9.2. Intersection and morphisms 397

Proof Let Γ ⊆ X×S Z be the graph of ϕ. This is an S-scheme and the projection p :Γ → X is projective birational. Let us show that Γ admits a desingularization. This is Theorem 8.3.44 if dim S = 0. Let us therefore suppose that dim S = 1. Let K = K(S). Then pK :ΓK → XK is proper birational with XK normal, and hence pK is an isomorphism. We can therefore apply Theorem 8.3.50. Let g : Xe → Γ be a desingularization morphism, and f : Xe → X the composition p ◦ g. It now suffices to apply Theorem 2.2 to the morphism f.

9.2.2 Projection formula Definition 2.8. Let X,Y be Noetherian schemes, and let f : X → Y be a proper morphism. For any prime cycle Z on X (Section 7.2), we set W = f(Z) and ½ [K(Z): K(W )]W if K(Z) is finite over K(W ) f Z = ∗ 0 otherwise.

By linearity, we define a homomorphism f∗ from the group of cycles on X to the group of cycles on Y . This generalizes Definition 7.2.17. It is clear that the construction of f∗ is compatible with the composition of morphisms. Remark 2.9. We can interpret the intersection of two divisors in terms of direct images of cycles. Let C,D be two Cartier divisors on a regular fibered surface X → S, of which at least one is vertical. Let us suppose that C is effective and has no common component with Supp D. Let h : C → S denote the morphism induced by X → S. Then C · D = h∗[D|C ]. This generalizes Lemma 1.29. See also Remark 2.13. Lemma 2.10. Let f : X → Y be a projective birational morphism of Noetherian integral schemes such that OY → f∗OX is an isomorphism. Then the following properties are true. (a) There exists an open subset V of Y such that f −1(V ) → V is an isomor- phism, and that codim(Y \ V,Y ) ≥ 2.

(b) Let Z be a cycle of codimension 1 in X. Then f∗Z is a cycle of codimen- sion 1 in Y . Proof (a) Let V be the open subset of Y defined by Proposition 4.4.2. Let y be a point of codimension 1 in Y . We have to prove that y ∈ V . Let x be a generic point of Xy. Then dim OX,x 6= 0 because f is dominant. By Theorem 8.2.5, we have trdegk(y) k(x) = 0, hence Xy is finite and y ∈ V . (b) We can suppose that Z is a prime cycle. Let x be the generic point of Z and y = f(x). If y ∈ V , then y has codimension dim OY,y = dim OX,x = 1, and f∗Z = {y} is a cycle of codimension 1. Let us suppose y∈ / V . Let F be the irreducible component of Xy containing x. Then dim F ≥ 1 because Xy has no isolated point by definition of V . Hence trdegk(y) k(x) = dim F ≥ 1, and f∗Z = 0. We can now generalize Theorem 7.2.18. See also [37], Proposition 1.4(b). 398 9. Regular surfaces

Proposition 2.11. Let f : X → Y be a surjective projective morphism of Noetherian integral schemes. We suppose that [K(X): K(Y )] = n is finite. Then for any Cartier divisor D on Y , we have

∗ f∗[f D] = n[D].

Proof The morphism f factors into a projective birational morphism 0 0 X → Y := Spec f∗OX followed by a finite surjective morphism Y → Y (Exercise 5.3.11). It suffices to show that the proposition is true for each of these morphisms. Finite morphism are dealt with in Theorem 7.2.18. We can therefore ∗ suppose that f is birational and f∗OX = OY . Then f∗[f D] − [D] is a cycle of codimension 1 (Lemma 2.10(b)), and its restriction to some open subset V ⊆ Y ∗ with codim(Y \ V,Y ) ≥ 2 is zero (Lemma 2.10(a)). Hence f∗[f D] − [D] = 0. Theorem 2.12. Let f : X → Y be a dominant morphism of regular fibered surfaces over S. Let C (resp. D) be a divisor on X (resp. on Y ). Then the following properties are true. (a) For any divisor E on X such that f(Supp E) is finite, we have E·f ∗D = 0. (b) Let us suppose that C or D is vertical. Then

∗ C · f D = f∗C · D (Projection formula), (2.4)

where f∗C is the Cartier divisor on Y such that [f∗C] = f∗[C]. (c) The extension K(X)/K(Y ) is finite. Let F be a vertical divisor on Y . Then f ∗F is vertical and we have

f ∗F · f ∗D = [K(X): K(Y )]F · D.

