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ma 4.6. Using Lemmas 4.5–4.6 we deﬁne the intersection number D1 · · · Dn of any n divisors on a nonsingular n-dimensional variety, without requiring them to be in general position. For this, we ﬁnd any divisor D 1,..., D n satisfying the assumptions of Lemma 4.5, so that the intersection number D · · · D n is deﬁned, and deﬁne 1 D1 · · · Dn by D1 · · · Dn = D n. We need to verify that this deﬁnition is inde1 pendent of the choice of the auxiliary divisors D n, but this is exactly what Lemma 4.6 guarantees. 1 For example, we can now speak of the selﬁntersection number CC of a curve C on a surface X. This number is also denoted by C2. We give some examples of how C2 is computed. Example 4.5 Let X = P2, and let C ⊂ P2 be a line. By deﬁnition C2 = CC where C ∼ C ∼ C and C and C are in general position. We can, for example, take C and C to be two distinct lines. These intersect in a single point x, and since they are transverse at x, we have CC = (CC)x = 1. Hence C2 = 1. Example 4.6 Let X ⊂ PN be an n-dimensional nonsingular projective variety. Write E for the intersection of X with a hyperplane of PN. Obviously E ∈ Div X. Our aim is to give an interpretation of the number En. (We have seen in Section 1.4, 244 4 Intersection Numbers Chapter 3 that all hyperplanes deﬁne linearly equivalent divisors, so that this number does not depend on the choice of the hyperplane E.) By deﬁnition En = E(1) · · · E(n), where E(i) for i = 1,..., n are hyperplane sections of X in general position. By Section 6.2, Chapter 1, these always exist. Then the points xi ∈ E(1) ∩ · · · ∩ E(n) are the points of
intersection of X with a (N − n)dimensional projective linear subspace L = PN −n in general position with X. Since E(1) · · · E(n) = xi ∈X∩L(E(1) · · · E(n))xi and each (E(1) · · · E(n))xi > 0, it follows that En is ≥ the number of points of X ∩ L. Now if L is transversal to X at every = 1 for each xi, and En is equal to the point of intersection then (E(1) · · · E(n))xi number of points of X ∩ L. We check that such a subspace L does exist, which gives us the following interpretation of the number En: it is the maximum number of points of intersection of X with a projective linear subspace PN −n of complimentary dimension in general position with respect to X. This number is called the degree of X and denoted by deg X. For the case of a hypersurface see Example 4.8. The existence of the required subspace L is proved by the traditional method of dimension counting (compare Section 6.4, Chapter 1). Consider the variety of projective linear subspaces L = PN −n ⊂ PN ; this is the Grassmannian G = Grass(N − n + 1, V ) (see Example 1.24), where PN = P(V ), that is, dim V = N + 1. In the product X × G, consider the subvariety Γ of pairs (x, L) such that the subspace L is not in general position with ΘX,x. This is obviously a closed subspace (for example, the conditions that x ∈ L and 0 L ∩ ΘX,x ⊂ ΘPN,x can be written as the vanishing of minors of matrixes made up by the linear equations of the subspace L). The ﬁbre of the ﬁrst projection Γ → X above x ∈ X consists of subspaces L ∈ Grass(N − n, ΘPN,x) that are in nongeneral position with respect to ΘX,x. Its dimension is at most dim Grass(N − n, ΘPN,x) − 1 = (N − n)n − 1. Hence
dim Γ ≤ (N − n)n − 1 + n. A fortiori the projection of Γ to G has dimension ≤(N − n)n − 1 + n. But dim G = (N − n + 1)n, and hence there exists a point of G not contained in the projection of Γ. Example 4.7 Let X ⊂ P3 be a nonsingular surface of degree m and L ⊂ X a line; we calculate L2. Consider a plane of P3 containing L and not tangent to X at at least one point of L, and let E be the hyperplane section of X by this plane. Then L is contained in E as a component of multiplicity 1 E = L + C, with C = kiCi and ki deg Ci = m − 1. We compute ﬁrst C2. For this, we observe that the curve E is singular at a point of intersection of L and C, which means that the plane cutting out E equals the tangent plane to X at this point. Consider another plane through L distinct from the tangent planes at all the points of L ∩ C. This plane deﬁnes a divisor E = L + C, and the points of L ∩ C and L ∩ C are all distinct. This means that C ∩ C = ∅; hence C2 = CC = 0. Now using EL = 1 (since L ⊂ P3 is a line), we get m = E2 = E(L + C) = EL + EC = 1 + EC, 1 Deﬁnition and Basic Properties 245 and hence EC = m − 1; since we have just proved that C2 = 0, m − 1 = EC = (L + C)C = LC + C2 = LC; and ﬁnally, 1 = EL = L2 + LC = L2 + m − 1, therefore L2 = 2 − m. Note that L2 < 0 if m > 2. Lines can indeed lie on a nonsingular surface of + arbitrary degree, for example the line x0 = x1, x2 = x3 on the surface xm 0 xm 2 − xm 1 − xm 3 = 0. 1.5 Exercises to Section 1 1 Let X be a surface, x ∈ X a nonsingular point, u, v local parameters at
x, and f a local equation of a curve C in a neighbourhood of x. If f = (au + bv)(cu + dv) + g x, and the linear forms au + bv and cu + dv are not proportional then with g ∈ m3 we say that x is a double point with distinct tangent directions or node and the lines in Θx with equations au + bv = 0 and cu + dv = 0 are called the tangent lines to C at x (compare Exercise 12 of Section 3.3, Chapter 2). Under the stated assumptions, let C be a nonsingular curve on X passing through x. Prove that (CC)x > 2 if and only if ΘC,x is one of the tangent directions to C at x. 2 Let C = V (F ) and D = V (G) be two plane curves in A2 and x a nonsingular point on both of them. Let f be the restriction of F to the curve D and vx(f ) the order of zero of f at x. Prove that this number is unchanged if F and G are interchanged. 3 Let Y be a nonsingular irreducible codimension 1 subvariety of an n-dimensional nonsingular variety X. Prove that if D1,..., Dn−1 are divisors on X in general position with Y at x then (D1 · · · Dn−1Y )x = (ρY (D1) · · · ρY (Dn−1))x, where the right-hand side is the intersection number computed in Y. 4 Find the degree of the surface vm(P2), where vm is the Veronese embedding. 5 Let X ⊂ Pn be a nonsingular projective surface and L = Pn−2 ⊂ Pn a projective linear subspace of dimension n − 2. Suppose that L and X intersect in a ﬁnite number of points, and that at k of these points, L intersects the tangent plane ΘX,x in a line. Prove that the number of points of intersection of L and X is at most deg X − k. 6 Let X ⊂ Pn be a nonsingular projective surface not contained in any Pn−1 and L = Pn−m a (n − m
)-dimensional projective linear subspace such that X ∩ L is ﬁnite. Suppose that at k of these points, L intersects the tangent plane ΘX,x in a line. Prove that the number of points of intersection of L and X is ≤ deg X − k − m + 2. 246 4 Intersection Numbers [Hint: Find a suitable projective linear subspace passing through L and satisfying the assumptions of Exercise 5. Start from the case k = 0.] 7 Prove that if D1,..., Dn−1 are effective divisors in general position on a nonsingular n-dimensional variety and C ⊂ Supp D1 ∩ · · · ∩ Supp Dn−1 is an irreducible component then (D1 · · · Dn−1)C = min(D1 · · · Dn−1D)x, where the minimum is taken over all x ∈ C and all effective divisors with x ∈ Supp D. 8 Compute (D1D2)C, where D1, D2 ⊂ A3 are given by x = 0 and x2 + y2 + xz = 0 respectively, and C is the line x = y = 0. 2 Applications of Intersection Numbers 2.1 Bézout’s Theorem in Projective and Multiprojective Space Theorems 4.1 and 4.2 allow us to compute intersection numbers of any divisors on a variety X provided that we know the divisor class group Cl X well enough. We illustrate this with two examples. Example 4.8 (X = Pn) We know that Cl X ∼= Z, and that we can take a hyperplane divisor E as generator. Any effective divisor D is the divisor of a form F, and if deg F = m then D ∼ mE. It follows that if Di ∼ miE for i = 1,..., n then D1 · · · Dn = m1 · · · mnEn = m1 · · · mn, (4.19) since obviously En = 1. If the divisors Di are effective, that is, they correspond to forms Fi of degree mi, Supp Di are exactly the nonzero and are in general position, then the points of solutions of the system of equations F1(x0,
..., xn) = · · · = Fn(x0,..., xn) = 0. Here we only consider nonzero solutions, and consider proportional solutions to be the same. For such a point x (or solution), it is natural to call the local intersection number (D1 · · · Dn)x the multiplicity of the solution. Then (4.19) says that the number of solutions of a system of n homogeneous equations in n + 1 unknowns is either inﬁnite, or is equal to the product of the degrees of the equations, provided that solutions are counted with their multiplicities. This result is called Bézout’s theorem in projective space Pn. In particular, if D2,..., Dn are hyperplanes then we see that DEn−1 = deg F, where F = 0 is the equation of D. If D is a nonsingular hypersurface then by definition the intersection number DEn−1 in Pn is equal to the intersection number En−1 on D. Therefore deg F = deg D in the sense of the deﬁnition in Example 3.2. 2 Applications of Intersection Numbers 247 Example 4.9 X = Pn × Pm. In this case Cl X = Z ⊕ Z, since by Theorem 1.9, any effective divisor D is deﬁned by a polynomial F that is homogeneous separately in the two sets of variables x0,..., xn and y0,..., ym (the coordinates in Pn and Pm respectively). If F has degree of homogeneity k and l then D → (k, l) deﬁnes an isomorphism Cl X ∼= Z ⊕ Z. In particular, we can take as generators of Cl X the divisors E and F deﬁned respectively by linear forms in the xi and yi ; then D ∼ kE + lF. Suppose that Di ∼ kiE + liF for i = 1,..., n + m. Then D1 · · · Dn+m = (kiE + liF ) = n+m i=1 ki1 · · · kir lj1 · · · ljs Er F s, (4.20) where the sum runs over all perm
utations (i1... ir j1... js) of {1,..., n + m} with i1 < i2 < · · · < ir and j1 < j2 < · · · < js. We now compute the intersection number Er F s. If r > n then we can ﬁnd r linear forms E1,..., Er having no common zeros in Pn, so that Er F s = E1 · · · Er F s = 0. The same thing happens if s > m. Since r + s = n + m, the intersection number Er F s can only be nonzero for r = n and s = m. In this case we can take E1,..., En and F1,..., Fm to be the divisors deﬁned by the forms x1,..., xn and y1,..., ym. These divisors have a unique common point, (1 : 0 : · · · : 0). They intersect transversally there, as one checks easily on passing to the open subset x0 = 0, y0 = 0, which is isomorphic to An+m. Thus D1 · · · Dn+m = (kiE + liF ) = n+m i=1 ki1 · · · kinlj1 · · · ljm, (4.21) where the sum runs over all permutations (i1... inj1... jm) of {1,..., n + m} with i1 < i2 < · · · < in and j1 < j2 < · · · < jm. This result is called Bézout’s theorem in Pn × Pm. One common feature of the two examples just treated is that Cl X is ﬁnitely generated. It is natural to ask whether this holds for any nonsingular variety X. This is not so; a counterexample is provided by the nonsingular plane cubic curve, which has a subgroup Cl0 X ⊂ Cl X with Cl X/ Cl0 X ∼= Z, and the elements of Cl0 X are in one-to-one correspondence with points of X. Hence, for example, if k = C then Cl0 is not even countable. This big sub
group Cl0 X has, however, no effect on intersection numbers, since deg D = 0 for D ∈ Cl0 X. The same thing also holds for an arbitrary nonsingular projective variety. Namely, one can prove14 that if a divisor D is algebraically equivalent to 0 (see Section 4.4, Chapter 3 for the deﬁnition), then D1 · · · Dn−1D = 0 for 14See Fulton [29, Chapter 10] for a proof (in much more advanced terms). 248 4 Intersection Numbers any divisors D1,..., Dn−1. Thus intersection numbers depend only on the classes of divisors in Div X/ Diva X (the Néron–Severi group NS X). Theorem D asserts that this group is ﬁnitely generated. Obviously, if E1,..., Er are generators of this group, then in order to know any intersection numbers of divisors on X, it is enough to know the ﬁnitely many numbers Ei1 · · · Eir r with i1 + · · · ir = dim X, by analogy 1 with what we saw in Examples 4.8–4.9. In other words, an analogue of Bézout’s theorem holds for X. 2.2 Varieties over the Reals The different versions of Bézout’s theorem proved in Section 2.1 have some pretty applications to algebraic geometry over R. We return to Example 4.8, and suppose that the equations Fi = 0 for i = 1,..., n have real coefﬁcients, and that we are interested in real solutions. If deg Fi = mi and the divisors Di are in general position then D1 · · · Dn = m1 · · · mn, as proved in Example 4.8. By deﬁnition, D1 · · · Dn = (D1 · · · Dn)x, where the sum runs over solutions x of the system of equations F1 = · · · = Fn = 0. In this we must of course consider both real and complex solutions x. However, since the Fi have real coefﬁcients, whenever x is a solution then so is the complex conjugate x. By deﬁnition of the intersection number it follows
at once that (D1 · · · Dn)x = (D1 · · · Dn)x, and hence D1 · · · Dn ≡ (D1 · · · Dn)y mod 2, where now the sum takes place only over real solutions. In particular if D1 · · · Dn is odd (which holds if and only if all the degrees deg Fi = mi are odd), then we deduce that there exists at least one real solution. This assertion is proved under the assumption that the Di are in general position. But the following simple argument allows us to get rid of this restriction. The point is that in our case the theorem on moving the support of a divisor can be proved very simply and in a more explicit form. Namely, we can choose a linear form l nonzero at all the points x1,..., xr we want to move the support of the divisor away from. If D is deﬁned by a form F of degree m then the divisor D deﬁned by the form Fε = F + εlm will satisfy all the conditions in the conclusion of the theorem if F (xi) + εl(xj )m = 0 for j = 1,..., r. These conditions can be satisﬁed for arbitrarily small values of ε. We now show how to get rid of the general position restriction in the assertion we proved above on the existence of a real solution of a system of equations of odd degrees. Let F1 = · · · = Fn = 0 (4.22) be any such system. By what we have said above we can ﬁnd linear forms li and arbitrarily small values of ε such that the divisors deﬁned by the forms Fi,ε = Fi + εlmi are in general position. Now we proved above that the system F1,ε = · · · = i Fn,ε = 0 has a real solution xε. Because projective space is compact, we can ﬁnd a sequence of numbers εm → 0 such that the sequence xεm converges to a point x ∈ Pn. Now Fj,εm → Fj as εm → 0, so that x is a solution of the system (4.22). We state the result we have just proved. 2 Applications of Inter
section Numbers 249 Theorem 4.3 A system of n homogeneous real equations in n + 1 variables has a nonzero real solution if the degree of each equation is odd. Entirely analogous arguments apply to the variety Pn × Pm (see Example 4.9). We get the following result. Theorem 4.4 A system of real equations Fi(x0 : · · · : xn; y0 : · · · : ym) = 0 for i = 1,..., n + m · · · ljm is odd. Here ki and has a nonzero real solution if the number li are the degrees of homogeneity of Fi in the two sets of variables, and we consider a solution to be zero if either x0 = · · · = xn = 0 or y0 = · · · = ym = 0. · · · kinlj1 ki1 Theorem 4.4 has interesting applications to algebra. One of these is concerned with the question of division algebras over R. If an algebra over R has rank n then it has a basis e1,..., en, and the algebra structure is determined by a multiplication table eiej = n k=1 ck ij ek for i, j = 1,..., n. (4.23) We do not assume that the algebra is associative, so that the structure constants ck ij can be arbitrary. The algebra is called a division algebra if the equation ax = b (4.24) has a solution for every a = 0 and every b. It is easy to see that this is equivalent to the nonexistence of zerodivisors in the algebra. For this it is enough to consider the linear map ϕ given by ϕ(x) = ax in the real vector space formed by elements of the algebra. The condition that (4.24) has a solution means that the image of ϕ is the whole space, and this is equivalent to ker ϕ = 0, as is well known. This condition means just that the algebra has no zerodivisors, that is xy = 0 implies either x = 0 or y = 0. If x = n j =1 yj ej, then (4.23) gives n i=1 xiei and y = xy = n k=1 zkek, where zk = n n i=
1 j =1 ck ij xiyj for k = 1,..., n. Thus the algebra is a division algebra if the system of equations Fk(x, y) = n n i=1 j =1 ck ij xiyj = 0 for k = 1,..., n (4.25) has no real solutions with (x1,..., xn), (y1,..., yn) = (0,..., 0). These equations very nearly satisfy the conditions of Theorem 4.4. The difference is that the Fk 250 4 Intersection Numbers deﬁne divisors in Pn−1 × Pn−1, the number n of which is not equal to the dimension 2n − 2 of the variety. We therefore choose any integer r with 1 ≤ r ≤ n − 1 and set xr+2 = · · · = xn = 0 and yn−r+2 = · · · = yn = 0. The equations Fk((x1,..., xr+1, 0,..., 0), (y1,..., yn−r+1, 0,..., 0)) = 0 for k = 1,..., n are now deﬁned in Pr × Pn−r, and a fortiori have no nonzero real roots. According to Theorem 4.4 this is only possible if the sum ki1 · · · kir lj1 · · · ljn−r (4.26) is even, and this must moreover hold for all r = 1,..., n − 1. In our case the forms Fk are bilinear, so that ki = li = 1, and the sum (4.26) equals the number of summands, which is. We see that if (4.25) has no nonzero real solutions then all the are even for r = 1,..., n − 1. This is only possible if n = 2k. Indeed, integers can be expressed as follows: over the ﬁeld with 2 elements F2 our condition on we have (T + 1)n = T n + 1. If n = 2lm with m odd and m > 1 then over F2, n r n r n r (
T + 1)2l m = T 2l + 1 m = T 2l m + mT 2l (m−1) + · · · + 1 = T 2l m + 1. We have proved the following result: Theorem 4.5 The rank of a division algebra over R is a power of 2. It can be proved that a division algebra over R exists only for n = 1, 2, 4 and 8. The proof of this fact uses rather delicate topological arguments. Applying analogous arguments, one can investigate for which values of m and n the system of equations n m k=1 j =1 ck ij xkyj = 0 for i = 1,..., n, (4.27) does not have nonzero real solutions. Based on the interpretation of the tangent space to P(V ) given in Section 1.3, Chapter 2, one can easily show that under the stated assumption, (4.27) deﬁnes (m − 1) linearly independent tangent vectors at each point of Pn−1, that is, (m − 1) everywhere linearly independent vector ﬁelds on Pn−1. In this form, the question of the possible values of m and n has been completely answered by topological methods. The question is interesting because it is equivalent to that of knowing whether the system of partial differential equations n m k=1 j =1 ck ij ∂uj ∂xk = 0 for i = 1,..., m is elliptic. 2 Applications of Intersection Numbers 251 2.3 The Genus of a Nonsingular Curve on a Surface The following formula plays an enormous role in the geometry on a nonsingular projective surface X. It is usually called the adjunction formula or the genus formula, and expresses the genus of a nonsingular curve C ⊂ X in terms of certain intersection numbers: gC = 1 2 C(C + K) + 1; (4.28) here gC is the genus of C and K the canonical class of X. This formula can be proved using only the methods we already know. However, a clearer and more transparent geometric proof follows from the elementary properties of vector bundles. This is given in Theorem 6.4 of Section 1.4, Chapter 6. Here we only discuss a number of applications. Example 4.10 (The projective
