a ij x i y j = y p xy p + 1 X i + j ≤ p − a ij x i (cid:18) y (cid:19) p − j = y p xy p + 1 X i + j ≤ p − a ij x i (cid:18) − xy p + 1 x p +1 (cid:19) p − j = X i + j ≤ p − ( − j +1 a ij y p x p + p − jp x i + j ( xy p + 1) p − − j X i + j ≤ p − ( − j +1 a ij y p x p + p − jp x i + j p − − j X k =0 (cid:18) p − − jk (cid:19) x k y kp = X i + j ≤ p − p − − j X k =0 ( − j +1 (cid:18) p − − jk (cid:19) a ij ( x jp − p − p y ( k +1) p ) x i + j + k = X i + j ≤ p − p − − j X k =0 w pijk x i + j + k where w ijk = (cid:18) ( − j +1 (cid:18) p − − jk (cid:19) a ij (cid:19) /p x j − p − y k +1 . Hence, u p + u p x + · · · + u pp − x p − = X i + j ≤ p − p − − j X k =0 w pijk x i + j + k . The term of degree p − x on the right side is X i + j ≤ p − w pijk x p − where k = p − − j − i . It means that X i + j ≤ p − w pijk = 0 X i + j ≤ p − w ijk = 0 X i + j ≤ p − (cid:18) ( − j +1 (cid:18) p − − jk (cid:19) a ij (cid:19) /p x j − p − y k +1 = 0 X i + j ≤ p − (cid:18) ( − j +1 (cid:18) p − − jp − − j − i (cid:19) a ij (cid:19) /p x j − p − y p − ( i + j ) = 0 X i + j ≤ p − (cid:18) ( − j +1 (cid:18) p − − jp − − j − i (cid:19) a ij (cid:19) /p x j y p − ( i + j ) = 0 . Since the last equation has degree at most p , it must be equal to zero. Thus all coefficients a ij are equalto zero, and therefore u ( x, y ) = 0. This shows that w = 0, that is, the Cartier operator C : ∆ (1)Σ → ∆ (1)Σ hastrivial kernel.Let V be the space of all w ∈ ∆ (1)Σ such that C i ( w ) = 0 for some i ≥
1. Note that if C i ( w ) = 0,then C i − ( w ) ∈ ker( C ). Hence V = { } . This implies that the Hasse-Witt matrix ( h ij ) over K of C hasmaximum rank equal to g , and consequently the matrix M = ( h ij )( h pij ) · · · ( h p g − ij )has rank g . Therefore, γ = g . 10 cknowledgment :The author thanks G´abor Korchm´aros for his guidance and helpful discussions about the topic of the presentresearch which was carried out when the author was visiting the Universit`a degli Studi della Basilicata (Italy)with a financial support of FAPESP-Brazil (grant 2015/10181-8). References [1] V. Abatangelo and G. Korchm´aros, Irreducible hypersurfaces of minimal degree containing all points ofa finite projective space over a finite field, in
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Codes, curves, andsignals (Urbana, IL, 1997), 9-20, Kluwer Internat. Ser. Engrg. Comput. Sci., 485, Kluwer Acad. Publ.,Boston, MA, 1998.[8] G. Tallini, Sulle ipersuperfici irriducibili d’ordine minimo che contengono tutti i punti di uno spazio diGalois S r,q , Rend. Mat. e Appl. (5) (1961), 431-479.[9] G. Tallini, Le ipersuperficie irriducibili d’ordine minimo che invadono uno spazio di Galois, Atti Accad.Naz. Lincei Rend. Cl. Sci. Fis. Mat. Nat. (8) (1961), 706-712. Authors’ addresses :Gregory Duran CunhaInstituto de Ciˆencias Matem´aticas e de Computa¸c˜aoUniversidade de S˜ao PauloAvenida Trabalhador S˜ao-carlense, 40013566-590 - S˜ao Carlos - SP (Brazil).E–mail: [email protected]@usp.br