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⊂ PN is an irreducible quasiprojective variety, with X = PN, and that X is nonsingular at ξ. Then the inverse image σ −1(X) of X under the blowup of PN centred at ξ is reducible, consisting of two components σ −1(X) = ξ × PN −1 ∪ Y. (2.31) 118 2 Local Properties The restriction of σ to the component Y deﬁnes a regular map σ : Y → X, which is an isomorphism of some neighbourhood U of x if x = ξ and a local blowup σ −1(U ) → U with centre ξ if x = ξ. Proof Let Y denote the closure σ −1(X \ ξ ) of σ −1(X \ ξ ). Since σ −1 is an isomorphism on PN \ ξ, it follows that σ −1(X \ ξ ) is isomorphic to X \ ξ, and is hence irreducible. Hence also Y is irreducible. (2.31) is obvious by deﬁnition: if x ∈ X \ ξ then σ −1(x) ∈ Y, and σ −1(ξ ) = ξ × PN −1. The fact that σ : Y → X is an isomorphism in a neighbourhood of any points x ∈ X other than x = ξ has already been noted. It remains to study σ : Y → X over a neighbourhood of ξ. Now we use the fact that the blowup can be described as a local blowup for an afﬁne space AN containing ξ, together with the independence of the local blowup on the choice of local parameters. Namely, by Theorem 2.14, we can choose a system of local parameters u1,..., uN at ξ ∈ PN such that in some neighbourhood of ξ, the subvariety X has local equations un+1 = · · · = uN = 0, (2.32) and the functions u1,..., un deﬁne a system of local parameters on X at ξ. We can choose a neighbourhood U ⊂ PN of ξ such that u1,
..., uN satisfy conditions (a) and (b) of Lemma of Section 4.2, so that the proof of the theorem reduces to the special case when X is given by (2.32). From conditions (a) and (b) and uitj = uj ti we get that tn+1(x) = · · · = tN (x) = 0 for x = ξ. Hence Y is contained in the subspace Y ⊂ X × PN −1 deﬁned by the equations tn+1 = · · · = tN = 0 uitj = uj ti for i, j = 1,..., n. (2.33) (2.34) If we write Pn−1 for the subspace of projective space PN −1 of points satisfying (2.33) then we see that Y ⊂ X × Pn−1 and is deﬁned there by (2.34). Thus Y is the same thing as the variety obtained as a result of the local blowup. We have proved that Y = σ −1(X \ ξ ). Hence Y = Y, which proves the theorem. Now we can give the most general deﬁnition of blowup centred at a point. If X ⊂ PN is a quasiprojective variety, ξ a nonsingular point of X and Y the variety introduced in the statement of Theorem 2.15 then σ : Y → X is called the blowup of X with centre ξ. From what we proved concerning the local blowup, it follows that Y is irreducible if X is, and σ −1(ξ ) ∼= ξ × Pn−1, with all points of σ −1(ξ ) nonsingular points of Y. Notice that a blowup is an isomorphism if X is a curve. Thus nontrivial blowups are a typical phenomenon of higher dimensional algebraic geometry. 4 The Structure of Birational Maps 119 4.4 Exceptional Subvarieties The example of a blowup shows a difference of principle between algebraic curves and varieties of dimension n > 1: whereas for nonsingular projective curves a birational map is an isomorphism, a blowup shows that this is not always the case in higher dimensions. Notice one peculi
arity of a blowup: it is a regular map, and only fails to be an isomorphism because the rational map σ −1 is not regular at a point ξ. In this section we study a map f : X → Y where f is a regular map and is birational, that is, g = f −1 : Y → X is a rational, but not regular, map. In the example of a blowup we saw that a codimension 1 subvariety in Y is contracted to the point ξ. We prove that the same property always holds in this situation. Theorem 2.16 Let f : X → Y be a regular birational map. For x ∈ X, assume that y = f (x) is a nonsingular point of Y and that the inverse map g = f −1 is not regular at y. Then there exists a subvariety Z ⊂ X with Z x such that codim Z = 1, but codim f (Z) ≥ 2. Proof We can if necessary replace X by an afﬁne neighbourhood of x, and thus assume that X is afﬁne. Suppose that X ⊂ AN, with coordinates t1,..., tN, and that g = f −1 is the map given by ti = gi for i = 1,..., N, with gi ∈ k(Y ). Obviously gi = g∗(ti); since g is not regular at y, at least one of the functions gi is not regular at y. Suppose this is g1, so that g1 /∈ Oy. We can write g1 in the form g1 = u/v with u, v ∈ Oy and v(y) = 0, and since Oy is a UFD (because we assume that y is nonsingular), we can suppose that u and v have no common factors. Since g = f −1, we have t1 = f ∗(g1) = f ∗(u/v) = f ∗(u)/f ∗(v), and hence f ∗(v)t1 = f ∗(u). (2.35) Now f ∗(v)(x) = v(y) = 0, so that x ∈ V (f ∗(v)). Set Z = V (f ∗(v)). By the theorem
on dimension of intersection, codim Z = 1, and since x ∈ Z it is nonempty. It follows from (2.35) that f ∗(u) = 0 on Z, so that t1 is a regular function. Hence also u = 0 on f (Z), and thus f (Z) ⊂ V (u) ∩ V (v). It remains to check that codim(V (u) ∩ V (v)) ≥ 2. But if V (u) ∩ V (v) contained a component Y with y ∈ Y and codim Y = 1 then by Theorem 2.10, Y would have a local equation h. This means that u, v ∈ (h), which contradicts the assumption that u and v have no common factor in Oy. The theorem is proved. Deﬁnition Let f : X → Y be a regular birational map. A subvariety Z ⊂ X is exceptional for f if codim Z = 1, but codim f (Z) ≥ 2. Corollary 2.5 If f : X → Y is a regular birational map between nonsingular varieties, not an isomorphism, then f has an exceptional subvariety. 120 2 Local Properties Corollary 2.6 Let f : X → Y be a regular birational map between curves X and Y, and suppose that Y is nonsingular; then f (X) is open in Y and f deﬁnes an isomorphism from X to f (X). Proof f (X) open in Y follows from the fact that X and Y have isomorphic open subsets U and V (Proposition 1.1); indeed, since f (U ) = V is obtained by discarding a ﬁnite number of points, a fortiori so is f (X), and therefore it is open in Y. If f : X → f (X) were not an isomorphism, we would get a contradiction to Theo- rem 2.16, since in our case only the empty set has codimension ≥2. 4.5 Isomorphism and Birational Equivalence Consider a birational equivalence class of quasiprojective varieties, that is, a class consisting of all quasiprojective varieties birationally equivalent to one another. A representative of this class is called a model. In Section 5.3
below, we prove that there exists a nonsingular projective model X0 in every birational equivalence class of algebraic curves. Theorem 2.12, Corollary 2.4 asserts that there is at most one such model up to isomorphism in every birational equivalence class. Therefore, sending a birational equivalence class of curves to the unique nonsingular projective model contained in it reduces the question of the classiﬁcation of curves up to birational equivalence to that of the classiﬁcation of nonsingular projective curves up to isomorphism. The function ﬁeld k(X) of an algebraic curve X is an extension ﬁeld of k generated over k by ﬁnitely many elements and of transcendence degree 1. Hence we can establish a one-to-one correspondence between such ﬁelds K and nonsingular projective curves X. This correspondence is given by K = k(X). We will also say that X is a model of K. One can attempt to ﬁnd the nonsingular projective model X directly from algebraic properties of K. We make this question more precise by asking how the local rings Ox of points x ∈ X are characterised within K. It is easy to see that every local ring Ox of a point x ∈ X has the following properties: (1) O is a subring of K with k O K; (2) O is a local ring, and its maximal ideal m is principal, that is m = (u); (3) K equals the ﬁeld of fractions of O. It can be proved (see Exercises 7–9) that any subring O ⊂ K satisfying (1)–(3) is the local ring Ox of some point x ∈ X. Thus the nonsingular projective model X is universal: it contains all the local rings of K satisfying the natural conditions (1)–(3). What about these questions in dimension n > 1? Things turn out reasonably well as far as the existence of a nonsingular projective model goes: it is proved for n = 2 or 3 (Walker, Zariski over ﬁelds of characteristic 0, Abhyankar over ﬁelds of characteristic p > 5), and for arbitrary n in characteristic 0 (Hironaka). For arbitrary
ﬁelds 4 The Structure of Birational Maps 121 and arbitrary n the existence of a nonsingular projective model seems extremely plausible. The uniqueness of the nonsingular projective model, on the contrary, is a wholly exceptional feature of the case n = 1. This can be seen already in the example of the projective plane P2 and the nonsingular quadric Q ⊂ P3, which are birational, but not isomorphic. One might ask for the existence, in each birational equivalence class, of a model X that would be universal in the sense that the local rings Ox of points x ∈ X exhaust all the local subrings of the ﬁeld K = k(X) that satisfy conditions (1), (2) and (3), as in the case n = 1, where m = (u) in condition (2) is replaced by the appropriate n-dimensional version m = (u1,..., un). However, no such model can exist, for the same reasons. Namely if σ : X → X is the blowup of X with centre in ξ, then the local rings of points y ∈ σ −1(ξ ) are not equal to any of the local rings Ox with x ∈ X. The reader can easily prove this as an exercise. Admittedly, putting together all the nonsingular points of all models of a birational equivalence class one does obtain a certain object, the so-called Zariski Riemann surface with this universal property, but this object is not a ﬁnite dimensional algebraic variety. Some information about this “inﬁnite model” can be found in Zariski and Samuel [81, Section 17 of Chapter VI, Vol. 2]. Given that there does not exist a distinguished model, the problem arises of studying the relations between the nonsingular projective models in each birational equivalence class. We describe here without proof the main results in this area known to date. From now on, all varieties considered will be assumed to be irreducible, nonsingular and projective. We start with two deﬁnitions. We say that a model X dominates X if there exists a regular birational map f : X → X. A variety X is a relatively minimal model if it does not dominate any variety not isomorphic to itself. For example, a nons
ingular projective curve is always a relatively minimal model. By Theorem 2.16, a variety is a relative minimal model if it does not contain any exceptional subvarieties. It can be proved that every variety dominates at least one relatively minimal model. Thus every birational equivalence class contains at least one relatively minimal model. The question thus arises of the uniqueness of a relatively minimal model. If every birational equivalence class had such a unique relatively minimal model then this would again reduce the birational classiﬁcation of varieties to the classiﬁcation up to isomorphism. However, for n > 1 this does not work. An example is provided by the projective plane P2 and the nonsingular quadric Q ⊂ P3, which, as we know, are birational, that is, they both belong to the same birational equivalence class. We prove that P2 and Q are both relatively minimal models, that is, they do not contain any exceptional curves. Since P2 and Q are not isomorphic (compare the discussion after Corollary 1.8), this gives the required example. In our case, an irreducible exceptional curve C ⊂ X must be contracted to a point by a regular birational map f : X → Y, that is, f (C) = y ∈ Y. Here X and Y are projective surfaces. Curves of this type have a series of very special properties (hence the name “exceptional”). We discuss just one of these. By Theorem 2.12, the map f −1 is not regular at only ﬁnitely many points yi ∈ Y. Suppose that U is a afﬁne neighbourhood of y, sufﬁciently small that f −1 is regular 122 2 Local Properties at all points of U other than y. Set V = f −1(U ) and C = f −1(y). Obviously V is an open subset of X and V ⊃ C. We prove that V does not contain any irreducible curve C that is closed in X and not contained in C. Indeed, C is a projective curve and its image f (C) is again projective. But f (C) ⊂ U, which is afﬁne. According to Theorem 1.11, Corollary 1.2, this is only possible if
f (C) = y is a point. If y = y then, since f −1 is an isomorphism, C would also have to be a point. If y = y then C ⊂ f −1(y) = C. Thus C is isolated in X, in the sense that there exists a neighbourhood V of C that does not contain any irreducible projective curve except for those contained in C. In other words, it is impossible to “wiggle C slightly”. We can deduce from this that many surfaces do not contain any exceptional curves. For example, let X = P2 and let C be an exceptional curve with C ⊂ V = P2 \ D. Then dim D = 0, since otherwise, by the theorem on the dimension of intersection, C and D would intersect. But if dim D = 0, that is, D is a ﬁnite set of points, then V contains any number of curves C not intersecting D, for example lines. Now let X = Q be the nonsingular quadric of P3. Here we make use of the existence of a group G of projective transformations taking Q to itself. Recall that transformations of G are given by 4 × 4 matrixes satisfying the relation A∗FA = F, where F is the matrix of the quadratic form deﬁning Q. It follows that G is an algebraic variety in the space of all 4 × 4 matrixes. Hence we will from now on assume that G is an algebraic afﬁne variety. It is easy to see that G acts transitively on Q. If C is a curve and C ⊂ Q \ D, then we construct a transformation ϕ ∈ G such that ϕ(C) ⊂ C and ϕ(C) ⊂ Q \ D, which contradicts the property of exceptional curves obtained above. For this, it is enough to prove that the set of ϕ ∈ G such that ϕ(C) ∩ D = ∅ is closed. Then we have at our disposal a whole neighbourhood of the identity transformation e ∈ G consisting of elements with the required property. In order to describe the set S of elements ϕ ∈ G such that ϕ(C) ∩ D = ∅ we consider the direct product G × Q and the subset Γ ⊂ G × Q of pairs (ϕ
, x) such that x ∈ C and ϕ(x) ∈ D. Obviously Γ is closed. If f : G × Q → G is the projection then S = f (Γ ), and f (Γ ) is closed by Theorem 1.10. This complete the proof the Q is a relatively minimal model, and hence the existence of two different relatively minimal model. Thus it is all the more surprising that uniqueness of minimal models does hold for algebraic surfaces, provided only that we exclude some special types of surfaces. Namely, as proved by Enriques, a birational equivalence class of surfaces contains a unique relatively minimal model provided that it does not contain a surface of the form C × P1, with C an algebraic curve. (A surface birational to C × P1 is called a ruled surface.) The proof of Enriques’ theorem is treated in Shafarevich [69, Chapter II], or Barth, Peters and van de Ven [9]. There has recently been signiﬁcant progress in the direction of constructing a theory of minimal models in dimension ≥3. In this case, minimal models cannot exist in the class of nonsingular varieties, but there is reason to hope that the theory can be generalised if we allow a certain class of rather well-controlled singularities. For this, see for example the surveys Kawamata, Matsuda and Matsuki [46] and Shokurov [70]. 4 The Structure of Birational Maps 123 4.6 Exercises to Section 4 1 Suppose that dim X = 2 and that ξ ∈ X is a nonsingular point. Let C1, C2 ⊂ X be two curves passing through ξ and nonsingular there, σ : Y → X the blowup centred at ξ, and set C ∩ Z if and i only if C1 and C2 touch at ξ. = σ −1(Ci \ ξ ) and Z = σ −1(ξ ). Prove that C 1 ∩ Z = C 2 2 Suppose that dim X = 2 and that ξ ∈ X is a nonsingular point. Let C be a curve passing through ξ and f the local equation of C in a neighbourhood of ξ. In local parameters u, v at ξ, suppose that f ≡ Π r, where k = li and the forms αiu + β
iv are not proportional. i=1(αiu + βiv)li modulo mk+1 ξ As in Exercise 1, σ : Y → X and C = σ −1(C \ ξ ). Prove that C ∩ Z consists of r points. 3 Use the notation of Exercise 2, and suppose also that f ≡ (α1u+β1v)(α2u+β2v) ξ, where the linear forms α1u + β1v and α2u + β2v are not proportional. modulo m3 Prove that both the points of C ∩ Z are nonsingular on C. 4 Consider the rational map ϕ : P2 → P4 given by ϕ(x0 : x1 : x2) = x0x1 : x0x2 : x2 1 : x1x2 : x2 2. Prove that ϕ is a birational map to a surface ϕ(P2), and that the inverse map ϕ(P2) → P2 coincides with the blowup. 5 In the spirit of Exercise 4, study the map ϕ : P2 → P6 deﬁned by all the monomials of degree 3 except for x3 0, x3 1 and x3 2. 6 For any n ≥ 2, construct an example of a regular birational map f : X → Y between n-dimensional nonsingular varieties having an exceptional codimension 1 subvariety Z whose image f (Z) ⊂ Y has codimension 2. 7 Let X be a nonsingular projective curve and O ⊂ k(X) a local subring of the function ﬁeld k(X) satisfying the conditions Section 4.5, (1)–(3). Prove that for any u ∈ k(X) either u ∈ O or u−1 ∈ O. Suppose that X ⊂ Pn with x0,..., xn homogeneous coordinates of Pn. Prove that there exists an i such that xj /xi ∈ O for j = 0,..., n. 8 Use the notation of Exercise 7. Let X be the afﬁne curve X = X ∩ An i. Prove that k[X] ⊂ O, and that k[X] ∩ m is the ideal of some
point x ∈ X with Ox ⊂ O. 9 Prove that if O1 and O2 are two subrings satisfying the conditions Section 4.5, (1)–(3). and O1 ⊂ O2 then O1 = O2. Deduce from this, using the results of Exercises 7–8, that O = Ox (in the notation of Exercise 8). 124 2 Local Properties 10 Let V ⊂ A3 be the quadratic cone deﬁned by xy = z2; let X → A3 be the blowup of A3 with centre in the origin, and V the closure of σ −1(V \ 0) in X. Prove that V is a nonsingular variety and that the inverse image of the origin under σ : V → V is a nonsingular rational curve. 5 Normal Varieties 5.1 Normal Varieties We start by recalling a notion of algebra: a ring A with no zerodivisors is integrally closed if every element of its ﬁeld of fractions K that is integral over A (Section 5.3, Chapter 1) is in A. Deﬁnition An irreducible afﬁne variety X is normal if k[X] is integrally closed. An irreducible quasiprojective variety X is normal if every point has a normal afﬁne neighbourhood. We will prove presently (Theorem 2.17) that a nonsingular variety is normal. Here is an example of a nonnormal variety: on the curve X deﬁned by y2 = x2 + x3, the rational function t = y/x ∈ k(X) is integral over k[X], since t 2 = 1 + x, but t /∈ k[X]. (See Exercise 7 of Section 3.4, Chapter 1.) This example shows that the condition that a variety is normal is somehow related to singular points of a variety. We now give an example of a variety that is normal although it has a singular point. This is the cone X ⊂ A3 given by x2 + y2 = z2 (we assume that the ground ﬁeld k has characteristic = 2). Let us prove that k[X] is integrally closed in k(X). For this we use the simplest properties of integral elements
