INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

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32 Mezzettiis closed in U P . The proposition then follows from:8.3. Lemma. Let T be a topological space, T = ∪ i∈I U i be an open covering ofT, Z ⊂ T be a subset. Then Z is closed if and only if Z ∩ U i is closed in U i for alli.Proof. Assume that U i = X \ C i and Z ∩ U i = Z i ∩ U i , with C i and Z i closed inX.Claim: Z = ⋂ i∈I (Z i ∪ C i ), hence it is closed.In fact: if P ∈ Z, then P ∈ Z ∩ U i for a suitable i. Therefore P ∈ Z i ∩ U i , soP ∈ Z i ∪ C i . If P /∈ Z j ∩ U j for some j, then P /∈ U j so P ∈ C j and thereforeP ∈ Z j ∪ C j .Conversely, if P ∈ ⋂ i∈I (Z i ∪ C i ), then ∀ i, either P ∈ Z i or P ∈ C i . Since ∃jsuch that P ∈ U j , hence P /∈ C j , so P ∈ Z j , so P ∈ Z j ∩ U j = Z ∩ U j .□8.4. Corollary.1. Let φ ∈ O(X): then φ −1 (0) is closed. It is denoted V (φ) and called theset of zeroes of φ.2. Let X be a quasi–projective variety and φ, ψ ∈ O(X). Assume that thereexists U, open non –empty subset such that φ| U = ψ| U . Then φ = ψ.Proof. φ − ψ ∈ O(X) so V (φ − ψ) is closed. By assumption V (φ − ψ) ⊃ U, whichis dense, because X is irreducible. So V (φ − ψ) = X.□If X ⊂ A n is locally closed, we can use on X both homogeneous and non–homogeneous coordinates. In the second case, a regular function is locally representedas a quotient F/G, with F and G ∈ K[x 1 , . . ., x n ]. In particular allpolynomial functions are regular, so, if X is closed, K[X] ⊂ O(X).If α ⊂ K[X] is an ideal, we can consider V (α) := ⋂ φ∈αV (φ): it is closedinto X. Note that α is of the form α = α/I(X), where α is the image of α inthe canonical epimorphism, it is an ideal of K[x 1 , . . ., x n ] containing I(X), henceV (α) = V (α) ∩ X = V (α).If K is algebraically closed, from the Nullstellensatz it follows that, if α isproper, then V (α) ≠ ∅. Moreover the following relative form of the Nullstellensatzholds: if f ∈ K[X] and f vanishes at all points P ∈ X such that g 1 (P) = . . . =g m (P) = 0 (g 1 , . . ., g m ∈ K[X]), then f r ∈ 〈g 1 , . . ., g m 〉 ⊂ K[X], for some r ≥ 1.8.5. Theorem. Let K be an algebraically closed field. Let X ⊂ A n K be closedin the Zariski topology. Then O(X) ≃ K[X]. It is an integral domain if and onlyif X is irreducible.Proof. Let f ∈ O(X).
Introduction to algebraic geometry 33(i) Assume first that X is irreducible. For all P ∈ X fix an open neighbourhoodU P of P and polynomials F P , G P such that V P (G P ) ∩ U P = ∅ andf| UP = F P /G P . Let f P , g P be the functions of K[X] defined by F P and G P . Theng P f = f P holds on U P , so it holds on X (by Corollary 8.3). Let α ⊂ K[X] bethe ideal α = 〈g P 〉 P ∈X ; α has no zeroes on X, because g P (P) ≠ 0, so α = K[X].Therefore there exists h P ∈ K[X] such that 1 = ∑ P ∈X h Pg P (sum with finitesupport). Hence in O(X) we have the relation: f = f ∑ h P g P = ∑ h P (g P f) =∑hP f P ∈ K[X].(ii) Let X be reducible: for all P ∈ X, there exists R ∈ K[x 1 , . . ., x n ] such thatR(P) ≠ 0 and R ∈ V (X \ U P ), so r ∈ O(X) is zero outside U P . So rg P f = f P ron X and we conclude as above by replacing g P with g P r and f P with f P r.□The characterization of regular functions on projective varieties is completely different:we will see in §12 that, if X is a projective variety, then O(X) ≃ K, i.e.the unique regular functions are constant.This gives the motivation for introducing the following weaker concept.8.6. Definition. Let X be a quasi–projective variety. A rational function on Xis a germ of regular functions on some open non–empty subset of X.Precisely, let K be the set {(U, f)|U ≠ ∅, open subset of X, f ∈ O(U)}. Thefollowing relation on K is an equivalence relation:(U, f) ∼ (U ′ , f ′ ) if and only if f| U∩U′ = f ′ | U∩U ′.Reflexive and symmetric properties are quite obvious. Transitive property: let(U, f) ∼ (U ′ , f ′ ) and (U ′ , f ′ ) ∼ (U ′′ , f ′′ ). Then f| U∩U′ = f ′ | U∩U′ and f ′ | U ′ ∩U ′′ =f ′′ | U ′ ∩U ′′, hence f| U∩U ′ ∩U ′′ = f ′′ | U∩U ′ ∩U ′′. U ∩ U ′ ∩ U ′′ is a non–empty opensubset of U ∩ U ′′ (which is irreducible and quasi–projective), so by Corollary 8.4f| U ′ ∩U ′′ = f ′′ | U ′ ∩U ′′.Let K(X) := K/ ∼: its elements are by definition rational functions on X.K(X) can be given the structure of a field in the following natural way.Let 〈U, f〉 denote the class of (U, f) in K(X). We define:〈U, f〉 + 〈U ′ , f ′ 〉 = 〈U ∩ U ′ , f + f ′ 〉,〈U, f〉〈U ′ , f ′ 〉 = 〈U ∩ U ′ , ff ′ 〉(check that the definitions are well posed!).There is a natural inclusion: K → K(X) such that c → 〈X, c〉. Moreover, if〈U, f〉 ≠ 0, then there exists 〈U, f〉 −1 = 〈U \ V (f), f −1 〉: the axioms of a field areall satisfied.There is also an injective map: O(X) → K(X) such that φ → 〈X, φ〉.
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Page 41 and 42: circle. We define φ : X → P 1 by
Page 45 and 46: Then v n,d (X) ≃ X: it is the set
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