plane) If X = P2 then Cl X = Z, with generator L, the class containing all the lines of P2. If C ⊂ P2 has degree n then C ∼ nL. In view of K = −3L and L2 = 1, in this case (4.28) gives g = n(n − 3) 2 + 1 = (n − 1)(n − 2) 2. We obtained the same result in Section 6.4, Chapter 3 by a different method. Example 4.11 (The nonsingular quadric surface) Let X ⊂ P3 be a nonsingular quadric surface in P3. Let’s see how to classify nonsingular curves on X in terms of their geometric properties. The algebraic classiﬁcation is clear. Since X ∼= P1 × P1, any curve on X is deﬁned by an equation F (x0 : x1; y0 : y1) = 0, where F is a polynomial homogeneous in the two sets of variables x0, x1 and y0, y1; write m and n for the degrees of homogeneity. F has (m + 1)(n + 1) coefﬁcients, and hence the curves of bidegree (m, n) correspond to points of PN, where N = (m + 1)(n + 1) − 1. There exists a nonsingular irreducible curve of any bidegree (m, n) with m > 0, n > 0, for example the curve given by 2xm 0 yn 0 + xm 0 yn 1 + xm 1 yn 0 + xm 1 yn 1 = 0; thus nonsingular irreducible curves correspond to points of a nonempty open set of PN. We saw in Section 2.1 that Cl X = Z ⊕ Z, and that a curve C given by a polyno- mial of bidegree (m, n) satisﬁes C ∼ mE + nF, (4.29) where E = P1 × x and F = x × P1. Thus the curves corresponding to the given bidegree (m, n) are the effective divisors of the class mE + nF. 252 4 Intersection Numbers The classes E and F correspond to
the two families of line generators on X. It is easy to ﬁnd the intersection number of curves given in the form (4.29): if C ∼ mE + nF and C ∼ mE + nF then In particular CC = mn + nm. m = CF and n = CE. (4.30) (4.31) (4.32) This shows the geometric meaning of m and n: just as the degree of a plane curve equals the number of points of intersection with a line, so m and n are the two degrees of C with respect to the two families of line generators E and F on X. Taking account of the embedding X ⊂ P3 provides a new geometric invariant of a curve, its degree. We know that the families of curves on X are simply classiﬁed by the invariants m and n. Our aim at present is to recover this classiﬁcation in terms of the invariants deg C and gC. We know that where H is a hyperplane section of X. Now note that deg C = CH, H ∼ E + F (4.33) (4.34) which follows at once from (4.32) and from the fact that H intersects both E and F transversally in one point. Substituting in (4.33) and using (4.31) gives deg C = m + n. (4.35) Note that, except for the case C a line, any irreducible curve C has m > 0, n > 0. Indeed, if C is not, say, in the ﬁrst family of line generators, then taking any point x ∈ C and the line E of the ﬁrst family through x, we see that C and E are in general position and CE = n ≥ (CE)x > 0. We proceed to calculate gC. To apply (4.28), we need to know the canonical class of X, which we now determine. We use the fact that X ∼= P1 × P1. It is easy to solve the even more general question, to ﬁnd the canonical class of a surface X = Y1 × Y2 which is the product of two nonsingular projective curves Y1 and Y2. We write π1 : X → Y1 and π2 : X → Y2 for the two projections,
consider arbitrary 1-forms ω1 ∈ Ω 1(Y1) and ω2 ∈ Ω 1(Y2) and the pullback 1-forms π ∗ 2 (ω2) on X. Then ω = π ∗ 2 (ω2) is a 2-form on X and its divisor div ω belongs to the canonical class. We now calculate this divisor. 1 (ω1) and π ∗ 1 (ω1) ∧ π ∗ Let x = (y1, y2) ∈ X where y1 ∈ Y1 and y2 ∈ Y2, and write t1, t2 for local parameters on Y1 and Y2 in a neighbourhood of y1 and y2. An obvious veriﬁcation then shows that π ∗ 2 (t2) is a local system of parameters at x ∈ X. Write ω1 and ω2 in the form ω1 = u1dt1 and ω2 = u2dt2. Then div(ω1) = div(u1) and 1 (t1), π ∗ 2 Applications of Intersection Numbers 253 1 (t1) ∧ dπ ∗ div(ω2) = div(u2) in a neighbourhood of y1 and y2. Obviously ω = π ∗ dπ ∗ 2 (t2), and it follows that in some neighbourhood of x, + π ∗ div(ω2) 2 π ∗ 1 (u1) div(ω1) π ∗ 2 (u2) div(ω) = div + div = π ∗ 1 1 (u1)π ∗. 2 (u2) · Since this holds for any point x ∈ X it follows that div(ω) = π ∗ π ∗ 2 (div(ω2)), or in other words, 1 (div(ω1)) + KX = π ∗ 1 (KY1 ) + π ∗ Now return to the case X = P1 × P1. We know that KP1 = −2y for a point y ∈ P1. Thus in our case, (4.36) gives KX = −2π ∗ 1 (y1) = E and π ∗ 2 (y2) = F,
we ﬁnally get 2 (y2). Since π ∗ 1 (y1) − 2π ∗ 2 (KY2 ). (4.36) KX = −2E − 2F. (4.37) Now for the genus of a curve C ∼ mE + nF, we substitute this formula into (4.28) and use (4.31). We get gC = (m − 1)(n − 1). (4.38) The numbers m and n are thus determined uniquely up to permutation by the degree and genus of C. We see that curves on X of given degree d form d + 1 families M0,..., Md. Curves in Mk have genus (k − 1)(d − k − 1); curves in Mk and Ml have the same genus if and only if k = l or k + l = d, that is, the two families are obtained from one another by the automorphism of P1 × P1 that interchanges the two factors. The dimension of Mk is (k + 1)(d − k + 1) − 1, which in terms of the degree and genus is g + 2d − 1. In his “Vorlesungen über die Entwicklung der Mathematik im 19. Jahrhundert,” Chapter VII, p. 319, Felix Klein gives the classiﬁcation of curves of degree 3 and 4 on the hyperboloid as an example of the application of ideas of birational geometry. We take pictures of curves with d = 4 from this reference: Figure 16, (b) has m = 1, n = 3 and Figure 16, (a) has m = n = 2. Example 4.12 (Curves on the cubic surface) As a further application of (4.28), we determine the possible negative values for the selﬁntersection of curves C on a cubic surface of P3. According to the result of Section 6.4, Chapter 3, in this case K = −E, where E is the hyperplane section. Thus (4.28) takes the form C2 − deg C = 2g − 2. Obviously C2 < 0 only if g = 0 and deg C = 1, that is, C is a line of the cubic surface. In this case C2 = −1. 2.4 The Riemann–Roch
Inequality on a Surface Recall from Section 1.5, Chapter 3 that we write (D) for the dimension of the vector space associated with a divisor D. Another fundamental relation involving inter- 254 4 Intersection Numbers Figure 16 Curves on a quadric surface section numbers on an irreducible nonsingular projective surface X is the Riemann– Roch inequality: (D) + (K − D) ≥ 1 2 D(D − K) + χ(OX); (4.39) here D is an arbitrary divisor on X and χ(OX) is an invariant depending only on X and not on D; in the case of a ﬁeld of characteristic 0 we have χ(OX) = 1−h1(X)+ h2(X), where hr = dim Ω r [X] are as deﬁned in Section 6.1, Chapter 3. Inequality (4.39) is obtained by omitting one term from the Riemann–Roch equality, which we do not treat here. The Riemann–Roch equalities for curves and surfaces generalise to varieties of arbitrary dimension. We illustrate the usefulness of the Riemann–Roch inequality by discussing one of its consequences. As mentioned in Section 2.1, the intersection number of divisors D1, D2 ∈ Div X depends only on their images in Div X/ Diva X, which is a ﬁnitely generated group. We can also pass to the quotient by the torsion subgroup, since torsion elements, of course, give zero intersection numbers. As a result we get a group isomorphic to Zm, and if u1,..., um is a basis, intersection numbers deﬁne a symmetric integral matrix (uiuj ), that is, an integral quadratic form. This is an extremely important invariant of a surface. We now determine the crudest invariant of this quadratic form, its index of inertia. It certainly takes positive values, since E2 = deg X > 0, where E is a hyperplane section. It turns out that on reducing the quadratic form to a sum of squares, all but one of the nonzero diagonal entries are negative. We prove this result in a form that does not use the result that Div X/ Diva X is ﬁnitely generated
. 2 Applications of Intersection Numbers 255 Hodge Index Theorem If D is a divisor on a surface X and DE = 0, where E is the hyperplane section, then D2 ≤ 0. Proof Suppose that D2 > 0. We prove that for all sufﬁciently large n > 0, either (nD) > 0 or (−nD) > 0. The theorem will follow from this: if, say, (nD) > 0, then nD is linearly equivalent to an effective divisor, that is; nD ∼ D > 0; therefore nDE = DE > 0, because every curve intersects a hyperplane. Hence nDE > 0 and so also DE > 0, which contradicts the assumption. Using (4.39), the assumption D2 > 0 implies that (nD) + (K − nD) ≥ c(n) and (−nD) + (K + nD) ≥ c(n), (4.40) where c(n) grows with n without bound. If (nD) = (−nD) = 0, we get (K − nD) ≥ c(n) and (K + nD) ≥ c(n). But now if (D1) > 0, we always have (D1 + D2) ≥ (D2); thus we would deduce that (2K) ≥ c(n), which is an ob- vious contradiction. The theorem is proved. 2.5 The Nonsingular Cubic Surface Let X ⊂ P3 be a nonsingular cubic surface. X contains a line L, by Theorem 1.28. Through L, pass two distinct planes E1 and E2 with equations ϕ1 = 0 and ϕ2 = 0, and consider the rational map ϕ : X → P1 given by ϕ(x) = (ϕ1(x) : ϕ2(x)). The linear system λ1ϕ1 + λ2ϕ2 corresponding to this map has L as a ﬁxed component: if Eλ1,λ2 is the section of X by the plane with equation λ1ϕ1 + λ2ϕ2 = 0 then Eλ1,λ2 = L + Fλ1,λ2, where Fλ1,λ2 is a plane conic. The linear system Fλ
1,λ2 obviously deﬁnes the same map ϕ. We prove that ϕ is regular. For this it is enough to prove that = ∅ if (λ1 : λ2) = (μ1 : μ2). Note that Fλ1,λ2 cannot contain L Fλ1,λ2 = 3L or 2L + L would contradict the relations as a component: an equality Eλ1,λ2 L2 = −1, Eλ1,λ2 L = 1 and LL ≥ 0 (see Example 4.7). Moreover, Fλ1,λ2 and Fμ1,μ2 cannot have a common component; indeed, this would have to be a line distinct from L, and would determine the plane containing it. Thus Fλ1,λ2 and Fμ1,μ2 are = 0, that is, F 2 = 0, in general position, and it is enough to prove that Fλ1,λ2 Fμ1,μ2 where F = E − L. This follows from E2 = 3, EL = 1 and L2 = −1. ∩ Fμ1,μ2 If L is the line given by ξ0 = ξ1 = 0 then the equation of X can be written as A(ξ0, ξ1)ξ 2 2 + 2B(ξ0, ξ1)ξ2ξ3 + C(ξ0, ξ1)ξ 2 3 + 2D(ξ0, ξ1)ξ2 + 2E(ξ0, ξ1)ξ3 + F (ξ0, ξ1) = 0, (4.41) where A, B, C, D, E and F are forms in ξ0, ξ1 of degrees deg A = deg B = deg C = 1, deg D = deg E = 2 and deg F = 3. We see from this that our map ϕ : X → P1 represents an open set V = ϕ−1(A1) ⊂ X as a pencil of conics V → U over the afﬁne line A1 ⊂ P1; we always choose the coordinates so that the ﬁbre over the point at inﬁnity P1 \ A1 is a nondegenerate conic. From Example 2.7,
it follows 256 Figure 17 Lines on the cubic surface 4 Intersection Numbers that the degenerate ﬁbres correspond to zeros of the discriminant, each zero has multiplicity 1, and each degenerate ﬁbre is a pair of distinct lines. Then the number of degenerate ﬁbres equals the degree of the discriminant Δ = det A C D C B E D E F, which is 5. The next result follows from this. Proposition Every line L on a nonsingular projective cubic surface X meets exactly 10 other lines on X, which break up into 5 pairs of intersecting lines. It follows from Tsen’s theorem and Corollary 1.12 that a nonsingular cubic surface is rational: Δ is not identically zero, since it has only simple roots. The rationality of X can also be proved otherwise: consider any line L intersecting L, and apply Proposition to it. Then L meets 10 lines, of which only L and one further line meet L. Therefore there exists a line M not intersecting L, and the rationality of X follows by Example 1.23. The line M just found obviously satisﬁes MF = 1, where F is the ﬁbre of the conic bundle, since ME = 1, ML = 0 and E ∼ L + F. Hence M intersects F in exactly one point, and, in particular, it intersects exactly one of each pair of lines meeting L. Write L i for this one, and Li for the other, for i = 1,..., 5. Then LiM = 0, L iM = 1. The conﬁguration of lines obtained thus is illustrated in Figure 17. It follows from Theorem 3.4 that the group Cl X has generators the classes deﬁned by the divisors L1, L2, L3, L4, L5, F, S, where S is some section of the conic bundle X → P1. We prove that S can be taken to be the line M found above. Indeed, since M ∩ L = ∅, the equations of M can be written ξ2 = aξ0 + bξ1, ξ3 = cξ0 + dξ1; that is, passing to inhomogeneous coordinates, x2 = ξ2/ξ0 and x3 = ξ3/ξ
0 can be expressed as rational functions of x1 = ξ1/ξ0, the coordinate of P1, and these expressions satisfy (4.41). We thus obtain the following result. 2 Applications of Intersection Numbers 257 Proposition Cl X is a free group with 7 generators, the classes of the lines L1, L2, L3, L4, L5, M and F. The intersection numbers of L1, L2, L3, L4, L5, M and F are easily determined; they are tabulated as follows: L1 L1 −1 L2 0 L3 0 L4 0 L5 0 M 0 F 0 L2 0 −1 0 0 0 0 0 L3 0 0 −1 0 0 0 0 L4 0 0 0 −1 0 0 0 L5 1 0 0 −1 1 0 0 1 0 The group Cl X to a signiﬁcant extent determines the geometry of X. In particular, we now show how to use it to ﬁnd all the lines on X. A line C on X satisﬁes C2 = −1. We know L and a further 10 lines intersecting it. We now try to ﬁnd the lines disjoint from L. These satisfy CL = 0, and therefore CF = 1. Suppose that 5 1 xiLi + yM + zF. Then CF = 1 implies y = 1, and C2 = −1 and CL = 0 C ∼ give 5 − 1 x2 i + 2z = 0, 5 1 xi + 2z = 0; (4.42) 5 1(x2 i + xi) = 0, that is, each xi = 0 or −1. Moreover, (4.42) implies It follows that also that the number of i for which xi = −1 is even, so that either (a) all xi = 0; or (b) all xi = −1 except one; or (c) xi = xj = −1, and the three remaining xk = 0. Of these possibilities, (a) gives the class of the line M, (b) and (c) 5 and 10 cases, that is, 16 classes altogether. Each class contains at most one line: for if C and C are distinct lines then CC = 0 or 1, whereas if C ∼ C are in the same class then CC = C
2 = −1. Thus it remains to exhibit at least one line in each class. In case (a), this is M. In case (b), if xi = 0 and xj = −1 for j = i we get the class Ci = − j =i Lj + i and M lie in the same plane, in which there must αkLk + βM + γ F and + M = E. Setting E ∼ Lk + 2M + 3F. Substituting this expression M + 2F. We note that the lines L be a third line L + L i arguing as before, we get that E ∼ − for E and L i i, so that L i ∼ F − Li for L i, we ﬁnd easily that L In case (c) we get a class Dij = −Li − Lj + M + F. Note that L i Lj = CiLj = 1, so that the lines L i and Lj for i = j intersect, and hence there is a third line Lij in the plane through them. Arguing exactly as before we show that Lij ∼ Dij. Thus we have found 1 line in case (a), 5 in case (b) and (10) in case (c), altogether 1 + 5 + 10 = 16. Together with L and the 10 lines meeting it, this gives 27 lines. This proves the next result. ∼ Ci. i 258 4 Intersection Numbers Theorem A nonsingular cubic surface of P3 contains exactly 27 lines. 2.6 The Ring of Cycle Classes The theory of divisors and their intersection numbers is a particular case of a general theory that deals with subvarieties of any dimension. The notion of divisor is replaced by that of k-cycle. A k-cycle is an element of the free Abelian group generated by irreducible k-dimensional subvarieties. Two irreducible subvarieties Y1 and Y2 are in general position, by deﬁnition, if every irreducible component Zi of the intersection Y1 ∩ Y2 has the same dimension, and codim Zi = codim Y1 + codim Y2. Two k-cycles are in general position if all components of the ﬁrst are in general position with those of the second. The foundation of the theory is a method of assigning to each component Zi of Y1 ∩ Y
2 a positive integral multiplicity ni(Y1, Y2). The cycle Y1 · Y2 = ni(Y1, Y2)Zi is called the product of the subvarieties Y1 and Y2. The notion extends by additivity to any two cycles in general position. The reader can learn about this theory from Fulton [29]. The basic property of this product is that it is invariant under an equivalence relation that we now describe. It generalises the algebraic equivalence of divisors introduced in Section 4.4, Chapter 3, and is deﬁned in an entirely similar way. Namely, let T be an arbitrary irreducible nonsingular variety and Z ⊂ X × t a cycle such that Z and the ﬁbre X × T are in general position for every t ∈ T. The set of cycles Ct = Z · (X × t) is called an algebraic family of cycles. Two cycles C1 and C2 are algebraically equivalent if there exists a family of cycles Ct with t ∈ T such = C2 for two points t1, t2 ∈ T. The set of cycle classes under that Ct1 algebraic equivalence forms a group. = C1 and Ct2 The product of cycles on a nonsingular projective variety is invariant under algebraic equivalence. There is a theorem on reducing to general position (the so-called moving lemma), according to which for any two cycles C1 and C2 there exist cycles C 1 and C 2 equivalent to C1 and C2 respectively, and in general position. These two results allow us to deﬁne the product of any two cycle classes. Now let X be a nonsingular n-dimensional projective variety, and write Ar to denote the group of cycle classes (under algebraic equivalence) of codimension r on X. The group A = n! r=0 Ar is a ring, where the product is deﬁned on the individual summands as described above, and on arbitrary elements by additivity. This ring is commutative and associative. By the formula for the dimension of intersections (1.49), Ar · As ⊂ Ar+s, 2 Applications of Intersection Numbers 259 where we set Am = 0 for m > n. That is, A is a graded ring. It is easy to prove that all points of X, viewed as 0-