(see Section 5.3, Chapter 1 and Atiyah and Macdonald [8, Chapter 5]). The ﬁeld k(X) consists of elements of the form u + vz with u, v ∈ k(x, y), where x and y are independent variables. Similarly, k[X] consists of elements u + vz = k(X) with u, v ∈ k[x, y]; hence k[X] is a ﬁnite module over k[x, y], and hence all elements of k[X] are integral over k[x, y]. If α = u + vz ∈ k(X) is integral over k[X] then it must be also integral over k[x, y]. Its minimal polynomial is T 2 − 2uT + u2 − x2 + y2 v2; hence 2u ∈ k[x, y], so that u ∈ k[x, y]. Similarly, u2 − (x2 + y2)v2 ∈ k[x, y], and hence also (x2 + y2)v2 ∈ k[x, y]. Now since x2 + y2 = (x + iy)(x − iy) is the product of two coprime irreducibles, it follows that v ∈ k[x, y], and thus α ∈ k[X]. We prove some simple properties of normal varieties. Lemma If X is a normal variety then its local ring OY at any irreducible subvariety Y ⊂ X (see the end of Section 1.1 for the deﬁnition) is integrally closed. Conversely, if X is irreducible and the local ring Ox at each point x ∈ X is integrally closed then X is normal. 5 Normal Varieties 125 Proof Since the deﬁnition of normal is local in nature, we can restrict to the case that X is afﬁne. Suppose that X is normal, and let Y ⊂ X be an irreducible subvariety. We prove that OY is integrally closed. Suppose that α ∈ k(X) is integral over OY, that is, αn + a1αn−1 + · · · + an = 0 with ai ∈ OY. Since ai
∈ OY we have ai = bi/ci with bi, ci ∈ k[X] and ci /∈ aY. Set d0 = c1 · · · cn, and multiply (2.36) by d0. We get that (2.36) d0αn + d1αn−1 + · · · + dn = 0 with di ∈ k[X] and d0 /∈ aY. (2.37) Multiplying (2.37) through again by d n−1 and setting d0α = β, we get that β is integral over k[X]. By assumption k[X] is integrally closed, and hence d0α = β ∈ k[X]. Then α = β/d0 ∈ OY, because d0 /∈ aY. This proves that OY is integrally closed. 0 Conversely, suppose that all the local rings Ox are integrally closed. We prove that k[X] is also. If α ∈ k(X) is integral over k[X] then αn + a1αn−1 + · · · + an = 0 with ai ∈ k[X]. But then a fortiori ai ∈ Ox for every x ∈ X, and since Ox is integrally closed by assumption, it follows that α ∈ Ox. Therefore α ∈ Ox. Now Ox = k[X], and hence α ∈ k[X]. The lemma is proved. by Theorem 1.7, x∈X x∈X Theorem 2.17 A nonsingular variety is normal. Proof By the lemma, it is enough to show that if x is a nonsingular point then Ox is integrally closed. We know by Theorem 2.11 that Ox is a UFD. Any element α ∈ k(X) can be represented in the form α = u/v, were u, v ∈ Ox and have no common factors. If α is integral over Ox then αn + a1αn−1 + · · · + an = 0 with ai ∈ Ox. Hence un + a1un−1v + · · · + anvn = 0, and we see that v \| un. But now, since u, v have no common factors and Ox is a UFD,
it follows that α ∈ Ox. The theorem is proved. Theorem 2.17 shows that the deﬁnition of normal is a certain weakening of the notion of nonsingularity. This is also reﬂected in the properties of normal varieties. In particular, we show that one of the basic properties of nonsingular varieties (Theorem 2.10) extends in a weak form to normal varieties. Theorem 2.18 If X is a normal variety and Y ⊂ X a codimension 1 subvariety then there exists an afﬁne open set X ⊂ X with X ∩ Y = ∅ such that the ideal of Y = X ∩ Y in k[X] is principal. Proof We can of course assume that X is afﬁne. Moreover, it is enough to prove that the maximal ideal mY is principal in the local ring OY. Indeed, if mY = (u) with u ∈ OY then u = a/b with a, b ∈ k[X] and b /∈ aY. Suppose that aY = (v1,..., vm). Since aY ⊂ mY, we can write vi = uwi, where wi = ci/di, with ci, di ∈ k[X] and 126 2 Local Properties di /∈ aY. Then the ideal aY of the variety Y = X ∩ Y is the principal ideal (u) in k[X], where we set X = X \ V (b) V (di). m i=1 Suppose that 0 = f ∈ k[X] and f ∈ aY ⊂ OY. Then Y ⊂ V (f ), and since both of these are codimension 1 subvarieties (by assumption and by the theorem on dimension of intersection), Y consists of components of V (f ). Suppose that V (f ) = Y ∪ Y1 with Y ⊂ Y1. Setting X1 = X \ Y1, we get that Y ∩ X1 = ∅ and Y ∩ X1 = V (f ) ∩ X1. Thus we can assume from the start that Y = V (f ). By the Nullstellensatz, Y = V (f ) in X implies that ak Y ⊂ (f ) for some k > 0, and �
�� (f ) in OY. Suppose that k is the minimal number having this property. hence mk Y Then there exist α1,..., αk−1 ∈ mY such that α1 · · · αk−1 /∈ (f ) but α1 · · · αk−1mY ⊂ (f ). That is, setting g = α1 · · · αk−1 we have g /∈ (f ) but gmY ⊂ (f ), or in other words, u = f/g satisﬁes u−1 /∈ OY, but u−1mY ⊂ OY. Now we use the fact that, by the lemma, OY is integrally closed. It follows from this that u−1mY ⊂ mY ; for otherwise, by the basic relation between ﬁnite modules and integral elements, the “determinant trick” (Atiyah and Macdonald [8, Proposition 2.4]), u−1 would be integral over OY and therefore contained in it, which is not the case. But mY is the maximal ideal of OY, so that u−1mY ⊂ OY but u−1mY ⊂ mY implies that u−1mY = OY. This means that mY = (u). The theorem is proved. Theorem 2.19 The set of singular points of a normal variety has codimension ≥2. Proof Suppose that X is normal, with dim X = n, and let S be the set of singular points of X. We have seen in Section 1.4 that S is closed in X. Suppose that S contains an irreducible component Y of dimension n − 1. Let X be the open subset whose existence we established in Theorem 2.18 and Y = Y ∩ X. There is at least one point y ∈ Y that is a nonsingular point of the variety Y (but not of X, by assumption). Let OY,y be the local ring of Y at y, and u1,..., un−1 local parameters. By Theorem 2.18, aY = (u) is a principal ideal of k[X], and hence k[Y ] = k[X]/(u). Similarly OY,y = OX,y/(
u), and obviously mX,y is equal to the inverse image of mY,y under the natural map OX,y → OY,y. Choose arbitrary inverse images v1,..., vn−1 ∈ OX,y of the local parameters u1,..., un−1 ∈ OY,y. Then ≤ n, and hence that mX,y = (v1,..., vn−1, u). This proves that dim mX,y/m2 y is a nonsingular point of X, which contradicts the assumption y ∈ Y ⊂ S. The theorem is proved. X,y Corollary For algebraic curves, normal and nonsingular are equivalent conditions. Example Let X be a normal afﬁne variety and G a ﬁnite group of automorphisms of X. We prove that the quotient variety Y = X/G (see Example 1.21) is normal. 5 Normal Varieties 127 Suppose that h ∈ k(Y ) is integral over k[Y ]. Then h is a fortiori integral over k[X], and hence h ∈ k[X]. But h ∈ k(Y ), so that g∗(h) = h for any g ∈ G, and hence h ∈ k[X]G = k[Y ]. In particular, suppose that X = A2 and G = {1, g}, where g(x, y) = (−x, −y). It is easy to check that k[X]G = k[x, y]G is generated by w = xy, u = x2 and v = y2. In other words, Y is the quadratic cone deﬁned by uv = w2 constructed at the start of this section. Since X is normal by Theorem 2.17, we get another proof that Y is normal. We now compare the nonsingular and normal properties of varieties we have introduced. We ﬁrst note that the proof of Theorem 2.17 did not make full use of the nonsingularity of X; we only used that Ox is a UFD. In this connection, it is natural to distinguish the class of varieties with the property that each local ring Ox is a UFD; these are called factorial varieties. Thus nonsingular varieties are factorial, and fact
orial varieties normal; in essence, that is what is proved in Theorem 2.17. One can show that these three classes of varieties are really different. For example, it is known that for n ≥ 5, a hypersurface X ⊂ An with just one singular point is factorial (Grothendieck [34, (SGA2), Section 3.14, Chapter XI]). A beautiful example of a surface that is singular, but factorial, is the surface given by x2 + y3 + z5 = 0. An example of a variety that is normal, but not factorial, is given by the quadratic cone considered above: z2 = (x + iy)(x − iy) are two different factorisations into irreducibles of the same element. Theorem 2.19 focuses attention on a new property of varieties: the set of singular points has codimension ≥2. A variety with this property is said to be nonsingular in codimension 1. Theorem 2.19 asserts that this class includes, in particular, normal varieties. These two classes of varieties are also distinct. Constructing a counterexample is a bit more complicated here; the point is that normal is equivalent to nonsingular in codimension 1 for a hypersurface. Hence the simplest possible example would be a surface in A4. An irreducible variety X is not normal if there exists an afﬁne variety Y and a surjective regular map f : Y → X, not an isomorphism, that restricts to an isomorphism of open subsets V ⊂ Y and U ⊂ X, and such that k[Y ] is a ﬁnite module over f ∗(k[X]). A ﬁrst approximation to the counterexample is thus given by X = L1 ∪ L2, where L1 and L2 are two planes of A4 meeting in a point (deﬁned by xz = xt = yz = yt = 0 in coordinates x, y, z, t of A4) and Y = L1 L2 the disjoint union of L1 and L2 (for example, in A5). But this is a reducible variety, and our deﬁnition of normal assumes irreducibility. We therefore construct an example that imitates this situation near the
singular point. For this, it is enough to construct a ﬁnite regular map f : A2 → A4 birational onto its image X = f (A2), with X ⊂ A4 closed in A4 such that two points, y1, y2, say, have the same image z ∈ X and f : A2 \ {y1, y2} → X \ {z} is an isomorphism. Thus f is very similar to the parametrisation (1.3) of the curve (1.2). The existence of the map f means that X is not normal, and z will be the unique singular point of X. Writing ξ, η for coordinates in A2 and x, y, z, t for coordinates in A4, we deﬁne f by x = ξ(1 − η), y = η(η − 1)2, z = ξ η, t = η2(η − 1). 128 2 Local Properties One sees easily that the ideal AX is generated by the four equations xz = −(t − y)(x + z)2, xt = −yz = (t − y)2(x + z), yt = (t − y)3. The relations ξ = x + z and η2 − η = t − y prove that ξ and η are integral over f ∗(k[X]), so that f is ﬁnite. The remaining properties of f we need are very easy to check. It is easy to see that the tangent cone to X at origin (see Section 1.5) is L1 ∪ L2. 5.2 Normalisation of an Afﬁne Variety Consider the simplest possible example of a nonnormal variety, the curve X deﬁned by y2 = x2 + x3. Its parametrisation, using the parameter t = y/x, deﬁnes a map f : A1 → X, or equivalently, an inclusion k[X] → k[t]. Since f is birational, we have k[X] ⊂ k[t] ⊂ k(X) = k(t). The line A1 is normal, of course, corresponding to that fact that k[t] is integrally closed. Moreover, the ring k
[t] can be characterised as the set of all elements u ∈ k(X) that are integral over k[X]. Indeed, t 2 = 1 + x, hence t and all elements of k[t] are integral over k[X]; moreover, if u is integral over k[X] then it is also integral over k[t], and hence u ∈ k[t] since k[t] is integrally closed. Finally, in geometric terminology, k[t] integral over k[X] says that f is a ﬁnite map. We show that for any irreducible afﬁne variety X, there exists a variety X and a map X → X having the same properties. We start with a deﬁnition that relates to arbitrary irreducible varieties. Deﬁnition A normalisation of an irreducible variety X is an irreducible normal variety Xν, together with a regular map ν : Xν → X, such that ν is ﬁnite and birational. (If X = Xi is a reducible variety then one can deﬁne Xν = Xν i.) Theorem 2.20 An afﬁne irreducible variety has a normalisation that is also afﬁne. Proof Let A be the integral closure of k[X] in k(X), that is, the set of elements u ∈ k(X) that are integral over k[X]. It follows from elementary properties of integral elements that A is a ring, and is integrally closed. Suppose that we can ﬁnd an afﬁne variety X such that A = k[X]. Then X is normal and the inclusion k[X] → k[X] deﬁnes a regular birational map f : X → X. Obviously X is a normalisation of X. By Theorem 1.3, the required afﬁne variety X exists if and only if A is ﬁnitely generated over k and has no zerodivisors. We will prove more, that A is a ﬁnite module over k[X]. If A = k[X]w1 + · · · + k[X]wm then w1,..., wm, together with the generators of the algebra k[X
] over k, provide a ﬁnite system of generators of A as a k-algebra. To prove that A is a ﬁnite k[X]-module, we use Noether normalisation, Theorem 1.18. By this theorem, there exists a subring B ⊂ k[X] such that B is isomorphic to a polynomial ring B ∼= k[T1,..., Tr ] and k[X] is integral over B. We write 5 Normal Varieties 129 out all the current rings and ﬁelds: k(T1,..., Tr ) ⊂ k(X) = K B ⊂ k[X] ⊂ A From this diagram and from basic properties of integral elements, one sees that A is equal to the integral closure of B in k(X). Moreover, K = k(X) is a ﬁnite ﬁeld extension of k(T1,..., Tr ), since T1,..., Tr is a transcendence basis of k(X). Finally, B is integrally closed, since Ar is normal, indeed, nonsingular. Thus the ﬁnal result we are aiming for, that A is a ﬁnite k[X]-module, follows from the fact that if B = k[T1,..., Tr ], L = k(T1,..., Tr ), and K is any ﬁnite extension ﬁeld of L, then the integral closure of B in K is a ﬁnite B-module. For the proof of this assertion, see Proposition A.15. The theorem is proved. Theorem 2.21 (i) If g : Y → X is a ﬁnite regular birational map, then there exists a regular map h : Xν → Y such that the diagram Xν h ν Y −→ X g is commutative. (ii) If g : Y → X is a regular map, g(Y ) is dense in X and Y is normal then there exists a regular map h : Y → Xν such that the diagram Y h g Xν −→ ν X is commutative. Proof of (i) By assumption we have inclusions k[X] ⊂ k[Y ] �
� k(X), with k[Y ] integral over k[X]. Now by deﬁnition of integral closure, k[Y ] ⊂ k[Xν], which provides the required regular map h : Xν → Y. Proof of (ii) An element u ∈ k[Xν] is integral over k[X] and contained in k(X) ⊂ k(Y ); since k[X] ⊂ k[Y ], it is a fortiori integral over k[Y ], and thus, since k[Y ] is integrally closed, u ∈ k[Y ]. Hence k[Xν] ⊂ k[Y ], which provides the regular map h : Y → Xν with the required properties. The theorem is proved. Corollary The normalisation of an afﬁne variety X is unique. More precisely, if ν : Xν → X and ν : Xν → X are two normalisations of X then there exists an 130 2 Local Properties isomorphism g : Xν ∼→ Xν such that the diagram g−→ Xν Xν ν ν X is commutative. This follows from either of the assertions of Theorem 2.21. We do not prove the existence of the normalisation for arbitrary quasiprojective varieties; the proof is discussed in Section 1.1, Chapter 6 in a more general context. Note that for those varieties for which the normalisation is known to exist, it has the properties established in Theorem 2.21, as follows at once from considering afﬁne covers. 5.3 Normalisation of a Curve Theorem 2.22 An irreducible quasiprojective curve X has a normalisation Xν, and Xν is again quasiprojective. Proof Let X = malisation of Ui, which exists by Theorem 2.20, and fi : U ν i regular map, which is birational and ﬁnite. Ui be a cover of X by afﬁne open sets. Write U ν i for the nor→ Ui for the natural We embed the afﬁne space containing U ν i into projective space, and write Vi for its closure in projective space. Note that all the varieties appearing so far are birational to X: for Ui ⊂ X is open
, f : U ν ⊂ i → Vj for the Vi is also open. Therefore U ν corresponding birational map. i and Vj are birational; write ϕij : U ν i → Ui is a birational map, and U ν i By Theorem 2.19, Corollary of Section 5.1, U ν is a nonsingular curve, and, since i → Vj is a regular map by Theorem 2.12, Corollary 2.3. Vj is projective, ϕij : U ν i Set W = → W, that is, ϕi(u) = (ϕi1(u), ϕi2(u),... ). j Vj and ϕi = Write X = i ). We claim that X = i ) ⊂ W for the union of all the ϕi(U ν ϕi(U ν Xν. For this we have to show that X is (a) quasiprojective, (b) irreducible and (c) normal, and (d) that it has a ﬁnite birational map ν : X → X. j ϕij : U ν i To prove these statements, set U0 = struction of ϕi it follows easily that U ν 0 ϕ for their common restriction to U ν 0. Then U ν i U ν 0 ⊂ ϕi ϕ ⊂ ϕ U ν 0, Ui ; this is an open subset of X. By coni, and all the ϕi coincide on U ν 0. Write ⊂ U ν where ϕ(U ν quasiprojective curve and ϕ(U ν 0 ) is the closure of ϕ(U ν 0 ) in W. Obviously ϕ(U ν 0 ) is an irreducible 0 ) consists of a ﬁnite number of points. By 0 ) \ ϕ(U ν 5 Normal Varieties 131 construction, ϕ(U ν number of points. This proves (a) and (b). 0 ) ⊂ X ⊂ ϕ(U ν 0 ) and hence ϕ(U ν 0 ) \ X also consists of a ﬁnite Let x
∈ X; then x ∈ ϕi(U ν → ϕi(U ν i ) for some i, and ϕi(U ν i ) is a neighbourhood of x. i ) ⊂ W is an isomorphism; since U ν We prove that ϕi : U ν is normal, it i i follows that X is normal, proving (c). For this, note that ϕii is an embedding of −1 U ν ii (ui) is an inverse to ϕi, i which proves that ϕi is an isomorphic embedding. to its projective closure Vi. Hence (u1, u2,... ) → ϕ Finally for the proof of (d) we construct the map −1 gi = fi ◦ ϕ i : ϕi U ν i → Ui ⊂ X. 0. If g : U ν 0 By what we have said above, all the gi are ﬁnite maps. We prove that all the gi deﬁne on X a single ﬁnite map f : X → X. For this, note that all the gi coincide → U0 is the normalisation map then gi = g on U ν on U ν 0. Hence the maps i ) ∩ ϕj (U ν gi and gj coincide on the open set of ϕ(U ν 0 ) contained in ϕi(U ν j ). But two regular maps that coincide on an nonempty open set coincide everywhere. This follows from the same statement for functions. Thus gi and gj coincide at all points at which they are both deﬁned, so that they all deﬁne a regular map ν : X → X. Obviously ν is birational. The theorem is proved. Theorem 2.23 The normalisation of a projective curve is projective. Proof Let X be a projective curve and Xν its normalisation, with ν : Xν → X the normalisation map. Suppose that Xν is not projective, and write Y for its closure in projective space. Choose a point x ∈ Y \ Xν ; let U be an afﬁne neighbourhood of x in Y and U ν its normalisation, with ν : U ν → U
the normalisation map. We have a diagram ν−→ X h−→ Xν ⏐ ϕ U ν ν⏐ U −→ Y ψ where ϕ : Xν → Y and ψ : U → Y are the inclusions of open sets. The composite map ν ◦ ϕ− is birational, and since U ν is nonsingular, it is regular by Theorem 2.12, Corollary 2.3. By Theorem 2.21, there exists a regular map h : U ν → Xν as in the diagram, making the square commute, ϕ ◦ h = ψ ◦ ν. However, the existence of h leads to a contradiction: ϕ(h(U ν)) ⊂ Xν, and ψ(ν(U ν)) x, since the normalisation map is ﬁnite, and hence surjective by Theorem 1.12. This contradiction proves the theorem. Corollary An irreducible algebraic curve is birational to a nonsingular projective curve. This is a combination of Theorem 2.19, Corollary of Section 5.1 and Theo- rem 2.23. Normalisation allows us to study properties of curves in more detail. 132 2 Local Properties Theorem 2.24 A regular map ϕ : X → Y from an irreducible nonsingular projective curve X is ﬁnite (Section 5.3, Chapter 1) if Y = ϕ(X) is a variety with dim Y > 0. Proof Write V for an afﬁne neighbourhood of a point y ∈ Y, and B = k[V ]. We view k(Y ) as a subﬁeld of k(X) under the inclusion ϕ∗. In particular, B ⊂ k(X); let A be the integral closure of B in k(X). In the proof of the existence of the normalisation of an algebraic variety, we proved that A is a ﬁnite B-module, and hence A = k[U ], where U is an afﬁne normal curve. Since U is birational to X, by Theorem 2.16, Corollary 2.6, we can assume that U is an open subset of X. Let us prove that U = ϕ−1
