HARTSHORNE EXERCISES


 Tamsyn Flowers
 3 years ago
 Views:
Transcription
1 HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing up Y at O = (0, 0) is nonsingular. (b) We define a node (also called ordinary double point) to be a double point (i.e., a point of multiplicity 2) of a plane curve with distance tangent directions. If P is a node on a plane curve Y, show that ϕ 1 (P ) consists of two distinct nonsingular points on the blownup curve Ỹ. We say that blowing up P resolves the singularity at P. (c) Let P Y be the tacnode x 2 = x 4 + y 4. If ϕ : Ỹ Y is the blowing up at P, show that ϕ 1 (P ) is a node. Using (b) we see that the tacnode can be resolved by two successive blowingsup. (d) Let Y be the plane curve y 3 = x 5, which has a higher order cusp at O. Show that O is a triple point; that blowing up O gives rise to a double point (what kind?) and that one further blowing up resolves the singularity. Proof. (a) We first work with the cusp. Let [t : u] be homogeneous coordinates of P 1. The variety Ỹ is covered by two open affine sets, those where either t or u doesn t vanish. First, consider t 0. Then we may assume that t = 1 and treat u as an affine coordinate, so that we obtain an affine variety in A 3. This variety is given by the polynomial equations: x u 2 x 2 x 2 u 4 = 0 xu y = 0 Therefore, the Jacobian of Ỹ at any point (x, y, u) of this affine piece is ( ) 1 2x 2xu 4 0 2u 4x 2 u 3 u 1 x This matrix has less than full rank if and only if 1 2x 2xu 4 = 2u 4x 2 u 3 = 0. Hence, to show that each point is nonsingular, we must show that the system of equations (1) x u 2 x 2 x 2 u 4 = 0 (2) 1 2x 2xu 4 = 0 (3) 2u 4x 2 u 3 = 0 has no solution. Notice that if u = 0 or x = 0 or char k = 2, then the system has no solution. So we can assume that neither u nor x are 0 and that the characteristic is not Date: Spring
2 2. Also, notice that the 2 (1) x (2) shows that x = 2u 2. This reduces us to the following system of equations: (4) 1 4u 2 4u 6 = 0 (5) 1 + 8u 6 = 0 2 (4) + (5) shows us that 3 = 8u 2 = 4x. Plugging this into (5) gives the equality = 0 which is only true if the characteristic of k is 7 or 13. If the characteristic of k is 7, the triple (6, 6u, u) is a singular point for any value of u k satisfying u 2 = 3. If the characteristic of k is 13, the triple (4, 4u, u) is a singular point for any value of u k satisfying u 2 = 2. In fact, the point (6, 6u) is singular on Y if char k = 7 and the point (4, 4u) is singular on Y if char k = 13, so it makes sense that they remain singular under the blowup. If u 0, we may assume that u = 1 and treat t as an affine coordinate. The resulting variety in A 3 is given by the equations: yt 3 1 y 2 t 4 y 2 = 0 x yt = 0 This leads to the Jacobian (0 ) t 3 2yt 4 2y 3yt 2 4y 2 t 3 1 t y The only point we need to check on this affine piece of Ỹ is the point t = 0, whose Jacobian is: ( ) 0 2y y If t = 0, then x = 0 and y satisfies the equation y = 0. Hence the point (0, y, 0) is singular on Ỹ if and only if 2y = 0 = y If char k = 2, then we have the singular points (0, 1, 0). In summary, Ỹ is nonsingular if and only if the characteristic of the field is not 2, 7, or 13. Proceeding as above, it can be seen that the blowup of the node at O is nonsingular no matter the characteristic of the field. This follows from the fact that the only singularity of the node is the origin for all fields k, and the singularity is a double point, so it is resolved by the blowup (part (b) below). (b) Suppose P = (a, b). If Y is defined by the polynomial f(x, y), then Y is isomorphic to the polynomial defined by f(x + a, y + b) under the map x x a, y y b, so we may assume that P = (0, 0). We can also rotate Y to assume that the tangent directions at the origin are not vertical. Decompose f(x, y) into its homogeneous components by f(x, y) = f 2 (x, y) + f 3 (x, y) f d (x, y), where f 2 (x, y) = (ax + by)(a x + b y), with [a : b] [a : b ] P 1 because P must have distinct tangent directions. Also, since the tangent directions are not vertical, 2