cycles, are algebraically equivalent, and the 0cycle x (for x ∈ X) is not algebraically equivalent to 0. Therefore An = Zu, with the standard generator u, the class of a point x ∈ X. The classes of divisors under algebraic equivalence form a group A1. For n elements α1,..., αn ∈ A1 the product α1 · · · αn ∈ An = Zu, that is, α1 · · · αn = ku with k ∈ Z. The number k equals the intersection number α1 · · · αn deﬁned in Section 1. The ring A is a very interesting invariant of X, and is still not well studied. A0 is isomorphic to Z, with generator X itself. As already pointed out, An ∼= Z. The group A1 is ﬁnitely generated, as asserted in Theorem D. However, already A2 may have an inﬁnite number of generators. The structure of these groups is quite mysterious. 2.7 Exercises to Section 2 1 Determine deg vm(Pn) where vm is the Veronese embedding (Section 4.4, Chapter 1). 2 Suppose that a nonsingular plane curve C of degree r lies on a nonsingular surface of degree m in P3. Determine C2. (This generalises Example 4.7.) 3 Suppose that a form of degree l on a nonsingular projective surface of degree m in P3 has divisor consisting of one component of multiplicity 1 that is a nonsingular curve. Find its genus. k i=1 ni simultaneous 4 Consider k sets of variables x(i) equations 0,..., x(i) ni and a system of x(1) 0,..., x(1) n1 fi ;... ; x(k) = 0, 0,..., x(k) nk 0,..., x(i) that are linear in each set of variables x(i) ni. Prove that the number of solutions in Pn1 × · · · × Pnk of the system equals the multinomial coefﬁcient n1+···+nk ni!. Here the number of solutions is taken as usual in the n1,...,nk!/ ni
= sense of the corresponding intersection number. 5 Let X ⊂ P3 be a nonsingular surface of degree 4 and C ⊂ X a nonsingular curve. Prove that if C2 < 0 then C2 = −2. 6 Prove that the selﬁntersection number of a nonsingular curve on a nonsingular surface in P3 of even degree is always even. 7 Let X be a nonsingular curve and D the diagonal of X × X (that is, the set of points of the form (x, x)). Prove that D2 = − deg KX. [Hint: Use the fact that D and X are isomorphic.] 260 4 Intersection Numbers 8 Generalise the result of Exercise 7 to the case that D ⊂ C1 × C2 is the graph of a map ϕ : C1 → C2 of degree d. 9 If D ⊂ C1 × C2 is a divisor, prove the inequality D2 ≤ 2(C1 × c2)D · (C2 × c1)D for c1 ∈ C1 and c2 ∈ C2. [Hint: Cook up α and β such that D = D − α(C1 × c2) − β(C2 × c1) satisﬁes (C1 × c2)D = (C2 × c1)D = 0, then apply the Hodge index theorem to D.] 10 In the notation of Exercises 8–9, suppose that C1 = C2 = C is a curve of genus g. Let ϕ : C → C be a map of degree d with graph Γϕ ⊂ C × C, and write Δ ⊂ C × C for the diagonal. Prove that \|ΓϕΔ − d − 1\| ≤ 2g √ d. [Hint: Set D = mΔ + nΓϕ and view D2 − 2(C × c)D · (C × c)D as a quadratic form in m and n; then write out the condition for this to be negative deﬁnite.] Here ΓϕΔ is the number of ﬁxed points of ϕ. This inequality, applied to the case that ϕ is the Frobenius map, generalises the inequality
(3.47) to curves of arbitrary genus. 3 Birational Maps of Surfaces This section treats an application of intersection numbers to the proof of some basic properties of birational maps of algebraic surfaces. We start by deriving the elementary properties of blowups of algebraic surfaces. 3.1 Blowups of Surfaces Let X be an algebraic surface, ξ ∈ X a nonsingular point, x, y local parameters at ξ and σ : Y → X the blowup centred at ξ. By Theorem 2.15 there exists a neighbourhood U of ξ in X such that V = σ −1(U ) is the subvariety of U × P1 deﬁned by t0y = t1x, where (t0 : t1) are coordinates on P1. In the open set where t0 = 0 the blowup is given by the simple equations x = u and y = uv, where v = t1/t0. (4.43) Set L = σ −1(ξ ). A local system of parameters at any point η ∈ L is given by u and v − v(η). The local equation of L is obviously u = 0. Let C be an irreducible curve on X passing through ξ. By analogy with Theorem 2.15, the inverse image σ −1(C) consists of two components: the exceptional 3 Birational Maps of Surfaces 261 curve L and a curve C that can be deﬁned as the closure in Y of σ −1(C \ ξ ). The curve C is called the birational15 transform of C. We denote it by C = σ (C). Now consider C as an irreducible divisor on X. Then σ ∗(C) = σ (C) + kL, (4.44) where σ (C) appears with coefﬁcient 1, because σ is an isomorphism of Y \ L and X \ ξ. We now determine the coefﬁcient k in (4.44). Suppose for this that C has ξ as an r-fold point. This means that the local equation f of C in a neighbourhood of ξ \ mr+1. Then σ ∗(C) has local equation σ ∗(f ) in a
neighbourhood satisﬁes f ∈ mr ξ ξ of any point η ∈ L. Set f = ϕ(x, y) + ψ with ϕ a form of degree r and ψ ∈ mr+1 ξ. (4.45) Substituting the formulas (4.43) for σ in (4.45), we see that (σ ∗(f ))(u, v) = ϕ(u, uv) + σ ∗(ψ). Since ψ ∈ mr+1, it follows that we can write ψ = F (x, y) with F a form of degree r + 1 in x, y with coefﬁcients in Oξ. Therefore σ ∗(ψ) = (σ ∗(F ))(u, uv), and ﬁnally σ ∗(f ) σ ∗(F ) ϕ(1, v) + u ; (1, v) (u, v) = ur (4.46) ξ since ϕ(1, v) is not divisible by u it follows that k = r in (4.44). We state the result we have proved. Theorem 4.6 If C is a prime divisor on X passing through the centre ξ of a blowup σ then the inverse image σ ∗(C) of C is given by σ ∗(C) = σ (C) + kL, where σ (C) ⊂ Y is a prime divisor, L = σ −1(ξ ) and k is the multiplicity of C at ξ. 3.2 Some Intersection Numbers We start from general properties of a birational regular map f : Y → X between nonsingular projective surfaces. Theorem 4.7 (i) If D1 and D2 are divisors on X then f ∗(D1)f ∗(D2) = D1D2. (4.47) (ii) If D is a divisor on Y all of whose components are exceptional curves of f and D is any divisor on X then f ∗(D)D = 0. (4.48) 15The terms strict transform and proper transform are also widely used in the literature. 262 4
Intersection Numbers Proof We write S ⊂ X for the ﬁnite set of points at which the inverse map f −1 is not regular, and set T = f −1(S) for the set-theoretic inverse image. Then f deﬁnes an isomorphism Y \ T ∼→ X \ S. (4.49) If Supp D1 ∩ S = Supp D2 ∩ S = ∅ and D1 and D2 are in general position then (4.47) is obvious from (4.49). Otherwise we use Theorem 3.1 on moving the support of a divisor away from points. Suppose that D ∼ D1 and D ∼ D2 are divisors with 1 2 Supp D 2 in general position. Then D1D2 = 1 D 1D 2). Since f ∗(D 2, by Lemma 4.6, and by what we said above D i) ∼ f ∗(Di), the required equality (4.47) holds. Equality (4.48) is likewise obvious if Supp D ∩ S = ∅. The general case reduces to this by an entirely similar argument. The theorem is proved. ∩ S = Supp D 2 ∩ S = ∅ and D 1 and D 1)f ∗(D = f ∗(D 1D 2 We now give a corollary that relates directly to blowups. We use the notation of Section 3.1. Corollary 4.1 L2 = −1. (4.50) Proof Consider the curve C ⊂ X with local equation y. By Theorem 4.6 σ ∗(C) = σ (C) + L, and moreover, it is clear from (4.43) that the local equation of σ (C) is v. Since u is the local equation for L, it follows that σ (C)L = 1 and (4.50) follows from (4.48): 0 = σ ∗(C)L = The corollary is proved. σ (C) + L L = 1 + L2. Corollary 4.2 If C ⊂ X is a curve with multiplicity k at ξ then σ (C)L = k. This follows at once from (4.48) and (4.50) and from (4.44). Cor
ollary 4.3 σ (C1)σ (C2) = C1C2 − k1k2, (4.51) where k1, k2 are the multiplicities of C1, C2 at ξ. Proof C1C2 = σ ∗(C1)σ ∗(C2) = σ (C1) + k1L = σ (C1)σ ∗(C2) = σ (C1) = σ (C1)σ (C2) + k1k2; σ (C2) + k2L σ ∗(C2) here the 3rd equality comes from (4.48) and the ﬁnal one from Corollary 4.2. This gives (4.51). The corollary is proved. 3 Birational Maps of Surfaces 263 3.3 Resolution of Indeterminacy We can now prove an important property of rational maps from algebraic surfaces. Theorem 4.8 Let X be a nonsingular projective surface and ϕ : X → Pn a rational map. Then there exists a chain of blowups Xm → · · · → X1 → X such that the composite rational map ψ = ϕ ◦ σ1 ◦ · · · ◦ σm : Xm → Pn is regular; in other words, there is a commutative diagram Xm σm ↓... σ2 ↓ X1 σ1 ↓ − − − − − − − ψ − − − − − − → X −−−−−−−−→ ϕ Pn in which the vertical column is a chain of blowups, and the diagonal arrow ψ is a regular map. Remark The assumption that X is projective is used in the proof of Theorem 4.8 in order to be able to use intersection numbers D1D2. However, the result itself holds for noncomplete surfaces or for complete nonprojective surfaces, and can be proved without difﬁculty by reducing to the statement of Theorem 4.8. Proof By Theorem 2.12, we know that ϕ only fails to be regular at a ﬁnite number of points; Theorem 3.2 gives a more precise description of this set, which we now recall. Suppose that ϕ = (f
0 : · · · : fn), and set div(f0),..., div(fn) and Di = div(fi) − D. D = hcd Then the set of points of irregularity of ϕ is exactly m i=0 Supp Di. We introduce an invariant d(ϕ) of a rational map ϕ as follows. All the divisors Di are obviously linearly equivalent. Hence we can set d(ϕ) = D2 i. Let us prove that d(ϕ) ≥ 0. For this, let λ = (λ0,..., λn) ∈ kn+1, and deﬁne Dλ = n i=0 λifi) − D; obviously Dλ is an effective divisor linearly equivalent to the div( Di. It is enough to prove that there exists λ such that D0 and Dλ have no common components, since then d(ϕ) = D0Dλ ≥ 0. By construction no curve is a common component of all the Di. Hence for every irreducible component C ⊂ D0 there exists some Di such that vC(Di) = 0. If gi are 264 4 Intersection Numbers local equations for the Di in a neighbourhood of some point c ∈ C then vC(Dλ) > 0 ⇐⇒ λi(gi \|C) = 0. Therefore the set of λ such that vC(Dλ) > 0 is a strict vector subspace of kn+1. A vector space (over an inﬁnite ﬁeld) is not the union of ﬁnitely many strict vector subspaces, and hence there exists λ such that vC(Dλ) = 0 for every component C ⊂ D0. Then D0 and Dλ have no common components for this λ. If x0 ∈ Supp Di then x0 ∈ Supp Dλ for every λ. Hence d(ϕ) > 0 if Supp Di = ∅, that is, if ϕ is not regular. If this happens, write σ : X → X for the blowup Supp Di, and set ϕ = ϕ ◦ σ : X → Pn. We prove that centred at a point x0 ∈ d(ϕ) < d(ϕ). The
orem 4.8 of course follows from this. We deﬁne the multiplicity of any divisor D = liCi at a point ξ to be k = liki, where ki are the multiplicities of the Ci at ξ. Obviously if D ≥ 0 then k ≥ 0, with k = 0 if and only if ξ /∈ Supp D. In the same way, we deﬁne the birational liσ (Ci); then Theorem 1 remains true for any divisor transform of D by σ (D) = D, that is, σ ∗D = σ (D) + kL. We now write νi for the multiplicity of Di at x0 and set ν = min νi. The map ϕ is given by the functions f i f div i = σ ∗ div(fi) = σ ∗(fi), and = σ (Di) + (νi − ν)L + νL + σ ∗(D), = σ (Di) + (νi − ν)L for i = 0,..., n have no common com- where the divisors D i ponents. Choose some i such that νi = ν. Then by deﬁnition ϕ σ (Di) 2 = 2. = d D i Now applying Theorem 4.7 to the equality σ ∗(Di) = σ (Di) + νL gives σ (Di) 2 = σ ∗(Di) − νL 2 = σ ∗(Di) 2 − ν2 = D2 i − ν2, and hence d(ϕ) = d(ϕ) − ν2. This proves Theorem 4.8. The simplest example of Theorem 4.8 is the map (occurring in the deﬁnition of the projective line) f : A2 → P1 given by f (x, y) = (x : y), which is not regular at ξ = (0, 0). Substituting (4.43) we see that at points of σ −1(ξ ) with t0 = 0 we have f (x, y) = (1 : v), and hence f ◦ σ is regular. 3
.4 Factorisation as a Chain of Blowups We now have all we need for the proof of the main result on birational maps of surfaces. Theorem 4.9 Let ϕ : X → Y be a birational map of nonsingular projective surfaces. Then there exist a surface Z, surfaces Xi and Yj with X0 = X, Y0 = Y, 3 Birational Maps of Surfaces Xk = Yl = Z, and maps 265 σi : Xi → Xi−1 for i = 1,..., k and τj : Yj → Yj −1 for j = 1,..., l such that each σi and τj is a blowup, and ϕ ◦ σ1 ◦ · · · ◦ σk = τ1 ◦ · · · ◦ τi. In other words, there is a commutative diagram Z...... σ2 X1 τ2 Y1 σ1 X −−−−−−−−−−−−−−−−−−−→ ϕ τ1 Y in which the diagonal arrows σi and τj are blowups. Theorem 4.9 is an obvious corollary of Theorem 4.8 together with the next result. Theorem 4.10 Let ϕ : X → Y be a regular map between nonsingular projective surfaces which is birational. Then there exists a chain of surfaces and blowups σi : Yi → Yi−1 for i = 1,..., k such that Y0 = Y, Yk = X and ϕ = σ1 ◦ · · · ◦ σk. We precede the proof of Theorem 4.10 with some general remarks on birational maps of surfaces. First of all, for any rational map ϕ : X → Y from a nonsingular surface X to a projective variety Y, it makes sense to talk of the image ϕ(C) of a curve C ⊂ X. Indeed, ϕ is regular at all points of C except possibly a ﬁnite set S. Thus by ϕ(C) we understand the closure in Y of ϕ(C \ S). Moreover, Theorem 2.16 on the existence of exceptional subvarieties remains valid in this setup. Lemma Let ϕ : X → Y be a birational
map of nonsingular projective surfaces, and suppose that ϕ−1 is not regular at some point y ∈ Y. Then there exists a curve C ⊂ X such that ϕ(C) = y. Proof Consider open sets U ⊂ X and V ⊂ Y such that ϕ : U → V is an isomorphism, and let Z be the closure in X × Y of the graph of the isomorphism ϕ : U → V. The projections to X and Y deﬁne regular birational maps p : Z → X and q : Z → Y. Obviously ϕ−1 = p ◦ q−1, so that, because we are assuming that ϕ−1 is irregular at y, the same is true of q−1. We can now apply Theorem 2.16 on the existence of exceptional subvarieties to the regular map q : Z → Y. This theorem shows that there exists a curve D ⊂ Z such that q(D) = y. We set p(D) = C and verify that C satisﬁes the conclusion 266 4 Intersection Numbers of the lemma. We really only need prove that dim C = 1, that is, dim C = dim D. Now otherwise, p(D) would be a point x ∈ X, so that p(D) = x and q(D) = y would imply D ⊂ (x, y) ∈ X × Y, which contradicts that D is a curve. The lemma is proved. We now proceed to the proof of Theorem 4.10. Suppose that ϕ is not an isomorphism, that is, ϕ−1 is not regular at some point y ∈ Y. Consider the blowup σ : Y → Y with centre in y and deﬁne ϕ : X → Y by ϕ = ϕ ◦ σ −1, so that the diagram below is commutative4.52) Auxiliary notation introduced in the course of the proof is summarised in the following diagram: X ⊃ Z x ϕ ↓ ↓ &ψ Y y ←−− σ L ⊂ Y. The theorem will be proved if we prove that ϕ is a regular map. Indeed, from the commutative diagram (4.52) it then follows that ϕ maps the subvariety ϕ
−1(y) to σ −1(y) = L ∼= P1. Since ϕ maps X onto the whole of Y, it follows that it maps ϕ−1(y) onto the whole of L. Thus not every component of ϕ−1(y) maps to a point. Therefore, for any y ∈ L the number of components of (ϕ)−1(y) is less than the number of components of ϕ−1(y). Hence after a ﬁnite number of blowups we arrange that X does not contain any exceptional subvarieties, that is, our regular map becomes an isomorphism. It remains to prove that ϕ is regular. Suppose otherwise. Then by the lemma, ψ = (ϕ)−1 maps some curve on Y to a point x ∈ X. It follows from the commutative diagram (4.52) that this curve can only be L, hence ψ(L) = x. Now according to Theorem 2.12, there exists a ﬁnite subset E ⊂ L such that ψ is regular at all point y ∈ L \ E. Since σ (y) = y, it follows from the commutativity of (4.52) that ϕ(x) = y. We now prove that dxϕ : ΘX,x → ΘY,y (4.53) is an isomorphism. For this, it is enough to prove that it is onto. Suppose that dxϕ(ΘX,x) ⊂ l ⊂ ΘY,y for some line l in the plane ΘY,y. Then from the commutativity of the diagram (4.52) it follows that also dy σ (ΘY,y ) ⊂ l (4.54) for every point y ∈ L \ E. However, this contradicts the most elementary property of blowups. Indeed, suppose that C is a nonsingular curve on Y with y ∈ C and 3 Birational Maps of Surfaces 267 ΘC,y = l, for example, the curve given by αu + βv = 0 where u and v are local parameters at y. Then by (4.44) we have σ (σ (C)) = C, where σ (C) intersects L in one point y
which has coordinates (−β : α) on L, and σ (C) is nonsingular with σ : σ (C) → C an isomorphism. We can choose α and β so that y /∈ E and then already dyσ (Θσ (C),y ) ⊂ l. The fact that (4.53) is an isomorphism contradicts the assumption that ϕ−1 is irregular at y. Indeed, using Theorem 2.16 on the existence of exceptional subvarieties, we ﬁnd a curve Z ⊂ X with Z x such that ϕ(Z) = y. Then ΘZ,x ⊂ ΘX,x (recall that the tangent space is deﬁned even if x is a singular point of Z). Since ϕ(Z) = y, we have dxϕ(ΘZ,x) = 0, and hence (4.53) has a kernel. This contradiction proves Theorem 4.10. 3.5 Remarks and Examples Consider a regular birational map f : X → Y between nonsingular projective surfaces. Suppose that f −1 fails to be regular at only one point η ∈ Y, and that the curve C = f −1(η) is irreducible. By Theorem 4.10, f is a composite f = σ1 ◦ · · · ◦ σk of blowups, and since every blowup gives rise to its own exceptional curve, C is irreducible only if k = 1, that is, f is itself a blowup. Then C is the curve L, concerning which we proved in Sections 3.1–3.2 that L ∼= P1 and L2 = −1. (4.55) Such a curve is called16 a −1-curve. The converse statement is also true: if a nonsingular projective surface X contains a −1-curve C, then there exists a regular birational map f : X → Y such that Y is nonsingular, f (C) = η ∈ Y, and f coincides with the blowup of η ∈ Y. Thus the conditions (4.55) are necessary and sufﬁcient for the curve C to be contracted to a point in the sense just described. This result was proved by Castel
nuovo, and is known as Castelnuovo’s contractibility criterion. We will not give the proof, which can be found, for example, in Shafarevich [69, Chapter II] or Hartshorne [37, Chapter V, Section 5]. Example (Standard quadratic transformation) We conclude this section with a construction in a simple example of a factorisation of a birational map into blowups, as in the conclusion of Theorem 4.9. This example is the birational map f from P2 to itself given by f (x0 : x1 : x2) = (y0 : y1 : y2), where y0 = x1x2, y1 = x0x2, y2 = x0x1; (4.56) it is called the standard quadratic transformation. 16−1-curves are called exceptional curves of the ﬁrst kind in the older literature. 268 4 Intersection Numbers We consider f as a birational map between two copies P2 and P2 of the projective plane, the ﬁrst with homogeneous coordinates (x0 : x1 : x2), and the second (y0 : y1 : y2). Obviously f fails to be regular at the 3 points ξ0 = (1, 0, 0), ξ1 = (0, 1, 0), ξ2 = (0, 0, 1). According to Theorem 4.8, we must start by performing the blowups σ0, σ1, σ2 in the three points ξ0, ξ1, ξ2. We arrive at a surface X with a regular map ϕ = σ2 ◦ σ1 ◦ σ0 : X → P2. We now prove that already the map ψ = f ◦ ϕ : X → P2 is regular. Indeed, ψ is regular at a point z if ϕ(z) = ξi. At points ζ ∈ σ (ξ0) the map f ◦ σ1 is already regular. To verify this, it is enough to set x = x1/x0, y = x2/x0 and substitute (4.43) in (4.56). We see that −1 1 f (1, x, y) = (xy : x : y) = = (uv :