(V ). This will guarantee the ﬁniteness of ϕ. Suppose that for some point y0 ∈ V there is a point x0 /∈ U with ϕ(x0) = y0. Consider a function f such that f /∈ Ox0 but f ∈ Oxi for all xi ∈ U with ϕ(xi) = y0 and xi = x0. Such a function can easily be constructed by putting x0 and xi in one afﬁne open set. If f has poles at points x ∈ U, then ϕ(x) = y = y0, and hence we can ﬁnd a function h ∈ B such that h(y0) = 0 and f h ∈ Ox, that is, f h ∈ A; for this, we need only take a sufﬁciently high power of a function that vanishes at y. Then f1 = f h is integral over B, that is + b1f n−1 1 + · · · + bn = 0 with bi ∈ B, f n 1 so that f1 = −b1 − b2/f1 − · · · − bn/f n−1 ∈ mx0. 1 Hence the ﬁnal equality is a contradiction: the right-hand side is regular at x0, but the left-hand side is not. The theorem is proved.. Since f1 /∈ Ox0, we get f −1 1 Another application concerns curve singularities: the existence of the normalisa- tion allows us to introduce some useful invariants of singular points of curves. Let X be a curve and p ∈ X a point, possibly singular; write ν : Xν → X for the normalisation and q1,..., qk for the inverse images of p in Xν. The points qi are called branches of X at p. The terminology is explained in that if k = C (or R) and Ui are sufﬁciently small complex (or real) neighbourhoods of the qi, then some neighbourhood of p ∈ X is the union of the branches ν(Ui). Write Θi for the tangent line to Xν at qi. The differential dqi ν of ν maps Θi onto a linear subspace of the tangent space to
X at p. Obviously (dqi ν)(Θi) is either the point p or a line; in the second case, we say that qi is a linear branch, and (dqi ν)(Θi) the tangent line to the branch.,..., ν∗(tn) + m2 A branch qi is linear if and only if ν∗ takes mp/m2 ) = 1, and therefore a branch is linear if and only if ν∗(ts) /∈ m2 p onto the whole of mqi /m2 qi. Suppose that p is the origin in An with coordinates t1,..., tn. Then ν∗(mp/m2 p) is generated by ν∗(t1) + m2 qi. Since qi is nonsingular we have qi dim(mqi /m2 qi for at qi least one s = 1,..., n. In other words, ν∗(ts) should be a local parameter at qi. Since mp = (t1,..., tn), in invariant form the condition for qi to be a linear branch takes the form ν∗(mp) ⊂ m2 qi. As a measure of how far qi fails to be a linear branch, we can take the number k such that ν∗(mp) ⊂ mk. This is called the multiplicity of the branch qi. qi but ν∗(mp) ⊂ mk+1 qi 5 Normal Varieties Figure 11 The Newton polygon 133 The point (0, 0) of y2 = x2 + x3 gives an example of two linear branches with the tangent lines y = x and y = −x, and the point (0, 0) of the cusp y2 = x3 an example of a nonlinear branch of multiplicity 2. If x is the centre of a single branch, and this is linear, then x is a nonsingular point. This is a corollary of a lemma that we prove at the end of this section. Thus the simplest invariants measuring how singular a point is are its number of branches, and their multiplicities. We say that a point of an algebraic curve in the plane is an ordinary singularity, or
a singular point with distinct tangent lines if it has only linear branches and all its branches have distinct tangent lines. Suppose that X is given by F (x, y) = 0, and that char k = 0. Let (0, 0) = p ∈ X and let q ∈ Xν be one of the branches corresponding to p. If t is a local parameter at q then there are formal power series expansions x = ant n + an+1t n+1 + · · ·, y = bmt m + am+1t m+1 + · · ·, with n, m > 0 and an, bm = 0. (2.38) There exists a formal power series τ = r1t + r2t 2 + · · · with r1 = 0 such that τ n = x. This is easy to check: we have ﬁrst to set r1 = a1/n, and from then on, for each i > 1 we get a linear equation for ri, to solve which we must divide by n, which is possible under the assumption char k = 0. On the other hand, t can also be expressed in terms −1 1 τ + s2τ 2 + · · ·, as can also be checked at once of τ as a formal power series, t = r by equating coefﬁcients. Finally, substituting this expression in (2.38), we get a parametrisation x = τ n, y = cmτ m + cm+1τ m+1 + · · ·, that can be rewritten n y = cmxm/n + cm+1xm+1/n + · · · (2.39) A parametrisation of a branch of this type is called a Puiseux expansion of y. This is particularly useful in problems of analysis, where y is viewed as a function of x. To ﬁnd explicitly the Puiseux expansions corresponding to different branches, there is an extremely convenient method using the Newton polygon of a polynomial F. Suppose that F (x, y) = Aij xiyj. In the plane, we draw the points with coordinates (i, j ) for which Aij = 0 (Figure 11). A necessary condition for the expansion (2.39) to satisfy F (x, y) = 0 is that after substituting (2.39) in F,
the lowest powers of x arising from the various monomials Aij xiyj must cancel out. In order for this to be possible, at least two monomials Aij xi must give terms of the same degree d in x, and and Aij xi yj yj 134 2 Local Properties other monomials terms of degree ≥d. In other words, the exponent α = m/n should satisfy the condition for all (i, j ) with Aij = 0. In Figure 11 this is expressed by saying that α is minus the slope of the line through points (i, j ) and (i, j ), with all other points drawn in the picture either on or above the line. In other words, the only exponents α that can appear are minus the slopes of the lower convex boundary of the convex hull of the set of points drawn in the picture. We rewrite the expansion (2.39) in the form y = cνi xνi, where νi are increas= 0. Certain of these exponents play an especially ing rational exponents, and cνi important role as invariants of the singularity. Suppose that the ﬁrst nonintegral exponent is m1/n1. Obviously, n1 \| n, and if n1 = n then there must be exponents with denominators strictly divisible by n1. Suppose that the ﬁrst of these beyond m1/n1 is m2/(n1n2); then suppose that m3/(n1n2n3) is the ﬁrst exponent with denominator strictly divisible by n1n2, and so on, up to mk/(n1 · · · nk), where n1 · · · nk = n. The pairs (m1, n1), (m2, n2),..., (mk, nk) are called the characteristic pairs of the branch. We state in its simplest form a result that illustrates the signiﬁcance of characteristic pairs. Consider only singularities having a single branch. For any sequence of characteristic pairs there exists a natural number l such that the singularities with given characteristic pairs are uniquely determined up to formal analytic equivalence (see Section 2.2 for the deﬁnition) by the ﬁrst l terms of the expansion (2.39). Thus singularities with given sequence of characteristic
pairs form a ﬁnite dimensional space. For a simple proof and various generalisations, see Hironaka [39] or Walker [77, Sections 2–3, Chapter IV]. 5.4 Projective Embedding of Nonsingular Varieties The nonsingular projective model of an algebraic curve constructed in the preceding section is contained in some projective space Pn. The natural question arises as to how small n we can take to be. We answer this by proving a general result on varieties of arbitrary dimension. Theorem 2.25 A nonsingular projective n-dimensional variety is isomorphic to a subvariety of P2n+1. Let X ⊂ PN be a nonsingular projective variety. Theorem 2.25 will be proved if for N > 2n + 1 we can choose a point ξ ∈ PN \ X such that the projection from ξ is an isomorphic embedding of X into PN −1. We therefore start by elucidating when a regular map is an isomorphic embedding. Lemma A ﬁnite map f from a variety X is an isomorphic embedding if and only if it is one-to-one and dxf is an isomorphic embedding of the tangent space Θx for every x ∈ X. 5 Normal Varieties 135 Proof Set f (X) = Y and ϕ = f −1. The lemma will be proved if we show that ϕ is a regular map. The assertion is local in nature. For y ∈ Y, let x ∈ X be such that f (x) = y. Write U and V for afﬁne neighbourhoods of x and y with f (U ) = V and such that k[U ] is integral over k[V ], and continue to write f : U → V for the restriction of f. It is enough to prove that f is an isomorphism for suitable choice of U and V, since then ϕ = f −1 is a regular map at y. Recall that the tangent space Θx is the dual vector space to mx/m2 x, where mx is the maximal ideal of the local ring Ox. The second hypothesis of the lemma is that f ∗ : my/m2 x is surjective. In other words, if my = (u1,..., u
k), y then the elements f ∗(ui) + m2 x generate mx/m2 x. We apply Nakayama’s lemma (Proposition A.11) to mx as an Ox -module. It then follows from this that mx = (f ∗(u1),..., f ∗(uk)), or in other words → mx/m2 mx = f ∗(my)Ox. (2.40) We check that Ox is a ﬁnite module over f ∗(Oy). Since k[U ] is a ﬁnite k[V ]module, it is enough to prove that each element of Ox can be expressed in the form ξ/f ∗(a) with ξ ∈ k[U ] and a /∈ my. For this, it is enough to check that for α ∈ k[U ] with α /∈ mx there exists an element a ∈ k[V ] with a /∈ my such that f ∗(a) = αβ with β ∈ k[U ]. By Theorem 1.12 the set f (V (α)) is closed, and since f is one-to-one, y /∈ f (V (α)). Hence there exists a function c ∈ k[V ] such that c = 0 on f (V (α)) and c(y) = 0. Then f ∗(c) = 0 on V (α) and f ∗(c)(x) = 0. By the Nullstellensatz, f ∗(c)n = αβ for some n > 0 and β ∈ k[U ]. We can set a = cn. Now we can apply Nakayama’s lemma to Ox as a f ∗(Oy)-module. The equality (2.40) shows that Ox/f ∗(my)Ox = Ox/mx = k, and hence is generated by the single element 1. It now follows by Nakayama’s lemma that Ox = f ∗(Oy). Let u1,..., ul be a basis of k[U ] as a module over k[V ]. By what we have proved, ui ∈ Ox = f ∗(Oy). Write V = V \ V (h)
for a principal afﬁne neighbourhood of y such that all (f ∗)−1(ui) are regular in U = U \ V (f ∗(h)). Then k[U ] = f ∗k[V ]ui. By assumption ui ∈ f ∗(k[V ]), and it follows that k[U ] = k[V ], which means that f : U → V is an isomorphism. The lemma is proved. Corollary 2.7 Let X ⊂ PN be a variety and ξ ∈ PN \ X. Suppose that every line through ξ intersects X in at most one point, and ξ is not contained in the tangent space to X at any point then the projection from ξ is an isomorphic embedding X → PN −1. It is enough to apply the lemma, together with Theorem 1.15. Proof of Theorem 2.25 It is enough to prove that if X ⊂ PN is a nonsingular ndimensional variety and N > 2n + 1 then there exists ξ as in Corollary 2.7; this is a standard dimension count. Let U1 and U2 ⊂ PN be the sets of points ξ ∈ PN not satisfying the two assumptions of Corollary 2.7. In PN × X × X consider the set Γ of triples (a, b, c) with a ∈ PN, b, c ∈ X such that a, b, c are collinear. Γ is obviously a closed subset of PN × X × X. The 136 2 Local Properties projections of PN × X × X to PN and to X × X deﬁne regular maps ϕ : Γ → PN and ψ : Γ → X × X. Obviously if y ∈ X × X with y = (b, c), and b = c then ψ −1(y) consists of points (a, b, c) where a is any point of the line through b and c. Hence dim ψ −1(y) = 1 and it follows by Theorem 1.25 that dim Γ = 2n + 1. By deﬁnition U1 = ϕ(Γ ), and the same theorem gives dim U1 ≤ dim Γ = 2n + 1.
In a similar way, to study the set U2 we consider in PN × X the set Γ consisting of points (a, b) such that a ∈ Θb. In exactly the same way we have projections ψ : Γ → X and ϕ : Γ → PN. For b ∈ X we have dim ψ −1(b) = n since X is nonsingular, and hence dim Γ = 2n, and since U2 = ϕ(Γ ), also dim U2 ≤ 2n. We see that dim U1 ≤ dim Γ = 2n + 1 and dim U2 ≤ 2n; therefore if N > 2n + 1 then U1 ∪ U2 = PN, which was what we wanted. The theorem is proved. Corollary 2.8 Any nonsingular quasiprojective curve is isomorphic to a curve in P3. We will see later that not every curve is isomorphic to a curve in the projective plane. That is, not every algebraic curve has a nonsingular plane projective model. However, it can be proved that, continuing the process of projection used in the proof of Theorem 2.25, we can obtain a plane curve all of whose singular points are ordinary double points (we assume here that char k = 0). By Theorem 2.25, every nonsingular projective surface is isomorphic to a surface in P5; in general, it cannot be projected isomorphically into P4. However, one can choose a projection to P4 so that it is an isomorphism outside ﬁnitely many points. In this way we easily arrive at isolated surface singularities that are not normal, one example of which was constructed at the end of Section 5.1. 5.5 Exercises to Section 5 1 Let X be an afﬁne variety and K a ﬁnite extension of k(X). Prove that there exists an afﬁne variety Y and a map f : Y → X with the properties (1) f is ﬁnite; (2) Y is normal; (3) k(Y ) = K with f ∗ : k(X) → k(Y ) = K the given inclusion. Prove that Y is uniquely determined by these properties. It is called the normalisation of X in K. √ 2 Let X
be the cone z2 = xy. Prove that the normalisation of X in the ﬁeld x) equals the afﬁne plane, and the normalisation map is of the form x = u2, k(X)( y = v2, z = uv. 3 Prove the assertions analogous to those of Exercise 1 for an arbitrary quasiprojective curve X. Prove that if X is projective then so is Y. 4 How is the normalisation of X × Y related to those of X and Y? 6 Singularities of a Map 137 5 Prove that x is a normal point of X if the completed local ring Ox of Section 2.2 has no zerodivisors and is integrally closed. [Hint: Extend Exercise 7 of Section 3.3 to singular points and apply.] 6 Prove that the cone X ⊂ An given by x2 1 + · · · + x2 n = 0 is normal. 7 In Exercise 13 of Section 3.3, prove that the origin is a normal point of the hypersurface X. 8 Is the Steiner surface of Exercise 15 of Section 1.6 normal? 9 Prove that any algebraic curve has a plane projective model all of whose singularities have only linear branches. 6 Singularities of a Map When studying a regular map f : X → Y, the following question arises: to what extent do the ﬁbres f −1(y) over points y ∈ Y inherit properties of X. As a rule, there are relations that do not hold everywhere, but hold over “most” points y ∈ Y, that is, over points of some dense open set U ⊂ Y. Over other points y /∈ U, the ﬁbres f −1(y) may suffer some kind of degeneration, or acquire singularities not present on X. The situation should be compared with that of Theorem 1.25 on the dimensions of ﬁbres. 6.1 Irreducibility Of course, even if X is irreducible, we cannot hope that almost all the ﬁbres of f : X → Y are irreducible. For example, if f is ﬁnite, its ﬁbres are ﬁnite collections of points. We now formulate a restriction that allows us to guarantee the ir
reducibility of “most” ﬁbres. Suppose that X and Y are irreducible, and that f (X) is dense in Y. A variety X deﬁned over k can also be viewed as a variety over the bigger ﬁeld k(Y ) ⊃ k. Since all our considerations so far related to algebraically closed ﬁelds, we have to view it over an even bigger ﬁeld, the algebraic closure k(Y ) of k(Y ). Now over k(Y ), our variety X may no longer be irreducible. For example, let X be the pencil of conics 2 i,j =0 aij (t)ξiξj = 0 in P2 × A1. Set D(t) = det \|aij (t)\|. If D(t) is not deﬁned by 2 i,j =0 aij (t)ξiξj = 0 is nondegenerate, and X is irreducible identically 0, the conic over k(t). But if D(t) ≡ 0, then over k(t), the equation of the conic can be reduced to a(t)ξ 2 1 is 0 irreducible over k(t), but nevertheless reducible over k(t). = 0. If −b(t)/a(t) is not a square in k(t) then a(t)ξ 2 0 + b(t)ξ 2 1 + b(t)ξ 2 138 2 Local Properties In the general case, it can be shown that a variety X is irreducible over k(Y ) if and only if the map f : X → Y cannot be factored as a composite X → Y → Y where k(Y ) is a nontrivial ﬁnite extension ﬁeld of k(Y ), or in other words, if and only if f ∗ : k(Y ) → k(X) embeds k(Y ) as an algebraically closed subﬁeld of k(X). This is a purely algebraic fact, see Zariski and Samuel [81, Theorem 38 of Section 11, Chapter VII]. Theorem 2.26 (The ﬁrst Bertini theorem) Let X and Y
be irreducible varieties deﬁned over a ﬁeld of characteristic 0, and f : X → Y a regular map such that f (X) is dense in Y. Suppose that X remains irreducible over the algebraic closure k(Y ) of k(Y ). Then there exists an open dense set U ⊂ Y such that all the ﬁbres f −1(y) over y ∈ U are irreducible. Remark 2.1 The theorem also holds over a ﬁeld of characteristic p; the proof just becomes slightly more complicated. Remark 2.2 By the remark just before the statement of the theorem, the only reason for “most” ﬁbres of f : X → Y to be reducible is the existence of a factorisation X → Y → Y where Y → Y is a generically ﬁnite maps. Proof We can replace Y by an afﬁne open subset Y1, and so by Theorem 1.25, we can assume that for y ∈ Y1 all the components of the ﬁbres f −1(y) have the same dimension r = dim X − dim Y. In this situation, we can also replace X by any open subset X1. Indeed, set X \ X1 = Z and let Z = Zi be its decomposition into irreducible components. By passing to a smaller open set Y2 ⊂ Y1 if necessary, we can discard the components Zi for which f (Zi) = Y2. If f (Zi) is dense in Y2, possibly shrinking Y2 still further, we can once more assume that all components of ﬁbres of f : Zi → Y2 have the same dimension equal to dim Zi − dim Y2 < r. Therefore they meet the ﬁbres of f : X → Y2 in subsets of smaller dimension, and since all components of these ﬁbres have equal dimension, discarding subsets of smaller dimension from them does not affect their irreducibility. We now make use of the fact that our ﬁelds have characteristic 0. We can ﬁnd r + 1 elements u1,..., ur, ur+1 ∈ k(X) such that u1,..., ur are algebraically independent over k(
Y ), and such that ur+1 is a primitive element for the ﬁeld extension k(Y )(u1,..., ur ) ⊂ k(X), and is integral over k[Y2][u1,..., ur ]. Let X2 be the afﬁne variety for which k[X2] = k[Y2][u1,..., ur, ur+1]. By construction X2 is birational to X, and hence they contain isomorphic open subsets, so that it is enough to prove the theorem for X2 in place of X, with the map f : X2 → Y2 deﬁned by the inclusion k[Y2] ⊂ k[Y2][u1,..., ur, ur+1]. Let F = T k + a1(u1,..., ur )T k−1 + · · · + ak(u1,..., ur ) be the irreducible polynomial with ai ∈ k[Y2][u1,..., ur ] of which ur+1 is a root. The assumption that X is irreducible over the ﬁeld k(Y ) means that F is irreducible over the ring k(Y )[T, u1,..., ur ]. Now the thing that we have to prove is that there exists an open subset U ⊂ Y2 such that F remains irreducible on making the substitution ai → ai(y) for each y ∈ U, that is, replacing each coefﬁcient ai ∈ k[Y2] of F by its 6 Singularities of a Map 139 value ai(y) at y. But this follows at once from Proposition of Section 5.2, Chapter 1 and Remark 1.6, according to which the reducibility of a polynomial is expressed by polynomial relations Rj (a1,..., ak) = 0 between its coefﬁcients. At least one of these relations fails for F, say Rj (a1,..., ak) = R = 0 ∈ k[Y2]; but then for any point y ∈ Y2 with R(y) = 0, the polynomial obtained by
