MIT OpenCourseWare http://ocw.mit.edu 18.727 Topics in Algebraic Geometry: Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� �� ��������� � �� ��������� � �ALGEBRAIC SURFACES, LECTURE 4 LECTURES: ABHINAV KUMAR We recall the theorem we stated and lemma we proved from last time: Theorem 1. Let f : X ��� S be a birational morphism of surfaces s.t. f−1 is not defined at a point p ∈ S. Then f factors as f : X g S˜π S where g is a→ →birational morphism and π is the blowup of S at p. Lemma 1. Let S be an irreducible surface, possibly singular, and S� a smooth surface with a birational morphism f : S S�. Suppose f−1 is undefined at p ∈ S. Then f−1(p) is a curve on S. → Lemma 2. Let φ : S ��� S� be a birational map s.t. φ−1 is undefined at a point p ∈ S�. Then there is a curve C on S s.t. φ(C) = {p}. Proof. φ corresponds to a morphism f : U S� where U is some open set in S.→Let Γ ⊂ U × S� be the graph of f, and let S1 denote its closure in S × S�. S1 is irreducible but may be singular. S1 q� S ��S�φ The projections q, q� are birational morphisms and the diagonal morphism com-mutes. Since φ−1(q) is not defined, (q�)−1(p) is not defined either, so ∃C1 ⊂ S an irreducible curve s.t. q�(C1) = {p}. Moreover, q(C1) = C is a curve in S: if not, since S1 ⊂ S × S�, q(C1) a point = C1 ⊂ {x} × S� for some x ∈ S; but such a⇒C1 can only intersect the graph of f in {(x, f(x))} so the closure of the graph of f can’t contain the curve C1. By construction, C contracts to {p} under φ. � Proof of theorem. Let g = π−1 f be the rational map in question. We need to show that g is a morphism. Let◦ s = g−1, and suppose that g is undefined at a point q ∈ X. (1) q X s(2) q f S˜π ��S 1� �� ��������� 2 LECTURES: ABHINAV KUMAR Applying the second lemma, we obtain a curve C ⊂ S˜s.t. s(C) = {q}. Then π(C) = f(q) by composing s(C) = {q} with f. So we must have C = E, the exceptional divisor for π, and f(q) = p. Let Ox,q be the local ring of X at q, and let mq be its maximal ideal. We claim that there is a local coordinate y on S at p s.t. f∗y ∈ m2 q . To see this, let (x, t) be a local system of coordinates at p. If f∗t ∈ m2 q then we are done. If not, i.e. f∗t ∈/ m2 q , then f∗t vanishes on f−1(p) with multiplicity 1, so it defines a local equation for f−1(p) in OX,q. So f∗(x) = u f∗t for some u ∈ Ox,q. Let y = x − u(q)t; then· (3) f∗y = f∗x − u(q)f∗t = uf∗(t) − u(q)f∗(t) = (u − u(q))f∗(t) ∈ m 2 q Next, let e be any point on E where s is defined. Then we have s∗f∗y = (f ◦ s)∗y = π∗y ∈ m2 . This holds for all e outside a finite set. But π∗y is a localecoordinate at every point of E except one, by construction, giving the desired contradiction. � This proves the universal property of blowing up. Here is another: Proposition 1. Every morphism from S˜to a variety X that contracts E to a point must factor through S. Proof. We can reduce to X affine, then to X = An, then to X = A1 . Then f defines a function on ˜= S � {p} which extends. �S � E ∼Theorem 2. Let f : S S0 be a birational morphism of surfaces. Then ∃ a→ sequence of blowups πk : Sk → (k = 1, . . . , n) and an isomorphism S ∼ SnSk−1 →s.t. f = π1 ◦ · · · ◦ πn ◦ u, i.e. f factors through blowups and an isomorphism. Proof. If f−1 is a morphism, we’re done. Otherwise, ∃ a point p of S0 where f−1 is not defined. Then f = π1 ◦ f1, where π1 is the blowup of S0 at p. If f1−1 is a morphism, we are done, otherwise we keep going. We need to show that this process terminates. Note that the rank of the Neron-Severi group rk NS(Sk) = 1+rk NS(Sk−1): since rk (S) is finite, this sequence must terminate. More simply, since f contracts only finitely many curves, it can only factor through finitely many distinct blowups. � Corollary 1. Any birational map φ : S ��� S� is dominated by a nonsingular surface S with birational morphisms q, q� : S S, S� which are compositions of→blow-up maps, i.e. S q� (4) q S ��S�φ Proof. First resolve the indeterminacy of φ using S and then note that q� is a birational morphism, i.e. a composition of blowups by the above. �3 ALGEBRAIC SURFACES, LECTURE 4 1. Minimal Surfaces We say that a surface S1 dominates S2 if there is a birational morphism S1 →S2. A surface S is minimal if it is minimal up to isomorphism in its birational equivalence class with respect to this ordering. Proposition 2. Every surface dominates a minimal surface. Proof. Let S be a surface. If S is not minimal, ∃ a birational morphism S S1→that is not an isomorphism, so rk NS(S) > rk NS(S1). If S1 is minimal, we are done: if not, continue in this fashion, which must terminate because rk NS(S) is finite. � Note. We say that E ⊂ S is exceptional if it is the exceptional curve of a blowup π : S S�. Clearly an exceptional curve E is isomorphism to P1 and satisfies E2 = −→1 and E KS = −1 (since −2 = 2g − 2 = E (E + K).· · Theorem 3 (Castelnuovo). Let S be a projective surface and E ⊂ S a curve ∼P1 with E2 = Then ∃ a morphism S → s.t. it is a blowup and E is = −1. S� the exceptional curve (classically called an “exceptional curve of the first kind”). Proof. We will find S� as the image of a particular morphism from S to a pro-jective space: informally, we need a “nearly ample” divisor which will contract E and nothing else. Let H be very ample on S s.t. H1(S, OS (H)) = 0 (take any hyperplane section H˜, then H = nH˜will have zero higher cohomology by Serre’s theorem). Let k = H E > 0, and let M = H + kE. Note that · M E = (H + kE) E = k + kE2 = 0. This M will define the morphism · · S → P(H0(S, OS (M))) (i.e. some Pn). Now, OS(H)|E ∼OE (k) since E ∼P1 = = and deg OS (H)|E = H · E = k …