v : 1). (4.57) u2v : uv : u −1 i = σ Since σ1 and σ2 both induce isomorphisms in a neighbourhood of ζ, also ψ is regular at points z for which ϕ(z) = ξ0. The same holds for ξ1 and ξ2. By Theorem 4.9, ψ is a composite of blowups ψ = τ1 ◦ · · · ◦ τk. We now determine which curves C ⊂ X can map to points under ψ. Obviously any such curve is either one of the M (ξi) for i = 0, 1 or 2, or the birational transform in i X of a curve L ⊂ P2 mapped to a point by f. It is easy to see that f deﬁnes an isomorphism of P2 \ (L0 ∪ L1 ∪ L2) and P2 \ (M0 ∪ M1 ∪ M2) where Li is the line xi = 0 in P2 and Mi the line yi = 0 in P2. Hence the only curves ψ can contract to a point are M 2, where the L 2, L 1, L i are the birational transform in X of the lines Li. But we see from (4.57) that M 0, given by the local equation u = 0, maps onto the whole curve M0 given by y0 = 0. In the same way, M i maps onto Mi for i = 1, 2. Thus the only curves ψ can contract are the L i. Moreover, ψ −1 is not regular at the points η0 = (1 : 0 : 0), η1 = (0 : 1 : 0), η2 = (0 : 0 : 1), since otherwise f −1 would be regular at one of them, and f −1 is given by the same formulas as f, as one sees from (4.56). Thus on the one hand, a factorisation ψ = τ0 ◦ · · · ◦ τk of ψ can have at most 3 blowups as factors, and on the other hand, it must include the blowups at η0, η1 and η2. We deduce that 0, M 1, M 0, L f = τ2 ◦ τ1 ◦ τ
0 ◦ σ −1 0 ◦ σ −1 1 −1 2 ◦ σ. 1, M 0, M It is easy to visualise the conﬁguration of the curves M 2 on the surface X; see Figure 18, where the arrows indicate which curves contract to which points. 0, L 1, L 2, L Of course, the standard quadratic transformation depends on the choice of the coordinate system in P2, or, what is the same thing, the choice of ξ0, ξ1, ξ2. Composing different standard quadratic transformations gives different birational maps of the plane to itself. M. Noether proved the theorem that any birational maps of the plane to itself can be written as a composite of quadratic transformations and projective linear transformations. We will not give the proof of this theorem, which is very delicate; it can be found in Shafarevich [69, Chapter V]. A description of the relations between these generators has been obtained comparatively recently: see Gizatullin [31] and Iskovskikh [42]. 3 Birational Maps of Surfaces 269 Figure 18 The standard quadratic transformation 3.6 Exercises to Section 3 1 For every integer k (positive, negative or 0), construct a nonsingular projective surface X and a curve C on it with C2 = k. [Hint: Construct X by blowing up a number of points on P2.] 1, C 2 Let X be a nonsingular projective surface, and C1, C2 two curves on X. Suppose that x ∈ C1 ∩ C2 is a nonsingular point of C1 and C2. Let σ : Y → X be the blowup of x and C 2 the birational transforms of C1, C2. Prove that C 2 intersect at a point y ∈ σ −1(x) if and only if C1 and C2 are tangent at x. Moreover, then σ −1(x) ∩ C 2 and 1 y is 1 less than that of C1 and C2 at x. = y is a single point, and the order of tangency of C 1 and C 1 and C ∩ C 2 3 Suppose that f : P2 → P1 is given by f (x0 : x1 : x2) = P (x0, x1, x2
) : Q(x0, x1, x2), where P and Q are forms of degree n. How many blowups does one have to perform to get a surface ϕ : X → P2 such that f ◦ ϕ : X → P1 is regular? 4 Let X ⊂ P3 be a nonsingular surface of degree 2 and f : X → P2 the birational map consisting of projection from a point x ∈ X. Factor f as a composite of blowups. 5 Let f : P2 → P2 be the birational map given in inhomogeneous coordinates by x = x, y = y + x2. Factor f as a composite of blowups. 6 Let L ⊂ P2 be a line, and x, y two points of L. Write X → P2 for the composite of the blowups at x and y, and L for the birational transform of L. Prove that (L)2 = −1. According to Castelnuovo’s contractibility theorem stated in Section 3.5, there is a regular map f : X → Y that is birational and contracts L to a point. Construct f in the given special case. [Hint: Try to ﬁnd it among the preceding exercises.] 270 4 Intersection Numbers 7 Let f : X → Y be a birational regular map of nonsingular n-dimensional projective varieties. Prove that f ∗(D1) · · · f ∗(Dn) = D1 · · · Dn for D1,..., Dn ∈ Div Y. 8 Let σ : X → Y be a blowup with centre in a point y ∈ Y and Γ = σ −1(y). For D1 ∈ Div(Y ), and D2,..., Dn−1 ∈ Div(X), prove that σ ∗(D1)D2 · · · Dn−1Γ = 0. 9 In the notation of Exercises 7–8, calculate Γ n for any n > 1. 10 Prove that if a curve of degree n passes through k of the points ξ0, ξ1, ξ2 (for k = 0, 1 or 2) deﬁning the standard quadratic transformation f (see Example of Section 3.5), and is not singular
there, then its image under f has degree 2n − k. 11 Let ϕ be the transformation of inversion with respect to a circle with centre O and radius 1, that is, ϕ(P ) = Q, where P, Q and O are collinear and \|OP \| · \|OQ\| = 1. Taking coordinates with O as the origin, write out the formulas for ϕ in coordinates x, y and u = x + iy, v = x − iy. Prove that after composing with the reﬂection (u, v) → (u, −v), ϕ becomes the standard quadratic transformations deﬁned by O and the two circular points at inﬁnity. Deduce from this that under inversion circles through O transform to lines, and other circles to circles. 4 Singularities 4.1 Singular Points of a Curve Theorem 4.11 Let C be an irreducible curve on a nonsingular surface X; then there exists a surface Y and a regular map f : Y → X, such that f is a composite of blowups Y → X1 → · · · → Xn → X and the birational transform C of C on Y is nonsingular. Proof We can consider separately each singularity of C. Indeed, if we can construct a map f : Y → X for one point x ∈ C, with f a composite of blowups above x and the birational transform C of C on Y nonsingular at all point of f −1(x), then we can subsequently apply the same argument to the remaining singular points of C; the number of these equals the number of singularities of C outside x. Thus let x ∈ C be a singular point. We blow up x; if some points of the inverse image of x are singular points of the birational transform of C, we blow these up too, and so on. We have to prove that this process stops after ﬁnitely many steps. Write μx(C) for the multiplicity of a singular point x ∈ C; let σ : X → X be the blowup, C the birational transform of C, and L = σ −1(x). By Corollary 4.2, μx(C) = CL. On the other hand, CL = σ (x)=x(CL)x, where the sum takes
place over all points x ∈ C with σ (x) = x. Since (CL)x ≥ μx(C), we get μx(C) ≥ C. μx σ (x)=x 4 Singularities 271 Therefore, if there is more than one point x, each of them must have multiplicity μx(C) < μx(C), so our process must stop after ﬁnitely many steps. It remains to consider the case when C has only one singular point x with σ (x) = x, and the same continues to hold after each blowup. Write Ox for the local ring of x ∈ C, and Ox for its integral closure in the ﬁeld k(C). Then Ox is a ﬁnite module over Ox ; this follows because by Theorem 2.20, x has an afﬁne neighbourhood U for which the normalisation k[U ]ν of the coordinate ring k[U ] is a ﬁnite module over k[U ]. Suppose that k[U ]ν = α1k[U ] + · · · + αmk[U ]. Then Ox = α1Ox + · · · + αmOx ; in fact if f ∈ Ox then f n + a1f n−1 + · · · + an = 0 with ai ∈ Ox, that is, ai = bi/c with bi, c ∈ k[U ] and c(x) = 0. Then cf is integral over k[U ], hence cf = α1r1 + · · · + αmrm with ri ∈ k[U ] and f = α1r1/c + · · · + αmrm/c. Since αi is in the ﬁeld of fractions of Ox (or even in that of the smaller ring k[U ]), there exists a nonzero element d ∈ Ox such that dαi ∈ Ox for each i, and hence dOx ⊂ Ox. It follows from this that Ox/Ox is a ﬁnite dimensional vector space. In fact its dimension is at most the dimension of the vector space Ox/dOx, which is generated by the m subspaces αi(Ox/dOx); but Ox/dOx is ﬁnite dimensional, ⊂ dOx. since C is a curve,
and so for any function d = 0 there exists k such that mk x Obviously Ox ⊂ Ox after one blowup. Moreover, we now prove that Ox ⊂ Ox. Indeed, let ν : Cν → C be the normalisation and (ν)−1(x) = {yi}. Then ν = σ ◦ ν : Cν → C coincides with the normalisation of C and ν−1(x) = {yi}. Obviously Ox ⊂ = Ox. Again, Oyi. Since ν is a ﬁnite map, we can assume that C and Cν are obviously Ox ⊂ afﬁne. If u ∈ Oyi then all the poles of u on Cν are distinct from the yi, and it follows that there exists a function v ∈ k[C] such that v(x) = 0 and uv ∈ k[Cν] (for this, it is sufﬁcient that ν∗(v) has zeros of sufﬁciently high degree at each pole of u). Then uv is integral over k[C], and it follows easily that u is integral over Ox, that is, u ∈ Ox. Oyi, so that we will be home if we check that Hence (Ox/Ox ) ≤ (Ox/Ox). If (Ox/Ox) = 0 then Ox = Ox, so that Ox is integrally closed, and then x is nonsingular and our process has stopped. It now only remains to prove that (Ox/Ox ) < (Ox/Ox), since then our process must stop after at most (Ox/Ox) steps. But (Ox/Ox) = (Ox/Ox) implies that Ox = Ox. Let u, v be local parameters at x ∈ X, so that the local parameters at x ∈ X are u and t = v/u. Since t restricted to C is an element of Ox, from Ox = Ox it follows that t ∈ Ox and mx = (u, v) = (u, ut) = (u). It follows from this that mx/m2 = x (u)/(u2) ∼= Ox/(u) ∼= k, that is, x ∈ C was already nonsingular. The theorem is proved. Oyi Theorem 4.11 enables us to de�
��ne an important characteristic of a singular point of a curve on a nonsingular surface, its tree of inﬁnitely near points. This is the diagram consisting of the singular point, the singular points arising out of it after one blowup, the singular points arising out of these after blowing them up again, and so on. All these points are said to be inﬁnitely near to the original point of the curve. We write the multiplicity of each point. Once we get to a point of multiplicity 1, we don’t carry out any further blowups there. Some examples are illustrated in Figure 19. 272 4 Intersection Numbers Figure 19 Resolutions of some curve singularities The genus of the normalisation of a singular curve lying on a nonsingular projective surface is expressed in terms of these invariants: for this, by (4.28), we have to determine how the expression C(C + K) changes on replacing C by σ (C) and KX by KX, where σ : X → X is the blowup of a point x ∈ C of multiplicity k. According to Theorem 4.6, σ (C) = σ ∗(C) − kL. To compute KX, consider a differential form ω ∈ Ω 2(X) such that x /∈ Supp(div ω); this exists by Theorem 3.1 (on moving the support of a divisor away from a point). Then since σ : X \ L → X \ x is an isomorphism, obviously div(σ ∗ω) = σ ∗(div ω) over X \ L. If x, y are local parameters at x then ω = f dx ∧ dy, where f ∈ Ox and f (x) = 0. If x = u, y = uv are as in (4.43) then σ ∗(ω) = σ ∗(f )vdu ∧ dv on X, and since σ ∗(f ) = 0 on L, we get div(σ ∗(ω)) = σ ∗(div ω) + L, that is, KX = σ ∗(KX) + L. Substituting in (4.28) gives σ (C) σ (C) + KX σ
∗(C) − kL = σ ∗(C) + σ ∗(KX) − (k − 1)L = C(C + KX) − k(k − 1). Now using Theorem 4.11, we get C(C + KY ) = C(C + KX) − ki(ki − 1), for the nonsingular curve C ⊂ Y, where ki are the multiplicities of all the inﬁnitely near points. It follows from (4.28) that g(C) = 1 2 C(C + KX) + 1 − ki(ki − 1) 2. (4.58) In particular if X = P2 and C is a curve of degree n then g = (n − 1)(n − 2) 2 − ki(ki − 1) 2. A corollary of (4.58) that is often used is that since g(C) ≥ 0, C(C + KX) ≥ −2, (4.59) and equality holds if and only if C is nonsingular (that is, all ki = 1), and g(C) = g(C) = 0, so that C ∼= P1. 4 Singularities 4.2 Surface Singularities 273 The theorem on resolution of singularities has been proved for algebraic surfaces over a ﬁeld of arbitrary characteristic; we can suppose that X is normal, so has only ﬁnitely many singular points. Resolution of singularities asserts that there exists a nonsingular projective surface Y birational to X. Using the theorem on the resolution of indeterminacies (Theorem 4.8), we can assume given a birational regular map f : Y → X. It is often convenient to consider the situation locally, dropping the assumption that X and Y are projective, thus replacing them by open subsets U ⊂ X and f −1(U ) ⊂ Y. Then the map f : Y → X will be proper (see Remark after Theorem 1.11). It can be shown that Theorem 2.16 remains true in this case, and f contracts a bunch of projective curves C1,..., Cr ⊂ Y to each singular point x ∈ X. Moreover, using Castelnuovo’s contractibility criterion discussed in Section 3.5, one
can prove that Y can be chosen so that there are no −1-curves among the Ci. In this case Y is a minimal resolution of singularities of X. We will not prove all these assertions, and will not make use of them: they only serve as motivation for the questions that we now discuss. If x ∈ X is a surface singularity, the bunch of curves C1,..., Cr ⊂ Y on the nonsingular surface Y that are contracted by f : Y → X is an important geometric characteristic of the singularity, and it is interesting to see what can be said in general about such a bunch of curves. Theorem 4.12 Let f : Y → X be a regular map of algebraic surfaces, with Y nonsingular and C1,..., Cr ⊂ Y projective curves that are contracted to x ∈ X; sup∼→ X \ x is an isomorphism. Then the matrix of pose that f : Y \ (C1 ∪ · · · ∪ Cr ) intersection numbers {CiCj } is negative deﬁnite. Proof Consider a curve E on Y distinct from all the Ci but intersecting each of them (for example, a hyperplane section of Y ); set f (E) = H ⊂ X and choose a function u ∈ Ox vanishing along H. Set g = f ∗(u). Then div g = miCi + F, where all the mi > 0 and F Ci > 0 for i = 1,..., r. Since div g is a principal divisor, if we set D = miCi then the restriction of D to each of the Cj satisﬁes ∼ −F\|Cj ; hence DCj < 0 for j = 1,..., r. The theorem now follows from the D\|Cj following result of linear algebra, the proof of which can be found in Proposition A.2 of the Appendix. Proposition Let M be a Z-module with a scalar product ab ∈ Z deﬁned for a, b ∈ M, and e1,..., er a set of generators of M with eiej ≥ 0 for i = j ; suppose that there exists an element d = miei with mi > 0 such that dei < 0 for i = 1,..., r. Then
every nonzero m ∈ M satisﬁes m2 < 0 and e1,..., er is a free basis of M. The theorem is proved. It is interesting to note the analogy between Theorem 4.12 and the Hodge index theorem in Section 2.4. 274 4 Intersection Numbers If x ∈ X is a surface singularity, the bunch of curves C1,..., Cr contracted to x under the minimal resolution can be drawn as a graph: each curve Ci is represented by a node, and intersecting curves Ci and Cj are joined by an edge, marked by the intersection number CiCj if CiCj = 1, and left unmarked if CiCj = 1 (that is, when Ci and Cj intersect transversally at one point); the node corresponding to Ci is marked with C2 i. Interesting examples of singularities are provided by the quotients A2/G of the plane by a ﬁnite group G of linear transformations. Recall that these are normal varieties (Examples 1.21 and 1.29), and points which are images of x ∈ A2 for which g(x) = x for all g = G are nonsingular (Example of Section 2.1, Chapter 2). Suppose for example that G = g is a cyclic group of order n generated by g(x, y) = (εx, εq y), where ε is a primitive nth root of 1 and q is coprime to n. It can be shown that after excluding certain uninteresting cases, every action of a cyclic group reduces to this form. In this case G acts freely on A2 \ (0, 0), and hence A2/G has a single singularity, the image of (0, 0) ∈ A2. This is called a singularity of type (n, q). For example, if q = −1, the ring of invariants k[x, y]G is generated by u = xn, v = yn and w = xy, with the single relation uv = wn. (4.60) This is the equation of the surface A2/G. For q = 1 the generators of k[x, y]G are ui = xiyn−i for i = 0,..., n. The relations holding between these are the same as those between the coordinates of
the Veronese curve in Section 5.4, Chapter 1. Thus in this case A2/G is the cone over the Veronese curve. It is not hard to verify that the resolution graph corresponding to an arbitrary singularity of type (n, q) has the form of a chain −e1 • −e2 • −en−1 • · · · −en • The curves Ci and Ci+1 intersect transversally, and C2 i = −ei, where the ei ≥ 2, and are deﬁned by an expansion which is very close to a continued fraction expansion of n/q: n q = e1 − 1 e2 − 1 e3−···. See for example de la Harpe and Siegfried [24] for the proof. 4.3 Du Val Singularities An extremely important class of singularities is deﬁned by the following condition. 4 Singularities 275 Figure 20 The Dynkin diagrams An, Dn, E6, E7 and E8 Deﬁnition A point x ∈ X of a normal surface is called a Du Val singularity if17 there exists a minimal resolution f : Y → X contracting curves C1,..., Cr to x, such that KY Ci = 0 for all i, where KY is the canonical class of Y. The meaning of the Du Val singularities, as formulated by Du Val himself, is that they “do not affect the canonical class”. For example, it is easy to see that for a surface X ⊂ P3 of degree n with only Du Val singularities, the invariant h2 = dim Ω 2[Y ] of its minimal resolution Y is the same as that of a nonsingular surface of degree n. This is a sharp contrast between surfaces and curves, for which, according to (4.58), any singularity decreases the genus of the normalisation of the curve. The types of resolution graphs corresponding to Du Val singularities x ∈ X can be completely determined. Indeed, if Ci is one of the irreducible projective curves contracted to x by f : Y → X then KY Ci = 0, and according to the inequality (4.59), C2 i ≥ −2. = −2 and Ci i < 0, and C2 i = −1 by minimality of the resolution, it follows that Because C2 ∼= P1. From the fact
that (Ci + Cj )2 < 0 for i = j it now follows C2 i that CiCj ≤ 1, that is, Ci and Cj either do not intersect, or intersect transversally in one point. The purely algebraic question of classifying Z-modules Ze1 + · · · + Zer having a = −2 for i, j = 1,..., r negative deﬁnite scalar product satisfying eiej > 0 and e2 i occurs in a number of problems. It ﬁrst appeared in connection with the classiﬁcation of simple Lie algebras in the theory of root systems (see Bourbaki [19]). The answer is as follows: the basis e1,..., er breaks up into disjoint “connected components” with eiej = 0 for ei and ej belonging to different components, and the module decomposes as a direct sum of submodules corresponding to the different components. Thus the problem reduces to describing the “connected” modules, which can have only the graphs of Figure 20 (each vertex is marked with −2). It can be shown that the set of curves that appear on resolving a singularity is always connected. For k = C we will prove this in Theorem 7.2 of Section 2.5, 17There are many alternative names in the literature: Kleinian singularities, rational double points, simple singularities, etc. 276 4 Intersection Numbers Chapter 7. Thus the Du Val singularities correspond only to graphs of type An, Dn, E6, E7 and E8. It can be proved that a Du Val singularity is determined up to formal analytic equivalence by its graph. They can be given by equations An : Dn : E6 : E7 : E8 : x2 + y2 + zn+1 = 0 for n ≥ 1, x2 + y2z + zn−1 = 0 for n ≥ 4, x2 + y3 + z4 = 0, x2 + y3 + yz3 = 0, x2 + y3 + z5 = 0. One of the realisations of these singularities is as follows: Theorem 4.13 Suppose that char k = 0 and that G is a ﬁnite group of linear transformations of the plane A2, with det g = 1 for all g ∈ G.