3 we have b 0 b. As in part (a), let s first consider the affine piece of the blowup where t 0. This is given by the equations x 2 f(x, xu) = (a + bu)(a + b u) + x 2 (f 3 (x, xu) f d (x, xu)) = 0 xu y = 0 Then varphi 1 ((0, 0)) are points of the form (0, 0, u) where (a + bu)(a + b u) = 0. Since both b and b are nonzero, this equation has two distinct solutions given by u = ab 1 and u = a b 1. Notice the Jacobian at (0, 0, u) is then given by ( x 3 f 3 (x, xu) (0,0,u) 0 ) ab + a b + 2bb u u 1 x which has rank 2 because ab + a b + 2bb u 0 for both solutions u (this follows from our assumption that [a : b] [a : b ] P 1 ). (c) First suppose P = (0, 0). Checking the affine pieces of the blowup reveals that the only point in ϕ 1 (P ) is (x, y, t) = (0, 0, 0) when u 0. This affine piece in A 3 is defined by the equations t 2 y 2 y 2 t 4 = 0 x yt = 0 We can isomorphically project this onto the y, tplane, to obtain the plane curve defined by f(y, t) = t 2 y 2 y 2 t 4 = (t y)(t + y) y 2 t 4. Notice then that the point (0, 0) is a node so long as char k 2, otherwise it is not a node. The result is not true for any other P (0, 0), for no other point is singular on Y when char k 2. (d) O is a triple point because the smallest nonzero homogeneous component of f(x, y) = y 3 x 5 has degree 3. Blowing up at O gives a curve with an affine open piece isomorphic to the cusp u 3 = x 2, where the preimage of O is the point (0, 0). The point (0, 0) is a double point on the cusp defined by f(x, u) = x 2 u 3, but it is not ordinary because x 2 has the same linear factors. On a previous homework assignment, we showed that blowing up the cusp u 3 = x 2 at the origin resolves the singularity, so one further blowing up of (0, 0) will resolve the singularity of Y. Hartshorne, Exercise I.6.3. Show by example that the result of (6.8) is false if either (a) dim X 2, or (b) Y is not projective. Proof. (a) Let X = A 2, P = (0, 0), and Y = P 1 with ϕ : X P Y (x, y) [x : y] and suppose that ϕ extended to ϕ : X Y where ϕ((0, 0)) = [a : b] P 1. Then, consider any point [c : d] [a : b] P 1. This point is closed (defined by the homogeneous polynomial equation dx = cy). However, ϕ 1 ([c : d]) = {(x, y) dx = cy} (0, 0) is not closed, so the map ϕ is not continuous. 3
4 (b) Let X = P 1, P = [0 : 1], and Y = A 1 with ϕ : X P Y [x : y] y x and suppose that ϕ extended to ϕ : X Y. A morphisms of varieties into A 1 is a regular function, so that ϕ is a regular function of P 1, so by a previous homework, ϕ is constant. This contradicts the definition of ϕ, so no such extension exists. Hartshorne, Exercise I.6.7. Let P 1,..., P r, Q 1,..., Q s be distinct points of A 1. If A 1 {P 1,..., P r } is isomorphic to A 1 {Q 1,..., Q s }, show that r = s. Is the converse true? Proof. Let U = A 1 {P 1,..., P r } and V = A 1 {Q 1,..., Q s }, and view U and V as open subsets of P 1. Let ϕ : U V be an isomorphism, and i : V P 1 be inclusion. Then the morphism i ϕ : U P 1 can be uniquely extended (Hartshorne I.6.8) to a map ϕ : P 1 P 1. I claim that ϕ is an isomorphism. If so, The result follows because then ϕ must map the set {P 1,..., P r, } bijectively to the set {Q 1,..., Q s, }. To see that ϕ is an isomorphism, we will exhibit its inverse. Namely, begin with ϕ 1 : V U and complete to ϕ 1 : P 1 P 1 as