Curves on P 1 P 1. Peter Bruin 16 November 2005


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1 Curves on P 1 P 1 Peter Bruin 16 November Introduction One of the exercises in last semester s Algebraic Geometry course went as follows: Exercise. Let be a field and Z = P 1 P 1. Show that the Picard group PicZ is the free Abelian group generated by the classes of a horizontal and a vertical line. Here PicZ is to be interpreted as the divisor class group ClZ, to which it is naturally isomorphic for Noetherian integral separated locally factorial schemes [Hartshorne, Corollary 6.16]. We view the first P 1 as the result of glueing Spec([x]) and Spec([1/x]) via Spec([x, 1/x]), and similarly for the second P 1 with y instead of x. Then Z = P 1 P 1 is the result of glueing the spectra of [x, y], [x, 1/y], [1/x, y] and [1/x, 1/y] in the obvious way. To prove the claim (see [Hartshorne, Example II.6.6.1] for a different approach), let L x and L y be the vertical and horizontal lines x = and y =. More precisely, L x is determined by the coherent sheaf of ideals I Lx with {Ã for A = [x, y] and A = [x, 1/y] I Lx Spec A = 1/x Ã for A = [1/x, y] and A = [1/x, 1/y], and similarly for L y. If Y is a curve on Z different from L x and L y (curves are assumed to be integral), the intersection of Y with Spec([x, y]) is a plane curve defined by an irreducible polynomial f [x, y]. Let a be the degree of f as a polynomial in x and b is its degree as a polynomial in y; then the divisor of f as a rational function on Z equals so we see that the divisor class of Y is equal to (f) = Y a L x b L y, [Y ] = a[l x ] + b[l y ]. This shows that Cl Z is generated by [L x ] and [L y ]; because there are no rational functions f (x, y) with the property that (f) = a L x + b L y as a divisor on Z unless a = b = 0, the classes [L x ] and [L y ] are linearly independent. If Y is a divisor on Z and a, b are the unique integers with [Y ] = a[l x ] + b[l y ], we say that Y is of type (a, b). The isomorphism Cl Z Pic Z sends the class of a divisor Y of type (a, b) to the invertible sheaf O Z (Y ) = O Z (a L x + b L y ). Note that O Z (a L x ) is isomorphic to the pullbac p1( OP 1 (a )), where the invertible sheaf O P 1 (a ) on P1 is defined by O P 1 (a ) Spec [x] = ([x]) O P 1 (a ) Spec [1/x] = x a ([1/x]). On the other hand, there is the invertible sheaf O P 1(a) with which is clearly isomorphic to O P 1(a ), so O P 1 (a) Spec [x/y] = y a ([x/y]) O P 1 (a) Spec [y/x] = x a ([y/x]), O Z (a L x ) = p 1(O P 1 (a)). Something similar is true for the second projection. Using O Z (a L x + b Ly) = O Z (a L x ) OZ (b L y ) 1
2 we conclude that O Z (Y ) is isomorphic to the invertible sheaf O(a, b) on Z defined by O(a, b) = p 1( OP 1 (a)) OZ p ( 2 OP 1 (b)). The aim of this tal is to study the cohomology of the sheaves O(a, b) and to derive some consequences for the ind of curves that exist on Z. We will do the following: 1. Prove the Künneth formula: if X and Y are Noetherian separated schemes over a field, there is a natural isomorphism H(X Y, p 1 F O X Y p 2 G) = H(X, F) H(Y, G) for all quasicoherent sheaves F on X and G on Y. 2. Deduce a connectedness result for closed subschemes and a genus formula for curves on Z. 3. Prove Bertini s theorem: if X is a nonsingular subvariety of P n with an algebraically closed field, there exists a hyperplane H P n not containing X such that H X is a regular scheme. 