Then the image of the origin 0 ∈ A2 is a Du Val singularity y0 ∈ A2/G. The proof uses the following construction, which we discuss in complete generality in Section 4.1, Chapter 5. Let X, Y and S be three varieties, and f : X → S and h : Y → S regular maps. The ﬁbre product of X and Y over S is the closed subset in X × Y consisting of pairs (x, y) for which f (x) = h(y); it is denoted by X ×S Y. The maps f : X → S and h : Y → S deﬁne a map X ×S Y → S, and the projections X × Y → X and X × Y → Y deﬁne projections X ×S Y → X and X ×S Y → Y. Let h : X → A2/G = S be the minimal resolution of the singularity y0 ∈ A2/G. Consider the ﬁbre product Z = A2 ×S X and its normalisation Zν. (Here we are using the existence of the normalisation, proved in Theorems 2.20 and 2.22 only for afﬁne varieties and curves; in Section 1.1, Chapter 6, the normalisation will be constructed in sufﬁcient generality for our present purposes.) We have the diagram of maps q→ X Zν p ↓ A2 → f ↓ h A2/G. Consider the differential form ω = dx ∧ dy on A2. From the condition det g = 1 for all g ∈ G it follows that g∗(ω) = ω. Write ω in the form h ds ∧ dt with s, t ∈ k(A2/G) and h ∈ k(A2). Then from the fact that g∗(ω) = ω it follows that g∗(h) = h; writing h in the form P /Q with P, Q ∈ k[A2] we see that - h = g∗(Q) P g=e g∗(Q), g and it follows that h ∈ k(A2/G). Thus ω = f ∗(ω0) with ω0 ∈ Ω 2(A2/G). Write ω
1 = h∗(ω0) and ω = q∗(ω1) = p∗(ω). From the fact that ω = p∗(ω) it follows that 4 Singularities 277 ω is regular on the set of nonsingular points of the surface Zν. On the other hand, for any maps f : X → S and h : Y → S it is easy to check that if f is ﬁnite then so is X ×S Y → X. Thus Z → X is ﬁnite, and hence also Zν → X. We use the following fact: Lemma If ϕ : U → V is a ﬁnite map of nonsingular surfaces and ω1 a rational differential 2-form on V such that ϕ∗(ω1) is regular, then ω1 is also regular. Proof of Theorem 4.13 We leave the proof of the lemma until after that of Theorem 4.13. It follows from the lemma that ω1 is regular outside the image of the ﬁnite set of singular points of Zν, and hence is regular on the whole of X. Let us determine the divisor div(ω1) on X. At any point α ∈ A2 with α = (0, 0) we can ﬁnd local parameters of the form f ∗(u), f ∗(v) (see Example of Section 2.1, Chapter 2), and it follows that ω0 is regular and nonzero at all points y = y0 ∈ A2/G, and these points are nonsingular. In exactly the same way, h is an isomorphism on X \ f −1(y0), and ω1 is nonzero on X \ f −1(y0). Thus D = div(ω1) = riCi with ri ≥ 0; obviously D ∼ KX. From the inequality (4.59), and the minimality of the resolution, we get DCi = KXCi ≥ 0. But then D2 = riDCi ≥ 0, which by Theorem 4.11 is only possible if DCi = 0 for all i. Thus KXCi = 0, that is, y0 is a Du Val singularity. The theorem is proved. Proof of the Lemma It is enough to
prove that if vC(div(ω1)) < 0 for any irreducible curve C ⊂ V then also vC(ϕ∗ div(ω1)) < 0 for any component C of the inverse image of C. This can be checked on any open subset V ⊂ V meeting C. What makes the problem nontrivial is that ϕ∗(div(ω1)) = div(ϕ∗(ω1)) in gen∼→ ΘV,ϕ(α) of ϕ at a point α ∈ U is eral. However, if the differential dαϕ : ΘU,α an isomorphism on the tangent spaces then the inverse images ϕ∗(v1), ϕ∗(v2) of local parameters v1, v2 at ϕ(α) are local parameters at α. Thus if ω1 = f dv1 ∧ dv2 then ϕ∗(ω1) = ϕ∗(f )dϕ∗(v1) ∧ dϕ∗(v2), and in an neighbourhood of α this has divisor div(ϕ∗(f )) = ϕ∗(div(f )) = ϕ∗(div(ω1)). Thus we need only consider curves C ⊂ U such that dαϕ is degenerate at every point α ∈ C. Set ϕ(C) = C. Since the map ϕ : C → C is an isomorphism of the tangent spaces over an open set, we can assume that at α the local parameter along C is ϕ∗(v1), where v1, v2 are local parameters at ϕ(α), where v2 = 0 is a local equation of C and v1 restricts to a local parameter along C. Set w = v1 and let (w, t) be local parameters at α, with t a local equation of C. Suppose ϕ∗(v2) = t eh where vC(h) = 0, and set w1 = f dv1 ∧ dv2. Then ϕ∗(ω1) = ϕ∗(f )dw ∧ d ϕ∗(v2) et e−1hdw ∧ dt + t edw ∧ dh
= ϕ∗(f ), and it follows that vC(ϕ∗(div(ω1))) = vC(ϕ∗(div(f ))) + e − 1. But if C is in the divisor of poles of ω1 then vC(f ) = −l with l > 0, and then vC(ϕ∗(div(ω1))) = −le + e − 1, which is also < 0. In other words, an effective term gets added on to the divisor ϕ∗(div(ω1)), but not enough to compensate for the pole that arises. The lemma is proved. 278 4 Intersection Numbers The groups G appearing in Theorem 4.13 are well known. Write SL(2, k) for the group of linear transformations with determinant 1, and consider the homomorphism π : SL(2, k) → PSL(2, k) to the group of projective transformations of P1; the kernel of π is ±1. Then the ﬁnite subgroups G ⊂ SL(2, k) are the following: either the cyclic group of order n consisting of transformations (x, y) → (εx, ε−1y) for εn = 1, or the binary dihedral group of order 4n, generated by (x, y) → (εx, ε−1y) for ε2n = 1, and (x, y) → (−y, x), or the binary tetrahedral, binary octahedral or binary icosahedral groups, that is, the inverse image under π of the subgroups of PSL(2, k) isomorphism to the tetrahedral, octahedral or icosahedral groups. These groups have order n, 4n, 24, 48 and 120 respectively (see for example Springer [75]). It is not hard to ﬁnd the corresponding Du Val singularities, which turn out to be An−1 for the cyclic group of order n, Dn+2 for the binary dihedral group of order 4n, and E6, E7 and E8 for the binary tetrahedral, binary octahedral or binary icosahedral groups (see de la Harpe and Siegfried [24]). 4.4 Degeneration of Curves Let X be a nonsingular
projective irreducible surface and f : X → S a regular map to a curve S; ﬁx some point s0 ∈ S, and assume that for every s ∈ S with s = s0, the ﬁbre f −1(s) is a nonsingular irreducible curve. We can consider {f −1(s) \| s ∈ S \ s0} as a family of nonsingular curves, and f −1(s0) as a degeneration of this. By the Castelnuovo contractibility criterion mentioned in Section 3.5, any −1-curve among the components of f −1(s0) can be contracted to a point without affecting the nonsingularity of X. Hence we assume in what follows that there are no such components. Moreover, it can be proved that f −1(s0) is connected, that is, cannot be written as a union of two closed disjoint curves. In the case k = C this will be proved in Theorem 7.2 of Section 2.5, Chapter 7. Theorem 4.14 Under the above assumptions, consider the divisor s0 on S consisting of one point, and suppose that its inverse image f ∗(s0) decomposes as f ∗(s0) = riCi, where the Ci are irreducible components and ri > 0. Then any divisor D = liCi satisﬁes D2 ≤ 0, with D2 = 0 if and only if D is proportional to riCi. riCi = f ∗(s0) ∼ f ∗(Δ), where Δ is a divisor on C not conProof Obviously taining s0. Hence the restriction to any component Ci of these divisors satis= 0. It follows from this that f ∗(s0)Ci = 0, that is ﬁes f ∗(s0)\|Ci liCi. In particular (f ∗(s0))2 = 0. Theorem 4.14 ( now follows from the following result of linear algebra proved in Proposition A.3 of the Appendix: riCi)B = 0 for every B = ∼ f ∗(Δ)\|Ci Proposition Let M be a free Z-module with a scalar product ab ∈ Z for
a, b ∈ M. Suppose that M has a basis e1,..., er satisfying eiej ≥ 0 for each i = j, and that the {ei} cannot be split up into two components with eiej = 0 for ei, ej in different 4 Singularities 279 components; assume that there exists an element d = liei with li > 0 such that dei = 0 for i = 1,..., r. Then every m ∈ M satisﬁes m2 ≤ 0, with equality only if m is proportional to d. Theorem 4.14 is proved. It is interesting to note that Theorem 4.14 occupies an intermediate position between the Hodge index theorem of Section 2.4, and Theorem 4.12: the curves considered in Theorem 4.14 are contained in a ﬁbre of a map f : X → C to a curve C, those of Theorem 4.12 in a ﬁbre of a map f : X → Y to a surface Y, and those of the Hodge index theorem in a ﬁbre of a map f : X → z to a point z. We now study the simplest examples of the situation described in Theorem 4.14. If the genus of the curves f ∗(s) for s = s0 is 0, that is, if f ∗(s) ∼= P1 then under the above assumption one can show that there is no degeneration, that is, the curve f −1(s0) is also nonsingular and isomorphic to P1. See, for example, Shafarevich [69, Chapter V] or Grifﬁths and Harris [33, IV.5]. Next in order of difﬁculty is the case when the curves f −1(s) for s = s0 have genus 1, that is, they are isomorphic to nonsingular plane cubic curves. Consider the pencil X of elliptic curves of Example 2.8 given by the equation 2 ξ0 = ξ 3 ξ 2 1 + a(t)ξ1ξ 2 o + b(t)ξ 3 0 in P2 × A1. In the afﬁne part A2 × A1 this equation is y2 = x3 + a(t)x + b(t). The ﬁ
bre of f : X → A1 over a point c ∈ A1 will be nonsingular when Δ(c) = 0, where Δ = 4a3 + 27b2. We suppose that Δ does not vanish identically on A1, but that Δ(0) = 0, and study the ﬁbre f −1(0). In order to work with a projective surface we consider the closure of X in P2 × P1 ⊃ P2 × A1. The surface we obtain is in general singular: points of a ﬁbre f −1(c) will be nonsingular if Δ(c) = 0, but when Δ(c) = 0 this will only happen if c is a simple root of Δ (see Example 2.8). We consider the minimal resolution ϕ : Y → X, which maps to P1 by g = f ◦ ϕ : Y → P1; here at a point such that Δ(c) = 0, the ﬁbre of g is the same as that of the original pencil f. On Y, consider the differential 2-form ω = y−1dx ∧ dt. One sees easily that above points c ∈ A1 where Δ(c) = 0, this form is regular and nowhere vanishing; this comes from the fact that the 1-form y−1dx is regular and nowhere vanishing on the curve f −1(c). It follows from this that the canonical class KY contains a divisor consisting only of components of ﬁbres. Suppose that g∗(0) = riCi, where the Ci are components of the ﬁbre g−1(0) and ri > 0. We write KY in the form KY = niCi + D, where D consists of components of ﬁbres other than g−1(0). Since g∗(0) ∼ g∗(c) for c = 0, we can add in a multiple of g∗(0) − g∗(c) to arrange that all ni > 0. Since all ﬁbres g∗(c) are linearly equivalent, g∗(c)KY = 0. We consider two cases. Case A (g−1(0) is an irreducible curve C) Since in this case C2 = 0 and KY C = 0
, from (4.58) and the fact that g(C) ≥ 0 we get that ki(ki − 1) δ = ≤ 1, 2 280 4 Intersection Numbers Figure 21 The extended Dynkin diagrams An, Dn, E6, E7 and E8 that is, δ equals 0 or 1. If δ = 0 then C is a nonsingular curve. If δ = 1 then C has just one singularity of multiplicity 2 which is resolved by one blowup. It follows that the singular point is formally analytically equivalent to the singularity y2 = x2 or y2 = x3 (see Exercise 12 of Section 3.3, Chapter 2). These are exactly the types of singularities that can appear on irreducible plane cubics. Case B (g−1(0) is reducible) Then Theorem 4.14 implies that any component Ci nj Cj + D with i < 0. We write KY in the form KY = of the ﬁbre satisﬁes C2 all the nj > 0 and Supp D disjoint from g−1(0). Then if ( nj Cj )2 < 0 we would have KY Ci < 0 for at least one component Ci. But then, by the same argument as = −1, g(Ci) = 0 and ki = 0, so in Section 4.5, the inequality (4.58) implies that C2 i that Ci is a −1-curve, and we are assuming that there are no such components in nj Cj is ﬁbres. Hence ( proportional to the ﬁbre f ∗(0), and so KY Ci = 0. Now (4.58) gives that C2 = −2, i g(Ci) = 0, ki = 0, in other words, all the components of ﬁbres are isomorphic to P1 and have C2 i nj Cj )2 = 0, and so it follows from Theorem 4.14 that If the ﬁbre has just two components C1 and C2 and g∗(0) = n1C1 + n2C2 then (C1 + C2)2 ≤ 0 gives C1C2 ≤ 2, and (n1C1 + n2C2)2 = 0 implies that n2 = 1 n1n2
C1C2, which is only possible if C1C2 = 2. The two components can intersect transversally in 2 points or have a point of tangency. = −2. + n2 2 If the ﬁbre has more than two components, then (Ci + Cj )2 < 0 implies that CiCj = 0 or 1. Thus the curves Ci and Cj are either disjoint, or they meet transversally. We draw the system of curves C1,..., Cr as a graph with the same conventions as for the resolution of isolated singularities. We have seen that these deﬁne a basis of the Z-module ⊕ZCi satisfying the = −2. All such Zassumptions of Theorem 4.14, and the additions condition C2 i modules have been found in connection with root systems (see Bourbaki [19]); their graphs are presented in Figure 21. The relation with the theory of Du Val singularities is as follows. Suppose that the elliptic pencil is given by the equation y2 = x3 + a(t)x + b(t), (4.61) 4 Singularities 281 where a(t) and b(t) are polynomials. We will assume that a and b are not simultaneously divisible by a 4th and 6th power of any polynomial c(t), since in that case one could get rid of the factors by means of the birational transformation y = y1c3, x = x1c2. Then the surface given by (4.61) has a Du Val singularity on every ﬁbre f −1(c) with Δ(c) = 0, and the ﬁbre of the nonsingular surface consists of the curves appearing in the minimal resolution of this singularity, together with the birational transform of the ﬁbre. A singularity of type An gives a ﬁbre of type An, Dn a ﬁbre of type Dn, and E6, E7, E8 a ﬁbre of type E6, E7, E8. 4.5 Exercises to Section 4 1 Find the graph of inﬁnitely near points for the curve singularity y2 = xn. 2 Generalise the notion of class (see Exercise 22 of Section 7.7, Chapter 3) to
singular plane curves. Prove that the class of a plane projective curve of degree n with d ordinary double points is n(n − 1) − d. 3 What are the singularities of a curve lying on a nonsingular surface that can be resolved by a single blowup? Give a characterisation of these in terms of the local equation of the curve, or more precisely, in terms of all its terms of degree r and r + 1, where r is the multiplicity of the singularity. ri(ri − 1) ≤ (n − 4 Prove that an irreducible plane curve C of degree n satisﬁes 1)(n − 2), where ri are the multiplicities of the singular points. What happens in case of equality? 5 Find the resolution graph corresponding to the Du Val singularities A2/G of type (n, −1); for this, embed A2/G in afﬁne 3-space A3, and use a sequence of blowups of (0, 0, 0) and in the singularities of the variety arising after the blowup. Do the same for the quotient singularity A2/G where G ⊂ SL(2, C) is the binary dihedral group of order 4n, generated by α = ε 0 0 ε−1 and β =, 0 1 −1 0 where ε = exp(2πi/2n). 6 Suppose that X is a nonsingular projective surface such that, for some n > 0, the rational map ϕ corresponding to the class nKX is regular and a birational embedding with normal image. Prove that ϕ(X) has only Du Val singularities. 7 Find all the types of degenerate ﬁbres of a pencil of elliptic curves in Weierstrass normal form for which Δ = 4a3 + 27b2 has a double root at the point of degeneration. 282 4 Intersection Numbers 8 Resolve the singularity of the surface y2 = x3 + αt 2x + βt 3, where α, β ∈ k and 4α3 + 27β2 = 0. For this, blow up the ambient space at (0, 0, 0), then again at the new singular points, and so on. Verify that the singularity is a Du Val singularity of type D4, and that the singular �
�bre of the pencil of elliptic curves arising after resolving is of type D4. Algebraic Appendix 1 Linear and Bilinear Algebra Recall that a scalar product on an Abelian group M with values in an Abelian group B is a function (a, b) for a, b ∈ M with values in B satisfying the conditions (b, a) = (a, b), (a1 + a2, b) = (a1, b) + (a2, b). (A.1) (A.2) Proposition A.1 Let M be an arbitrary Abelian group and B an Abelian group in which division by 2 is possible and unique. A function f (a) on M with values in B can be expressed as f (a) = (a, a) for some scalar product (a, b) if and only if f (a + b) + f (a − b) = 2 f (a) + f (b). (A.3) Proof If f (a) = (a, a) then (A.3) follows at once from (A.1) and (A.2). Assume (A.3), and set (a, b) = 1 2 f (a + b) − f (a) − f (b). Then (A.1) is obvious and (A.2) is equivalent to (a + b, c) − (a, c) − (b, c) = 0. (A.4) (A.5) We write ψ(a, b, c) for the left-hand side of (A.5). It follows from (A.4) that 2ψ(a, b, c) = f (a + b + c) − f (a + b) − f (a + c) − f (b + c) + f (a) + f (b) + f (c), so that ψ(a, b, c) is a symmetric expression in a, b and c. Applying (A.3) with a = b = 0 implies that f (0) = 0, and with a = 0 that f (−b) = f (b). Now from (A.3) and (A.4) we deduce that (a, −b) = −(a, b), and so in view
of (A.1), (−a, b) = −(a, b). Putting this all together gives ψ(a, b, −c) = −ψ(a, b, c), and the same equality for a and b by symmetry. But (A.5) also gives ψ(−a, −b, c) = I.R. Shafarevich, Basic Algebraic Geometry 1, DOI 10.1007/978-3-642-37956-7, © Springer-Verlag Berlin Heidelberg 2013 283 284 Algebraic Appendix −ψ(a, b, c), whereas we have just proved that ψ(−a, −b, c) = ψ(a, b, c). Hence ψ(a, b, c) = 0. The proposition is proved. Proposition A.2 Let M be a Z-module having a scalar product (a, b) ∈ Z for a, b ∈ M, and suppose that e1,..., er is a set of generators of M satisfying (ei, ej ) ≥ 0 r for i = j. Assume that there exists an element d = i=1 miei with mi > 0 and (d, ei) < 0 for i = 1,..., r. Then (m, m) < 0 for all nonzero m ∈ M, and e1,..., er are linearly independent in M. Proof Write A for the symmetric r × r matrix with entries aij = (ei, ej ), and deﬁne a scalar product ϕ on Rr by ϕ(x, y) = aij xiyj. It is enough to prove that ϕ(x, x) < 0 for all nonzero x ∈ Rr ; because then the map Zr → M taking the basis element fi = (0,..., 1,..., 0) to ei is an isometry, hence an isomorphism, and so e1,..., er are linearly independent in M. The function ϕ(x, x)/\|x\|2 on Rr \ 0 achieves it supremum λ because the unit sphere Sr−1 is compact; moreover, it is easy to see that. * ϕ(u, u) \|u\|2 = ϕ
(x, x) \|x\|2 λ = sup 0=u∈Rr holds if and only if x is a nonzero eigenvector of A belonging to the maximum eigenvalue λ, so that Ax = λx. Now because aij ≥ 0 for i = j, we can assume that the coordinates xi of x = (x1,..., xr ) are all ≥ 0; for ϕ(x, x) = aij xixj ≤ aij \|xi\|\|xj \| = ϕ(y, y), where y = (\|x1\|,..., \|xr \|). Since x is an eigenvector of A, we have ximj = aij ximj = ϕ(ei, ej )ximj (∗) λ ij ij ij for any m = (m1,..., mr ) ∈ Rr. We apply this to m = (m1,..., mr ) ∈ Zr, where d = miei is the element given in the assumption. Then the right-hand side of (∗) equals xiϕ(ei, d), which is negative since xi ≥ 0 and ϕ(ei, d) < 0. Finally, ij ximj > 0 on the left-hand side of (∗), and therefore λ < 0; thus the matrix A has maximum eigenvalue λ < 0, and it follows that it is negative deﬁnite. Proposition A.3 Let M be a Z-module having a scalar product (a, b) ∈ Z for a, b ∈ M, and suppose that e1,..., er is a set of generators of M satisfying (ei, ej ) ≥ 0 r for i = j. Assume that there exists an element d = i=1 liei with li > 0 and (d, ei) = 0 for i = 1,..., r. Then (m, m) ≤ 0 for all m ∈ M. If in addition the elements e1,..., er cannot be partitioned into two components in such a way that (ei, ej ) = 0 for ei and ej in different components then (m, m) = 0 only for m proportional to d.
