INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

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26 Mezzettifor i ≠ j, then Y ′1 ⊂ Y 1 ∪ . . .Y r , so Y ′1 = ⋃ ri=1 (Y ′i, and we can assume i = 1. Similarly, Y 1 ⊂ Y j ′so j = 1 and Y 1 = Y ′1 ∪ Y i ), hence Y 1 ′ ⊂ Y i for some′, for some j, so Y 1 ⊂ Y 1 ⊂ Y ′1. Now let Z = Y − Y 1 = Y 2 ∪ . . . ∪ Y r = Y 2 ′ ∪ . . . ∪ Y s ′ andproceed by induction.□j ,6.9. Corollary. Any algebraic subset of A n (resp. of P n ) is in a unique way thefinite union of its irreducible components.□Note that the irreducible components of X are its maximal algebraic subsets.They correspond to the minimal prime ideals over I(X). Since I(X) is radical,these minimal prime ideals coincide with the primary ideals appearing in the primarydecomposition of I(X).6.10. Definition. An irreducible closed subset of A n is called an affine variety.Similarly, an irreducible closed subset of P n is a projective variety. A locally closedsubset in P n is the intersection of an open and a closed subset. An irreduciblelocally closed subset of P n is a quasi–projective variety.6.11. Proposition. Let X ⊂ A n and Y ⊂ A m be affine varieties. Then X × Yis irreducible, i.e. a subvariety of A n+m .Proof. Let X × Y = W 1 ∪ W 2 , with W 1 , W 2 closed. For all P ∈ X the map{P } × Y → Y which takes (P, Q) to Q is a homeomorphism, so {P } × Y isirreducible. {P } × Y = (W 1 ∩ ({P } × Y )) ∪ (W 2 ∩ ({P } × Y )), so ∃i ∈ {1, 2} suchthat {P } × Y ⊂ W i . Let X i = {P ∈ X | {P } × Y ⊂ W i }, i = 1, 2. Note thatX = X 1 ∪ X 2 .Claim. X i is closed in X.Let X i (Q) = {P ∈ X | (P, Q) ∈ W i }, Q ∈ Y . We have: (X × {Q}) ∩ W i =X i (Q) × {Q} ≃ X i (Q); X × {Q} and W i are closed in X × Y , so X i (Q) × {Q}is closed in X × Y and also in X × {Q}, so X i (Q) is closed in X. Note thatX i = ⋂ Q∈Y Xi (Q), hence X i is closed, which proves the Claim.Since X is irreducible, X = X 1 ∪ X 2 implies that either X = X 1 or X = X 2 ,so either X × Y = W 1 or X × Y = W 2 .□Exercises to §6.1. Let X ≠ ∅ be a topological space. Prove that X is irreducible if and onlyif all non–empty open subsets of X are connected.2*. Prove that the cuspidal cubic Y ⊂ A 2 C of equation x3 −y 2 = 0 is irreducible.(Hint: express Y as image of A 1 in a continuous map...)
Introduction to algebraic geometry 273. Give an example of two irreducible subvarieties of P 3 whose intersection isreducible.4. Find the irreducible components of the following algebraic sets over thecomplex field:a) V (y 4 − x 2 , y 4 − x 2 y 2 + xy 2 − x 3 ) ⊂ A 2 ;b) V (y 2 − xz, z 2 − y 3 ) ⊂ A 3 .5*. Let Z be a topological space and {U α } α∈I be an open covering of Z suchthat U α ∩ U β ≠ ∅ for α ≠ β and that all U α ’s are irreducible. Prove that Z isirreducible.7. Dimension.Let X be a topological space.7.1. Definition. The topological dimension of X is the supremum of the lengthsof the chains of distinct irreducible closed subsets of X, where by definiton thefollowing chain has length n:X 0 ⊂ X 1 ⊂ X 2 ⊂ . . . ⊂ X n .Example.1. dim A 1 = 1: the maximal length chains have the form {P } ⊂ A 1 .2. dim A n = n: a chain of length n is{0} = V (x 1 , . . ., x n ) ⊂ V (x 1 , . . ., x n−1 ) ⊂ . . . ⊂ V (x 1 ) ⊂ A n ;note that V (x 1 , . . ., x i ) is irreducible ∀i, because the ideal 〈x 1 , . . ., x i 〉 is prime.Indeed the quotient K[x 1 , . . ., x n ]/〈x 1 , . . ., x i 〉 is isomorphic to K[x i+1 , . . ., x n ].Therefore dim A n ≥ n. On the other hand, from every chain of irreducible closedsubsets of A n , passing to the ideals, we get a chain of the same length of primeideals in K[x 1 , . . ., x n ], therefore dim A n ≤ dim K[x 1 , . . ., x n ] = n.3. Let X be irreducible. Then dim X = 0 if and only if X is the closure ofevery point of it.We prove now some useful relations between the dimension of X and thedimensions of its subspaces.7.2. Proposition.1. If Y ⊂ X, then dim Y ≤ dimX. In particular, if dimX is finite, then alsodim Y is (in this case, the number dim X − dimY is called the codimension of Yin X).2. If X = ⋃ i∈I U i is an open covering, then dim X = sup{dim U i }.3. If X is noetherian and X 1 , . . ., X s are its irreducible components, thendim X = sup i dim X i .
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Page 45 and 46: Then v n,d (X) ≃ X: it is the set
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