4. Deduce that if is algebraically closed field, there exist nonsingular curves of type (a, b) on Z for all a, b > Tensor products of complexes Let A be a ring, (C, d) a complex of right Amodules and (C, d ) a complex of left Amodules, i.e. C and C are graded Amodules C = n Z C n and C = n Z and d, d are Amodule endomorphisms such that dd = 0 and d(c n ) C n+1 (similarly for d ). Let C A C be the usual tensor product, graded in such a way that (C A C ) n = C p A C q. p+q=n There is a group endomorphism D of C A C defined by C n D(x y) = dx y + ( 1) p x d y for x C p ; it fulfills D((C A C ) n ) (C A C ) n+1 and DD = 0, so ((C A C ), D) is a complex of Abelian groups. For any complex (C, d) of Abelian groups, we write Z(C) for the subgroup of cocycles, B(C) for the subgroup of coboundaries and H(C) for the cohomology of C: Z(C) = erd, B(C) = imd, H(C) = Z(C)/B(C). If x and y are cocycles in C and C, respectively, then x y is a cocycle in C A C, because D(x y) = dx y + ( 1) p x d y = 0 for x C p. This means that there is a natural Abilinear map Z(C) Z(C ) Z(C A C ) (x, y) x y. If either x B(C) or y B(C ), then the image of (x, y) under this map is in B(C A C ), because for example dx y = D(x y) for all x C, y Z(C ) This means that we can divide out by the coboundaries in each of the groups and get a natural Abilinear map H(C) H(C ) H(C A C ) and therefore (by the universal property of the tensor product) a natural group homomorphism γ C,C : H(C) A H(C ) H(C A C ). In the next section we will need the following result: 2
3 Lemma. Let A be a ring, (C, d) a complex of right Amodules and (C, d ) a complex of left A modules. Assume d = 0. Then H(C) = C and γ C,C induces a natural group homomorphism C A H(C ) H(C A C ) x ȳ x y. (1) If C is flat over A, then this map is an isomorphism. Proof. We only need to prove the last claim. Because C is flat, er(d) = er(1 d ) = C A er(d ), so the natural map C A Z(C ) Z(C A C ) is an isomorphism. Furthermore, the image of C A B(C ) in C A Z(C ) corresponds to the subgroup B(C A C ) under this isomorphism, since both are generated by elements of the form x d y with x C and y C. This implies the map defined above is an isomorphism. 3. The Künneth formula From now on we restrict our attention to the case where A is a field. Then all complexes have the structure of vector spaces, and all modules are flat. For a treatment without this restriction, see [Bourbai]. We will prove the following theorem (note that the previous lemma is a special case of this): Theorem (Künneth formula). Let (C, d) and (C, d ) be complexes over. Then the natural linear map γ C,C : H(C) H(C ) H(C C ) is an isomorphism. Proof. Write Z = Z(C), B = B(C), H = H(C) and H = H(C ). Consider the short exact sequence of complexes defining Z(C) and B(C): 0 Z j C d B(1) 0. Here B(1) denotes the complex B shifted one place to the left, i.e. B(1) n = B n+1. Taing the tensor product with C gives a short exact sequence of complexes 0 Z C j 1 C C d 1 (B C )(1) 0. We tae the cohomology sequence of this short exact sequence. The coboundary map will go from H(B C ) to H(Z C ). To find out what it does, we write down the following diagram with exact rows: 0 (Z C ) n 1 j 1 (C C ) n 1 d 1 (B C ) n 0 D 0 (Z C ) n j 1 D (C C ) n d 1 (B C ) n+1 0. D Because d = 0 on B and because B is flat over, the ernel of