Proof The proof is almost the same as for Proposition A.2. Arguing as there, we ﬁnd that the matrix A has maximum eigenvalue λ = 0, which proves that ϕ(x, x) ≤ 0 for all x ∈ Rr ; and moreover, ϕ(x, x) = 0 for x ∈ Rr only if Ax = 0. Suppose that there 2 Polynomials 285 exist two linearly independent vectors x(1) and x(2) ∈ Rr with ϕ(x(i), x(i)) = 0, hence Ax(i) = 0. Then there is a nonzero linear combination x of x(1) and x(2) which has some zero coefﬁcients. As in the proof of Proposition A.2, passing to the vector y = (\|x1\|,..., \|xr \|), we can again assume that all the coefﬁcients of x are ≥ 0, some = 0; and x still satisﬁes Ax = 0, that is, ϕ(x, fi) = 0 for i = 1,..., r, where fi = (0,..., 1,..., 0) ∈ Rr is s the standard basis. Suppose that x = i=1 xifi with xi > 0 and s < r. Then for s j > s we have 0 = ϕ(x, fj ) = i=1 xiϕ(fi, fj ), and since ϕ(fi, fj ) ≥ 0 for all i and j and xi > 0 for i ≤ s it follows that (ei, ej ) = ϕ(fi, fj ) = 0 for i ≤ s and j > s. This partitions the set of vectors {e1,..., er } into two components {e1,..., es} and {es+1,..., er } consisting of pairwise orthogonal vectors. The proposition is proved. 2 Polynomials Proposition A.4 Let an ∈ Q be a sequence of numbers, and suppose that there exists a polynomial g(T ) ∈ Q[T ] such that an+1 − an = g(n) for all sufﬁciently large n. Then there exists a polynomial f
(T ) ∈ Q[T ] such that an = f (n) for all sufﬁciently large n. Proof For any g(T ) ∈ Q[T ], there exists a polynomial h(T ) ∈ Q[T ] such that h(T + 1) − h(T ) = g(T ). This assertion can be proved by induction on n = deg g; for if g has leading term equal to aT n then setting h0(T ) = a/(n + 1)T n+1, we ﬁnd that h0(T + 1) − h0(T ) − g(T ) has degree < n, and then we can use induction. Note that h is determined up to an additive constant. For any choice of the polynomial h we get an+1 − an = h(n + 1) − h(n), that is, h(n + 1) − an+1 = h(n) − an for all sufﬁciently large n, that is, h(n) − an = c. The polynomial f = h − c satisﬁes the requirements of the proposition. The proposition is proved. 3 Quasilinear Maps Let L be a vector space over a ﬁeld K and ϕ : L → L a map. We say that ϕ is quasilinear if ϕ(x + y) = ϕ(x) + ϕ(y) for x, y ∈ L, and there exists an automorphism g of K such that ϕ(αx) = g(α)ϕ(x) for all α ∈ K and x ∈ L; then we say that g is the automorphism of K associated with ϕ. 286 Algebraic Appendix Proposition A.5 Let L be a ﬁnite dimensional vector space over a ﬁeld K, and G a ﬁnite group of quasilinear maps of L. Assume that every element e = ϕ ∈ G has associated automorphism g = idK. Then L has a basis consisting of elements invariant under G. Remark Obviously, in this basis, each map in G has the identity matrix; this does not mean, of course, that it is the identity map: it acts on coordinates
by the corresponding automorphism of K. We start with a well-known lemma. Lemma Any set {g1,..., gn} of distinct ﬁeld homomorphism gi : K → K is linearly independent over K; that is, there does not exist any nontrivial relation of the form n i=1 λigi(ξ ) = 0 for all ξ ∈ K with λi ∈ K. In other words, there exist α1,..., αn ∈ K such that gi(αj ) = 0. det (A.7) (A.8) Proof of the Lemma Among all relations of the form (A.7), choose one with the minimal number of nonzero coefﬁcients λi. There are obviously at least two such nonzero coefﬁcients, say λj = 0 and λk = 0 with j = k. Since by assumption gj = gk there exists α ∈ K such that gj (α) = gk(α). Substituting αξ for ξ in (A.7) gives n i=1 λigi(α)gi(ξ ) = 0 for all ξ ∈ K. (A.9) Subtracting gj (α) times (A.7) from (A.9) gives a relation gi(α) − gj (α) λi gi(ξ ) = 0, n i=1 (A.10) in which gj (ξ ) has coefﬁcient 0, but the coefﬁcient gk(ξ ) is gk(α) − gj (α) λk = 0. This contradicts the minimality of the choice of relation (A.7). The lemma is proved. Proof of Proposition By assumption the different maps in G have different associated homomorphisms. Thus we can index the elements of G by their associated homomorphisms. Write Ag ∈ G for the map with associated homomorphism g. 4 Invariants 287 Let LG be the set of vectors x ∈ L invariant under all Ag ∈ G; let us prove that Ag(x); obviously S(x) ∈ LG for any x ∈ L. LG generates L. For this, set S(
x) = We prove that already the vectors S(x) with x ∈ L generate L over K. For this, note that the space spanned by the S(x) contains the elements S(αx) = g(α)Ag(x) for all α ∈ K. By the lemma, we can choose elements α1,..., αn for which det \|gi(αj )\| = 0. We see that the Ag(x) can be expressed as linear combinations of the S(αix). In particular for g = e we get an expression for x itself, which was what we wanted. Now it is enough to choose vectors y1,..., yr ∈ LG that generate L, and a maximal linearly independent subset among these. This will be the required basis. The proposition is proved. 4 Invariants Proposition A.6 Let A be a ﬁnitely generated algebra over k and G a ﬁnite group of automorphisms of A. Assume that the order n of G is not divisible by char k. Write AG for the subalgebra of elements a ∈ A such that g(a) = a for all g ∈ G. Then AG is ﬁnitely generated as an algebra over k. Proof We write S for the averaging operator S(a) = 1 n g(a). g∈G For any a ∈ A, the coefﬁcients of the polynomial Pa(T ) = T − g(a) = T n + σ1T n−1 + · · · + σn g∈G belong to AG. The coefﬁcients σi are the elementary symmetric functions in g(a), that can be expressed in terms of the Newton sums S(ai) for i = 1,..., n. Let u1,..., um be a set of generators of A. Write B for the subalgebra of AG generated by the elements S(uj i ) for i = 1,..., m and j = 1,..., n. Then Pui (ui) = 0, and hence the un with i coefﬁcients in B. Therefore it follows by induction that any monomial ua1 · · · uam m 1 can be expressed as a linear combination of monomials of
the same kind with a1,..., am < n. Thus any element a ∈ A has an analogous expression i can be expressed as linear combinations of 1, ui,..., un−1 a = ai <n ϕa1...am ua1 1 · · · uam m with ϕa1...am ∈ B. In particular, let a ∈ AG. Applying the operator S to this, we get a = S(a) = ϕa1...am S ua1 1. · · · uam m 288 Algebraic Appendix It follows that AG is generated by elements S(ua1 1 The proposition is proved. · · · uam m ) with ai < n and S(un i ). 5 Fields Proposition A.7 Let k be an algebraically closed ﬁeld and k ⊂ K a ﬁnitely generated extension. Then there exist elements z1,..., zd+1 ∈ K with K = k(z1,..., zd+1), and such that z1,..., zd are algebraically independent over k and zd+1 is separable over k(z1,..., zd ). Proof Suppose that K is generated over k by a ﬁnite number of elements t1,..., tn, and let d be the maximal number of algebraically independent elements among the ti. Suppose that t1,..., td are algebraically independent. Then any element y ∈ K is algebraically dependent on t1,..., td, and moreover, there exists a relation f (t1,..., td, y) = 0 with f (T1,..., Td, Td+1) irreducible over k. Let f (T1,..., Td+1) be such a polynomial for t1,..., td+1. We assert that the partial derivative f (T1,..., Td+1) = 0 for at least one i = 1,..., d + 1. Indeed, if Ti not, then each Ti only occurs in f in powers that are multiples of the characteristic p ai1...id+1T pi1 of k; that is, f is of
the form f = · · · T id+1 bi1...id+1T i1 and g = 1 ducibility of f. · · · T pid+1 i1...id+1 d+1 ; then we get f = gp, which contradicts the irre- d+1. Set ai1...id+1 = bp 1 = 0, the d elements t1,..., ti−1, ti+1,..., td+1 are algebraically independent over k. Indeed, ti is algebraic over k(t1,..., ti−1, ti+1,..., td+1) because f = 0, so that Ti occurs in f. Thus if t1,..., ti−1, ti+1,..., td+1 were algebraically Ti dependent, the transcendence degree of k(t1,..., td+1) would be less than d, which contradicts the algebraic independence of t1,..., td. If f Ti Td+1 Thus we can always renumber the ti, so that t1,..., td are algebraically independent over k, and f = 0. This shows that td+1 is separable over k(t1,..., td ). Since td+2 is algebraic over k(t1,..., td ), by the primitive element theorem (see van der Waerden, [76, Section 46]), we can ﬁnd an element y such that k(t1,..., td+2) = k(t1,..., td, y). Repeating the process of adjoining elements td+1,..., tn, we express K as k(z1,..., zd+1), where z1,..., zd are algebraically independent over k and f (z1,..., zd, zd+1) = 0, with f an irreducible polynomial over k with f proved. Td+1 = 0. Proposition A.7 is Proposition A.8 Let k be an algebraically closed ﬁeld of characteristic p, and K a ﬁnitely generated ﬁeld extension of k, having transcendence degree 1 over k. Let K (p)
be the subﬁeld consisting of pth powers αp, with α ∈ K. Then [K : K (p)] = p. If L ⊂ K is a subﬁeld such that K is an inseparable extension of L then L ⊂ K (p). 6 Commutative Rings 289 Proof Recall that α → αp deﬁnes a ﬁeld homomorphism K → K, whose image is the subﬁeld K (p). Let t ∈ K be transcendental over k. The ﬁrst assertion follows from the diagram K (p) ⊂ K k(t)(p) ⊂ k(t). Indeed, this implies that K : k(t) K : k(t)(p) = k(t) : k(t)(p) K : K (p) = K (p) : k(t)(p). Since α → αp deﬁnes an isomorphic inclusion, it follows that [K : k(t)] = [K (p) : k(t)(p)], and therefore [K : K (p)] = [k(t) : k(t)(p)]. Finally, it is obvious that k(t)(p) = k(t p), and hence k(t) : k(t)(p) = p and K : K (p) = p. To prove the second assertion, write L for the set of all elements of K that are separable over L. It is very easy to prove that this is a subﬁeld. We can obviously replace L by L, and thus assume that any element of K that is separable over L belongs to L. Let α ∈ K and suppose that its minimal polynomial is of the form P (T ) = a0T pmr + a1T pm(r−1) + · · · + ar−1T pm + ar where Q(T ) = a0T r + a1T (r−1) + · · · + ar−1T + ar is a separable polynomial, that is Q(T ) = 0. Then β = αpm satisﬁes Q(β) = 0, that is, β is separable over L, and therefore belongs to L. It follows that K can be obtained
from L by successively adjoining pth roots; √ that is, there is a chain L = K1 ⊂ · · · ⊂ Km = K, with Ki = Ki−1( p αi), for some √ αi ∈ Ki−1. Set K = Km−1 and α = αm−1, so that K = K ( p α). We prove that K = K (p), and it is at this point that we use that K has transcendence degree 1 over k. Any element β ∈ K has an expression β = a0 + a1 1 α + · · · + αp √ α)p−1 α + · · · + αp−1( p with ai ∈ K, and hence βp = ap p−1αp−1, that is K (p) ⊂ K. 0 But [K : K ] = p, and we proved that [K : K (p)] = p in the ﬁrst part of the proof. Therefore K = K (p) and L ⊂ K (p). Proposition A.8 is proved. + ap √ p 6 Commutative Rings Proposition A.9 (The Hilbert Nullstellensatz) Let k be an algebraically closed ﬁeld and F1,..., Fm ∈ k[T1,..., Tn]. If the ideal (F1,..., Fm) = (1) then the system of equations F1 = · · · = Fm = 0 has a solution in k. Lemma If a system of equations F1 = · · · = Fm = 0 with Fi ∈ k[T1,..., Tn] has a solution in some ﬁnitely generated extension ﬁeld K of k, then it has a solution in k. 290 Algebraic Appendix Proof By Proposition A.7, K is of the form k(x1,..., xr, θ ) where x1,..., xr are algebraically independent over k, and θ is a root of a polynomial P (X, U ) = p0(X)U d + · · · + pd (X) ∈ k(X)[U ], with P (X, U ) irreducible
over k(X) = k(x1,..., xr ); here we write X = (x1,..., xr ). Suppose that Fj (ξ1,..., ξn) = 0 with ξi ∈ K. Write the ξi in the form ξi = Ci(X, θ ), with Ci(X, U ) ∈ k(X)[U ]. The relation Fj (ξ1,..., ξn) = 0 gives the identity C1(X, U ),..., Cn(X, U ) Fj = P (X, U )Qj (X, U ) (A.11) in X = (x1,..., xr ) and U, where Qj (X, U ) ∈ k(X)[U ]. Choose values xi = αi ∈ k for i = 1,..., n such that (α1,..., αn) is not a zero of the denominators of any coefﬁcient of P, Qi, C1,..., Cn ∈ k(X)[U ], nor a zero of the leading coefﬁcient of P. Now choose U = τ ∈ k to be one of the roots of P (α1,..., αn, τ ) = 0, and set Cj (α1,..., αn, τ ) = λj for j = 1,..., m. Then it follows from (A.11) that Fj (λ1,..., λn) = 0, that is, (λ1,..., λn) is a solution of the system F1 = · · · = Fm = 0. The lemma is proved. Proof of Proposition A.9 If the ideal (F1,..., Fm) = (1) then it is contained in some maximal ideal M ⊂ k[T1,..., Tn], and K = k[T1,..., Tn]/M is a ﬁeld. Write ξi for the image of Ti in K. Obviously K = k(ξ1,..., ξn) and (ξ
1,..., ξn) is a solution in K of the system F1 = · · · = Fm = 0. We get a solution in k by applying the lemma. This proves Proposition A.9. Corollary If G, F1,..., Fm ∈ k[T1,..., Tn] and G is 0 at all solutions of the system F1 = · · · = Fm = 0, then GN ∈ (F1,..., Fm) for some N ≥ 0. Proof It is enough to consider the case G = 0. We introduce a new variable U, and consider the polynomials F1,..., Fm and U G − 1 ∈ k[T1,..., Tn, U ]. By assumption these have no common solutions in k, and therefore by Proposition A.9 there exist polynomials P1,..., Pm, Q ∈ k[T1,..., Tn, U ] such that P1F1 + · · · + PmFm + Q(U G − 1) = 1. This identity is preserved if we set U = 1/G. Clearing denominators we get GN ≡ 0 mod (F1,..., Fm). The corollary is proved. Proposition A.10 Let A be a commutative ring with a 1. An element a ∈ A is nilpotent (that is, an = 0 for some n > 0) if and only if a belongs to every prime ideal of A. 6 Commutative Rings 291 Proof A nilpotent element is obviously contained in every prime ideal. Conversely, suppose that a is not nilpotent. We construct a prime ideal not containing a. Consider the ideals I ⊂ A not containing any power of a. By assumption, I = (0) has this property. Let a be a maximal element of this set of ideals, which exists by Zorn’s lemma. We prove that a is a prime ideal; then since a /∈ a, this will prove the proposition. For this, set B = A/a and write b for the image of a; we prove that B is an integral domain. By assumption, any nonzero ideal b ⊂ B contains some power of b, but b itself is not nilpot
ent. Suppose b1, b2 ∈ B with b1, b2 = 0. Then by assumption, bn1 ∈ (b1) and bn2 ∈ (b2) for some n1, n2 > 0. Hence bn1+n2 ∈ (b1b2), and therefore b1b2 = 0. Proposition A.10 is proved. Proposition A.11 (Nakayama’s Lemma) Let M be a ﬁnite module over a ring A and a ⊂ A an ideal. Suppose that for any element a ∈ 1 + a, aM = 0 implies M = 0. Then aM = M implies that M = 0. Proof Suppose that M = (μ1,..., μn). The assumption aM = M implies that there are equalities μi = n j =1 αij μj with αij ∈ a. n j =1(αij − δij )μj = 0 for i = 1,..., n, and by Cramér’s rule dμi = 0 for Thus i = 1,..., n, where d = det(αij − δij ); therefore dM = 0. Since d ∈ 1 + a, it follows by assumption that M = 0. The proposition is proved. Corollary A.1 If A ⊂ B are rings with B a ﬁnite A-module and a ⊂ A an ideal, then a = A implies aB = B. Proof Since B contains the unit element of A, aB = 0 only if a = 0, and if a = (1) then 0 /∈ 1 + a. This veriﬁes the assumptions of Proposition A.11, and so aB = B. The corollary is proved. Corollary A.2 If a ⊂ A is an ideal such that every element of 1 + a is invertible, M a ﬁnite A-module and M ⊂ M any submodule, then M + aM = M implies that M = M. Proof Apply Proposition A.11 to the module M/M. The corollary is proved. Remark It is easy to see that the assumption on the ideal a in Corollary A.2 holds if A/a is
a local ring. Corollary A.3 Under the assumptions of Corollary A.2, elements μ1,..., μn ∈ M generate M if and only if their images generate M/aM. 292 Algebraic Appendix Proof Apply Corollary A.2 to the submodule M = (μ1,..., μn). The corollary is proved. Proposition A.12 Let A be a Noetherian ring, and a ⊂ A an ideal such that every element of 1 + a is invertible in A. Then n>0(b + an) = b for any ideal b ⊂ A. (1) The case b = 0. Apply Proposition A.11 to M = (2) The general case. Set B = A/b and let a = (a + b)/b be the image of a in B. Then (a)n = (a + b)n/b = (an + b)/b is the image of an in B. By the case (1), an. n>0(a)n = 0, and hence n>0(b + an) = b. The proposition is proved. Proposition A.13 Suppose that a is an ideal of a Noetherian ring A such that every element of 1 + a is invertible in A. Then the property that a sequence of elements f1,..., fm ∈ a is a regular sequence (see Section 1.2, Chapter 4) is preserved under permutations of the fi. Proof It is enough to prove that permuting two adjacent elements fi, fi+1 of a regular sequence again gives a regular sequence. Set (f1,..., fi−1) = b and A/b = B, and write a, b for the images in B of fi, fi+1. Everything reduces to the proof of Proposition A.13 for a regular sequence a, b of B. We need to prove (1) that b is not a zerodivisor in B, and (2) that a is not a zerodivisor modulo b. (1) Suppose that xb = 0. We prove then that ak x ∈ for all k. (A.12) Since A is Noetherian, it follows by Proposition A.12 that x = 0. We verify (A.12) by induction.