substituting ai → ai(y) for the coefﬁcients of F is also irreducible. In other words, U = Y2 \ V (a). The theorem is proved. 6.2 Nonsingularity In the theory of differentiable manifolds, one proves that for a smooth map f : X → Y, the points y ∈ Y over which the ﬁbre f −1(y) is not a smooth manifold form a subset of measure 0 in Y (an analogue in differential topology of a subvariety of smaller dimension). This is called Sard’s theorem (see Lang [56, Section 1, Chapter VIII] or Abraham and Robbin, [2, Section 15]). Theorem 2.27 below is an algebraic geometric equivalent of this over a ﬁeld of characteristic 0. We will see in Section 6.4 that the same assertion in characteristic p is false. Theorem 2.27 (The second Bertini theorem) Let f : X → Y be a regular map of varieties deﬁned over a ﬁeld of characteristic 0, with f (X) dense in Y ; assume that X is nonsingular. Then there exists a dense open set U ⊂ Y such that the ﬁbre f −1(y) is nonsingular for every y ∈ U. Theorems 2.26 and 2.27 and their various generalisations are called the Bertini Theorems. Set dim X = n and dim Y = m. By Theorem 1.25, there exists a dense open subset of Y over which all components of the ﬁbres f −1(y) are of the same dimension n − m. We can assume that Y is the whole of this open set. In the same way, we can assume that Y is nonsingular. We prove ﬁrst two lemmas. Lemma 2.3 The ﬁbre f −1(y) is nonsingular if dxf : ΘX,x → ΘY,y is surjective for all points x ∈ f −1(y). Proof Note that the tangent space Θf −1(y),x to the ﬁbre f −1(y) is contained in the kernel of dxf. Indeed, the composite of Θf −1(y),x → Θ
X,x with dxf is 0. To check this, by duality we must check that the composite of the dual my/m2 x of y dxf with mx/m2 x is 0, where mx is the maximal ideal of x on the ﬁbre x f −1(y), and mx → mx the restriction from X to the ﬁbre. But this is obvious. Thus dxf surjective implies that → mx/m2 → mx/m2 dim Θf −1(y),x ≤ dim ker dxf = dim OX,x − dim ΘY,y ≤ n − m; (here we use the fact that X is nonsingular, that is, dim ΘX,x = n). Since all the components of the ﬁbre f −1(y) have dimension n − m, it follows that it is nonsin- gular. 140 2 Local Properties Lemma 2.4 There exists a nonempty open subset V ⊂ X such that dxf is surjective for x ∈ V. Proof The surjectivity of dxf : ΘX,x → ΘY,y is dual to the injectivity of my/m2 → y mx/m2 x, that is, if u1,..., um are local parameters at Y, to the linearly independence of dxu1,..., dxum. Using the inclusion of Oy to the formal power series ring as in Section 2.2, it is easy to see that u1,..., um are algebraically independent, and since f (X) is dense in Y, it follows that they are also algebraically independent as functions on X. We complete them to a system u1,..., un of n = dim X algebraically independent functions. Lemma 2.4 will be proved if we check that for any system u1,..., un of algebraically independent functions on X, the set of points at which u1,..., un are local parameters is open and nonempty. We can assume that X is afﬁne, X ⊂ AN, with coordinates x1,..., xN. We prove that for points x of a nonempty open set U ⊂ X all the dxxi can be expressed as linear combinations of dxu1
,..., dxun. If these were linearly dependent it would then follow that dim ΘX,x < n. Each xi is related to u1,..., un by a relation Fi(xi, u1,..., un) = 0, with Fi an irreducible polynomial, and hence (using that char k = 0), ∂Fi/∂xi is not identically 0. Suppose that Fi = a0xni + · · · + an, with aj ∈ k[u1,..., un]. Now dxaj i are linear combinations of dxu1,..., dxun. Using the basic properties (2.10) of the differential dx, it follows from Fi(xi, u1,..., un) = 0 that + a1xni −1 i ∂Fi ∂xi (x)dxxi + xni i dxa0 + · · · + dxan = 0 at any point x ∈ X. The points at which all ∂Fi/∂xi(x) = 0 form a nonempty open set, and at such points dxxi can be written as linear combinations of dxu1,..., dxun. Lemma 2.4 is proved. Proof of Theorem 2.27 It is now easy to complete the proof of Theorem 2.27. Let Z ⊂ X denote the subset of points x ∈ X at which dxf is not surjective. It is easy to see that it is a closed subset, since it is deﬁned by the vanishing of certain minors. We need to prove that f (Z) is contained in a proper closed subset of Y. If not, then f (Z) is dense in Y. Applying Lemma 2.4 to Z, we ﬁnd a nonempty open set V ⊂ Z such that ΘZ,x → ΘY,f (x) is surjective at all points of V. But ΘZ,x ⊂ ΘX,x, and thus a fortiori the map ΘX,x → ΘY,x must be surjective. This contradiction proves the theorem. 6.3 Ramiﬁcation We consider now an especially simple case of maps, those with 0-dimensional ﬁbres
. For a ﬁnite map f : X → Y, as we saw in Theorem 1.13, the inverse image f −1(y) of any point y ∈ Y is a ﬁnite number of points. Let us study this number. 6 Singularities of a Map 141 By analogy with the theorem on dimension of ﬁbres, it is natural to expect that it is constant for all y in some open set, deviating from this value only on some closed subset Z ⊂ Y. This is what happens in the simplest case f : A1 → A1 given by y = f (x) = x2. (2.41) To state in a general form the speciﬁc property of this example, we introduce the following notion. Deﬁnition Let X and Y be irreducible varieties of the same dimension and f : X → Y a regular map such that f (X) ⊂ Y is dense. The degree of the ﬁeld extension f ∗(k(Y )) ⊂ k(X), which is ﬁnite under these assumptions, is called the degree of f : k(X) : f ∗ k(Y ) deg f =. The map (2.41) has deg f = 2, and if char k = 2, every point y = 0 has two distinct inverse images, and the point y = 0 one only. Is it always true that the number of inverse images is ≤ the degree of a map? This is not so for the example of the parametrisation f : A1 → Y of the cubic curve with an ordinary double point (1.2)–(1.3): here deg f = 1, but the inverse image of the singular point consists of two points. It turns out that the reason here is that Y is not normal. Theorem 2.28 If f : X → Y is a ﬁnite map of irreducible varieties, and Y is normal, then the number of inverse images of any point y ∈ Y is ≤ deg f. Proof In view of the deﬁnition of a ﬁnite map, we can restrict to the case that X and Y are afﬁne. Set k[X] = A, k[Y ] = B, k(X) = K, k(Y
) = L, with [K : L] = deg f = n. Since Y is normal, B is integrally closed, and since f is ﬁnite, A is a ﬁnite Bmodule. Hence for any a ∈ A, the coefﬁcients of the minimal polynomial of a are in B. This is a simple property of integrally closed rings, whose proof can be found in Atiyah and Macdonald [8, Proposition 5.15]. Suppose that f −1(y) = {x1,..., xm}. Consider an element a ∈ A taking distinct values a(xi) at the points xi for i = 1,..., m; if X ⊂ AN, the point is to ﬁnd a polynomial on AN with this property, which is entirely elementary. Let F ∈ B[T ] be the minimal polynomial of a. Obviously deg F ≤ n. We replace all the coefﬁcients of F by their values at y, writing F (T ) for the resulting polynomial. Then this has m distinct roots a(xi). Thus m ≤ deg F = deg F ≤ n, so that m ≤ n, as asserted. The theorem is proved. In what follows, throughout this section, we consider a ﬁnite map f : X → Y between irreducible varieties, and assume that Y is normal. 142 2 Local Properties Deﬁnition f is unramiﬁed over y ∈ Y if the number of inverse images of y equals the degree of the map. Otherwise, we say that f is ramiﬁed at y, or that y is a ramiﬁcation point or a branch point of f. Theorem 2.29 The set of points at which a map is unramiﬁed is open, and is nonempty if f ∗(k(Y )) ⊂ k(X) is a separable ﬁeld extension. Proof We preserve the notation introduced in the proof of Theorem 2.28. If f is unramiﬁed at y then deg F = deg F = n, and F has n distinct roots. Write D(F ) for the discriminant of F. As we have seen, a sufﬁcient condition for f
to be unramiﬁed at a point y can be written D(F ) = D(F )(y) = 0. (2.42) But then D(F )(y) = 0 for points y in some neighbourhood of y. This is what we had to prove. Thus the set of branch point is a closed set; it is called the branch locus or ramiﬁcation locus of f. The question remains as to whether it is a strict subset. Suppose that f ∗(k(Y )) ⊂ k(X) is separable. In this case we also say that f is separable. We can again assume that X and Y are afﬁne, and use the previous notation. If a ∈ A is a primitive element for the ﬁeld extension f ∗(k(Y )) ⊂ k(X) and F (T ) its minimal polynomial, then deg F = n and D(F ) = 0. Therefore, there exist points y ∈ Y such that D(F )(y) = 0, so that f is unramiﬁed. This proves Theorem 2.29. Remark In the case of an inseparable map, every point is a ramiﬁcation point; the standard example of this is the map A1 → A1 deﬁned by x → xp. We see that if f : X → Y is ﬁnite and separable, with X and Y irreducible and Y normal, then the picture is as in the example (2.41): points of some nonempty open subset U ⊂ Y have deg f distinct inverse images, and points in the complement have fewer inverse images. See Theorem 7.3 of Section 3.1, Chapter 7, for a more concrete local description of the ramiﬁcation of a map f : X → Y between algebraic curves over C. Now suppose that Y is nonsingular. The preceding considerations allow us to describe ﬁnite unramiﬁed maps f : X → Y in a very explicit form. Consider a function a ∈ A = k[X] that takes distinct values at all the points of the inverse image f −1(y) of some point y ∈ Y. Then k(X) = k(Y )(a
). If B = k[Y ], and F = F (T ) ∈ B[T ] is the minimal polynomial of a, then by (2.42), the discriminant D(F )(y) = 0, and hence F (a)(x) = 0 for x ∈ f −1(y). From now on, we write Y for an afﬁne neighbourhood of y on which D(F ) is nonzero, and X for its inverse image. Set A = B[a] = B[T ]/(F (T )). Then A = k[X], where X ⊂ Y × A1 is deﬁned by the equation F (T ) = 0. We prove that, in view of the nonsingularity of Y, also X is nonsingular. But then X is normal, and therefore A is integrally closed; since A ⊂ A and the two rings have the same ﬁeld of fractions, we have A = A and X = X, that is, the explicit construction of X actually describes X. 6 Singularities of a Map 143 It remains to prove that X is nonsingular. Suppose that F (T ) = T n + b1T n−1 + · · · + bn with bi ∈ B. We prove that the map dxf : ΘX,x → ΘY,z is an inclusion for any point x ∈ X, where z = f (x). By duality, this is equivalent to mz/m2 x surjective. Let z u1,..., um be local parameters at z. We need to prove that dxu1,..., dxum generate x. By deﬁnition this space is generated by elements dxb for b ∈ B (which are mx/m2 linear combinations of dxu1,..., dxum) together with dxa. It remains to prove that dxa can be written as a linear combination of dxu1,..., dxum. For this, we use the fact F (a) = 0, and the properties (2.10) of differentials. We get → mx/m2 F (a)(x)dxa + an−1(x)dxb1 + · · · + dxbn = 0. Since F (a)(x
) = 0, this expresses dxa in terms of dxu1,..., dxum. Now recall that Y is nonsingular and dim X = dim Y = m. Hence dim ΘY,z = m, and in view of the inclusion dxf : ΘX,x → ΘY,z, also dim ΘX,x = m. Hence X is nonsingular and X = X. But we have proved a little more. We summarise what we have proved. Theorem 2.30 An unramiﬁed ﬁnite map f : X → Y to a nonsingular variety Y is locally described as the projection to Y of a subvariety X ⊂ Y × A1, where X is deﬁned by an equation F (T ) = 0 and D(F ) = 0 on Y. The differential dxf deﬁnes an isomorphism ΘX,x ∼→ ΘY,f (x) on the tangent spaces. In the case k = C, this theorem shows that as a map of topological spaces, f : X → Y is an unramiﬁed cover, that is, any point y ∈ Y has a neighbourhood U such that f −1(U ) decomposes as a disjoint union of open sets, each of which is mapped homeomorphically to U by f. Indeed, suppose that f −1(y) = {x1,..., xn}, and that u1,..., um are local parameters in a neighbourhood of y and ∼→ ΘY,y shows v(i) 1,..., v(i) m local parameters at xi. The isomorphism dxi that det \|∂v(i) k /∂uj \|(xi) = 0 for all i = 1,..., n. By the implicit function theorem it follows from this that there exist neighbourhoods Vi of xi and U of y such that f deﬁne a homeomorphism from each Vi to U. We can choose these neighbourhoods sufﬁciently small that Vi and Vj do not intersect for i = j. We check Vi. If y ∈ U then, since f is unramiﬁed, f −1(y) consists of that f −1(U
) = n = deg f points. But since y already has n inverse images in Vi, we have f −1(U ) = : ΘX,xi Vi. 6.4 Examples Example 2.7 (Pencil of quadrics) Assume that char k = 2, and consider the hyper- n i,j =0 aij (t)ξiξj = 0, where aij (t) ∈ surface X ⊂ Pn × A1 deﬁned by the equation 144 2 Local Properties k[A1] = k[t], and the projection X → A1 is induced by the projection Pn × A1 → A1. This is called a pencil of quadrics, and the polynomial D(t) = det \|aij (t)\| the discriminant of the pencil. Pencils of conics have already appeared in Example 1.35. We determine ﬁrst of all when X is nonsingular, and secondly, over what points α ∈ A1 the ﬁbre of X → A1 is singular. Set F = aij (t)ξiξj. If D(α) = 0 at a point t = α then the equations ∂F /∂ξi = 0 for i = 0,..., n and t = α have only the solution 0, so that points of the ﬁbre over t = α are nonsingular both as points of X and as points of the ﬁbre. It remains to consider the values t = α for which D(α) = 0. We will assume that α = 0. Write F for the quadratic form F = aij (0)ξiξj, and r for its rank. We can make a nondegenerate linear transformation with coefﬁcients in k to put F in the form ξ 2 r−1. Now we apply to F the standard method of completing the 0 square; we can make a linear transformation with coefﬁcients in the local ring O0 of the origin of A1 (that is, the coefﬁcients are rational function with no t in the denominators), and with determinant invertible in O0, to put F in the form + · · · + ξ 2 F = a0(t)ξ 2 0 +
· · · + ar−1(t)ξ 2 r−1 + tG(ξr,..., ξn), with ai(t) ∈ O0 and ai(0) = 1 for i = 0,..., r − 1. Any points with t = 0 and ξ0 = · · · = ξr−1 = 0 (and arbitrary ξr,..., ξn) lie on X, and there ∂F /∂ξi = 0 for all i. Suppose that G(ξr,..., ξn) = G(ξr,..., ξn) + tG1(ξr,..., ξn), with G ∈ k[ξr,..., ξn]. Then at our point, ∂F /∂t = G(ξr,..., ξn). If r < n then there exist ξr,..., ξn, not all 0, such that G(ξr,..., ξn) = 0, and the point is singular on X. For r = n, the equation looks like F = a0(t)ξ 2 0 + · · · + an−1(t)ξ 2 n−1 + t kan(t)ξ 2 n, with ai(0) = 0 for i = 0,..., n and some k ≥ 1. If k > 1 then ∂F /∂t = 0 at the point t = 0, (ξ0,..., ξn) = (0,..., 0, 1), and this is a singular point of X. There remains the case k = 1, when it is easy to see that no point of the ﬁbre over t = 0 is a singular point of X. Thus we have proved the following result. Proposition 2.1 The quadric bundle X is a nonsingular variety if and only if its discriminant has no repeated roots. The singular ﬁbres are precisely the ﬁbres over the roots of the discriminant. In particular, the number of singular ﬁbres of X → A1 equals the degree of the discriminant. Example 2.
8 (Pencil of elliptic curves) Assume that the characteristic of k is not 2 or 3, and consider the surface X ⊂ P2 × A1 deﬁned by the equation 2 ξ0 = ξ 3 ξ 2 1 + a(t)ξ1ξ 2 0 + b(t)ξ 3 0 with a(t), b(t) ∈ k A1 = k[t]. The projection P2 × A1 → A1 deﬁnes a map f : X → A1. The ﬁbre f −1(α) over a + a(α)ξ1ξ 2 point α is the cubic curve ξ 2 0. This cubic has a unique 0 2 ξ0 = ξ 3 1 + b(α)ξ 3 6 Singularities of a Map 145 point on the “line at inﬁnity” ξ0 = 0, the ﬂex (0 : 1 : 0); in the chart A2 with ξ0 = 0, it is given in afﬁne coordinates x = ξ1/ξ0 and y = ξ2/ξ0 by y2 = x3 + a(α)x + b(α). If the ﬁbre f −1(α) is nonsingular, then as in Example 2.7, X has no singular points on it. Suppose that f −1(α) is singular. It is easy to see that (0 : 1 : 0) is nonsingular. Thus there must be a simultaneous solution of y = 0, 3x2 + a(α) = 0 and y2 = x3 + a(α)x + b(α), from which it follows that 4a(α)3 + 27b(α)2 = 0. The polynomial D(t) = 4a(t)3 + 27b(t)2 is called the discriminant of the pencil X → A1. We will assume that D(t) is not identically 0. We have proved that if D(α) = 0 then all points of the ﬁbre f −1(α) are nonsingular both on the ﬁbre and on the surface X. If D(α) = 0 then the same argument shows that the ﬁbre f −1
(α) has a singular point, and it follows from the equations 3x2 + a(α) = 0 and x3 + a(α)x + b(α) = 0 that the x-coordinate of this point is given by 2a(α)x + 3b(α) = 0. In order for this to be a singular point of X, it must also satisfy a(α)x + b(α) = 0, whence 2ab − 3ba = 0. Since moreover 4a(α)3 + 27b(α)2 = 0, either a(α) = b(α) = 0 or a(α) = 0 and b(α) = 0. When a(α) = b(α) = 0 our relations are equivalent to b(α) = 0, and when a(α) = 0 and b(α) = 0, they can be expressed as (a3/b2)(α) = 0, or D(α) = (b2(4a3/b2 + 27))(α) = 0. This proves the following result. Proposition 2.2 The pencil of elliptic curves X → A1 is a nonsingular surface if the discriminant has simple roots or are common roots of a and b that are simple roots of b. Singular ﬁbres correspond to roots of the discriminant. 2 ξ0 = ξ 3 1 Example 2.9 (Pathologies in ﬁnite characteristic) We can construct examples in which the assertion of Theorem 2.27 fails in characteristic 2. For this, consider the ﬁnite part of the pencil of elliptic curves ξ 2 0, given by y2 = x3 + a(t)x + b(t). In characteristic 2 every ﬁbre y2 = x3 + a(α)x + b(α) is singular at the point x = a(α)1/2, y = b(α)1/2, and at no other point. In order for this to be a singular point of the surface, we must have a(α)x + b(α) = 0, that is, ((a)2a + (b)2)(α) = 0. Thus all the ﬁbres of X → A1 are singular, but the surface X itself only has singular points in the �
�bres f −1(α), where α is a root of (a)2a + (b)2. If S is the set of these roots, then the surface X \ f −1(S) is nonsingular, but all the ﬁbres of X \ f −1(S) → A1 \ S are singular. + a(t)ξ1ξ 2 0 + b(t)ξ 3 There is a similar example in characteristic 3, the pencil with equation y2 = x3 + a(t). It can be proved that such “pathological” pencils of cubic curves exist only in characteristic 2 or 3, although of course similar examples occur for all p for curves of higher degree, for example y2 = xp + a(t). An example of a ﬁnite map f : X → Y such that every point y ∈ Y is a branch point is given by the Frobenius map, Example 1.16 in characteristic p > 0. It has ϕ(α1,..., αn) = (αp n ), so that in characteristic p, every point x has a unique inverse image ϕ−1(x). 1,..., αp In the theory of curves, the Frobenius map particularly reﬂects the speciﬁc properties of ﬁnite characteristic. For this, we need to generalise somewhat. If C aij xiyj = 0, we let C be the curve g(x, y) = is the plane curve f (x, y) = 146 2 Local Properties ap ij xiyj = 0. In characteristic p, the map u = xp, v = xp obviously deﬁnes a rational map ϕ : C → C (in fact, a regular map). This is also called the Frobenius map, and coincides with that introduced in Example 1.16 if aij ∈ Fp, when ap ij = aij, and therefore C = C. Theorem 2.31 The Frobenius map of an algebraic curve has degree p. Every inseparable rational map of curves f : X → Y factors as a composite f = g ◦ ϕ where g : X → Y is some map and ϕ : X → X the Frobenius map. Proof This follows from general properties of ﬁeld