D: (B C ) n (B C ) n+1 equals er(d) = er(1 d ) = B er(d ), so erd is generated by elements of the form dx y with x y (C C ) n 1 such that y Z(C ). The image of x y (C C ) n 1 in (C C ) n is now D(x y) = dx y, which is in (Z C ) n. We see therefore that the coboundary map sends the class of dx y to that of (i 1)(dx y), where i: B Z is the inclusion. In other words, the coboundary map equals H(i 1). The long exact sequence is now H n (B C ) H(i 1) H n (Z C ) H(j 1) H n (C C ) H(d 1) H n+1 (B C ) H(i 1) H n+1 (Z C ). 3
4 We can also tae the tensor product with H of the short exact sequence defining H to obtain an exact sequence 0 B H i 1 Z H p 1 H H 0. We connect this sequence with the long exact sequence above via the natural maps γ B,C : B H H(C C ) γ Z,C : Z H H(C C ) γ C,C : H H H(C C ), the first two of which are the isomorphisms occurring in the lemma from Section 2. This gives a commutative diagram with exact rows (B H ) n i 1 (Z H ) n p 1 γ B,C γ Z,C (H H ) n 0 γ C,C H n (B C ) H(i 1) H n (Z C ) H(j 1) H n (C C ) H(d 1) H n+1 (B C ) H(i 1) H n+1 (Z C ) γ B,C γ Z,C 0 (B H ) n+1 i 1 (Z H ) n+1 The lower right part shows that H(i 1) is injective, so H(d 1) = 0 by exactness. From the rest of the diagram we now see that γ C,C is an isomorphism. 4. The cohomology of sheaves of the form F G Let X and Y be two compact separated schemes over a field. Consider the scheme Z = X Y together with its projection morphisms p 1 : Z X and p 2 : Z Y. Let F and G be quasicoherent sheaves on X and Y, respectively. Recall that the pullbacs p 1 F and p 2G of F and G to Z are defined by p 1 F = O Z p 1 p 1 1 OX 1 F p 2 G = O Z p 1 p 1 OY 2 G. 2 It is a general fact that the pullbac of a quasicoherent sheaf is quasicoherent. We use this for p 1 F and p 2G. Suppose U = Spec A and V = Spec B are affine opens of X and Y, respectively, M is an Amodule such that F U = M and N is a Bmodule such that F V = N. Then the restrictions of p 1F and p 2G to the affine open subscheme W = U V = Spec(A B) of Z are p 1F W = (p 1 F(W)) = ((A B) A F(U)) = (B M), p 2G W = (p 2 G(W)) = ((A B) B G(V )) = (A N). From this we get the following expression for the sheaf p 1 F O Z p 2 G: In particular, we see that p 1F OZ p 2G W = ((B M) A B (A N)) = (M N). p 1 F O Z p 2 G(U V ) = F(U) G(V ) for all open affine subschemes U of X and V of Y. It seems therefore useful to introduce the abbreviated notation F G = p 1F OZ p 2G for quasicoherent sheaves F on X and G on Y. (To prevent confusion, this notation should only be used if the sheaves are quasicoherent.) We are now going to compare the cohomology of the sheaf F G on Z to the cohomology of F on X and G on Y. This we will do using a variant of Čech cohomology with respect to finite affine coverings of X, Y and Z. 4
5 Definition. The unordered Čech complex of a sheaf F of Abelian groups on a topological space X with respect to an open covering U = {U i } i I is the complex defined by C n (U, F) = F(U i0,...,i n ) where, as usual, i 0,...,i n I U i0,...,i n = U i0... U in. The maps d: C n C n+1 are defined using the same formula as for the usual (alternating) Čech complex: d ( ) n+1 {s i0,...,i n } i0,...,i n I = ( 1) j s i0,...,î j,...,i n+1 Ui0,...,i n+1. j=0 i 0,...,i n+1 I Notice that, in contrast to the alternating Čech cohomology, all the Cn (U, F) are nonzero (unless X = ), but that the product occuring in the