If x = x1ak then x1akb = 0. Since a, b is a regular sequence, a is a non-zerodivisor, and hence x1b = 0. Again because a, b is a regular sequence, it follows that x1 ∈ (a), hence x ∈ (ak+1). (2) Suppose that xa = yb. Because a, b is a regular sequence, it follows that y = az with z ∈ A, and hence x = zb. The proposition is proved. 7 Unique Factorisation Proposition A.14 Suppose that a Noetherian local ring A is contained in a local ring A which is a UFD. Suppose that the maximal ideals m ⊂ A and m ⊂ A satisfy the following conditions: (a) m A = m; (b) (mn A) ∩ A = mn for n > 0; (c) for any α ∈ A and any integer n > 0 there exists an ∈ A such that α − an ∈ mn A. Then A is also a UFD. 8 Integral Elements 293 Proof (taken from Mumford [60, Section 1C]) The usual method of proving unique factorisation into prime factors deduces it from the statement that if a divides bc, and a and b have no common factors then a divides c. We need to establish this result in A, knowing that it holds in A. For this it is enough to prove the following two assertions: (1) for a, b ∈ A, a \| b in A =⇒ a \| b in A; (2) if a and b have no commons factors in A, then they have no common factors in A. Both these assertions are based on the following lemma. Lemma (a A) ∩ A = a for any ideal a ⊂ A. Proof It is enough to prove that (a A)∩A ⊂ a. Suppose that a = (a1,..., an), and let aiαi with αi ∈ A. By assumption (c), there exist elements x ∈ (a A)∩A. Then x = a(n) ξ (n) + ξ (n) i ai = i i a +ξ with a ∈ a and ξ ∈ mn. Hence ξ = x −a ∈ A
∩ mn = mn. Therefore x ∈ a+mn for all n > 0 and so x ∈ a by Proposition A.12. The lemma is proved. ∈ A such that αi = a(n) ∈ mn. Then x = i with ξ (n) a(n) i ai + i Proof of (1) If a divides b in A then b ∈ A ∩ (a) A, which by the lemma is equal to (a). This just means a divides b in A. Proof of (2) If a and b have a common factor in A then they can be written a = γ α, b = γβ where α, β ∈ A are proper divisors of a and b with no common factors. Then aβ − bα = 0. By assumption (c), there exist xn, yn ∈ A and un, vn ∈ mn such that α = xn + un, β = yn + vn. Hence ayn − bxn ∈ (a, b)mn = (a, b)mn A. By the lemma, ayn − bxn ∈ (a, b)mn, that is, ayn − bxn = atn + bsn with sn, tn ∈ mn. Hence a(yn − tn) = b(xn + sn) and so α(yn − tn) = β(xn + sn). From the assumption that α and β have no common factors in A it follows that xn + sn is divisible by α, that mk = 0, for sufﬁciently large n we have α, β /∈ mn−1. is, xn + sn = αλ. Since Then also xn + sn /∈ mn−1, and hence λ /∈ m, that is, λ is invertible in A. Hence A(xn + sn) = (α) A, and xn + sn divides α, and so divides a in A. By (1) it also divides a in A, that is, a = (xn + sn)h. But a(yn − tn) = b(xn + sn), and hence b = (yn − tn)h. Since a and b have no common factors in
A it follows that h is invertible in A, that is, (a) = (xn + sn) = (α), and this contradicts the assumption that α is a proper divisor of a in A. The proposition is proved. 8 Integral Elements Proposition A.15 Let B = k[T1,..., Tn] be the polynomial ring and L = k(T1,..., Tn) its ﬁeld of fractions, and suppose that L ⊂ K is a ﬁnite ﬁeld extension. Write A for the integral closure of B in L. Then A is a ﬁnite B-module. 294 Algebraic Appendix 1 + a1αpl (m−1) Proof The proof in the case that the extension L ⊂ K is separable, which is very simple, is given in Atiyah and Macdonald [8, Proposition 5.17]. We do not reproduce it here, but show how to reduce everything to the case of a separable extension. Suppose that K = L(α1,..., αs). If α1 is not separable over L then its minimal polynomial is of the form αpl m + · · · + am = 0, where ai ∈,..., T 1/pl k(T1,..., Tr ) and αpl ], L = r k(T 1/pl ), and let A be the integral clo1 sure of B in K. Now set ai = bpl, with bi ∈ L. Then K = L(α1,..., αs) and + · · · + bm = 0, so that α1 is separable over L. On the other hand αm 1 A ⊂ A, and if the proposition is proved for A then A is a ﬁnite B-module. But B · · · T ir /pl is itself a ﬁnite B-module: it has a basis consisting of monomials T i1/pl r with 0 ≤ ir,..., ir < pl. Therefore A is a ﬁnite B-module, and hence so is its submodule A. is separable over L. Write B = k[T 1/pl,..., T 1/pl r 1 ) and
K = K(T 1/pl + b1α(m−1),..., T 1/pl r 1 1 1 1 1 i We see that the proof of the proposition reduces to the case that α1 is separable. By the primitive element theorem, then L(α1,..., αs) = L(α 2, α3,..., αs). Applying the same argument s times we reduce the proof to the case of a separable extension. The proposition is proved. 9 Length of a Module Deﬁnition A module M over a ring A has ﬁnite length if there exists a chain of A-submodules M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = 0 with Mi = Mi+1, (A.13) such that each quotient Mi/Mi+1 is a simple A-module, that is, does not contain any proper submodule. By the Jordan-Hölder theorem, all such chains have the same length n; this common length n is called the length of M, and denoted by (M), or A(M) to stress the role of the ring A. Obviously, the quotient modules Mi/Mi+1 in (A.13) are isomorphic to A/m where m are maximal ideals of A. If M has ﬁnite length then the same holds for all its submodules and quotient modules. If a module M has a chain (A.13) such that each quotient Mi/Mi+1 has ﬁnite length then M has ﬁnite length, and (M) = (Mi/Mi+1). Proposition A.16 Let O be a Noetherian local ring, with maximal ideal m, and suppose that a ⊂ O is an ideal such that a ⊃ mk for some k > 0; then the module O/a has ﬁnite length. 9 Length of a Module 295 Proof It is enough to prove that O/mk is a module of ﬁnite length. By considering the chain of submodules Mi = mi/mk for i = 0,..., k, we see that it is enough to check that each module mi/mi+1 has ﬁnite length. But in the O-module structure of mi/mi+1, multiplication by
m kills every element. Therefore O/m = k acts on mi/mi+1, so that it is a vector space over the ﬁeld k, and its length equals the dimension of this vector space over k. Since O is a Noetherian ring, mi/mi+1 has a ﬁnite number of generators, that is, it is a ﬁnite dimensional vector space. This proves the proposition. If M is an A-module and p a prime ideal of A then we write Mp for the localisa- tion of M at p, that is the module M ⊗A Ap, where Ap is the local ring of p. Example If M = A/p then Mq = 0 if q ⊃ p. If q ⊃ p then Mq = (A/p)q where q = q/p is the image of q in A/p. Lemma A ﬁnite module M over a Noetherian ring A has a chain (A.13) of submodules such that Mi/Mi+1 ∼= A/pi, where pi ⊂ A is a prime ideal. Proof For an element m ∈ M with m = 0, write Ann m for the ideal of elements of a such that am = 0. Because A is Noetherian, a chain of ideals of the form Ann(m1) ⊂ Ann(m2) ⊂ · · · must terminate. Hence we can choose m ∈ M with the following property: Ann m ⊂ Ann(m) with m = 0 implies that Ann m = Ann(m). We prove that Ann m is then a prime ideal. Let ab ∈ Ann m with b /∈ Ann m. Then Ann m ⊂ Ann(bm) and bm = 0, but then by assumption Ann m = Ann(bm). But a ∈ Ann(bm), hence a ∈ Ann m. Set p = Ann m. Then the submodule Am ⊂ M is isomorphic to A/p. In M = M/Am we can again ﬁnd a submodule isomorphic to A/p where p is a prime ideal of A. In this way we construct a chain M (1) ⊂ M (2) ⊂ · · · such that M (i−1)/M (i) ∼= A/pi. By the
assumption that M is Noetherian, this chain terminates. The lemma is proved. Deﬁnition A local ring A with maximal ideal m is 1-dimensional if there exists a prime ideal p m, and every such prime ideal p is minimal, that is, does not contain any strictly smaller prime ideal. Proposition A.17 Let O be a 1-dimensional local ring having a ﬁnite number of minimal prime ideals p1,..., pn, and a ∈ A a non-zerodivisor of A not contained in any of the pi. Then O/(a) = n i=1 Opi (Opi ) × O O/(pi + aO). (A.14) Proof (taken from Fulton [29, A.1–3]) At the same time as (A.14) we prove a generalisation to an arbitrary ﬁnite O-module M. For this, we set e(M, a) = 296 Algebraic Appendix A(M/aM) − A(AnnM (a)), where AnnM (a) denotes the A-module {m ∈ M \| am = 0}. The generalisation of (A.14) is the following: e(M, a) = n i=1 Opi (Mpi ) × O O/(pi + aO). (A.15) The advantage of the invariant e(M, a) is that it is additive: if a ∈ O and 0 → M → M → M → 0 is an exact sequence, then e(M, a and the left-hand side is ﬁnite if both terms on the right-hand side are. This follows at once from the following exact sequence 0 → AnnM (a) → AnnM (a) → AnnM (a) → M /aM → M/aM → M /aM → 0, which is trivial to verify. By induction we get that for any chain (A.13), e(M, a) = e(Mi/Mi+1, a). It follows from these considerations and from the lemma that we need only prove (A.15) for modules M isomorphic to O/p, where p is a prime ideal of O. If p = m is the maximal ideal then M ∼= k (as an O-module), so that e(M
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. Harper and Rowe, New York (1965); MR 34 #1327 64. Mumford, D., Fogarty, J.: Geometric Invariant Theory, 2nd edn. Springer, Berlin (1982) 65. Pham, F.: Introduction à l’étude topologique des singularités de Landau. Mém. Sci. Math., Gauthier-Villar, Paris (1967); MR 37 #4837 66. Pontryagin, L.S.: Continuous Groups, Gos. Izdat. Teor.-Tekh. Lit, Moscow (1954). English translation: Topological Groups (Vol. 2 of Selected Works), Gordon and Breach, New York (1986) 67. Saltman, D.J.: Noether’s problem over an algebraically closed ﬁeld. Invent. Math. 77, 71–84 (1984) 68. Seifert, G., Threlfall, V.: Lehrbuch der Topologie. Chelsea, New York (1934). English trans- lation: Academic Press, New York (1980) 69. Shafarevich, I.R., et al.: Algebraic Surfaces. Proceedings of the Steklov Inst., vol. 75. Nauka, Moscow (1965). English translation: AMS, Providence (1967); MR 32 #7557 70. Shokurov, V.V.: Numerical geometry of algebraic varieties. In: Proc. Int. Congress Math., vol. 1, Berkeley, 1986, pp. 672–681. AMS, Providence (1988) 71. Siegel, C.L.: Automorphic Functions and Abelian Integrals. Wiley-Interscience, New York (1971) 72. Siegel, C.L.: Abelian Functions and Modular Functions of Several Variables. Wiley- Interscience, New York (1973) 300 References 73. Siu, Y.-T.: A general non-vanishing theorem and an analytic proof of the ﬁnite generation of the canonical ring. arXiv:math/0610740 74. Springer, G.: Introduction to Riemann Surfaces, 2nd edn. Chelsea, New York (1981) 75. Springer, T.: Invariant Theory. Springer, Berlin (1977) 76. van der Waerden, B.L.: Moderne Algebra, Bd. 1, 2,
Morgan In the previous three sections, we deﬁned probabilities, at least approximately, in situations with symmetry, or when we could repeat a random procedure under essentially the same conditions, or when there was a plausible agreed “fair price” for a resulting prize. However, there are many asymmetric and non-repeatable random events, and for many of these we would feel distinctly unsatisﬁed, or even unhappy, with the idea that their probabilities should be determined by gamblers opening a book on the question. For example, what is the probability that some accused person is guilty of the charge? Or the probability of life on another planet? Or the probability of you catching a cold this week? Or the probability that Shakespeare wrote some given sonnet of doubtful provenance? Or the probability that a picture called “Sunﬂowers” is by van Gogh? Or the probability that Riemann’s hypothesis is true? (It asserts that all the nontrivial zeros of a certain function 2.) Or the probability that π e is irrational? In this last question, would the have real part + 1 knowledge that eπ is irrational affect the probability in your judgment? In these questions, our earlier methods seem more or less unsatisfactory, and, indeed, in a court of law you are forbidden to use any such ideas in deciding the probability of guilt of an accused. One is led to the concept of probability as a “degree of belief.” 10 0 Introduction If we assign a probability to any of these eventualities, then the result must of necessity be a personal or subjective assessment. Your ﬁgure need not be the same as mine or anyone else’s, and any probability so obtained is called subjective or personal. In fact, there is a strong (although perhaps minority) body of opinion that maintains that all probabilities are subjective (cf the remark of A. de Morgan at the head of this section). They argue that appeals to symmetry, the long run, or fair value, merely add a spurious objectivity to what is essentially intuition, based on personal experience, logical argument, and experiments (where these are relevant). The examples above are then simply rather trivial special cases of this general deﬁnition. According to this approach, a probability is a measure of your “degree of belief” in the guilt of the accused, the truth of some assertion, or that a die will show a six. In
the classical case based on dice, etc., this belief rests on symmetry, as it does in the “long-run” relativefrequency interpretation. The “fair price” approach uses the beliefs of all those concerned in ﬁxing such a price. Furthermore, the approach via degrees of belief allows (at least in principle) the possibility that such probabilities could be determined by strictly logical statements relating what we know for sure to the uncertain eventuality in question. This would be a kind of inductive probability logic, an idea that was ﬁrst suggested by Leibniz and later taken up by Boole. During the past century, there have been numerous clever and intriguing books about various approaches to establishing such an axiomatic framework. This argument is clearly seductive, and for all the above reasons it is tempting to regard all types of probability as a “feeling of the mind,” or as a “degree of belief.” However, there are several drawbacks. First, in practice, it does not offer a wholly convincing and universally accepted way of deﬁning or measuring probability, except in the cases discussed above. Thus, the alleged generality is a little artiﬁcial because different minds feel differently. Second, setting this more general idea in a formal framework requires a great deal more effort and notation, which is undesirable for a ﬁrst approach. Finally, it is in any case necessary for practical purposes to arrange things so that the rules are the same as those obtained from the simpler arguments that we have already outlined. For these reasons, we do not further pursue the dream of a universal interpretation of probability; instead, we simply note this remark of William Feller: All deﬁnitions of probability fall short of the actual practice. 0.7 FAQs Neither physicists nor philosophers can give any convincing account of what “physical reality” is. G.H. Hardy “What is the meaning of it, Watson?” said Holmes, solemnly, as he laid down the paper.... “It must tend to some end, or else our universe is ruled by chance, which is unthinkable.” A. Conan Doyle, The Adventure of the Cardboard Box In the preceding sections, we agree to develop a mathematical theory of probability and discuss some interpretations of probability. These supply deﬁnitions of probability in 0.7 FAQs 11 some simple cases and,
in any case, we interpret probability as an extension of the idea of proportion. The net result of this preliminary reconnaissance is our convention (or rule) that probability is a number between zero and unity inclusive, where impossibility corresponds to 0, and certainty corresponds to 1. In Chapter 1, we probe more deeply into our ideas about probability, to discover more subtle and important rules that describe its behaviour, and generate our mathematical model. However, there are a few very common doubts and queries that it is convenient to dispose of here. (1) The ﬁrst, and perhaps in one way, key question is to ask if anything is really random? This is called the question of determinism, and its importance was realised early on. One point of view is neatly expressed by Laplace in these two extracts from his monumental work on probability: The path of a molecule is ﬁxed as certainly as the orbits of the planets: the only difference between the two is due to our ignorance. Probability relates partly to this ignorance and partly to our knowledge.... Thus, given a sufﬁciently great intelligence that could encompass all the forces of nature and the details of every part of the universe,... nothing would be uncertain and the future (as well as the past) would be present before its eyes. This is called “determinism” because it asserts that the entire future is determined by a complete knowledge of the past and present. There are many problems with this, not the least of them being the unattractive corollary that people do not have free will. Your every action is inevitable and unavoidable. However, no one actually believes this, unless wasting time on fruitless cerebration is also unavoidable, and so on. The difﬁculties are clear; fortunately, on one interpretation of quantum theory, certain activities of elementary particles are genuinely unpredictable, an idea expressed by Heisenberg’s uncertainty principle. But this view also leads to paradoxes and contradictions, not the least of which is the fact that some events must perforce occur with no cause. However, causeless events are about as unattractive to the human mind as the nonexistence of free will, as expressed by Einstein’s remark, “In any event, I am sure that God does not dice.” At this point we abandon this discussion, leaving readers to ponder it for as long as they desire. Note, ﬁrst,
that however long you ponder it, you will not produce a resolution of the problems and, second, that none of this matters to our theory of probability. (2) If you accept that things can be random, the second natural question is to ask: What is probability really? The answer to this is that our intuition of probability begins as a large portmanteau of empirical observations that the universe is an uncertain place, with many areas of doubt and unpredictability. In response to this, we form a theory of probability, which is a description of what we observe in terms of (mathematical) rules. In the sense in which the question is usually posed, we here neither know nor care what probability “really” is. (If we were engaged in a certain type of philosophical investigation, this question might concern us, but we are not.) It is the same in all sciences; we label certain concepts such as “mass,” “light,” “particle,” and describe their properties by mathematical rules. We may never know what light or mass “really” are, nor does it matter. Furthermore, for example, even if it turned out that mass is 12 0 Introduction supplied by the Higgs boson, the next question would be to ask what is a boson “really”? You might just as well ask what are numbers really? (3) When it is appreciated that probability is our constructed model, it is next natural to ask is it the only one or are there others? There are indeed many other theories devised to explain uncertain phenomena. For example, the quantum theory designed to explain the odd behaviour of elementary particles uses a probabilistic structure quite different from that seen here. Another area in which a different theoretical structure would be necessary is in applying probability to questions of legal proof “beyond reasonable doubt” or “on the balance of probabilities.” The answer in this case is still a matter for debate. At the other extreme, attempts have been made to construct more general theories with a smaller set of rules. So far, none of these is in general use. (4) Accepting that we need a theory, another natural question is; Does it have to be so abstract and complicated? The answer here is (of course) yes, but the reason is not just the perversity of professors. The ﬁrst point is that probability is not directly tangible in the same way as the raw
