INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

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18 MezzettiPart 2.(i) of 4.10. implies that, if I(X ∩Y ) ≠ I(X)+I(Y ), then I(X)+I(Y )is not radical.We move now to projective space. There exist proper homogeneous ideals ofK[x 0 , x 1 , . . ., x n ] without zeroes in P n , also assuming K algebraically closed: forexample the maximal ideal 〈x 0 , x 1 , . . ., x n 〉. The following characterization holds:4.12. Proposition. Let K be an algebraically closed field and let I be ahomogeneous ideal of K[x 0 , x 1 , . . ., x n ].The following are equivalent:(i) V P (I) = ∅;(ii) either I = K[x 0 , x 1 , . . ., x n ] or √ I = 〈x 0 , x 1 , . . ., x n 〉;(iii) ∃d ≥ 1 such that I ⊃ K[x 0 , x 1 , . . ., x n ] d , the subgroup of K[x 0 , x 1 , . . ., x n ]formed by the homogeneous polynomials of degree d.Proof.(i)⇒(ii) Let p : A n+1 − {0} → P n be the canonical surjection. We have:V P (I) = p(V (I)−{0}), where V (I) ⊂ A n+1 . So if V P (I) = ∅, then either V (I) = ∅or V (I) = {0}. If V (I) = ∅ then I(V (I)) = I(∅) = K[x 0 , x 1 , . . ., x n ]; if V (I) = {0},then I(V (I)) = 〈x 0 , x 1 , . . ., x n 〉 = √ I by the Nullstellensatz.(ii)⇒(iii) Let √ I = K[x 0 , x 1 , . . ., x n ], then 1 ∈ √ I so 1 r = 1 ∈ I(r ≥ 1). If√I = 〈x0 , x 1 , . . ., x n 〉, then for all variable x k there exists an index i k ≥ 1 suchthat x i kk∈ I. If d ≥ i 0 + i 1 + . . . + i n , then any monomial of degree d is in I, soK[x 0 , x 1 , . . ., x n ] d ⊂ I.(iii)⇒(i) because no point in P n has all coordinates equal to 0.□4.13. Theorem. Let K be an algebraically closed field and I be a homogeneousideal of K[x 0 , x 1 , . . ., x n ]. If F is a homogeneous non–constant polynomial suchthat V P (F) ⊃ V P (I) (i.e. F vanishes on V P (I), then F ∈ √ I.Proof. We have p(V (I) − {0}) = V P (I) ⊂ V P (F). Since F is non–constant, wehave also V (F) = p −1 (V P (F)) ∪ {0}, so V (F) ⊃ V (I); by the NullstellensatzI(V (I)) = √ I ⊃ I(V (F)) = √ (F) ∋ F.□4.14. Corollary (homogeneous Nullstellensatz). Let I be a homogeneousideal of K[x 0 , x 1 , . . ., x n ] such that V P (I) ≠ ∅, K algebraically closed. Then √ I =I h (V P (I)).□4.15. Definition. A homogeneous ideal of K[x 0 , x 1 , . . ., x n ] such that √ I =〈x 0 , x 1 , . . ., x n 〉 is called irrelevant.4.16. Corollary. Let K be an algebraically closed field. There is a bijectionbetween the set of projective algebraic subsets of P n and the set of radical
Introduction to algebraic geometry 19homogeneous non–irrelevant ideals of K[x 0 , x 1 , . . ., x n ].□Remark. Let X ⊂ P n be an algebraic set, X ≠ ∅. The affine cone ofX, denoted C(X), is the following subset of A n+1 : C(X) = p −1 (X) ∪ {0}. IfX = V P (F 1 , . . ., F r ), with F 1 , . . ., F r homogeneous, then C(X) = V (F 1 , . . ., F r ).By the Nullstellensatz, if K is algebraically closed, I(C(X)) = I h (X).Exercises to §4.1. Give a non-trivial example of an ideal α of K[x 1 , . . ., x n ] such that α ≠ √ α.2. Show that the following closed subsets of the affine plane Y = V (x 2 +y 2 −1)and Y ′ = V (y − 1) are such that equality does not hold in the following relation:I(Y ∩ Y ′ ) ⊃ I(Y ) + I(Y ′ ).3. Let α ⊂ K[x 1 , . . ., x n ] be an ideal. Prove that α = √ α if and only if thequotient ring K[x 1 , . . ., x n ]/α does not contain non–zero nilpotents.4*. Let I be a homogeneous ideal of K[x 1 , . . ., x n ] satisfying the followingcondition: if F is a homogeneous polynomial such that F r ∈ I for some positiveinteger r, then F ∈ I. Prove that I is a radical ideal.5. The projective closure of an affine algebraic set.Let X ⊂ A n be Zariski closed. Fix an index i ∈ {0, . . ., n} and embed A n into P nas the open subset U i . So X ⊂ A n φ i֒→ P n .5.1. Definition. The projective closure of X, X, is the closure of X in theZariski topology of P n .Since the map φ i is a homeomorphism (see Proposition 3.2.), we have: X ∩A n = X because X is closed in A n . The points of X ∩ H i , where H i = V P (x i ),are called the “points at infinity” of X in the fixed embedding.Note that, if K is an infinite field, then the projective closure of A n is P n :indeed, let F be a homogeneous polynomial vanishing along A n = U 0 . We canwrite F = F 0 x d 0 +F 1x0 d−1 +. . .+F d . By assumption, for every P(a 1 , . . ., a n ) ∈ A n ,P ∈ V P (F), i.e. F(1, a 1 , . . ., a n ) = 0 = a F(a 1 , . . ., a n ). So a F ∈ I(A n ). Weclaim that I(A n ) = (0): if n = 1, this follows from the principle of identity ofpolynomials, because K is infinite. If n ≥ 2, assume that F(a 1 , ..., a n ) = 0 forall (a 1 , ..., a n ) ∈ K n and consider F(a 1 , ..., a n−1 , x): either it has positive degreein x for some choice of (a 1 , ..., a n ), but then it has finitely many zeroes againstthe assumption; or it is always constant in x, so F belongs to K[x 1 , ..., x n−1 ] andwe can conclude by induction. So the claim is proved. We get therefore thatF 0 = F 1 = . . . = F d = 0 and F = 0.
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Page 45 and 46: Then v n,d (X) ≃ X: it is the set
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