definition of C n (U, F) is finite if I is finite. Let U = {U i } i I and V = {V j } j J be finite coverings by open affine subschemes of X and Y, respectively. Because X and Y are separated over, the intersection of any positive number of such affines is again affine [Hartshorne, Exercise II.4.3]. We loo at the unordered Čech complex of the sheaf F G on Z with respect to the affine open covering U V. By the property (1) of F G and because I and J are finite, C n (U V, F G) = F G ( ) U i0 V j0... U in V jn (i 0,j 0),...,(i n,j n) I J = i 0,...,i n I j 0,...,j n J F(U i0,...,i n ) G(V j0,...,j n ). Since the tensor product is distributive over direct sums, we see that C n (U V, F G) = F(U i0,...,i n ) i 0,...,i n I = C n (U, F) C n (V, G). j 0,...,j n J G(V j0,...,j n ) We tae the direct sum over all n and conclude that C(U V, F G) = C n (U, F) C n (V, G). (2) Fact. There exists a natural homotopy equivalence of complexes C n (U, F) C n (V, G) C(U, F) C(V, G). n=0 n=0 After applying this fact, which follows from the Eilenberg Zilber theorem [Godement, Théorème 3.9.1], to the righthand side of (2) and taing cohomology, we obtain a natural isomorphism Now the Künneth formula implies that H(C(U V, F G)) = H(C(U, F) C(V, G)). Ȟ(U V, F G) = Ȟ(U, F) Ȟ(V, G). If X and Y are Noetherian, then from the fact that the Čech cohomology is isomorphic to the derived functor cohomology for open affine coverings (the proof of [Hartshorne, Theorem III.4.5] also wors for the unordered Čech cohomology) we get the following theorem: Theorem. Let X and Y be Noetherian separated schemes over a field. For all quasicoherent sheaves F on X and G on Y, there is a natural isomorphism of vector spaces H(X, F) H(Y, G) = H(X Y, F G). 5
6 5. Application to the sheaves O(a, b) and curves on P 1 P 1 We have seen in Dirard s tal (see also [Hartshorne, Section III.5]) that for any ring A the cohomology of the sheaves O X (n) on X = P r A is given by H 0 (X, O X (n)) = S n H i (X, O X (n)) = 0 for 0 < i < r H r (X, O X (n)) = Hom A (S n r 1, A) for all n Z, where S n is the component of degree n in S = A[x 0,..., x r ]. In particular, for A equal to the field and for r = 1, H 0 (P 1, O P 1 (n)) = [x 0, x 1 ] n The dimensions are therefore equal to H 1 (P 1, O P 1 (n)) = [x 0, x 1 ] n 2 dim H 0 (P 1, O P 1(n)) = max{n + 1, 0} dim H 1 (P 1, O P 1(n)) = max{ n 1, 0}. It is now a matter of simple calculations and applying the Künneth formula to find the following table for the cohomology of the sheaves O(a, b) on Z = P 1 P 1 : dim H 0 (Z, O(a, b)) dim H 1 (Z, O(a, b)) dim H 2 (Z, O(a, b)) a 1, b 1 (a + 1)(b + 1) 0 0 a 1, b 1 0 (a + 1)( b 1) 0 a 1, b 1 0 ( a 1)(b + 1) 0 a 1, b (a + 1)(b + 1) We can now loo at a few applications of this. Let Y be a locally principal closed subscheme of Z, and let i: Y Z be the inclusion map, which is a closed immersion. Viewing Y as a divisor on Z, we have an exact sequence of coherent sheaves: 0 O Z ( Y ) O Z i O Y 0. The corresponding long exact cohomology sequence is 0 H 0 (Z, O Z ( Y )) H 0 (Z, O Z ) H 0 (Z, i O Y ) H 1 (Z, O Z ( Y )) H 1 (Z, O Z ) H 1 (Z, i O Y ) H 2 (Z, O Z ( Y )) H 2 (Z, O Z ) H 2 (Z, i O Y ) 0. Because i is a closed immersion, we now that H(Z, i O Y ) = H(Y, O Y ). Furthermore, the case a = b = 0 gives us that H 0 (Z, O Z ) =, H 1 (Z, O Z ) = 0 and H 2 (Z, O Z ) = 0, so the long exact sequence breas down into two exact sequences 0 H 0 (Z, O Z ( Y )) H 0 (Y, O Y ) H 1 (Z, O Z ( Y )) 0 and 0 H 1 (Y, O Y ) H 2 (Z, O Z ( Y )) 0. 6