material of many branches of physics. Electricity, heat, magnetism, and so on, will all register on the meters of appropriate instruments. There is no meter to record the presence of probability, except us. But we are imperfect instruments, and not infrequently rather confused. It is the case that most people’s intuition about problems in chance will often lead them grossly astray, even with very simple concepts. Although many examples appear later, we mention a few here: (a) The base rate fallacy. This appears in many contexts, but it is convenient to display it in the framework of a medical test for a disease that affects one person in 100,000. You have a test for the disease that is 99% accurate. (That is to say, when applied to a sufferer, it shows positive with probability 99%; when applied to a nonsufferer, it shows negative with probability 99%.) What is the probability that you have the disease if your test shows a positive result? Most people’s untutored intuition would lead them to think the chance is high, or at least not small. In fact, the chance of having the disease, given the positive result, is less than one in a 1,000; indeed, it is more likely that the test was wrong. (b) The Monty Hall problem. This is now so well-known as to hardly need stating, but for all our extraterrestrial readers here it is: You are a contestant in a game show. A nice car and two feral goats are randomly disposed behind three doors, one to each door. You choose a door to obtain the object it conceals. The presenter does not open your chosen door, but opens another door that turns out to reveal a goat. Then the presenter offers you the chance to switch your choice to the ﬁnal door. Do you gain by so doing? That is to say, what is the probability that the ﬁnal door conceals the car? Many people’s intuition tells them that, given the open door, the car is equally likely to be behind the remaining two; so there is nothing to be gained by switching. In fact, this is almost always wrong; you should switch. However, even this simple problem raises issues of quite surprising complexity, which require our sophisticated theory of probability for their resolution. (c) Coincidences. Twenty-three randomly selected people are listening to a lecture on chance. What is the probability that at least two
of them were born on the same 0.7 FAQs 13 day of the year? Again, untutored intuition leads most people to guess that the chances of this are rather small. In fact, in a random group of 23 people, it is more likely than not that at least two of them were born on the same day of the year. This list of counterintuitive results could be extended indeﬁnitely, but this should at least be enough to demonstrate that only mathematics can save people from their ﬂawed intuition with regard to chance events. (5) A very natural FAQ that requires an answer is: What have “odds” got to do with probability? This is an important question, because there is a longstanding link between chance and betting, and bookmakers quote their payouts in terms of these “odds.” The ﬁrst key point is that there are two kinds of odds: fair odds and pay-off odds. (a) Fair odds: If the occurrence of some event is denoted by A, then Ac denotes the nonoccurrence of A, (≡ the event that A does not occur). If the probability of A is p and the probability of Ac is q, then the odds against A are q : p; (pronounced q to p). The odds on A are p : q. In the case of n equally likely outcomes of which r yield A and n − r yield Ac, the odds on A are r : n − r. Thus, the odds against a six when rolling a fair die are 5 to 1, and so on. (b) Pay-off odds. These are the odds that are actually offered by bookmakers and casinos. In a sense, they could be called Unfair Odds because they are ﬁxed to ensure that the advantage lies with the casino. For example, suppose you bet on an even number arising in an American roulette wheel. We have noted above that the probability P(E) of an even number is 18 19, so the fair odds on an even 38 number are 9 : 10; the fair odds against are 10 : 9. However, the pay-off odds in the casino are 1 : 1, which is to say that you get your stake and the same again if you win. The fair value of this $1 bet (as discussed in section 0.5) is therefore × 2 = 18 19 = 1 − 1 19 9 19 =
9. The fact that this is less than your stake is an indication that these odds are stacked against you. Of course, if the casino paid out at the fair odds of 10 : 9, the value of a bet of $1 is 10 9 9 19 + 1 = 1, which is equal to your stake, and this is to be regarded as fair. The difference between the fair odds (which reﬂect the actual probabilities) and the pay-off odds is what guarantees the casino’s proﬁt in the long run. We conclude with a couple of less problematical questions that arise in more everyday circumstances. (6) “The weather forecast gave only a 5% chance of rain today, but then it did rain. What is the use of that?” One sympathizes with this question, but of course the whole point of probability is to discuss uncertain eventualities in advance of their resolution. The theory cannot hope to turn uncertainty into certainty; instead, it offers the prospect of discussing 14 0 Introduction these chances rationally, with the intention of making better-informed decisions as a result. It is always better to play with the odds than against them. (7) Will a knowledge of probability enable me to design a betting system to win at roulette? Absolutely not! [unless, of course, your betting system comprises the one simple rule: “Buy a casino.”] 0.8 History Errors using inadequate data are much less than those using no data at all. Charles Babbage The excitement that a gambler feels when making a bet is equal to the amount he might win multiplied by the chance of winning it. Blaise Pascal One of the most interesting features of the history of probability is how remarkably short it is. The Ancient Greeks and others had been prompted (by everyday problems and their curiosity) to develop an extensive knowledge of geometry, astronomy, and numbers. Other branches of mathematics made great strides during the Italian Renaissance. Nevertheless, it was left to Pascal, Fermat, and Huygens to inaugurate mathematical probability in 1654 to 1656, with their work on the Problem of the Points (which concerned the fair division of the prize or stakes in unﬁnished games of chance), and the Gamblers Ruin Problem (see Examples 2.11 and 4.12). This tardy start is all the more surprising when one notes that games of chance, using dice and other familiar randomisers, were certainly widespread throughout the classical era
, and were almost certainly common before that period. For example, the game known as Alea was widespread throughout the Roman Empire at all levels of society. At various times, laws proscribed or restricted it; Marcus Aurelius was so addicted to the game that he had a personal croupier to roll his dice. The 6 × 6 squared boards, often inscribed appropriately, are found across the imperial domain. Various numbers of dice were used, usually three; the most desirable outcome was called the Venus throw, the least desirable was the Dogs. Note that the Latin for die is alea (from which we get aleatory meaning random), and the game used counters, from which the word calculus has also entered our language. The game was also played by the Greeks, with the legend that it was invented by Palamedes to occupy the tedious years of the Trojan wars, particularly the celebrated siege. (Another legend attributes it to the Lydians, who used it to distract their minds from hunger during famine.) The Greeks called the game pessoi, after the counters, a die was tessera. It seems safe to say that the origins of the game are lost in early antiquity. Despite these ancient origins, there is no evidence that anyone attempted to calculate chances, and play accordingly, until the second millennium. A poem entitled “De Vetula,” dated to around 1250 ad, includes elementary calculations of chances involving dice. Similar ideas and calculations can be found sporadically throughout Europe over the next four centuries, but the spirit of the times and difﬁculties of communication hampered any Notation 15 serious development. Nevertheless, we ﬁnd Cardano writing a book around 1520, entitled On Games of Chance, in which the ideas of symmetry and long-run frequency are nearly clear. Galileo was certainly taking such notions for granted in the early 17th century when he calculated the odds on rolling 9 or 10, respectively, with three dice. Following the ground-breaking correspondence of Pascal and Fermat on the problem of the points (which had been known but unsolved for two centuries previously), there was a comparatively swift and substantial development. Christiaan Huygens (1657) wrote a book on numerous problems in probability, followed by the books of James Bernoulli (1713), Pierre de Montmort (1708), and Abraham de Moivre (1718, 1738, 1756). The ﬁnal edition of de Moivre’
s book includes a law of large numbers, a central limit theorem (see Sections 7.5.11 and 8.9), and quite sophisticated solutions of problems using generating functions (see Section 3.6 and Chapter 6). The pace of development accelerated during the 19th century, until the structure that is most in use today was rigorously codiﬁed by A. Kolmogorov in his book of 1933, Grundbegriffe der Wahrscheinlichkeitsrechnung. (We note, however, that other systems have been, are being, and will be used to model probability.) Review In this chapter, we discuss our intuitive ideas about chance and suggest how they can help us to construct a more formal theory of probability. In particular, we exploit the interpretations of chances as a simple proportion in situations with symmetry and as relative frequency in the long run. These suggest that: Any probability should be a number between zero and one, inclusive. Things that are impossible should have zero probability. Things that are certain should have probability one. Thus, we can picture probabilities as lying on a scale between zero and one, where the more unlikely eventualities have their probabilities nearer to zero and the more likely eventualities have their probabilities nearer to one. The following chapters use similar arguments to develop more complicated rules and properties of probability. Appendix: Review of Elementary Mathematical Prerequisites It is difﬁcult to make progress in any branch of mathematics without using the ideas and notation of sets and functions. Indeed, it would be perverse to try to do so because these ideas and notation are helpful in guiding our intuition and solving problems. (Conversely, almost the whole of mathematics can be constructed from these few simple concepts.) We therefore give a brief synopsis of what we need here for completeness, although it is likely that the reader will already be familiar with this. Notation We use a good deal of familiar standard mathematical notation in this book. The basic notation for sets and functions is set out below. More specialized notation for probability theory is introduced as required throughout the book, and recorded in Chapter Reviews and the Index of Notation. 16 0 Introduction We take this opportunity to list some fundamental notation that you are likely to see soon: e log x log2x π n! \|x\| [x] R Z x ∧ y = min{x, y} x ∨ y = max{x, y} x + = x ∨ 0 n ar = a1 + ·
· · + an the base of natural logarithms; Euler’s number the logarithm of x to base e, unless otherwise stated the logarithm of x to base 2 the ratio of circumference to diameter for a circle the universal set n(n − 1)... 3.2.1; factorial n modulus or absolute value of x the integer part of x the real line the integers the smaller of x and y the larger of x and y summation symbol r =1 n r =1 ar = a1 a2... an product symbol Sets A set is a collection of things that are called the elements of the set. The elements can be any kind of entity: numbers, people, poems, blueberries, points, lines, and so on, endlessly. For clarity, upper case letters are always used to denote sets. If the set S includes some element denoted by x, then we say x belongs to S and write x ∈ S. If x does not belong to S, then we write x ∈ S. There are essentially two ways of deﬁning a set, either by a list or by a rule. Example If S is the set of numbers shown by a conventional die, then the rule is that S comprises the integers lying between 1 and 6 inclusive. This may be written formally as follows: S = {x : 1 ≤ x ≤ 6 and x is an integer}. Alternatively, S may be given as a list: S = {1, 2, 3, 4, 5, 6}. One important special case arises when the rule is impossible; for example, consider the set of elephants playing football on Mars. This is impossible (there is no pitch on Mars) and the set therefore is empty; we denote the empty set by φ. We may write φ as { }. If S and T are two sets such that every element of S is also an element of T, then we say that T includes S and write either S ⊆ T or S ⊂ T. If S ⊂ T and T ⊂ S, then S and T are said to be equal and we write S = T. Venn Diagrams 17 Note that φ ⊂ S for every S. Note also that some books use the symbol “⊆” to denote inclusion and reserve “⊂” to denote strict inclusion, that is to say
, S ⊂ T if every element of S is in T and some element of T is not in S. We do not make this distinction. Combining Sets Given any nonempty set, we can divide it up, and given any two sets, we can join them together. These simple observations are important enough to warrant deﬁnitions and notation. Let A and B be sets. Their union, denoted by A ∪ B, is the set of elements Deﬁnition that are in A or B, or in both. Their intersection, denoted by A ∩ B, is the set of elements in both A and B. Note that in other books the union may be referred to as the join or sum; the intersection may be referred to as the meet or product. We do not use these terms. Note the following. Deﬁnition If A ∩ B = φ, then A and B are said to be disjoint. We can also remove bits of sets, giving rise to set differences, as follows. Let A and B be sets. That part of A that is not also in B is denoted by A\B, Deﬁnition called the difference of A from B. Elements that are in A or B but not both, comprise the symmetric difference, denoted by AB. Finally, we can combine sets in a more complicated way by taking elements in pairs, one from each set. Deﬁnition Let A and B be sets, and let C = {(a, b) : a ∈ A, b ∈ B} be the set of ordered pairs of elements from A and B. Then C is called the product of A and B and denoted by A × B. Let A be the interval [0, a] of the x-axis, and B the interval [0, b] of the Example y-axis. Then C = A × B is the rectangle of base a and height b with its lower left vertex at the origin, when a, b > 0. Venn Diagrams The above ideas are attractively and simply expressed in terms of Venn diagrams. These provide very expressive pictures, which are often so clear that they make algebra redundant (see Figure 0.1). In probability problems, all sets of interest A lie in a universal set, so that A ⊂ for all A. That part of that is not in A is called
the complement of A, denoted by Ac. 18 0 Introduction Figure 0.1 Venn diagrams. Formally, Ac = \A = {x : x ∈, x ∈ A}. Obviously, from the diagram or by consideration of the elements A ∪ Ac =, A ∩ Ac = φ, (Ac)c = A. Clearly, A ∩ B = B ∩ A and A ∪ B = B ∪ A, but we must be careful when making more intricate combinations of larger numbers of sets. For example, we cannot write down simply A ∪ B ∩ C; this is not well deﬁned because it is not always true that (A ∪ B) ∩ C = A ∪ (B ∩ C). We use the obvious notation n r =1 n r =1 Ar = A1 ∪ A2 ∪ · · · ∪ An, Ar = A1 ∩ A2 ∩ · · · ∩ An. Deﬁnition If A j ∩ Ak = φ for j = k and n r =1 Ar =, then the collection (Ar ; 1 ≤ r ≤ n) is said to form a partition of. Functions Size 19 When sets are countable, it is often useful to consider the number of elements they contain; this is called their size or cardinality. For any set A, we denote its size by \|A\|; when sets have a ﬁnite number of elements, it is easy to see that size has the following properties. If sets A and B are disjoint, then \|A ∪ B\| = \|A\| + \|B\|, and more generally, when A and B are not necessarily disjoint, \|A ∪ B\| + \|A ∩ B\| = \|A\| + \|B\|. Naturally, \|φ\| = 0, and if A ⊆ B, then \|A\| ≤ \|B\|. Finally, for the product of two such ﬁnite sets A × B, we have \|A × B\| = \|A\| × \|B\|. When sets are inﬁnite or uncountable, a great deal more care and subtlety is required in dealing with the idea of size. However, we intuitively see that we can consider the length of subsets of a line, or areas of sets in a plane, or volumes in
space, and so on. It is easy to see that if A and B are two subsets of a line, with lengths \|A\| and \|B\|, respectively, then in general \|A ∪ B\| + \|A ∩ B\| = \|A\| + \|B\|. Therefore \|A ∪ B\| = \|A\| + \|B\| when A ∩ B = φ. We can deﬁne the product of two such sets as a set in the plane with area \|A × B\|, which satisﬁes the well-known elementary rule for areas and lengths \|A × B\| = \|A\| × \|B\| and is thus consistent with the ﬁnite case above. Volumes and sets in higher dimensions satisfy similar rules. Functions Suppose we have sets A and B, and a rule that assigns to each element a in A a unique element b in B. Then this rule is said to deﬁne a function from A to B; for the corresponding elements, we write b = f (a). Here the symbol f (·) denotes the rule or function; often we just call it f. The set A is called the domain of f, and the set of elements in B that can be written as f (a) for some a is called the range of f ; we may denote the range by R. Anyone who has a calculator is familiar with the idea of a function. For any function key, the calculator will supply f (x) if x is in the domain of the function; otherwise, it says “error.” 20 0 Introduction Inverse Function If f is a function from A to B, we can look at any b in the range R of f and see how it arose from A. This deﬁnes a rule assigning elements of A to each element of R, so if the rule assigns a unique element a to each b this deﬁnes a function from R to A. It is called the inverse function and is denoted by f −1(·): a = f −1(b). Example: Indicator Function : Let A ⊂ and deﬁne the following function I (·) on I (ω) = 1 I (ω) = 0 if ω ∈ A, if ω ∈ A. Then I is a function from to {0, 1}; it is called the indicator of A because
by taking the value 1 it indicates that ω ∈ A. Otherwise, it is zero. This is about as simple a function as you can imagine, but it is surprisingly useful. For example, note that if A is ﬁnite you can ﬁnd its size by summing I (ω) over all ω: \|A\| = I (ω). ω∈ Series and Sums Another method I have made use of, is that of Inﬁnite Series, which in many cases will solve the Problems of Chance more naturally than Combinations. A. de Moivre, Doctrine of Chances, 1717 What was true for de Moivre is equally true today, and this is therefore a convenient moment to remind the reader of some general and particular properties of series. Note that n! = n(n − 1)... 3.2 − 1, and that 0! = 1, by convention. Also, and Consider the series n r = n! r!(n − r )! M(a, b, c) = (a + b + c)! a!b!c!. Finite Series sn = n r =1 ar = a1 + a2 + · · · + an. The variable r is a dummy variable or index of summation, so any symbol will sufﬁce: Limits 21 n r =1 ar ≡ n i=1 ai. n r =1 (axr + byr ) = a n r =1 xr + b n r =1 yr. In general, In particular, 1 = n; r = 1 2 n(n + 1), the arithmetic sum(n + 1)(2n + 11 = 1 4 n2(n + 1)2; x r yn−r = (x + y)n, the binomial theorem; n r =1 n r =1 n r =1 n r =1 n r =0 n r M(a, b, c)x a ybzc = ybzc a + b + c = n a, b, b, c ≥ 0 = (x + y + z)n, the multinomial theorem; x r = 1 − x n+1 1 − x, the geometric sum. n r =0 Limits Often, we have to deal with inﬁnite series. A fundamental and extremely useful concept in this

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