s of characteristic p and transcendence degree 1; see Proposition A.8. It is proved there that [k(X) : ϕ∗(k(X))] = p, and this means that deg ϕ = p. Moreover, f ∗(k(Y )) ⊃ k(X)p, but k(X)p = k(X). The inclusion of ﬁelds f ∗(k(Y )) ⊂ ϕ∗(k(X)) and the isomorphism ϕ∗ : k(X) → ϕ∗(k(X)) deﬁne an inclusion (ϕ∗)−1(f ∗(k(Y ))) ⊂ k(X), that is, a rational map g : X → Y such that f = g ◦ ϕ. The theorem is proved. 6.5 Exercises to Section 6 1 Classify singular points of pencils of quadrics over the point t = 0 up to formal analytic equivalence, under the assumption that the rank of the quadric drops by 1 at t = 0. 2 i,j =0 aij (s, t)ξiξj = 2 Consider the net of conics X on P2 deﬁned in P2 × A2 by 0. Assume that the rank of a conic over every point α ∈ A2 drops by at most 1. Prove that X is nonsingular if and only if the discriminant curve det \|aij (s, t)\| = 0 is nonsingular. 3 Prove that if a pencil of elliptic curves (Example 2.8) is a nonsingular surface then its singular ﬁbre is irreducible. Is this true for any family of cubics? 4 Determine the branch locus of the map X → Pn, where X is the normalisation of Pn in the quadratic extension of k(Pn) = k(x1,..., xn) deﬁned by the equation y2 = f (x1,..., xn), where f is a polynomial of degree m. [Hint: The answer depends on the parity of m.] 5 Prove that if char k = p then the curve yp + y = f (x) where f is a polynomial
is an unramiﬁed cover of the line A1 with coordinate x. 6 Prove that for the surfaces y2 = x3 + a(t)x + b(t) over a ﬁeld of characteristic 2 and y2 = x3 + a(t) over a ﬁeld of characteristic 3, the singular points of ﬁbres form a nonsingular curve having projection to the line A1 with coordinate t of degree p = 2 or 3 respectively. Chapter 3 Divisors and Differential Forms 1 Divisors 1.1 The Divisor of a Function A polynomial in one variable is uniquely determined up to a constant factor by specifying its roots and their multiplicities; that is by specifying a set of points x1,..., xr ∈ A1 with multiplicities k1,..., kr. A rational function ϕ(x) = f (x)/g(x) with f, g ∈ k[A1] is determined by the zeros of f and g, that is, by the points at which it is 0 or is irregular. To distinguish the roots of g from those of f, we take their multiplicities with a minus sign. Thus the function ϕ is given by points x1,..., xr with arbitrary integer multiplicities k1,..., kr. The task we set ourselves here is to ﬁnd a similar way of specifying a rational function on an arbitrary algebraic variety. The starting point is that, according to the theorem on the dimension of intersections, the set of points at which a regular function is 0 is a codimension 1 subvariety. Thus the object we associate with a function is a collection of irreducible codimension 1 subvarieties, together with assigned multiplicities; the multiplicities we assign are integers, both positive and negative. Deﬁnition Let X be an irreducible variety. A collection of irreducible closed subvarieties C1,..., Cr of codimension 1 in X with assigned integer multiplicities k1,..., kr will be called a divisor on X. A divisor is written D = k1C1 + · · · + kr Cr. (3.1) If all the ki = 0, we
write D = 0. If all ki ≥ 0 and some ki > 0 then we write D > 0; in this case D is said to be effective. An irreducible codimension 1 subvariety Ci taken with multiplicity 1 is called a prime divisor. If all the ki = 0 in (3.1) then the variety C1 ∪ · · · ∪ Cr is called the support of D and denoted by Supp D. I.R. Shafarevich, Basic Algebraic Geometry 1, DOI 10.1007/978-3-642-37956-7_3, © Springer-Verlag Berlin Heidelberg 2013 147 148 3 Divisors and Differential Forms We deﬁne an addition operation on divisors. For this, note that, provided we also allow the coefﬁcients to take the value 0 in (3.1), any two divisors D and D can be written D = k1C1 + · · · + kr Cr and D = k 1C1 + · · · + k r Cr, with the same collection of prime divisors C1,..., Cr. Then by deﬁnition, D + D = k1 + k 1 C1 + · · · + kr + k r Cr. Thus divisors on X form a group, equal to the free Z-module with the irreducible codimension 1 subvarieties C of X as generators. This group is denoted by Div X.7 We now describe the map taking a nonzero function f ∈ k(X) into its divisor div f. Let C be a prime divisor; ﬁrst of all, to each nonzero f ∈ k(X), we assign an integer vC(f ). If X = A1 then vC(f ) is the order of zero or pole of a function at a point. This can be done only under one restriction on X. Namely, we assume that X is nonsingular in codimension 1 (see Section 5.1, Chapter 2); in other words, we assume that the set of singular points of X has codimension ≥2. Let C ⊂ X be an irreducible codimension 1 subvariety, and U some afﬁne open set intersecting C, consisting of
nonsingular points, and such that C is deﬁned in U by a local equation. Such an afﬁne set U exists by the assumption on X and by Theorem 2.10. Thus aC = (π) in k[U ]. We prove that for any 0 = f ∈ k[U ], there exists an integer k ≥ 0 such that f ∈ (π k) and f /∈ (π k+1). If this were not the case, that is, if f ∈ (π k) for (π k); the same then holds in the local ring OC at an irreducible every k, then f ∈ subvariety C. Hence f = 0 by Theorem 2.8 and Proposition A.12. The number k just determined is denoted by vC(f ). It has the properties vC(f1f2) = vC(f1) + vC(f2), vC(f1 + f2) ≥ min vC(f1), vC(f2), and if f1 + f2 = 0, (3.2) as follows easily from the deﬁnition and the irreducibility of C. In the case of a nonsingular plane curve, we have already deﬁned this function in Theorem 1.1. If X is irreducible, then any function f ∈ k(X) can be written in the form f = g/ h with g, h ∈ k[U ]. If f = 0 we set vC(f ) = vC(g) − vC(h). It follows at once from (3.2) that vC(f ) does not depend on the representation of f in the form g/ h, and that (3.2) holds for all f ∈ k(X) with f = 0. Our deﬁnition of vC(f ) depends at present on the choice of an open set U, and hence we temporarily write vU C (f ) is independent of U. Suppose ﬁrst that V ⊂ U is an afﬁne open set with V ∩ C = ∅. Then π is a local equation for C also in V, and obviously then vU C (f ). However, if V is any open set satisfying the same conditions
as U then U ∩ C and C (f ) in place of vC(f ). Let us show that in fact vU C (f ) = vV 7The current literature is inconsistent, some authors using Div X for the group of “ordinary” diviki Ci (Weil divisors) described here, some for locally principal divisors (Cartier divisors) sors (see Section 1.2). In case of ambiguity, one can write WDiv or CDiv. 1 Divisors 149 V ∩ C are open in C and nonempty, and since C is irreducible they have nonempty intersection. Taking W to be an afﬁne neighbourhood in U ∩ V of some point x ∈ C (f ) = U ∩ V ∩ C, by the preceding remark, we get that vU C (f ) = vV C (f ), and hence vU vW C (f ). Thus we have justiﬁed that the notation vC(f ) is well deﬁned. C (f ) and vV C (f ) = vW Notice that if X = A1 and C = x is the point α then vx(f ) equals the multiplicity of α as a root of f for any nonzero f ∈ k[T ]; the general deﬁnition essentially copies this particular case. If vC(f ) = k > 0 then we say that f has a zero of order k along C; if vC(f ) = −k < 0 that f has a pole of order k along C. Note that these notions are deﬁned for codimension 1 subvarieties, rather than for points. For example, if f is the function f = x/y on A2 then the point (0, 0) is contained both in the locus of zeros (x = 0) and the locus of poles (y = 0) of f. We now prove that for a given function f ∈ k(X), there are only a ﬁnite number of irreducible codimension 1 subvarieties C such that vC(f ) = 0. Consider ﬁrst the case that X is an afﬁne variety and f ∈ k[X]. Then it follows from the deﬁnition that if C is
not a component of the subvariety V (f ) then vC(f ) = 0. If X is still afﬁne, but f ∈ k(X) then f = g/ h with g, h ∈ k[X], and we see that vC(f ) = 0 if C is not a component of V (g) or V (h). Finally, in the general case, let X = Ui be a ﬁnite cover of X by afﬁne open sets. Then any subvariety C intersects at least one Ui, so that vC(f ) = 0 only for C that is the closure of an irreducible codimension 1 subvariety C ⊂ Ui for some i, with vC(f ) = 0 in Ui. Since there are only ﬁnitely many Ui and ﬁnitely many C in each Ui, there are only ﬁnitely many C with vC(f ) = 0. Thus we can consider the divisor vC(f )C, (3.3) where the sum takes place over all the irreducible codimension 1 subvarieties C for which vC(f ) = 0. This divisor is called the divisor of f, and denoted8 by div f. A divisor of the form D = div f for some f ∈ k(X) is called a principal divisor. If div f = kiCi then the divisors div0 f = {i\|ki >0} kiCi and div∞ f = {i\|ki <0} −kiCi are called respectively the divisor of zeros and divisor of poles of f. Obviously div0 f, div∞ f ≥ 0 and div f = div0 f − div∞ f. Notice a number a simple properties: div(f1f2) = div(f1) + div(f2), and div f = 0 for f ∈ k; and div f ≥ 0 for f ∈ k[X]. Let us prove that for a nonsingular irreducible variety the converse also holds, that is, if div f ≥ 0 then f is regular on X. The same thing holds if X is only normal, but we omit the proof. Let x �
� X be a point at which f is not regular. Then f = g/ h with g, h ∈ Ox but g/ h /∈ Ox. It follows from the fact that Ox is a UFD 8The divisor div f is also traditionally denoted by (f ) in the literature. 150 3 Divisors and Differential Forms (Theorem 2.11) that we can choose g and h without common factors. Let π be a prime element of Ox that divides h but not g. In some afﬁne neighbourhood U of x, the variety V (π) is irreducible and of codimension 1. Write C for its closure in X. Then obviously vC(f ) < 0. This proves that div f ≥ 0 =⇒ f regular. Since an everywhere regular function on an irreducible projective variety X is a constant (Theorem 1.11, Corollary 1.1), it follows from the result just proved that on a nonsingular projective variety X, if div f ≥ 0 then f = α ∈ k. In particular, on a nonsingular projective variety, a rational function is uniquely determined up to a constant factor by its divisor: if div f = div g then div(f/g) = 0, so that f = αg with α ∈ k. Example 3.1 (X = An) By Theorem 1.21, any irreducible codimension 1 subvariety C is deﬁned by one equation, AC = (F ) with F ∈ k[X]. It follows that C = div F, that is, every prime divisor, hence every divisor, is principal. Example 3.2 (X = Pn) Any irreducible codimension 1 subvariety C is deﬁned by a single homogeneous equation F, and moreover, if F has degree k then in the afﬁne chart Ui, we have aC = (F /T k i ). From this we get a method of constructing the divisor of a function f ∈ k(Pn) as follows: represent f as f = F /G with F and G forms of the same degree, and factor F and G into irreducibles: F = i and G =. Then H ki L mj j mj Dj,
div f = kiCi − where Ci and Dj are the irreducible hypersurfaces deﬁned by Hi = 0 and Lj = 0. Write deg F for the degree of the form F. Since deg F = deg G it follows that mj deg Lj. Deﬁne the degree of a divisor D = ki deg Hi = kiCi as the integer deg D = ki deg Hi. We have proved that if D is a principal divisor then deg D = 0. The converse is also easy to prove: if ki deg Ci = 0 and Ci is deﬁned by a form Hi then f = is homogeneous of degree 0 and div f = kiCi. (3.4) H ki i Example 3.3 (X = Pn1 × · · · × Pnk ) This case is treated similarly. A codimension 1 subvariety C is again given by one equation H = 0 by Theorem 1.21, although now H is homogeneous separately in each of the k sets of coordinates of Pni, and correspondingly has k different degrees degi H for i = 1,..., k. In the same way as in Example 3.2 one can introduce the degrees degi D of a divisor D on X and prove that a divisor D is principal if and only if degi D = 0 for i = 1,..., k. Principal divisors form a subgroup P (X) of the group Div X of all divisors. The quotient group Div X/P (X) is called the divisor class group of X, and is denoted by Cl X. A coset of Div X/P (X) is called a divisor class. Divisors in the same coset of Div X/P (X) are said to be linearly equivalent: D1 ∼ D2 if D1 − D2 = div f for some nonzero f ∈ k(X). In the three examples just worked out, we have respectively = 0, Pn Cl = Z and Cl Pn1 × · · · × Pnk = Zk. An Cl 1 Divisors 151 1.2 Locally Principal Divisors Suppose that the variety X is nonsingular. In this case, for any prime divisor C ⊂ X and any point x �
� X there exists an open set U x in which C is deﬁned by a local equation π. If D = kiCi is any divisor, and each of the Ci is deﬁned in U by a local equation πi, then we have D = div f in U, where f =. Thus every point x has a neighbourhood in which D is principal. From among all such neighbourhoods we can choose a ﬁnite cover X = Ui, and D = div(fi) on Ui. π ki i Obviously, the functions fi cannot be chosen arbitrarily: fi is not identically 0, and in Ui ∩ Uj the divisors div(fi) and div(fj ) coincide. As we saw in Section 1.1, it follows from this that fi/fj is regular in Ui ∩ Uj and nowhere 0 there. We say that a system {fi} of functions corresponding to the open sets Ui of a cover X = Ui is compatible if the fi satisfy these conditions, that is, fi/fj is regular in Ui ∩ Uj and nowhere 0 there. Conversely, any compatible system of functions deﬁnes a divisor on X. Indeed, for any prime divisor C we set kC = vC(fi) if Ui ∩ C = ∅, where fi and C are considered as a function and a prime divisor for the variety Ui. From the compatibility of the system of functions it follows that this number is independent of the choice of Ui. Hence we can consider the divisor D = kCC. Obviously the given {fi} are then a compatible system corresponding to D. Finally, it is easy to determine when a system of functions {fi} corresponding to the open sets of a cover X = Ui deﬁnes the same divisor as another system {gj } corresponding to the open sets of a cover X = Vj. For this, a necessary and sufﬁcient condition is that fi/gj should be regular in Ui ∩ Vj and nowhere 0 there. We leave the simple veriﬁcation to the reader. Specifying divisors in terms of compatible systems of functions allows us to study their behaviour under regular maps. Let ϕ : X → Y be a regular map of nonsingular irreduc
ible varieties, and D a divisor on Y. Suppose that ϕ(X) ⊂ Supp D. We prove that, under this restriction, one can deﬁne the pullback ϕ∗(D) of a divisor D by analogy with the deﬁnition of the pullback of a regular function. First we determine when we can construct the pullback of a rational function f on Y, and when it will not be identically 0 on X. For this it is sufﬁcient that there is at least one point y ∈ ϕ(X) at which f is regular and f (y) = 0; these points then form a nonempty open set V, and f is regular and nowhere 0 on V. Therefore ϕ∗(f ) deﬁnes a regular function on ϕ−1(V ) that is not identically 0 (in fact, nowhere 0). Since ϕ−1(V ) is open in X, ϕ∗(f ) deﬁnes a rational function on X. In terms of divisors, our condition on the map ϕ and the function f boil down to ϕ(X) ⊂ Supp(div f ). Now suppose that D is given by a compatible system of functions {fi} with respect to a cover Y = Ui. We consider the Ui with ϕ(X) ∩ Ui = ∅, and prove that ϕ(X) ∩ Ui ⊂ Supp(div fi). Indeed, the irreducibility of X implies that ϕ(X) is irreducible in Y. If we assume that ϕ(X) ∩ Ui ⊂ Supp(div fi) then since ϕ(X) is irreducible and ϕ(X) ∩ Ui = ∅, it would follow that ϕ(X) ⊂ Supp(div fi). Finally, from the fact that Supp(div fi) ∩ Ui = Supp D ∩ Ui, the irreducibility of ϕ(X) and the fact that ϕ(X) ∩ Ui is nonempty, it would follow that ϕ(X) ⊂ Supp D, contradicting our assumption. 152 3 Divisors and Differential Forms Therefore, for every i such that ϕ(X) �
� Ui = ∅, the rational function ϕ∗(fi) is deﬁned on Vi = ϕ−1(Ui). Then X = Vi is an open cover of X, with respect to which {ϕ∗(fi)} is a compatible system of functions deﬁning a divisor on X. This divisor is obviously unchanged if we deﬁne D using a different system of functions. The divisor obtained in this way is called the pullback or inverse image of D and denoted by ϕ∗(D). Example Suppose that X and Y are two curves, and f : X → Y a map taking X to a point a ∈ Y. If a = b ∈ Y and D = b is the divisor consisting of b with multiplicity 1, then 1 is a local equation of D in a neighbourhood of a, so that f ∗(D) = 0. In particular if ϕ(X) is dense in Y then the pullback of any divisor D ∈ Div Y is deﬁned. If D and D are two divisors on Y deﬁned by systems of functions {fi} and {gj } Vj then the divisor D + D is deﬁned with respect to covers Y = by the system of functions {figj } with respect to the cover Y = (Ui ∩ Vj ). It follows at once that ϕ∗(D + D) = ϕ∗(D) + ϕ∗(D), so that if ϕ(X) is dense in Y, the pullback ϕ∗ deﬁnes a homomorphism Ui and Y = ϕ∗ : Div Y → Div X. The principal divisor div f is given by the system of functions fi = f, and hence ϕ∗(div f ) = div(ϕ∗(f )). Therefore ϕ∗ maps P (Y ) to P (X), and so deﬁnes a homomorphism ϕ∗ : Cl Y → Cl X. As an application of the idea of a divisor deﬁned by a compatible system of functions, we show how one can associate a divisor not with a function, but with a form in the coordinates on a nonsingular
projective variety. Suppose that X ⊂ PN and let F be a form in the coordinates of PN that is not identically 0 on X. For any x ∈ X, consider a form G of the same degree d = deg F, but with G(x) = 0; such forms exist, of course: if, say, x = (α0 : · · · : αN ) and αi = 0 then we can take G = T d i. Then f = F /G is a rational function on X and is regular on the open set where G = 0. It is easy to see that there exist forms Gi such that the open sets Ui = X \ XGi form a cover of X. One checks just as easily that the functions fi = F /Gi form a compatible system of functions with respect to the open sets Ui, and therefore deﬁne a divisor on X. A different choice of the forms Gi does not change this divisor, which therefore depends only on F. It is called the divisor of F, and denoted by div F. Since the functions fi are regular in the Ui, it follows that div F ≥ 0. If F1 is another form with deg F1 = deg F then div F − div F1 = div(F /F1) is the divisor of the rational function F /F1. Therefore deg F1 = deg F implies that div F ∼ div F1. In particular, all the divisors div L, where L is a linear form, are linearly equivalent. Obviously Supp(div L) = XL is the section of X by the hyperplane L = 0. These divisors are thus called hyperplane section divisors of X. Taking F1 = Ld as the form in the above argument, where d = deg F, we get div F ∼ d div L, where div L is a hyperplane section divisor. 1 Divisors 153 All the arguments concerned with using compatible systems of functions to specify divisors generalise to arbitrary, possibly singular, varieties. However, for this, we must take the speciﬁcation by compatible systems of functions as the deﬁnition of divisor. The object we get is called a locally principal divisor. More precisely, we have the following deﬁnition. Deﬁnition A locally principal
divisor or Cartier divisor on an irreducible variety X is a system of rational functions {fi} corresponding to the open sets Ui of a cover X = Ui satisfying the conditions: (1) the fi are not identically 0; (2) fi/fj and fj /fi are both regular on Ui ∩ Uj. Here functions {fi} and open sets Ui deﬁne the same divisor as functions {gj } and open sets Vj if fi/gj and gj /fi are regular on Ui ∩ Vj. Every function f ∈ k(X) deﬁnes a locally principal divisor div f if we set fi = f. Divisors of this form are said to be principal. The product of the two locally principal divisors deﬁned by functions {fi} corresponding to open sets Ui and functions {gj } corresponding to open sets Vj is the divisor deﬁned by functions {figj } and open sets Ui ∩ Vj. All locally principal divisors form a group, and principal divisors a subgroup. The quotient group is called the Picard group of X, and denoted by Pic X. Any locally principal divisor has a support. This is the closed subset which in Ui consists of points at which fi is either not regular, or equal to 0. Just as for divisors on nonsingular varieties, one can deﬁne the pullback of a locally principal divisor D on Y under a regular map ϕ : X → Y if ϕ(X) is not contained in Supp D. We note an important special case. If X is a nonsingular variety and Y a possibly singular subvariety of X, then any divisor D on X with Supp D ⊃ Y deﬁnes a locally principal divisor D on Y. For this, we need to consider the inclusion map ϕ : Y → X and set D = ϕ∗(D). We call D the restriction of D to Y, and denote it by ρY (D). From the deﬁnition it follows that for a principal divisor div f we have ρY (div f ) = div( f ), where f is the restriction of f to Y. Of