Proof (a) We can suppose that E is a vertical prime divisor. Let y = f(E). ∗ Then OY (D) is free on an open neighborhood V of y. It follows that OX (f D) = ∗ −1 ∗ f OY (D) is free on f (V ) ⊃ E, and hence OX (f D)|E 'OE, which, in particular, implies that f ∗D · E = 0. (b) We can suppose that C is a prime divisor. If f(C) is a point, then f∗[C] = 0 and equality (2.4) is true by (a). Let us therefore suppose that dim f(C) = 1. Using the moving lemma 1.10, we can suppose that Supp D does not contain f(C). Hence Supp f ∗D ⊆ f −1(Supp D) does not contain C. Let π : X → S, π0 : Y → S be the structural morphisms. With the notation of Remark 2.13,

∗ ∗ 0 ∗ 0 C · f D = π∗([C].f D) = π∗f∗([C].f D) = π∗(f∗[C].D) = f∗C · D.

(c) The morphism of the generic fibers Xη → Yη is a dominant morphism of algebraic curves over K(S). This implies that K(Y ) → K(X) is finite. We have ∗ f∗[f F ] = [K(X): K(Y )][F ] by Proposition 2.11. It now suffices to apply (b) to the pair (f ∗F,D). 9.2. Intersection and morphisms 399

Remark 2.13. We have a more precise version of Part (b) of Theorem 2.12. Let us first define intersection cycles on an integral Noetherian scheme X of dimension 2. Let Z ∈ Z1(X) be a cycle of codimension 1 on X and let D be a Cartier divisor on X such that Supp D does not contain any irreducible component of Z. We denote by Z0(X) the subgroup of Z(XP) of 0-cycles on X. We define Z.D ∈ Z0(X) in the following way. Write Z = i niZi with Zi irreducible. Then D|Zi is a Cartier divisor on Zi (Lemma 7.1.29). We let X

Z.D := ni[D|Zi ] ∈ Z0(X) i

where [D|Zi ] is the 0-cycle on Zi (hence on X) associated to D|Zi (Definition 7.2.12). The 0-cycle Z.D is obviously additive in Z and in D. In the case when Z is the cycle [C] associated to some effective Cartier divisor C, we have [C].D = [D|C ]. P Indeed, if [C] = i niZi with Zi irreducible of generic point ξi, then ni = multξi (C) = length OC,ξi is equal to the multiplicity of Zi in the scheme C. InP the course of the proof of Proposition 7.5.7,P we saw that multx(D|C ) = i ni multx(D|Zi ). Hence the equality [D|C ] = i ni[D|Zi ] = [C].D. If π : X → S is a regular fibered surface and Z = [C] for some Cartier divisor C on X, then we have clearly X π∗([C].D) = C · D, is(C,D) = multx([C].D)[k(x): k(s)]

x∈Xs the second equality being true only if C or D has support contained in Xs. Let f : X → Y be a proper surjective morphism of integral Noetherian schemes of dimension 2. Let Z ∈ Z1(X), D ∈ Div(Y ). Suppose that f −1(Supp D) does not contain any irreducible component of Z. Then we have in Z0(Y ):

∗ f∗(Z.f D) = (f∗Z).D (Projection formula).

To prove this formula, we can suppose that Z is irreducible. If f(Z) = {y} is a point, then we see easily that both sides of the above formula vanish. Suppose now that V := f(Z) is one-dimensional. Then g = f|Z : Z → V is a surjective finite morphism. By Proposition 7.1.38, we have

∗ ∗ ∗ f∗(Z.f D) = g∗[(f D)|Z ] = g∗[g (D|V )] = [k(Z): k(V )]V.D = (f∗Z).D.