above. Notice that ϕ ϕ 1 restricts to the identity on V, so it must be the unique extension of the identity morphism on V, which is the identity on P 1. It follows that ϕ 1 = ϕ 1. The converse is true only for 0 r 2. For r = 0, A 1 = A 1. For r = 1, A 1 \ P = A 1 \ Q via the map x x P + Q. For r = 2, A 1 \ {P 1, P 2 } = A 1 \ {Q 1, Q 2 } via the map x (Q 1 Q 2 )x Q 2 P 1 + Q 1 P 2 P 2 P 1 To see that the converse is false for r > 3, we use without proof the result that an automorphism of P 1 is determined by where it sends three points (Hartshorne, exercise I.6.6). Let P 1,..., P r, Q 1,..., Q r be distinct points such that no automorphism of P 1 bijectively maps the set {P 1,..., P r, } to the set {Q 1,..., Q r, }. If A 1 \ {P 1,..., P r, } = A 1 \ {Q 1,..., Q r, }, as shown above, there must be an automorphism of P 1 which maps {P 1,..., P r, } to the set {Q 1,..., Q r, } bijectively, a contradiction. Hartshorne, Exercise II.2.3. Reduced Schemes. A scheme (X, O X ) is reduced if for every open set U X, the ring O X (U) has no (nonzero) nilpotent elements. (a) Show that (X, O X ) is reduced if and only if for every P X, the local ring O X,P has no (nonzero) nilpotent elements. (b) Let (X, O X ) be a scheme. Let (O X ) red be the sheaf associated to the presheaf U O X (U) red, where for any ring A, we denote by A red the quotient of A by its ideal of nilpotent elements. Show that (X, (O X ) red ) is a scheme. We call it the reduced scheme associated to X, and denote it by X red. Show that there is a morphism of schemes X red X, which is a homeomorphism on the underlying topological spaces. (c) Let f : X Y be a morphism of schemes, and assume that X is reduced. Show that there is a unique morphism g : X Y red such that f is obtained by composing g with the natural map Y red Y. 4
5 Proof. (a) Suppose (X, O X ) is reduced, P X, and s, U O X,P for some open U X containing P and some s O X (U). Suppose that s, U n = s n, U = 0 O X,P for some n N. This means there is some open V U containing P such that (s n ) V = (s V ) n = 0 O X (V ). Since (X, O X ) is reduced O X (V ) has no nilpotent elements, so s V = 0. It follows that s, U = s V, V = 0 so that the local ring O X,P has no nilpotent elements. Next, suppose that for every P X, O X,P has no nilpotent elements. For any U X, let s O X (U), and suppose that s n = 0 for some n N. For each P U, let s P = s, U O X,P. By our hypothesis, it follows that s P = 0 for all P U, that is, for each P there exists open V P U containing P such that s VP = 0. The V P cover U, so by the sheaf axiom, it follows that s = 0 O X (U), and thus, (X, O X ) is reduced. (b) First, assume that X = Spec A is affine. We will show that X red = Spec Ared. The canonical projection A A red induces a map of schemes ϕ : Spec A red Spec A. If N is the nilradical of A, then since all prime ideals in A contain N, there is a onetoone correspondence between prime ideals p A and prime ideals p A red. The map ϕ is a homeomorphism sending p to p. Now, ϕ # U : O Spec A(U) ϕ O Spec Ared (U) must vanish on nilpotent elements of O Spec A (U) because Spec A red is a reduced scheme (its localizations are of the form (A red ) p = (Ap ) red, so they are reduced, so by part (a), Spec A red is reduced). We thus have a map of presheaves (O Spec A (U)) red ϕ O Spec Ared (U) which by the universal property extends to a map of sheaves (O Spec A ) red ϕ O Spec Ared. To see that this is an isomorphism, we will see that the induced map on stalks is an isomorphism. The induced map on stalks is the middle map in the following sequence of compositions: (6) (A p ) red ((O Spec A ) red ) p (ϕ O Spec Ared ) p (A red ) p ( a ) a a b b, D(b) b, D ( b ) a b Here a A, b A \ p, D(b) = {q Spec A b / q}, D(b) = {q Spec A red b / q}, and for any commutative ring R, prime ideal p R, and element r R s p, we view r as s a function on D(s) via: D(s) p D(s) p r s (6) is the canonical isomorphism between (A p ) red and (A red ) p, so that the map of stalks is also an isomorphism. We have thus shown that as affine schemes, (Spec A) red = Spec A red. We have also exhibited the desired morphism of affine schemes Spec A red Spec A which is just the map induced by the projection A A red. Now, if X is any scheme, let x X, and choose some open affine neighborhood of x, U = Spec A. This means that (U, O X U ) = (Spec A, O Spec A ) as affine schemes. Thus, (U, (O X U ) red ) = (Spec A red, O Spec Ared ) as affine schemes. For any open V U, (O X U ) red and (O X ) red U are both the sheaf associated to the presheaf V O X (V ), they are isomorphic. It follows that (Spec A red, O Spec Ared ) is an open affine set of x X red so that X red is indeed a scheme. 5 R p
6 To obtain the desired map X red X, let ϕ : X X be the identity. Then, for any open U X, define ϕ # U : O X(U) (O X ) red (U) to be the composition of the quotient map O X (U) ((O X (U)) red with the natural map from the presheaf (O X (U)) red to its associated sheaf (O X ) red (U). (c) The natural map (ϕ, ϕ # ) from Y red to Y is the identity on topological spaces. Thus we can define g(x) = f(x). Next, for any open U Y, consider the map f # U : O Y (U) O X (f 1 (U)). Since X is reduced, f # U must factor through (O Y (U)) red. This gives a map of presheaves (O Y (U)) red O X (f 1 (U)). By the universal property of sheafification, there is a unique map of sheaves g # U : (O Y ) red (U) O X (g 1 (U)). The pair (g, g # ) is the desired morphism of schemes X Y red. Hartshorne, Exercise II.2.7. Let X be a scheme. For any x X, let O x be the local ring at x, and m x its maximal ideal. We define the residue field of x on X to be the field k(x) = O x /m x. Now let K be any field. Show that to give a morphism of Spec K to X it is equivalent to give a point x X and an inclusion map k(x) K. Proof. Let Y = Spec K, and suppose (f, f # ) : Y X is a morphism of schemes, that is f : Y X is a continuous map and f # : O X f O Y is a morphism of sheaves of rings on X. As a topological space Y consists of one point (the zero ideal in K), so we take x X to be the image of this point. Now let U be any open set in X containing x, so that we have a map f # U : O X(U) O Y (Y ) = K. Since f # is a morphism of sheaves, the maps f # U are compatible with restrictions, so that the map f x # : O x K s, U f # U (s) is welldefined, where s O X (U). Being a map of schemes, we require this induced morphism to be a local homomorphism of local rings. In this case, this condition means that ker f x # = m x so that f x # induces an inclusion k(x) K. Next, suppose we fix a point x X and an inclusion i : k(x) K. Define a continuous map f : Y X by f((0)) = x. For any open U X not containing X, f 1 (U) = so define f # U : O X(U) O Y ( ) = 0 to be the trivial ring map. If U contains x, then define : O X(U) O Y (Y ) = K as follows: f # U f # U : O X(U) K s i( s, U + m x ) In other words, f # U is the composition of maps O X(U) O x k(x) K. To see that this defines a map of sheaves, we must verify that for any V U, the diagram: O X (U) O Y (f 1 (U)) O X (V ) O Y (f 1 (V )) 6