course, the distinction between divisors and locally principal divisors, and between the groups Cl X and Pic X, occurs only for singular varieties. 1.3 Moving the Support of a Divisor away from a Point Theorem 3.1 For any divisor D on a nonsingular variety X, and any ﬁnite number of points x1,..., xm ∈ X, there exists a divisor D with D ∼ D such that xi /∈ Supp(D) for i = 1,..., m. Proof We can assume that D is a prime divisor, since otherwise we need only apply the assertion to each component separately. Choose an open afﬁne subset of X containing x1,..., xm; it is enough to prove the theorem for this, so that we can assume that X is afﬁne. By induction, we can assume that x1,..., xm−1 /∈ Supp D 154 3 Divisors and Differential Forms but xm ∈ Supp D, and it is enough to ﬁnd a divisor D such that D ∼ D and x1,..., xm /∈ Supp(D). Consider some local equation π of the prime divisor D in a neighbourhood of xm. We prove that we can choose a local equation π for D with π ∈ k[X] (by assumption X is afﬁne). Indeed, π is regular at xm, so that, if π has divisor of poles div∞(π ) = klFl, then xm /∈ Fl. Thus for each l there exists a function fl ∈ k[X] that vanishes along Fl and such that fl(xm) = 0. Then the function π = π is obviously regular on X and is a local equation of D in a neighbourhood of xm. For each i = 1,..., m − 1, since xi /∈ D ∪ {x1,..., xi−1, xi+1,..., xm} by assumption, there exists a function gi ∈ k[X] such that gi(xi) = 0, but gi = 0 on that set. Now adjust the constants αi ∈
k such that the function f kl l f = π + m−1 i=1 αig2 i, satisﬁes f (xi) = 0 for i = 1,..., m − 1. (3.5) For this it is sufﬁcient to take αi = −π(xi)/(gi(xi)2). We claim that D = D − div f satisﬁes the conclusions of the theorem. First, (3.5) shows that xi /∈ div f, and hence xi /∈ Supp(D) for i = 1,..., m − 1. Now since by construction the gi vanish on D, we get π \| gi in the local ring Oxm, so that = π 2h with h ∈ Oxm, and therefore f = π(1 + πh). Since (1 + πh)(xm) = 1, it follows that f is a local equation of D in a neighbourhood of xm. Therefore div f = D + rsDs, with no prime divisor Ds passing through xm. This means that xm /∈ Supp(D). The theorem is proved. αig2 i The same holds for a locally principal divisor on a singular X (the proof is very similar). Here is a ﬁrst application of Theorem 3.1. In Section 1.2 we deﬁned the pullback f ∗(D) of a divisor D on a variety X by a regular map f : Y → X under the assumption that f (Y ) ⊂ Supp D. Theorem 3.1 allows us to replace D by a linearly equivalent divisor D so that Supp(D) x, where x is an arbitrarily chosen point of f (Y ). Then automatically f (Y ) ⊂ Supp(D), so that the pullback f ∗(D) is deﬁned. This shows that we can deﬁne the pullback of a divisor class C ∈ Cl X without any restriction on f. For this, we must choose a divisor D in the class C such that f (Y ) ⊂ Supp D and consider the divisor class on Y containing the divisor f ∗(D). One checks easily that we
thus obtain a homomorphism f ∗ : Cl X → Cl Y. In other words, Cl X is a functor from the category of irreducible nonsingular algebraic varieties to the category of Abelian groups. Example Let f : X → Y be the constant map f (X) = a ∈ Y (see Example of Section 1.2). Then by Theorem 3.1, the divisor a is linearly equivalent to ribi with bi = a, and if Ca is the divisor class containing a then again f ∗(Ca) = 0. 1 Divisors 155 1.4 Divisors and Rational Maps Associating divisors with functions is useful for studying rational maps of varieties to projective space. Let X be a nonsingular variety and ϕ : X → Pn a rational map. We determine the points of X at which ϕ is not regular. A rational map is deﬁned by the formulas ϕ = (f0 : · · · : fn), with fi ∈ k(X), (3.6) and we can assume that none of the fi is identically 0 on X. Suppose that div(fi) = m j =1 kij Cj, with the Cj prime divisors; here we allow some of the kij to be 0. To determine whether ϕ is regular at a point x ∈ X, write πj for a local equation of Cj at x. Then kij π j fi = ui with ui ∈ Ox and ui(x) = 0. Since Ox is a UFD, there exists a highest common factor d of the elements f0,..., fn, that is, an element d ∈ k(X) such that fi/d ∈ Ox, and if d1 ∈ k(X) is some element for which fi/d1 ∈ Ox then d1 \| d, that is, d/d1 ∈ Ox. Since the local equations πj of prime divisors are prime elements of Ox, we have d = lj j, where lj = min π 0≤i≤n kij. Now ϕ is regular at x if there exists a function g ∈ k(X) such that fi/g ∈ Ox for all i = 0,..., n, and not all the (fi
/g)(x) are 0 at x. By deﬁnition of the highest common factor d it follows that g \| d. If d = gh with h ∈ Ox and h(x) = 0 then h \| (fi/g), and hence all the (fi/g)(x) = 0. Thus the required conditions can only be satisﬁed if d = gh with h(x) = 0. Then fi/g = (fi/d)h, that is fi/g = kij −lj π j uih, j and ϕ is regular at x if and only if not all the functions To translate this answer into the language of divisors, we deﬁne quite generally kij Cj for i = 1,..., n to be the highest common divisor of given divisors Di = the divisor are zero there. kij −lj j π j hcd{D1,..., Dn} = lj Cj, where lj = min 1≤i≤n kij. Obviously D i nents. We set in particular D = hcd{div(f0),..., div(fn)} and D i = Di − hcd{D1,..., Dn} ≥ 0, and the D i have no common compo- = div(fi) − D. 156 3 Divisors and Differential Forms Then in some neighbourhood of x we have div j kij −lj π j = D i, and we can say that ϕ is regular at x if and only if not all the subvarieties Supp(D i) pass through x. We have proved the following result. Theorem 3.2 The rational map (3.6) fails to be regular precisely at the points of = div(fi) − hcd{div(f0),..., div(fn)} ≥ 0 for i = 0,..., n. Supp(D i), where D i Since the D i have no common irreducible components, i) is a subvariety of codimension ≥2. Thus Theorem 3.2 is a more precise version of Theorem 2.12. Supp(D Remark The divisors D Supp D xi = 0 under the map ϕ : X → Pn. Indeed
, if x /∈ neighbourhood U of x, then in U the regular map is deﬁned by i can be interpreted as the pullbacks of the hyperplanes i and D = div h in a ϕ =. f0 h : · · · : fn h The pullback of the hyperplane xi has local equation fi/ h, that is, it is equal to D i. λixi = 0, More generally, if λ = (λ0 : · · · : λn) and Eλ ⊂ Pn is the hyperplane then ϕ∗(Eλ) = div λifi − D. 1.5 The Linear System of a Divisor The fact that all the polynomials f (t) of degree ≤n form a ﬁnite dimensional vector space has the following interpretation in terms of divisors. Write x∞ for the point at inﬁnity on the projective line P1 with coordinate t. A polynomial in t of degree k has pole of order k at x∞, and no other poles. Hence the condition deg f ≤ n can be expressed as the statement that div f + nx∞ is effective. In the same way, for an arbitrary divisor D on a nonsingular variety X, we consider the set consisting of 0 together with the nonzero functions f ∈ k(X) such that div f + D ≥ 0. (3.7) This set is a vector space over k under the usual algebraic operations on functions. Indeed, if D = niCi then (3.7) is equivalent to vCi (f ) ≥ −ni and vC(f ) ≥ 0 for C = Ci, and because of this, our assertion follows at once from (3.2). 1 Divisors 157 The space of functions satisfying (3.7) is called the associated vector space of D, or the Riemann–Roch space of D, and denoted by L(D) or L(X, D). The analogue of the ﬁnite dimensionality of the vector space of polynomials of degree ≤n is the fact that L(D) is ﬁnite dimensional if X is a projective variety and D any divisor. We prove this theorem in Theorem 3.9 for the case of algebraic curves. The proof
in the general case can be deduced from this without special difﬁculty using induction of the dimension. However, the status of the theorem becomes clearer if it is obtained as a particular case of a much more general assertion on coherent sheaves; we prove it in this form in Corollary 6.1 of Section 3.4, Chapter 6. The dimension of L(D) is also called the dimension of D, and denoted by (D). Theorem 3.3 Linearly equivalent divisors have the same dimension. Proof Suppose that D1 ∼ D2. This means that D1 − D2 = div g, with g ∈ k(X). If f ∈ L(D1) then div f + D1 ≥ 0. It follows that div(fg) + D2 = div f + D1 ≥ 0, that is, fg ∈ L(D2), so that g · L(D1) = L(D2). Thus multiplying functions f ∈ L(D1) by g deﬁnes an isomorphism of the vector spaces L(D1) and L(D2), and the theorem follows. Thus we see that it makes sense to speak of the dimension (C) of a divisor class C, that is, the common dimension of all the divisors of this class. This number has the following meaning. If D ∈ C and f ∈ L(D) then the divisor Df = div f + D is effective. Obviously, since Df ∼ D also Df ∈ C. Conversely, any effective divisor D ∈ C is of the form Df, for f ∈ L(D). Obviously, if X is projective, f is uniquely determined by Df up to a constant factor. Thus we can set up a one-to-one correspondence between effective divisors in the class C and points of the ((C) − 1)-dimensional projective space P(L(D)) corresponding to a divisor D (recall that the projective space P(L) of a vector space L consists of all the 1dimensional vector subspaces of L). The space L(D) is useful when specifying rational maps in terms of divisors, as described in Section 1.4. If ϕ = (f0 : · · · : fn) : X → Pn (3.
8) is a rational map, and, in the notation of Section 1.4, D = hcd div(f0),..., div(fn) with Di = div(fi) − D, (3.9) then Di ≥ 0 and hence all the fi ∈ L(−D). The choice of the functions fi depended on the choice of the projective coordinate system in Pn. Thus in an invariant way, ϕ corresponds to the set of all functions n i=0 λifi that are linear combinations of the functions fi. These functions form a vector subspace M ⊂ L(−D). From now on we assume that ϕ(X) is not contained λifi = 0 on X, provided that not all the in any proper linear subspace of Pn. Then λi = 0. The set of effective divisors that correspond to these functions, that is, the divisors div g − D with g ∈ M, is called a linear system of divisors. If M = L(−D) 158 3 Divisors and Differential Forms then we have a complete linear system. The meaning of the divisors div f − D for f ∈ M is very simple: they are the pullbacks of the hyperplane divisors of Pn under ϕ. In this way we can construct all rational maps of a given nonsingular variety X into different projective spaces. For this, we need to take an arbitrary divisor D, and a ﬁnite dimensional subspace M ⊂ L(−D). If f0,..., fn is a basis of M then (3.8) gives the required map. Note that the divisors Di have an additional property: they have no common components. Since multiplying all the fi through by a common factor g ∈ k(X) does not change the map ϕ, and replaces the divisor D by the linearly equivalent divisor div g + D, the class of the divisor D is an invariant of a rational map. Thus we have the following method of constructing all rational maps ϕ : X → Pm such that ϕ(X) is not contained in any proper linear subspace of Pm: take an arbitrary divisor class on X, and for any divisor D in this class, a ﬁnite dimensional vector sub
space M ⊂ L(−D) such that the effective divisors div f − D for f ∈ M have no common components. If f0,..., fn is a basis of M then our map is given by (3.8). Of course, it can happen that L(−D) = 0, or that all the divisors div f − D for f ∈ L(−D) have common components, and then this divisor class does not lead to any map. We observe one interesting feature of the picture we obtain. Among all the rational maps corresponding to a divisor class C, there is a maximal one: that obtained by taking M to be the whole space M = L(−D) with D ∈ C. (Here we take on trust the so far unproved theorem that L(−D) is ﬁnite dimensional.) All other maps corresponding to this class are obtained by composing this map X → PN with the various projection maps PN → Pn. Indeed, if ϕ = (f0 : · · · : fN ), and, say, ψ = (f0 : · · · : fn) with n < N then ψ = π ◦ ϕ, where π(x0 : · · · : xN ) = (x0 : · · · : xn) is the projection, viewed as a rational map. Let’s see how this scheme of things works if we take X to be projective space Pm. We know that Cl(Pm) ∼= Z, and the class Ck corresponding to an integer k consists of hypersurfaces of degree k. If k > 0 and D ∈ Ck then obviously L(−D) = 0. If k ≤ 0 then we set k = −k ≥ 0, and we can take −D to be the divisor kE, where E is the divisor of the hyperplane at inﬁnity x0 = 0. Then L(kE) consists of polynomials of degree ≤k in the inhomogeneous coordinates x1/x0,..., xm/x0 (see Exercise 15). If we multiply the formula for the resulting map through by xk 0 we get the Veronese − 1 (see Example 1.28). Thus embedding vk : Pm → PN
where N = νk,m = we see that any rational map from Pm is obtained by composing the Veronese map with a projection. k+m m Example Suppose that X ⊂ Pn+1 is an irreducible n-dimensional hypersurface deﬁned by F = 0, with deg F = k. We ﬁnd the space L(D), where D = div H, with H a form of degree m. Since div H ∼ mE, where E = div(x0) is the hyperplane section divisor, we can assume that D = mE. Obviously if Φ is any form of degree ∈ L(mE). We prove that these functions exhaust L(mE). m then Φ/xm 0 If ϕ ∈ L(mE) then ϕ ∈ k[U0], where U0 ⊂ X is the afﬁne open set given by x0 = 0. Let yi = xi/x0 for i = 1,..., n + 1 be inhomogeneous coordinates. We see that ϕ = P (y1,..., yn+1), where P is a polynomial, which can be altered by 0 of the hypersurface U0 ⊂ adding multiples of the deﬁning equation F0 = F /xk 1 Divisors 159 An+1. Our claim is that after adding such a multiple we get a polynomial P of degree deg P ≤ m. By contradiction, suppose that deg P = l > m, and that the degree of P cannot be reduced by adding a multiple of F0. We choose the coordinate system in such a way that the intersection of X with x0 = x1 = 0 has dimension n − 2. This means that if fk is the homogeneous component of F0(y1,..., yn+1) of top degree then fk is not divisible by y1. We pass to the open subset U1 ⊂ X with x1 = 0, and set z1 = x0/x1 = 1/y1 and zi = xi/x1 = yi/y1 for i > 1. Then y1 = 1/z1, yi = zi/z1 for i > 1, and P (z1,.
.., zn+1), where P is a polynomial of degree −l ϕ = P (y1,..., yn+1) = z 1 P = l. By assumption, m div z1 + div ϕ > 0 in U1, that is, zm Q(z1,..., zn+1) on U1, where Q is a polynomial. Let deg Q = r. By assumption, zm−l 1 is the equation of U1, and A is a rational 1 function whose denominator does not have F1 as a factor. Returning to U0, we get P = Q + AF1, where F1 = F /xk 1 ϕ ∈ k[U1], or zm−l 1 y −m 1 P = y −r 1 Q + BF0, (3.10) Q, where now C is a polynomial. Since deg(ym−r where Q(y1,..., yn+1) is a polynomial of degree r, and the denominator of B does not have F0 as a factor. If m ≥ r then multiplying (3.10) by ym 1 gives P − CF0 = Q) = m < l, this contradicts ym−r 1 the assumption that the degree of P cannot be reduced. If r ≥ m then similarly, we P − CF0 = Q. Write pl, qr, fk and c for the homogeneous components of get yr− > deg Q, we have top degree in P, Q, F0 and C. Since deg(yr−m 1 yr−m pl = cfk. By the choice of coordinates, fk is not divisible by y1, and hence 1 pl is divisible by fk, say pl = gl−kfk. Then deg(P − gl−kF0) < l, which again contradicts the assumption on P. 1 This proves the following result. Proposition Let X ⊂ Pn+1 be an irreducible hypersurface deﬁned by F = 0, with deg F = k. Then L(X, mE) is the vector space of forms of degree m, modulo the n+m+1 subspace of multiples of F by forms of degree m − k. Therefore (mE) = m if m < k or
if m ≥ k. − n+m−k+1 m−k n+m+1 m 1.6 Pencil of Conics over P1 We conclude this section with an example that is very pretty, and will be useful later. Let X be a nonsingular projective surface and ϕ : X → P1 a regular map. Suppose that the point ∞ ∈ P1 is chosen so that the inverse image ϕ−1(∞) is nonsingular, P1 \ ∞ = A1, and the map ϕ−1(A1) → A1 deﬁnes a pencil of conics as in Example 1.35 and Example 2.7. In this situation, X, together with its map ϕ : X → P1, is called a pencil of conics over P1. The open set ϕ−1(A1) is deﬁned in P2 × A1 by 160 the equation 3 Divisors and Differential Forms 2 i,j =0 aij (t)ξiξj = 0, (3.11) where t is a coordinate on A1. In Proposition 2.1, we saw that the singular ﬁbres of ϕ correspond to the roots t = α1,..., αm of the discriminant Δ(t) = det \|aij (t)\|, that these roots are simple and that the corresponding singular ﬁbres F1,..., Fm are of the form Fi = Li + L i, where Li and L i are distinct lines. Since Δ(t) has only simple roots, it is not identically 0, and the conic (3.11) is nondegenerate. In Proposition of Section 6.2, Chapter 1 and Corollary 1.6, we saw that ϕ has a section s : A1 → ϕ−1(A1), a regular map such that s(α) is contained in the ﬁbres ϕ−1(α) for each α ∈ A1, that is, ϕ ◦ s = id. This map extends from A1 to P1, and gives a regular map s : P1 → X. Write S for the curve s(P1). We choose some ﬁxed nonsingular ﬁbre F. Theorem 3.4 The
divisor class group Cl X is a free Abelian group with m + 2 generators, the classes deﬁned by L1,..., Lm, F and S. Proof Let C be a prime divisor on X. Then C ⊂ X is an irreducible curve, and ϕ either maps it to a point γ ∈ P1 or onto the whole of P1. In the ﬁrst case, C is contained in a ﬁbre ϕ−1(γ ). Suppose that ϕ(C) = P1. Then the map ϕ : C → P1 deﬁnes an inclusion k(P1) ⊂ k(C) of the function ﬁelds, and a nonzero function u ∈ k(P1) does not vanish on C; here we identify u and its pullback ϕ∗(u) ∈ k(X). In other words, vC(u) = 0 for any 0 = u ∈ k P1. (3.12) Hence vC deﬁnes a function v : (k(X) \ 0) → Z that satisﬁes (3.12) and is a valuation in the sense of (3.2). Proposition of Section 6.2, Chapter 1 and Corollary 1.6 prove that the conic (3.11) is rational over the ﬁeld K = k(P1) = k(t), that is, k(X) = K(T ); the birational map X → P1 K uses the point of the conic (3.11) corresponding to a section s, and in particular, it can be chosen so that T has a pole of order 1 at this point. Thus v is a function on K(T ) \ 0 that satisﬁes (3.12) and (3.2). It is easy to determine all such functions. Suppose that v(T ) ≥ 0. Then it follows from (3.12) that v(H ) ≥ 0 for every H ∈ K[T ], and if v is not identically 0 then v(H ) > 0 for some H. Therefore v(P ) > 0 for some irreducible factor P of H. But then v(Q) = 0 for every irreducible po