Example 2.14. Let f : X → Y be a finite dominant morphism of regular Noetherian schemes of dimension 2, of degree n = [K(X): K(Y )]. Let C and D be distinct prime divisors on Y . Let us suppose that C0 := f −1(C) 0 −1 0 0 and D := f (D) are irreducible. We have f∗C = [K(C ): K(C)]C. Let eC 0 denote the ramification index of OY,ξ → OX,ξ0 , where ξ (resp. ξ ) is the generic 0 ∗ 0 point of C (resp. of C ). Let us define eD in a similar way. Then f C = eC C 400 9. Regular surfaces

0 (Exercise 7.2.3(b)), and n = eD[K(D ): K(D)] (Lemma 7.1.36(c)). The projec- tion formula described in Remark 2.13 then gives X 0 0 niy(C,D) = eC eD ix(C ,D )[k(x): k(y)].

x∈Xy

Base change Let X → S be a regular fibered surface. Let λ : S0 → S be a dominant morphism 0 0 of Dedekind schemes with dim S = dim S. For example, S can be Spec OS,s or Spec ObS,s for a closed point s ∈ S, or a scheme that is finite and surjective 0 over S. Let us suppose that X ×S S is integral and admits a desingularization 00 0 00 X → X ×S S . In general, X → S is not projective, nor even of finite type. We therefore cannot use Theorem 2.12 directly with the morphism f : X00 → X 00 0 0 (composition of X → X ×S S and the projection X ×S S → X), except when S0 → S is finite. For any s0 ∈ S0, if s = λ(s0), then

e 0 := length (O 0 0 /m O 0 0 ) s /s OS0,s0 S ,s s S ,s is finite because OS0,s0 /msOS0,s0 is of dimension 0. Proposition 2.15. Let X → S be a regular fibered surface. Let λ : S0 → S be a dominant morphism of Dedekind schemes with dim S0 = dim S. Let us suppose 0 0 00 0 that X := X ×S S is integral and admits a desingularization g : X → X . Let f : X00 → X be the natural morphism. Let C,D be Cartier divisors on X, of which at least one is vertical. Then for any closed point s ∈ S and s0 ∈ λ−1(s), we have ∗ ∗ is0 (f C, f D) = es0/sis(C,D). (2.5) Proof We can suppose that S, S0 are local. Using the moving lemma 1.10, and by bilinearity, we reduce to the case when C and D are two distinct prime divisors. Let p : X0 → X denote the projection morphism and let C0 = p∗C, D0 = p∗D. We are in position to apply the projection formula of Remark 2.13 to the birational morphism g. We then get

∗ ∗ ∗ 0 ∗ 0 ∗ 0 0 0 0 0 g∗([f C].f D) = g∗([g C ].g D ) = g∗[g C ].D = [C ].D = [D |C0 ] Let π, π0, π00 be respectively the structural morphisms of X,X0 and X00. Then 00 ∗ ∗ 0 0 π∗ ([f C].f D) = π∗[D |C0 ]. Let J = OX (−C) + OX (−D). Let x ∈ Supp C. −1 Then multx([D|C ]) = length(OC,x/JOC,x) = [k(x): k(s)] lengthOS (OX /J )x (Lemma 7.1.36(a)). Therefore

0 is(C,D) = mults(π∗[D|C ]) = lengthOS H (X, OX /J ) because OX /J is a skyscraper sheaf on X. Similarly,

∗ ∗ 0 0 0 0 i 0 (f C, f D) = mult 0 (π [D | 0 ]) = length H (X , O 0 /(J )). s s ∗ C OS0 X

0 0 0 ∗ ∗ 0 0 0 As H (X , OX /(J )) = H (X, OX /J ) ⊗OS OS , we have is (f C, f D) = es0/sis(C,D) by Exercise 7.1.8(b).