Ample Subvarieties of Algebraic Varieties

Hartshorne, Robin

doi:10.1007/bfb0067839

Cited by 579 publications

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“…We reduce the proof of the first case to that of the second one as follows. By a theorem of Hartshorne (Theorem 4-2 in [7]), the linear system \nB\ is free of base points if B 2 > 0 and n is sufficiently large. Thus, using Bertini's theorem, we can choose a surface B n as in the proof of Theorem 5-1 to be a smooth algebraic curve supported near B.…”

Section: Remark If One Uses the Classes Fi(a) Instead Of 2/i{a) As mentioning

confidence: 99%

Embedded surfaces in four-manifolds, branched covers, and SO(3)-invariants

Kotschick

Matić

1995

Math. Proc. Camb. Phil. Soc.

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IntroductionOne of the outstanding problems in four-dimensional topology is to find the minimal genus of an oriented smoothly embedded surface representing a given homology class in a smooth four-manifold. For an arbitrary homology class in an arbitrary smooth manifold not even a conjectural lower bound is known. However, for the classes represented by smooth algebraic curves in (simply connected) algebraic surfaces, it is possible that the genus of the algebraic curve, given by the adjunction formula There are a number of results on this question in the literature. They can be divided into two classes, those proved by classical topological methods, which apply to topologically locally flat embeddings, including the G-signature theorem, and those proved by methods of gauge theory, which apply to smooth embeddings only.On the classical side, there is the result of Kervaire and Milnor[10], based on Rokhlin's theorem, which shows that certain homology classes are not represented by spheres. A major step forward was made by Rokhlin[20] and Hsiang and Szczarba [8] who introduced branched covers to study this problem for divisible homology classes. If the integral homology class represented by an embedded surface £ c: X is divisible by k, then the corresponding cover of X of order k branched along £ is again a smooth 4-manifold, so there is an obvious inequality, simply from the existence of the covering, asserting that its second Betti number is at least the absolute value of its signature. In the case k = 2, this gives

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Section: Remark If One Uses the Classes Fi(a) Instead Of 2/i{a) As mentioning

confidence: 99%

Embedded surfaces in four-manifolds, branched covers, and SO(3)-invariants

Kotschick

Matić

1995

Math. Proc. Camb. Phil. Soc.

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“…Pic (P N ) ~ Pic (A), where P N stands for Pξ and P» denotes the formal completion of P N along Z. The proof is in [3] Ch. IV (essentially Th.…”

Section: Cally Complete Intersection In Z Then C Is Actually a Complmentioning

confidence: 99%

“…Let y be the sheaf of ideals defining Z and call X n the scheme (X,Θ pN IZr n ). We can use the exact sequences where * denotes the multiplicative group of units and the first map sends x to 1 + x (for more details see [3] Ch. 4 p. 179 and [2] Exp II p. 124).…”

Section: Cally Complete Intersection In Z Then C Is Actually a Complmentioning

confidence: 99%

A problem of complete intersections

Robbiano

1973

Nagoya Mathematical Journal

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Let X be a non-singular projective surface in P\ (k an algebraically closed field of characteristic 0) and C an irreducible curve, which is a set-theoretically complete intersection in X is it true that C is actually a complete intersection in X ?In this paper we give a positive answer even in a more general hypothesis.We note that a similar question does not arise for a variety X with dim X ψ 2. In fact Lefschetz theorem says that, if X is a non-singular projective variety which is a complete intersection in Pξ and such that dim X > 3, any positive divisor on X is a complete intersection in X.On the other hand, if X is a non-singular conic in P\ and P SL point on X, then P is a set-theoretically complete intersection but not a complete intersection in X.As to the surfaces, it is a well known fact that on a "general" surface of degree > 4 in P| any curve is a complete intersection, but there are surfaces whose Picard group is different from Z (e.g. nonsingular quadric and cubic surfaces) (see [4]).Nevertheless no example is known of an irreducible curve on a nonsingular surface in P\, which is a set-theoretically complete intersection in X, but not a complete intersection in X (see [1]), and in fact we are going to prove that such an example cannot exist.For this we make use of the techniques developed by Grothendieck to prove Lef schetz theorem (see [2] and [3]).We now state the following THEOREM. Let k be an algebraically closed field of characteristic 0 and let X c Pf be a non-singular projective surface, which is a complete intersection. If C is an irreducible curve on X, which is a set-theoreti-

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“…(Lefschetz theory). According with the notations and the context of [7], let X be the formal completion of X along Y; by [7], chap. IV, theorem (1.5) and the proof of theorem (3.1), the map Pic(Z) ->Pic (Y) factorizes through Pic (X) -> Pic (X) -> Pic (Γ) and the map ( 6) Pic…”

Section: ) Ns(x) -> Ns(y)mentioning

confidence: 99%

“…Grothendieck's form (see [5], [7]), and on the theory of Picard schemes (see [6], [11]). We divide it into several steps.…”

Section: ) Ns(x) -> Ns(y)mentioning

confidence: 99%