lynomial in T not proportional to P : indeed, there exist polynomials U, V ∈ K[T ] such that P U + QV = 1; so if v(P ), v(Q) > 0 it would follow that v(1) > 0, whereas v(u) = 0 for u ∈ K. It follows that v(f ) = vP (f ) = m is the exponent of f when written in the form f = P mg, where P divides neither the numerator nor the denominator of g. In particular, for the divisor C we are considering, there exists an irreducible polynomial P ∈ k[T ] such that vP (f ) = vC(f ), and the divisor C is uniquely determined by P. 1 Divisors 161 Hence vC(P ) = 1, and since P determines C uniquely, div0 P does not contain any irreducible curve except for components of ﬁbres: div0 P = C + Gi, (3.13) where Gi are conics of the pencil or their components. If v(T ) < 0 then we set U = T −1 and ﬁnd that v corresponds in the same way to the polynomial U ∈ k[U ] ⊂ k(T ). In terms of K[T ], as we see easily, v(F ) = − deg H, where H ∈ K[T ], so that there is only one such function v. Since by assumption T has a pole at the point corresponding to the section s, we must have v = vS. As before, vS(H ) = − deg H, and S is the unique curve with this property, so that for any H ∈ K[T ], we have div∞ H = (deg H )S + G j, (3.14) where G i are conics of the pencil or their components. In particular, if P is the irreducible polynomial corresponding to a curve C = S, we have div∞ P = (deg P )S + G j and div P = C − (deg P )S + G j. Gi − rlG l, where G Hence C ∼ (deg P )S + l are components of conics of the pencil. It remains to consider these. They can either be nondegenerate con
ics, that is, ﬁbres ϕ∗(α) with α ∈ P1. But since all points of P1 are linearly equivalent, all ﬁbres of X → P1 are also linearly equivalent, therefore linearly equivalent to the chosen ﬁbre F, say. Or they can be components Li or L i of reducible ﬁbres. But since Li + L i in terms of Li and F. As a result, we see that i every irreducible divisor is linearly equivalent to a linear combination of S, F and L1,..., Lm. Hence the divisor classes of these curves generate Cl X. ∼ Fi ∼ F, we can express L It remains to check that the classes of S, F and L1,..., Lm are linearly indepenm dent in Cl X. Suppose that nF + lS + i=1 riLi ∼ 0. We consider the restriction of this divisor to various nonsingular curves. It must again be linearly equivalent to 0. Consider the restriction to an irreducible ﬁbre F = F. Since F ∩ F = ∅, Li ∩ F = ∅ and the restriction to S gives a point ξ, we must have lξ ∼ 0. This is only possible if l = 0. Considering the restriction to L i we get that ri = 0. The relation that remains is nF ∼ 0. If n = 0 then we can assume that n > 0. This is impossible: an effective divisor cannot be principal. The theorem is proved. 1.7 Exercises to Section 1 1 Determine the divisor of x/y on the quadric surface xy − zt = 0 in P3. 2 Determine the divisor of the function x − 1 on the circle x2 1 x = x1/x0. + x2 2 = x2 0, where 162 3 Divisors and Differential Forms 3 Determine the pullback f ∗(Da) where f (x, y) = x is the projection of the circle x2 + y2 = 1 to the x-axis, and Da = a is the divisor on A1 consisting of the point with coordinate a with multiplicity 1. 4 Let X be a nonsingular projective curve and
f ∈ k[X]. Viewing f as a regular function f : X → P1, prove that div f = f ∗(D), where D is the divisor D = 0 − ∞ on P1. 5 Let X be a nonsingular afﬁne variety. Prove that Cl X = 0 if and only if the coordinate ring k[X] is a UFD. 6 Suppose that X ⊂ PN is a nonsingular projective variety. Let k[S] be the polynomial ring in the homogeneous coordinates of PN and AX ⊂ k[X] the ideal of X. Prove that if k[S]/AX is a UFD then Cl X ∼= Z, and is generated by the class of a hyperplane section. 7 Find Cl(Pn × Am). 8 The projection p : X × A1 → X deﬁnes a pullback homomorphism p∗ : Cl X → Cl(X × A1). Prove that p∗ is surjective. [Hint: Use the map q∗ : Cl(X × A1) → Cl X, where q : X → X × A1 is given by q(x) = (x, 0).] 9 Prove that for any divisor D on X × A1 there exists an open set U ⊂ X such that D is a principal divisor on U × A1. [Hint: You can suppose that X is afﬁne, and that D is irreducible. Then it is deﬁned by a prime ideal of k[X × A1] = k[X][T ]. Use the fact that every ideal in k(X)[T ] is principal, and then replace X by some principal afﬁne open set.] 10 Prove that Cl(X × A1) ∼= Cl X. [Hint: Use the results of Exercises 8–9.] 11 Let X be the projective curve deﬁned in afﬁne coordinates by y2 = x2 + x3. Prove that every locally principal divisor on X is equivalent to a divisor whose support does not contain the points (0, 0). Using this, together with the normalisation map ϕ : P1 → X, for which ϕ−1(0
, 0) consists of two points x1, x2 ∈ P1, describe Pic X as D/P, where D is the group of all divisors on P1 whose support does not contain x1, x2, and P the group of principal divisors div f such that f is regular at x1, x2 and f (x1) = f (x2) = 0. Prove that Pic X is isomorphic to Z × k∗, where k∗ is the multiplicative group of nonzero elements of k. 12 Determine Pic X where X is the projective curve y2 = x3. 13 Let X be a quadratic cone. Using the map ϕ : A2 → X described in Exercise 2 of Section 5.5, Chapter 2, determine the image ϕ∗(Div X) ⊂ Div(A2). Prove that the principal divisor D = div F ∈ Div(A2) is contained in ϕ∗(Div X) if and only F (−u, −v) = ±F (u, v), that is, F is either an odd or an even function. Prove that the principal divisors on X correspond to even functions. Deduce that Cl X ∼= Z/2Z. 2 Divisors on Curves 163 14 Using Theorem 3.2, determine the points at which the birational map ϕ : X → P2 is not regular, where X is a surface of degree 2 in P3 and ϕ the projection from a point. The same for ϕ−1. 15 Prove that if E is the hyperplane x0 = 0 in Pn then the space L(kE) consists of polynomials of degree ≤k in the inhomogeneous coordinates x1/x0,..., xn/x0. ].] [Hint: f ∈ L(kE) implies that f ∈ k[An 0 16 Prove that any automorphism of Pn takes hyperplane divisors to one another. [Hint: The class of a hyperplane is determined in Cl(Pn) by intrinsic properties, and the hyperplane divisors are determined as the effective divisor in this class.] 17 Prove that any automorphism of Pn is a projective transformation. [Hint: Use the result of Exercise 16.] 18 Suppose
that Y is nonsingular, and let σ : X → Y be a blowup with centre y ∈ Y. Prove that Cl X ∼= Cl Y ⊕ Z. 2 Divisors on Curves 2.1 The Degree of a Divisor on a Curve ki. Consider a nonsingular projective curve X. A divisor on X is a linear combination kixi of points xi with coefﬁcients ki ∈ Z. The degree of D is the number D = deg D = The case n = 1 of Example 3.2 shows that when X = P1, a divisor D is principal if and only if it has degree 0. We prove that the equality deg D = 0 holds for a principal divisor on any nonsingular projective curve. For this we use the notion of the degree deg f of a map f introduced in Section 6.3, Chapter 2. Theorem 3.5 If f : X → Y is a regular map between nonsingular projective curves and f (X) = Y then deg f = deg(f ∗(y)) for any point y ∈ Y. In Theorem 3.5, f ∗(y) is the divisor on X obtained as the pullback of the divisor on Y consisting of y with multiplicity 1. Thus deg f equals the number of inverse images of any point y ∈ Y, taken with the right multiplicities. This makes the intuitive meaning of the degree of a map f easier to understand: it counts how many times X covers Y under the map f. Corollary The degree of a principal divisor on a nonsingular projective curve equals 0. 164 3 Divisors and Differential Forms Proof Indeed, any nonconstant function f ∈ k(X) deﬁnes a regular map f : X → P1. Moreover, we have f ∗(0) = div0 f, where the left-hand side is the pullback of the point 0 ∈ P1, as follows at once from the deﬁnition of the two divisors. Similarly, f ∗(∞) = div∞ f. By Theorem 3.5, deg(div f ) = deg(div0 f ) − deg(div∞ f ) f ∗(∞) f ∗(0) = deg
− deg = deg f − deg f = 0. If X and Y are varieties of the same dimension then a regular map f : X → Y with f (X) dense in Y deﬁnes an inclusion f ∗ : k(Y ) → k(X). We use this in what follows to view k(Y ) as a subﬁeld of k(X). (That is, for u ∈ k(Y ) we write u instead of f ∗(u) when this does not cause confusion.) Theorem 3.5 follows from two results. To state these, we introduce the following notation. Let x1,..., xr ∈ X be points of X, and set O = r i=1 Oxi. (3.15) Thus O consists of functions that are regular at all the points x1,..., xr. If {x1,..., xr } = f −1(y) for y ∈ Y then the ring Oy, viewed as a subring of k(X) according to the convention just explained, is contained in O. Theorem 3.6 O is a principal ideal domain with a ﬁnite number of prime ideals. There exist elements ti ∈ O such that vxi (tj ) = δij for 1 ≤ i, j ≤ r (Kronecker delta). (3.16) If u ∈ O and u = 0 then u = t k1 1 where ki = vxi (u) and v is invertible in O. · · · t kr r v, (3.17) Theorem 3.7 If {x1,..., xr } = f −1(y) then O is a free Oy -module of rank n = deg f, that is, O ∼= O⊕n y. Proof of Theorem 3.6 + Theorem 3.7 =⇒ Theorem 3.5 Let t be a local parameter on Y at y, and {x1,..., xr } = f −1(y). By Theorem 3.6, t = t k1 r v, where ki = 1 vxi (t) and v is invertible in O. Recalling the deﬁnition of the pullback of a divisor, we see that · · · t
kr f ∗(y) = r i=1 kixi and deg f ∗(y) = r i=1 ki. 2 Divisors on Curves 165 Since t1,..., tr are pairwise relatively prime in O, it follows that O/(t) ∼= O/ t ki i. r! i=1 Fix attention on one of the summands O/(t ki that any element w ∈ O can be written in a unique way in the form i ) in the direct sum. One sees easily w ≡ α0 + α1ti + · · · + αki −1t ki −1 with αi ∈ k. Indeed, if we already have an expression w ≡ α0 + α1ti + · · · + αs−1t s−1 i i mod t ki i, (3.18) mod t s i, then u = t −s i w − α0 − · · · − αs−1t s−1 i ∈ O ⊂ Oxi. Set u(xi) = αs ∈ k. Then vxi (u − αs) > 0, and it follows from Theorem 3.6 that u ≡ αs modulo ti, that is, w ≡ α0 + α1ti + · · · + αs−1t s−1 i + αst s i mod t s+1 i. This proves (3.18) by induction. It follows from (3.18) that dim O/(t ki i ) = ki. Hence dim O/(t) = r i=1 ki. (3.19) Now apply Theorem 3.7. It follows from this that O/(t) ∼= (Oy/(t))⊕n. But t is a local parameter at y, and hence Oy/(t) ∼= k dim O/(t) = n = deg f. (3.20) and The equalities (3.19) and (3.20) prove Theorem 3.5. Proof of Theorem 3.6 Write ui for a local parameter at xi. Then xi appears in the divisor div(ui) with multiplicity 1, that is, div(ui) = xi + D, where xi does not appear in D. By Theorem 3.1 we can move the support of D away from x1,
..., xr, that is, we can ﬁnd a function fi such that none of these points appear in D +div(fi). This means that the relations (3.16) are satisﬁed by ti = uifi. Let u ∈ O. Set vxi (u) = ki. By assumption, ki ≥ 0. The element v = ut −kr r satisﬁes vxi (v) = 0 for all i = 1,..., r, from which it follows that v ∈ O and v−1 ∈ O. This gives an expression (3.17) for u. It remains to check that O is a principal ideal ring. Let a be an ideal of O. Set r. Then ua−1 ∈ O for any u ∈ a, that is, a ⊂ (a). ki = infu∈a vxi (u) and a = t k1 1 We prove that a = (a). For this we denote by a the set of functions ua−1 with u ∈ a. · · · t kr −k1 1 · · · t 166 3 Divisors and Differential Forms Obviously a is an ideal of O and also infu∈a vxi (u) = 0. Hence for any i = 1,..., r, there exists ui ∈ a such that vxi (ui) = 0, that is ui(xi) = 0. An obvious veriﬁcation j =1 uj t1 · · ·tj · · · tr ∈ a satisﬁes vxi (c) = 0 for i = shows that the element c = 1,..., r. This means that c−1 ∈ O, and hence a = O, and a = (a). Theorem 3.6 is proved. r We proceed to the proof of Theorem 3.7. We ﬁrst prove that O is a ﬁnite Oy module. For this, recall that by Theorem 2.24, the map f is ﬁnite. Therefore the result we need follows from the next lemma. Lemma Let f : X → Y be a ﬁnite map of curves, with X nonsingular; for y ∈ Y, write f −1(y)
= {x1,..., xr } and O = Oxi. Then O is a ﬁnite Oy -module. Proof Since the assertion is local, we can assume that X and Y are afﬁne. Let k[X] = A and k[Y ] = B. Then B ⊂ A and A is a ﬁnite B-module. We prove that O = AOy. Indeed, if ϕ ∈ O and zi are the poles of ϕ on U then f (zi) = yi = y. There exists a function h ∈ B such that h(y) = 0 and h(yi) = 0, and moreover ϕh ∈ Ozi and hence ϕh ∈ A. Since h−1 ∈ Oy, we get ϕ ∈ AOy ; this proves that O ⊂ AOy. The converse inclusion is obvious. Obviously, generators of A over B = k[Y ] provide at the same time generators of AOy over Oy. Hence O is a ﬁnite Oy -module. The lemma is proved. Proof of Theorem 3.7 Now it is easy to complete the proof of Theorem 3.7. By the main theorem on modules over a principal ideal domain, O is the direct sum of a free module and a torsion module. However, both Oy and O are contained in the ﬁeld k(X), so that it follows that the torsion module is 0, and O ∼= O⊕m for some m. It remains to determine m, that is, the rank of O over Oy. It equals the maximal number of elements of O that are linearly independent over Oy. Since linear independence over a ring and over its ﬁeld of fractions is the same thing, and the ﬁeld of fractions of Oy is k(Y ), our m equals the maximal number of elements of O that are linearly independent over k(Y ). y By assumption [k(X) : k(Y )] = n, so that obviously m ≤ n. It remains to prove that O contains n elements that are linearly independent over k(Y ). Suppose that α1,..., αn is a basis of the ﬁeld extension k(Y ) ⊂ k(X). Let t be a
local parameter on Y at y, and write k for the maximum order of poles of the αi at the points xj. Then obviously the functions αit k are regular at these points, and are hence contained in O. Hence they are linearly independent over k[Y ]. Theorem 3.7 is proved. It follows from Corollary after Theorem 3.5 that on a nonsingular projective curve X, linearly equivalent divisors have the same degree. Hence it makes sense to talk about the degree of a divisor class. Thus we have a homomorphism deg : Cl X → Z, 2 Divisors on Curves 167 whose image is the whole of Z, and whose kernel consists of divisor classes of degree 0, and is denoted by Cl0 X. The role of this group is clear already from the following result. Theorem 3.8 A nonsingular projective curve X is rational if and only if Cl0 X = 0. Proof Indeed, if X ∼= P1 then we are in the case n = 1 of Example 3.2. We saw there that Cl(P1) = Z, and hence Cl0 X = 0. Conversely, suppose that Cl0 X = 0. This means that any divisor of degree 0 is principal. In particular, if x = y ∈ X are two points then there exists a function f ∈ k(X) such that x − y = div f. Viewing f as a map f : X → P1 we get from Theorem 3.5 that k(X) = k(f ), that is, f is birational. Since X and P1 are nonsin- gular projective curves, it follows that f is an isomorphism. 2.2 Bézout’s Theorem on a Curve We now indicate the simplest applications of the theorem on the degree of a principal divisor. They are very special cases of more general theorems, that we will prove in connection with the theory of intersection numbers in Chapter 4. However, it is convenient to treat these simple cases already at this stage, since we will ﬁnd them useful in Section 2.3. Suppose that X ⊂ Pn is a nonsingular projective curve and x ∈ X a point. Let F be a form in the coordinates of Pn, not identically 0 on X; we write Pn F for the
hypersurface deﬁned by F = 0. We introduced in Section 1.2 the divisor div F of F on X. Its degree deg(div F ) is also denoted by XF, and is called the intersection number of X and the hypersurface Pn F. An important corollary follows at once from Theorem 3.5: this number deg(div F ) is the same for all forms of the same degree. Indeed, if deg F = deg F1 then f = F /F1 ∈ k(X). From the deﬁnition of the divisor div F it follows at once that div F = div F1 + div f, and hence div F ∼ div F1. By Corollary after Theorem 3.5, deg(div F ) = deg(div F1). To determine how the number XF depends on the degree of F, it is enough to take F to be any form of degree m = deg F. In particular we can take F = Lm where L is a linear form. Then XF = mXL = (deg F )XL. (3.21) Finally, we explain the meaning of XL. Deﬁnition The degree of a curve X ⊂ PN, denoted by deg X, is the maximum number of points of intersection of X with a hyperplane not containing any component of X. 168 3 Divisors and Differential Forms Since XL = L(x)=0 vx(div L), we have deg X ≤ XL. Here we use the notation vx(D) for the multiplicity of a divisor D at x, that is, the coefﬁcient ki in the expression D = kixi. For any form F, we now determine when vx(div F ) = 1. Since the function vx is additive, it is enough to consider an irreducible form. Lemma Let X ⊂ Pn be a curve, F an irreducible form and Y = Pn ⊂ Pn the F hypersurface given by F = 0. Then vx(div F ) = 1 is equivalent to F (x) = 0 and ΘY,x ⊃ ΘX,x. Here we view both these spaces as vector subspaces of ΘPn,x. x, or equivalently, dxf
= 0. But dxf ∈ Θ ∗ Proof We obtain a proof by putting together a number of deﬁnitions from Chapter 2. Let G be a form such that G(x) = 0 and deg G = deg F. By deﬁnition, vx(div F ) = vx(f ), where f = (F /G)\|X. We know that vx(div f ) > 1 is equivalent to f ∈ m2 X,x, and is the restriction to ΘX,x of the differential dx(F /G) of the function F /G, which is a rational function on Pn, regular at x. Thus vx(div F ) > 1 is equivalent to dx(F /G) = 0 on ΘX,x. Furthermore, F /G is a local equation of the hypersurface Y in the neighbourhood of x given by G = 0. Hence dx(F /G) = 0 is the equation of ΘY,x, and dx(F /G) = 0 on ΘX,x if and only if ΘY,x ⊃ ΘX,x. The lemma is proved. We apply this to compute the intersection number XL. Since XL is the same for all linear forms L, the number of points x ∈ X with L(x) = 0 is a maximum when all the vx(L) = 1. By the lemma, this is equivalent to saying that the hyperplane L is not tangent to X at any point. Taking L to be such a linear form, we get deg X = XL. (3.22) We need only verify that linear forms with the required property actually exist. This is easy using the dimension counting argument that we have used many times: in the product X × Pn∗ (where Pn∗ is the dual projective space of hyperplanes of Pn), consider the set Γ of points (x, L) such that L is tangent to X at x. A standard application of the theorem on the dimension of ﬁbres of a map then shows that the image of Γ under the projection X × Pn∗ → Pn∗ has codimension ≥1. Putting together (3.21) and (3.22) we get the relation XF = (deg F )(deg X), (3.23
) which is called Bézout’s theorem. Thus we have ﬁnally proved this theorem, already stated in Section 1.6, Chapter 1. 2.3 The Dimension of a Divisor In Section 1.5, we associated with a divisor D on a nonsingular variety X a vector space L(D). 2 Divisors on Curves 169 Theorem 3.9 L(D) is ﬁnite dimensional for any effective divisor D on a nonsingular projective algebraic curve. Proof First of all, the assertion reduces easily to the case D ≥ 0. Indeed, let D = D1 − D2 with D1, D2 ≥ 0. Then L(D) ⊂ L(D1): indeed, f ∈ L(D) means that div f + D1 − D2 = D ≥ 0, and hence div f + D1 = D + D2 ≥ 0, that is, f ∈ L(D1). The required reduction follows from this. Let D ≥ 0 and let x be a point appearing in D with multiplicity r > 0, that is D = rx + D1. Set (r − 1)x + D1 = D, and let t be a local parameter on X at x. For a function f ∈ L(D), set λ(f ) = (t r f )(x). Then λ : L(D) → k is obviously a linear function, with kernel equal to L(D). Carrying out the same construction deg D times, we see that L(0) is a vector subspace of L(D) deﬁned by the vanishing of deg D linear forms. But we know that L(0) = k by Section 1.1 (just before Example 3.1). It follows from this that L(D) is ﬁnite dimensional, and in fact The theorem is proved. (D) ≤ deg D + 1. (3.24) Remark 3.1 Equality holds in (3.24) for X = P1. Indeed, in this case any divisor D is linearly equivalent to rx, where x ∈ P1 is the point at inﬁnity. Then L(D) equals the space of polynomials of degree ≤r and (D) = r + 1. Remark 3