7 If Y is of type (a, b) with a, b > 0, then O Z ( Y ) = O( a, b); for these sheaves we have by the bottom row of the table above Therefore, H 0 (Z, O Z ( Y )) = 0, H 1 (Z, O Z ( Y )) = 0, dim H 2 (Z, O Z ( Y )) = ( a + 1)( b + 1) = (a 1)(b 1). H 0 (Y, O Y ) = and dim H 1 (Y, O Y ) = (a 1)(b 1) if a, b > 0. The interpretation of this is that Y is connected, and if Y is a nonsingular curve it has genus (a 1)(b 1). 6. Bertini s theorem In this section we study intersections of projective varieties with hyperplanes. A hyperplane H P n is by definition the zero set of a single homogeneous polynomial f [x 0,...,x n ] of degree 1. Let V be the subspace of homogeneous elements of degree 1 in [x 0,..., x n ]. Form the projective space H = (V \ {0})/ = ([x 0,..., x n ] 1 \ {0})/ and view it as a projective variety over ; it is isomorphic to P n. Because two nonzero sections of O P n determine the same hyperplane if and only if one is a multiple of the other by an element of, there is a canonical bijection between H and the set of hyperplanes in P n. Theorem (Bertini). Let X be a nonsingular closed subvariety of P n, where is an algebraically closed field. Then there exists a hyperplane H P n, not containing X, such that the scheme H X is regular. Moreover, the set of all hyperplanes with this property is an open dense subset of H. Proof. Consider a closed point x of X. There is an i {0, 2,..., n} such that x is not in the hyperplane defined by x i ; after renaming the coordinates we may assume i = 0. Then f/x 0 is a regular function in a neighbourhood of x for all f V, so there is a linear map φ x : V O X,x f f/x 0, where O X,x is the local ring of X at x. If X is contained in the hyperplane H defined by f, then φ x (f) = 0; conversely, φ x (f) = 0 means that f vanishes on some open neighbourhood of x in X, hence on all of X since X is irreducible. We conclude that φ x (f) = 0 X H. Furthermore, φ x (f) m x x H. Assume X H but x X H, so that φ x (f) m x \ {0}. Then f = φ x (f)o X,x is a nonzero ideal of O X,x contained in m x. Now the local ring of H X at x is O X,x /f, and its maximal ideal is n = m x /f. The fact that O X,x is an integral domain and f is a nonzero principal ideal implies that dim(o X,x /f) = dim(o X,x ) 1. Furthermore, n 2 = (m 2 x + f)/f and n/n 2 = m x /(m 2 x + f). In particular, dim n/n 2 dim m x /m 2 x with equality if and only if f m 2. Recall that dim m x /m 2 x dim O X,x with equality if and only if O X,x is a regular local ring. Applying this also to O X,x /f we see that O X,x /f is regular if f m 2 (in which case dim n/n 2 = dim O X,x /f), and not regular if f m. Hence O X,x /f is a regular local ring if and only if φ x (f) m x \ m 2 x. Let B x H be the set of hyperplanes that are defined by an element f V for which φ x (f) m 2 x. In other words, if we put φ x : V O X,x /m 2 x f f/x 0 mod m 2 x 7
