AG for NT Week 8 We will use the language of schemes to study varieties 1 Blowing Up Varieties We will construct the blow up of a variety with respect to a non-singular closed subvariety. This tool/technique is the main method to resolve singularities of algebraic variety. Denition 1.1. The blowup of n at is constructed as follows: Take the product n n−1. If are A 0 A × P fx1; : : : ; xng ane coordinated for n, are the homogeneous coordinates of n−1, the blowup of n Bl n is the A fy1; : : : ; yng P A 0(A ) closed subset dened by Bl n 0(A ) := fxiyj = xjyij1 ≤ i; j ≤ n; i 6= jg We have the following commutative diagram Bl n n n−1 0(A ) / A × P φ projection & n A For the next few pages, we let φ be the morphism dened as above. Lemma 1.2. 1. If n, then −1 consist of one point. In fact gives an isomorphism of Bl n −1 ∼ p 2 A p 6= 0 φ (p) phi 0(A ) n φ (0) = An n f0g 2. φ−1(0) =∼ Pn−1 3. The points of φ−1(0) are in 1-1 correspondence with the lines of An through the origin 4. Bl n is irreducible 0(A ) Proof. n −1 aj 1. Let p = (a1; : : : ; an) 2 , assume ai 6= 0. So if p × (y1; : : : ; yn) 2 φ (p) then for each j, yj = yi. A ai So is uniquely determined as a point in n−1. By setting , we have (y1 : ··· : yn) P yi = ai (y1 : ··· : yn) = (a1 : ··· : an). Moreover setting (p) = (a1; : : : ; an) × (a1 : ··· : an) denes an inverse morphism to φ. n Bl n −1 A n f0g ! 0A n φ (0) 2. φ−1(0) consist of all points 0 × Q for Q 2 Pn−1 with no restrictions 3. Follows from 2. 4. Bl n Bl n −1 −1 . The rst component, by part . is irreducible, and each point in 0(A ) = 0(A ) n φ f0g [ φ f0g 1 −1 is contained in the closure of some line in Bl n −1 . Hence Bl n −1 is dense in Bl n φ (0) L 0A n φ f0g 0A n φ (0) 0(A ) and hence Bl n is irreducible 0(A ) 1 Denition 1.3. If n we dene Bl to be is −1 Y ⊂ A n 0 0Y Ye φ (Y n 0) We see from Lemma 1.2 φ induces a birational morphism of Ye to Y . Fact 1.4. Blowing up is independent of your choice of embedding. Example 1.5. (Node) Let x; y be coordinates in A2, and dene X :(y2 = x2(x + 1)). Let t; u be homogeneous coordinates for P1. Then Bl 2 2 2 1. On the ane piece , we have 2 2 and , 0X = fy = x (x + 1); ty = uxg ⊂ A × P t 6= 0 y = x (x + 1) y = ux hence u2x2 = x2(x + 1). This factors, hence we get a variety fx = 0g = E (this is the preimage of 0 under φ and is called the Exceptional Divisors) and the variety fu2 = x + 1g = Xe (This is called the proper transform of X). Note that Xe \ E consists of two points, u = ±1. Notice that this values for u are precisely the values of the slopes of X through the origin. Blowups separates points and tangent vectors) Exercise 1.6. (Tacnode) Let T :(y2 = x4(x + 1)). Blow this up at the origin and see what you get. Denition 1.7. Blowing up with respect to a subvariety. Let X ⊂ An be an ane variety. Let Z ⊂ X be a closed non-singular subvariety, dened by the vanishing of the polynomials in n. Let Z ff1; : : : ; fkg A (y1 : ··· : yk) be homogenous coordinates for k−1. Dene Bl n . As before, we get a P Z (A ) = fyifj = yjfij1 ≤ i; j ≤ k; i 6= jg birational map Bl n n k−1 Z (A ) / A × P φ projection & n A −1 It has a birational inverse, p = (a1; : : : ; an) 7! (a1; : : : ; an) × (f1(p): ··· : fk(p)). Also dene BlZ (X) = φ (X n 0). Exercise 1.8. Compare blowing up 2 2 in 3 with respect to the -axis. y = x (x + 1) A[x:y:z] z [Note: 0 the subvariety dened by the vanishing of polynomials fi = xi] For most purposes/classifying all surfaces only need to know about blowing up a point. Example 1.9. Let be the double cone dened by 2 2 2 3 and let be the line dened by X x + y = z ⊂ A[x:y:z] Z . Let be coordinates for 1, hence Bl 2 2 2 . So on the ane piece fy = z; x = 0g t; u P Z X = fx + y = z ; xt = (y − z)ug u 6= 0, we get xt = y − z hence x2 = xt(y + z). This factorises, so we get two pieces: fx = 0; y = z; t arbitraryg = E (the exceptional curve); fxt = y − zg := Xe (this should be nonsingular) 2 Invertible Sheaves Let X be a variety. Denition 2.1. • An invertible sheaf F on X is a locally free OX -module of rank 1. (That is, there exists an open covering ∼ fUig of X so that F(Ui) = OX (Ui)) • We will see soon that the Picard group is the group of isomorphism classes of invertible sheaves on X. • On varieties: Weil divisors are the same as Cartier divisors. A Cartier Divisor D = f(Ui; fi)g with fUig an open covering of X, and fi on Ui is an element of OX (Ui) (think of a rational function). Also on Ui \ Uj, we have fi is invertible. fj 2 Notation 2.2. Let be a divisors (Weil/Cartier), dene to be the sub- -module which is generated by −1 D L(D) OX fi fi −1 −1 on Ui. This is well dened since is invertible on Ui \ Uj, so f and f diers by a unit. This L(D) is called fj i j the sheaf associated to D = f(Ui; fi)g. Proposition 2.3. 1. For any divisors D, L(D) is an invertible sheaf on X and the map D 7! L(D) gives a 1-1 correspondence Pic(X) $Invertible sheaves on X. ∼ −1 2. L(D1 − D2) = L(D1) ⊗ L(D2) ∼ 3. D1 ∼ D2 (linearly equivalence) if and only if L(D1) = L(D2) Proof. 1. The map dened by −1 is the isomorphism, so is an invertible sheaf. Conversely, OUi !L(D)jUi 1 7! fi L(D) can be recovered from by on to be the inverse of a generator for . D L(D) fi Ui L(D)(Ui ) 2. If and , then on is generated by −1 . So ∼ D1 = f(Ui; fi)g D2 = f(Vi; gi)g L(D1 − D2) Ui \ Vj fi gj L(D1 − D2) = −1 L(D1) ⊗ L(D2) . ∼ 3. By part 2. it is sucient to show that D = D1 − D2 is principal if and only if L(D) = OX . If D is principal, dened by ∗ , then is globally generated by −1, so −1 is the isomorphism ∼ . f 2 Γ(X; OX ) L(D) f 1 ! f OX = L(D) So we have a 1-1 correspondence from Pic(X) !isomorphism classes of invertible sheaves. 3 Morphisms to Pn On n, the homogeneous coordinates give the standard cover and on , −1 is a local P x0 : ··· : xn fUi := (xi 6= 0)g Ui xi generator for the sheaf O(1). For any (projective) variety X, let φ : X ! Pn. Then L = φ∗(O(1)) is an invertible ∗ sheaf on X. The global sections s0; : : : ; sn (si := φ (xi)), si 2 Γ(X; L) generate the sheaf L. Conversely, L and si determines φ. Theorem 3.1. 1. If n is a morphism, then ∗ is an invertible sheaf generated by global sections ∗ φ : X ! P φ (O(1)) si = φ (xi) 2. Any invertible sheaf L on X determines a unique morphism φ : X ! Pn Proof. 1. From Above 2. Lengthy argument in Hartshorne, pg 150 Proposition 3.2. Let k be an algebraically closed eld. Let X be a variety, and φ : X ! Pn be a morphism ∗ corresponding to L and s0; : : : ; sn be as above. Let V ⊂ Γ(X; L) be a subspace spanned by si = φ (xi). Then φ is a closed immersion if and only if: 1. Elements of V separate points, i.e., for any P 6= Q on X, exists s 2 V with s 2 mP LP but s2 = mQLQ. 2. Elements of V separate tangent vectors, i.e., for each points P 2 X, the set of fs 2 V : sP 2 mpLpg span the vector space 2 . mpLp=mpLp 3 Proof. (Only proving ) If is a closed immersion, think of as a closed subvariety of n. So and the ) φ X P L = OX (1) vector space is spanned by the images of n . Given in , we can V ⊂ Γ(X; OX (1)) x0; : : : ; xn 2 Γ(P ; O(1)) P 6= Q X P P nd a hyperplane H containing P and not Q. If H = ( aixi = 0) for ai 2 k, then s = aixijX satises the rst property. For the second, each hyperplane passing through gives rise to sections which generate 2 . P mP LP =mpLp xi 2 Example. If P = (1 : 0 : ··· : 0), then U0 has local coordinates yi = , so P = (0;:::; 0) 2 Ui and mp=m is the x0 p vector space spanned by yi.