.2 If X is not rational then (2.22) can be improved. Namely, in this case, for any point x ∈ X we have L(x) = k. Indeed, if L(x) contains a nonconstant function, then we would have div∞ f = x. Then by Corollary after Theorem 2.4, deg(div0 f ) = 1, that is, div f = y − x, which contradicts the irrationality of X (see the proof of Theorem 3.8). Therefore in the process of proving (2.22) we already get to a divisor x after deg D − 1 steps, for which L(x) = 1, and hence (D) ≤ deg D if D > 0. (3.25) Thus rational curves are characterised by the fact that for them (D) = deg D + 1 for D > 0. Remark 3.3 The same argument shows that, quite generally, for divisors D1 and D2, D1 < D2 =⇒ (D2) ≤ (D1) + deg(D2 − D1). (3.26) The inequalities (2.22) and (2.23) are particular cases of this, with D1 = 0 and D1 = x respectively. 2.4 Exercises to Section 2 1 A line l is a double tangent or bitangent to a plane quartic curve X if l and X are tangent at any point of l ∩ X. Prove that the set of quartic curves having a given line 170 3 Divisors and Differential Forms l as a double tangent has codimension 2 in the space of all quartics. Prove that any irreducible quartic curve has a double tangent. 2 For a singular projective curve X, deﬁne the divisor of a form F on the normalisation Xν using the pullback of functions ν∗(F /G) as in Section 1.2, and the intersection number XF as the degree of this divisor on Xν. Prove that Bézout’s theorem continues to hold in this context. 3 Prove that the number of singular points of an irreducible plane curve of degree n is ≤ + 1 singular points, and as many nonsingular ones as possible. Then apply
Bézout’s theorem.]. [Hint: Pass a curve of degree n through n−1 2 n−1 2 4 If X is a nonsingular plane curve and l a line, and the multiplicity of tangency at x ∈ X is r ≥ 2, we say that r − 2 is the inﬂexion multiplicity of X at x. Prove that the sum of the inﬂexion multiplicities of a curve of degree n taken over all inﬂexion points it equal to 3n(n − 2). [Hint: Prove that the multiplicity of ﬂex points at x is equal to the multiplicity of the zero of the Hessian at x (Section 6.2, Chapter 1).] 5 Let X be a nonsingular curve and x1,..., xm ∈ X. Prove that we can take the functions ti in Theorem 2.5 to be the equations of hypersurfaces Ei such that Ei xi, Ei xj, for i = j, and Ei ⊃ ΘX,xi, that is, Ei is not tangent to X at xi. 6 Prove that a curve of degree n in Pn not contained in any hyperplane is rational. 3 The Plane Cubic 3.1 The Class Group We have seen in Theorem 3.8 that Cl0 X = 0 holds for rational curves X, and for them only. We now work out the simplest example for which Cl0 X = 0. This is the nonsingular plane cubic curve X, one of the most beautiful examples in algebraic geometry, with a wealth of unexpected properties. We proved in Section 6.2, Chapter 1 that X always has an inﬂexion point, and hence can be put in Weierstrass normal form. It follows from this, as we have seen in Section 1.6, Chapter 1, that X is irrational. Theorem 3.10 Pick any point α0 of a nonsingular plane cubic curve X, and consider the map X → Cl0 X that sends α ∈ X to the divisor class Cα containing α − α0. Then α → Cα deﬁnes a one-to-one correspondence between points α ∈ X and divisor classes C ∈ Cl0 X. Proof If Cα = Cβ
then α − α0 ∼ β − α0, so that α ∼ β. If α = β, it would follow from the proof of Theorem 3.8 that X is rational, whereas we know that it is not rational. 3 The Plane Cubic 171 It remains to prove that any divisor class C of degree 0 contains a divisor of the form α − α0. Suppose ﬁrst that D is any effective divisor. We show that there exists a point α ∈ X such that D ∼ α + kα0. (3.27) If deg D = 1, then (3.27) holds with k = 0. If deg D > 1 then D = D + β with deg D = deg D − 1 and D > 0. Using induction, we can assume that (3.27) is proved for D, that is, D ∼ γ + lα0. Then D ∼ β + γ + lα0. If we can ﬁnd a point α such that β + γ ∼ α + α0, (3.28) then (3.27) will follow. Suppose ﬁrst that β = γ. Pass the line given by L = 0 through β and γ. By Bézout’s theorem, LX = 3, hence div L = β + γ + δ for some δ ∈ X. (3.29) Suppose moreover that δ = α0 and pass the line given by L1 = 0 through δ and α0. In same way as for (3.29), we get div L1 = δ + α0 + α for some α ∈ X. Since div L ∼ div L1 we get β + γ + δ ∼ δ + α0 + α, and (3.28) follows. We still have to treat the cases with β = γ or δ = α0. If β = γ we pass the tangent to X at β; let L = 0 be the equation. According to 1.2, Lemma vβ (L) ≥ 2, and hence div L = 2β + δ. Thus (3.29) holds in this case. The case δ = α0 is treated similarly. Now suppose that D is any nonzero divisor with deg D = 0. Then D = D1 − D2 with D
1, D2 > 0 and deg D1 = deg D2. Applying (3.27) to both D1 and D2, we get D1 ∼ β + kα0 and D2 ∼ γ + kα0 with the same k, since deg D1 = deg D2. Hence D = D1 − D2 ∼ β − γ, and we need only ﬁnd a point α such that β − γ ∼ α − α0. This relation is equivalent to β + α0 ∼ α + γ, and is the same as (3.28) up to the notation. The theorem is proved. The proof of Theorem 3.10 allows us to determine explicitly the function (D) for divisors D on a nonsingular plane cubic. Theorem 3.11 Let X ⊂ P2 be a nonsingular cubic; then (D) = deg D for every effective divisor D > 0 on X. (3.30) Conversely, a curve for which (3.30) holds is isomorphic to a nonsingular cubic. (Compare also Corollary 3.4.) Proof By Theorem 3.9, Remark 3.2, (D) ≤ deg D for a divisor D > 0 on X, and it is enough to prove that (D) ≥ deg D. In the proof of Theorem 3.10 we proved that D ∼ α + mα0. Thus it is enough to prove that (α + mα0) > m (strict inequality!). If m = 1, then l(α + α0) > 1; because L(α + α0) contains the nonconstant function L1/L0, where L0 is deﬁning equation of the line through α and α0 and L1 any line through the third point of the intersection of L0 and X (see Figure 12, (a)). 172 3 Divisors and Differential Forms Figure 12 Constructing functions on a plane cubic Hence for m > 1 it is sufﬁcient to exhibit a function fm with div∞(f ) = mα0; indeed, then fm ∈ L(mα0) ⊂ L(α + mα0) and fm /∈ L(α + (m − 1)α0), whence (α + mα0) ≥ (α + (
m − 1)α0) + 1, and our assertion is proved by induction. It is easy to ﬁnd fm with this property for m = 2 or 3. Namely, f2 = L1/L0, where L0 is the tangent line to X at α0, and L1 is any line through the third point of the intersection of L0 and X (see Figure 12, (b)). Similarly, f3 = L1L3/L0L2, where L0 and L1 are as before (see Figure 12, (b)), L2 is the deﬁning equation of the line through α0 and one of the other points of intersection of L1 and X, and L3 = 0 a line through the third point of intersection of L2 and X. Finally, if m = 2r is even, then fm = f r 2. This proves the equality (3.30). 2 ; and if m = 2r + 3 is odd and ≥3 then fm = f3f r Conversely, suppose that X is a nonsingular projective curve X such that (3.30) holds for any divisor D > 0. Take any point p ∈ X. Since (2p) > 1 by (3.30), there exists a function x ∈ k(X) with div∞ x = 2p (note that div∞ x = p is impossible, since then the curve would be rational). By (3.30) L(3p) = L(2p), so that there exists a function y ∈ k(X) with dim∞ y = 3p. Finally, by (3.30), (6p) = 6. But we already know 7 functions belonging to L(6p), namely 1, x, x2, x3, y, xy, y2. Hence there must be a linear dependence relation between these a0 + a1x + a2x2 + a3x3 + b0y + b1xy + b2y2 = 0. (3.31) Thus the functions x and y deﬁne a rational, hence regular map from X to the plane cubic Y ⊂ P2 with (3.31) in inhomogeneous coordinates. This is the rational map deﬁned by the linear system L(3p). The map f de�
�nes an inclusion of function ﬁelds f ∗ : k(Y ) → k(X). Let us prove that f ∗(k(Y )) = k(X). For this, remark that k(Y ) ⊃ k(x) and k(Y ) ⊃ k(y), and the functions x and y each deﬁnes a map of X → P1. By assumption div∞ x = 2p, which means that the map g deﬁned by x satisﬁes g∗(∞) = 2p. From Theorem 3.5 it follows that deg g = 2, that is, [k(X) : k(f ∗(x))] = 2. Similarly, 3 The Plane Cubic 173 [k(X) : k(f ∗(y))] = 3. Since [k(X) : f ∗(k(Y ))] has to divide both these numbers, k(X) = f ∗(k(Y )), that is, f is birational. The cubic (3.31) cannot have singular points, since then it, and X together with it, would be a rational curve, which contradicts (3.30). Therefore Y is a nonsingular cubic, and hence f is an isomorphism. The theorem is proved. Thus nonsingular cubic curves in P2 are characterised by (3.30) in exactly the same way that rational curves are characterised by (D) = deg D + 1 for D > 0. 3.2 The Group Law Theorem 3.10 establishes a one-to-one correspondence between the points of a nonsingular cubic curve X ⊂ P2 and the elements of the group Cl0 X, under which a point α ∈ X corresponds to the class Cα of the divisor α − α0, where α0 is the ﬁxed point used to deﬁne the correspondence. Using this, we can transfer the group law from Cl0 X to X itself. The corresponding operation on points of X is called addition, and written ⊕, with subtraction denoted by. By deﬁnition, α ⊕ β = γ if Cα + Cβ = Cγ, that is α + β ∼ γ + α0. (3.32)
α0 is obviously the zero element. From now on we denote it by o, so that (3.32) can be rewritten α + β ∼ (α ⊕ β) + o. (3.33) The proof of Theorem 3.10 allows us to describe the operations ⊕ and in elementary geometric terms. Namely, if the tangent to X at o meets X at π and the line through π and α meets X in a third point α then 2o + π ∼ π + α + α so that α + α ∼ 2o, (3.34) which means that α = α is the inverse of α in the group law (Figure 13, a). If α = π, passing a line through α and π should be replaced by drawing the tangent line to X at α. Similarly, to describe ⊕, pass a line through α and β; let γ be the third point of intersection with X, and γ the third point of intersection of X with the line through o and γ (Figure 13, b). Then, that is, γ = α ⊕ β. (3.35) If α = β or γ = o then passing a secant through α and β, or through γ and o, should be replaced by drawing the tangent line to X at α or γ. 174 3 Divisors and Differential Forms Figure 13 The group law on a plane cubic Figure 14 The inverse map γ → γ1 This description becomes especially simple if we take o to be an inﬂexion point of X, which from now on we always assume. Then the section of X by a line is linearly equivalent to 3o: to see this, take the inﬂexional tangent line to X at o. If γ1 is the third point of intersection of X with the line through γ and o then γ + γ1 + o ∼ 3o, (3.36) that is, γ1 ∼ γ (Figure 14). To describe the operation ⊕, pass a line through α and β. Let γ be the third points of intersection of X, and γ the third point of intersection of X with the line 3 The Plane Cubic through γ and o. Then (Figure 13, α + β ∼ γ + o. 175 (3.37) If α =
β then the secant line through α and β should be replaced by the tangent line to X at α. Another form of relation (3.37) is that α, β, γ are collinear if and only if α ⊕ β ⊕ γ = o. In particular, β lies on the tangent line at α if and only if 2α ⊕ β = o (where 2α = α ⊕ α in the sense of the group law). Finally, also β = α if α is an inﬂexion point; then 3α = o. Thus the inﬂexion points of a cubic are precisely the elements of order 3 in the group law, together with the zero element. A cubic in Weierstrass normal form has an inﬂexion point at inﬁnity. We will assume that the characteristic of k is different from 2 or 3. (This is exclusively for the purpose of simplifying the formulas.) Then the equation of X can be written y2 = x3 + ax + b (3.38) and its point at inﬁnity o is on the lines x = c for all c ∈ k. Hence the minus operation of the group law is particularly simple to write down: (x, y) = (x, −y). (3.39) To write out the operation α ⊕ β, pass a line through α = (x1, y1) and β = (x2, y2): y − y1 = y2 − y1 x2 − x1 (x − x1). (3.40) The three points of intersection of this line with the cubic (3.38) are obtained from the equation that is, y1 + y2 − y1 x2 − x1 (x − x1) 2 = x3 + ax + b, x3 − 2 y2 − y1 x2 − x1 x2 + · · · = 0. We know two of the roots x1 and x2 of this equation. Therefore the third root is given by x3 = 2 y2 − y1 x2 − x1 − x1 − x2. (3.41) The coordinate y3 is given by (3.40), and ﬁnally α ⊕ β = (x3, −y3). 176 3 Divisors
and Differential Forms When α = β, we should take the tangent line to X at (x1, y1). Similar transformations give its third point of intersection as (x2, y2), where x2 = + a)2 (3x2 1 + ax1 + b) 4(x3 1 − 2x1, (3.42) and y2 is obtained by substituting for x2 from (3.42) into the equation of the tangent line. Then 2α = (x2, −y2). A remarkable property of the group law we have constructed is that it is given by rational formulas, that is, it deﬁnes a rational map X × X → X. We can even say more. Theorem 3.12 The maps ϕ : X → X given by ϕ(α) = α and ψ : X × X → X given by ψ(α, β) = α ⊕ β are regular. Proof For ϕ this is obvious from (3.39). Similarly, it follows from (3.40) that ϕ is regular at a point (α, β) provided that α = (x1, y1), β = (x2, y2) and x1 = x2; or in other words, since x2 = x1 implies that y2 = ±y1, provided that α = β and α = β. Now for any point γ ∈ X, consider the reﬂection map sγ, that takes a point α = γ into the third point of intersection of X with the line through α and γ. Obviously sγ (α) = (α ⊕ γ ). It can be seen explicitly from the formulas that this map is rational, hence regular by Theorem 2.12 and Corollary of Section 3.1, Chapter 2. = id, so that sγ is an automorphism. Let us prove that sγ (γ ) is the Moreover, s2 γ third point of intersection of X with the tangent line at γ. For this we apply sγ to the relation α + γ + sγ (α) ∼ 3o. Since sγ is an automorphism of X it obviously preserves linear equivalence of divisors, and moreover, sγ (o) = γ. Hence sγ (α) + sγ (γ ) + α ∼
3( γ ). Substituting in this the expression sγ = (α ⊕ γ ), we get sγ (γ ) = 2( γ ) (multiplication by 2 in the group law of X), and this is the third point of intersection of X with the tangent line at γ. Now we can consider the translation automorphism tγ (α) = α ⊕ γ for α = γ. Obviously tγ is the composite of the two reﬂections tγ = s0 ◦ sγ, from which it follows that if α = γ then tγ (α) = 2α. Finally, for any α, β ∈ X, we have ψ(α, β) = t −1 γ ⊕δψ tγ (α), tδ(β). Hence if ψ is regular at any point (α0, β0), then it is regular at any point (α, β) = (tγ (α0), tδ(β0)), where γ = α ⊕ ( α0) and δ = β ⊕ ( β0). But it is regular where α = β and α = β, and hence is regular everywhere. The theorem is proved. 3 The Plane Cubic 177 The map ψ : X × X → X has a differential at (α, β) ∈ X × X, d(α,β)ψ : Θ(α,β) → Θα⊕β. ∼= Θα ⊕ Θβ, and the linear map from the direct sum is determined Obviously Θ(α,β) by the map on the summands. Finally, the composite map Θα → Θα ⊕ Θβ → Θα⊕β comes from X → X × X → X, where the ﬁrst map is the inclusion γ → (γ, β), and the second the group law ψ. The composite map X → X is simply the translation tβ, and hence the restriction of dψ to Θα equals dtβ and ﬁnally dψ = dtα + dtβ. We have proved the next result. Lemma The differential dψ : Θ(α,β) → Θα⊕β of
the group law ψ : X × X → X is given by dψ = dtα + dtβ. In particular, it is surjective. 3.3 Maps We study regular maps λ : X → X of the cubic to itself. An example is the translation tγ given by tγ (α) = α ⊕ γ. If λ(o) = γ then t γ ◦ λ = λ ﬁxes o. From now on we always assume this, that is, λ(o) = o. We prove in Theorem 3.16 below that then λ is a homomorphism of the group law on X, but we do not use this at present. Just as with any maps to a group, we can add maps, deﬁning λ + μ by (λ + μ)(α) = λ(α) ⊕ μ(α). Obviously all regular maps λ : X → X with λ(o) = o form a group. If λ(X) is not just a point then λ(X) = X. Then the degree deg λ is deﬁned, and is positive; we write n(λ) for deg λ. If λ(X) = o then we set n(λ) = 0. The basic result is the following theorem, which has many applications. Theorem 3.13 There exists a scalar product (λ, μ) on the group of regular maps λ : X → X with λ(o) = o such that (λ, λ) = n(λ). Here by scalar product, we mean that a number (λ, μ) ∈ Q is deﬁned for any elements λ, μ, with the properties (λ, μ) = (μ, λ) and (λ1 + λ2, μ) = (λ1, μ) + (λ2, μ). For any Q-valued function n(λ) with n(λ) > 0 for λ = 0 and n(λ) = 0 for λ = 0, there exists a scalar product (λ, μ) with (λ, λ) = n(λ) if and only if n(λ + μ) + n(λ − μ) = 2 n(λ) + n(μ). (
3.43) This is an elementary and purely algebraic fact (see Proposition A.1). Thus to prove the theorem, it is enough to check the relation (3.43) for n(λ) = deg λ. We write Δ ⊂ X × X for the diagonal subvariety of pairs (α, α) with α ∈ X, and Σ ⊂ X × X for the set of pairs (α, α). Obviously these are both nonsingular irreducible subvarieties isomorphic to X. For Δ, compare Example 1.20; and Σ = 178 3 Divisors and Differential Forms ι(Δ) where ι = (id, ) is the involution (α, β) → (α, β). Consider the regular map ψ : X × X → X given by ψ(x, y) = x ⊕ y, and the divisor ψ ∗(o). By Lemma of Section 3.2, dψ is surjective. It follows from this that ψ ∗(o) = Σ is the prime divisor Σ with multiplicity 1. Indeed, if to is a local parameter on X at o then ψ ∗(to) is a local equation of Σ. Since dψ is surjective, the dual map mo/m2 α, α o is injective. Hence ψ ∗(to) /∈ m2 α, α, and it follows that ψ ∗(o) is nonsingular. In the same way, the map ψ1 : X × X → X deﬁned by ψ1(α, β) = α β differs from ψ by the involution ι, and a similar argument gives ψ ∗ 1 (0) = Δ. Finally, set p1(α, β) = α and p2(α, β) = β. Obviously p∗ 2(o) = X × o. The identity (3.43) follows easily from the next assertion. 1(o) = o × X and p∗ → mα, α/m2 Lemma The linear equivalence Δ + Σ ∼ 2(o × X + X × o) = 2 1(o) + p∗ p∗ 2(o) (3.44) holds on the
surface X × X. Proof To prove this, we exhibit a function f on X × X for which div0 f equals the left-hand side of (3.44) and div∞ f the right-hand side. We assume that an afﬁne piece of X is given in the Weierstrass normal form (3.38); then x deﬁnes a map x : X → P1 such that x(α) = x(β) only if α = β or α = β. Moreover, deg x = 2, and since x−1(∞) = {o}, we have div∞ x = x∗(∞) = 2o. The afﬁne product variety X × X is the subset of A4 (with coordinates = f (x2). Inside X × X, Δ is deﬁned = f (x1), y2 x1, y1, x2, y2) deﬁned by y2 1 2 by x1 = x2, y1 = y2, and Σ by x1 = x2, y1 = −y2. The function f we require to establish the linear equivalence (3.44) is f = x1 − x2. The veriﬁcation is an almost tautological calculation. The divisor deﬁned on the afﬁne surface by x1 − x2 is Δ + Σ. In fact (y1 − y2)(y1 + y2) = y2 1 − y2 2 = (x1 − x2)g(x1, x2), where g(x1, x2) = (f (x1)−f (x2))/(x1 −x2). Now (y1 −y2) is invertible outside Δ, so x1 −x2 is a local equation for Σ there, and similarly (y1 +y2) is invertible outside Σ, so x1 − x2 is a local equation for Δ there. Therefore div0(x1 − x2) = Δ + Σ. Now since div∞ x = 2o, it’s clear that on the projective variety X × X we have div∞(x1 − x2) = 2(o × X + X × o), so that �

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