8 then B x = (er φ x \ {0})/ H. This is a subvariety of H, the interpretation of which is as follows: a hyperplane H is in B x if and only if either H X or x H X and x is a singular point of H X. Let us tae a closer loo at B x. We put y i = x i /x 0 for 1 i n, so that Spec [y 1,...,y n ] is an affine open neighbourhood of x. Let g 1,..., g m [y 1,..., y n ] be local equations for X, and let (a 1,..., a n ) be the coordinates of the point x. Then O X,x is isomorphic to A p, where A = ([y 1,...,y n ]/(g 1,...,g m )), p = (y 1 a 1,..., y n a n ), and m x corresponds to pa p under this isomorphism. Furthermore, the vector space O X,x /m 2 x has dimension dim (O X,x /m 2 x) = dim (O X,x /m x ) + dim (m x /m 2 x) = 1 + dimx and is spanned over by the elements 1, y 1 a 1,..., y n a n (easy chec). This shows that φ x is surjective, and dimer φ x = dim V dim (O X,x /m 2 x) = (n + 1) (1 + dimx) = n dimx, from which we conclude that dimb x = n dimx 1. The polynomials g 1,..., g m which locally define X are modulo m 2 x congruent to the polynomials ḡ i = n (y j a j ) g i (a 1,...,a n ) y j j=1 (1 i m). Because φ x (f) is of the form b 0 + n j=1 b jy j, we see that φ x (f) m 2 x f/x 0 m ḡ i, i=1 or, equivalently, ( er φ m m ) x = x 0 ḡ i and B x = x 0 ḡ i \ {0} /. i=1 i=1 Consider the fibred product X H. Because of the above characterisation of er φ x, there is a closed subscheme B of X H such that the closed points of B are precisely the points of X H corresponding to the pairs (x, H) with x a closed point of X and H B x. We have seen that the fibre of B above each point of X has dimension n dimx 1, so B itself has dimension (n dimx 1) + dimx = n 1. Because X is proper over and proper morphisms are preserved under base extension, the projection p 2 : X H H is proper too. This implies that p 2 (B) is a closed subset of H of dimension at most n 1, and from this we conclude that H p 2 (B) is an open dense subset of H. For each H H \ p 2 (B), the scheme H X is regular at every point by the construction of B. 8
9 7. Application to the existence of nonsingular curves of type (a, b) Let be an algebraically closed field, and let a, b be positive integers. We want to show that there are nonsingular curves of type (a, b) on P 1 P 1. First we embed P1 P 1 into Pn, where n = ab+a+b, using the auple, buple and Segre embeddings: Recall that the auple embedding is defined by P 1 P 1 P a P a P n. (x 0 : x 1 ) (x a 0 : x a 1 0 x 1 :... : x a 1) and similarly for the buple embedding; the Segre embedding is defined by ((s 0 :... : s a ), (t 0 :... : t b )) (... : s i t j :...) in lexicographic order. Let j denote the composed embedding P 1 P 1 Pn. The image of j is a nonsingular surface X in P n that is isomorphic to P1 P 1. We apply Bertini s theorem to find a hyperplane H in P N such that H X is a onedimensional regular closed subscheme of X. This hyperplane is given by a homogeneous linear polynomial in the coordinates {z i,j : 0 i a, 0 j b} of P n. Now z i,j = j(x0 a i x i 1y b j 0 y j 1 ), so Y = j 1 (H X), viewed as a divisor on P 1 P 1, is of type (a, b). We have seen earlier that this implies that Y is connected. The local rings of Y are regular local rings, so in particular they are integral domains [Hartshorne, Remar II A]. This means that there cannot be two irreducible components of Y intersecting each other; therefore Y is irreducible, and hence a nonsingular curve. References [Bourbai] N. Bourbai, Algèbre, chapitre 10: Algèbre homologique. Masson, Paris, [Godement] R. Godement, Topologie algébrique et théorie des faisceaux. Actualités Sci. Ind Hermann, Paris, [Hartshorne] R. Hartshorne, Algebraic Geometry. Graduate Texts in Mathematics 52. Springer Verlag, New Yor Heidelberg,
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