7 commutes. If V does not contain x, then the diagram commutes. If V contains X, then the diagram becomes O X (U) O X (V ) K The commutativity of this diagram follows from that of O X (U) O X (V ) O x By definition, the map induced on stalks by f # is O x k(x) K, whose kernel is m x, so it is a local map of local rings, and thus (f, f # ) is a morphism of schemes. Hartshorne, Exercise II.2.8. Let X be a scheme. For any point x X, we define the Zariski tangent space T x to X at x to be the dual of the k(x)vector space m x /m 2 x. Now assume that X is a scheme over a field k, and let k[ε]/ε 2 be the ring of dual numbers over k. Show that to give a kmorphism of Spec k[ε]/ε 2 to X is equivalent to giving a point x X, rational over k (i.e., such that k(x) = k), and an element of T x. Proof. Let Y = Spec k[ε]/ε 2, and suppose (f, f # ) is a kmorphism from Y to X. This means the following diagram of schemes commutes: Y f Spec k X As a topological space, Y consists of one point (the ideal (ε)), so let x be the image of this point, and consider the induced commutative diagram of stalks: k[ε]/ε 2 f # x O x Each arrow above is a local homomorphism of local rings. In particular, the maps from k must be injections, and (f # x ) 1 ((ε)) = m x. If k[ε]/ε 2 k is the map obtained by dividing by (ε), then the map O x k[ε]/ε 2 k vanishes on m x so that we have a map k(x) = O x /m x k. This discussion results in the following commutative diagram of fields: k k f # x k(x) k 7
8 where the map k k is an isomorphism. This shows that k(x) = k so that x is rational. Notice also that f x # restricts to a kalgebra map α : m x (ε) = k. This map must vanish on m 2 x because ε 2 = 0, so α induces an element in T x. Next, suppose we are given a point x X such that k(x) = k and a klinear map α : m x /m 2 x k. To construct a morphism (f, f # ) from Y to X, first let f((ε)) = x. Next, for any open U X not containing x, define f # U : O X(U) O Y ( ) = 0 to be the zero map. Now suppose that U contains x. The kscheme structure on X gives O X (U) the structure of a kalgebra. Also, since x is a rational point, given any a O X (U), we can canonically decompose a into a = a(x) + (a a(x)), where a(x) = a + m x k(x) = k. Notice that a a(x), U m x O x. Now, define f # U : O X(U) O Y (f 1 (U)) = k[ε]/ε 2 by: O X (U) O x O x /m 2 x k[ε]/ε 2 a a, U a, U + m 2 x a(x) + α( a a(x), U + m 2 x)ε 2 That this is a map of rings follows from the klinearity of α and the fact that (a a(x)) 2, U m 2 x. f # is a map of sheaves because the natural maps O X (U) O x and O X (V ) O x commute with restriction for all open V U containing x. Hartshorne, Exercise II Let k = F p be the finite field with p elements. Describe Spec k[x]. What are the residue fields of its points? How many points are there with a given residue field? Proof. k[x] is a PID, so its nonzero prime ideals are maximal and are in one to one correspondence with monic, irreducible polynomials in k[x]. Let f(x) be a monic, irreducible polynomial of degree n in k[x], and let m = (f). Since quotienting and localizing commute, we can either view the residue field of f as k[x] m /(m k[x]), or equivalently as K = (k[x]/m) (0) = k[x]/m. K/k is the splitting field of f, and thus has degree equal n. By uniqueness of finite fields, it follows that K = F p n, so we see that the residue fields of the points of Spec k[x] are all of the form F p n for some n. For fixed n, to determine how many points there are with residue field F p n, we must determine how many monic, irreducible polynomials of degree n exist in k[x]. For this, we reference without proof a result of Gauss, which states that the number of monic, irreducible polynomials of degree n in k[x] is 1 ( n ) µ p d n d d n where µ is the Möbius function given by 0 if p 2 m for some prime p µ(m) = 1 if m = 1 or m is the product of an even number of distinct primes 1 if m is the product of an odd number of distinct primes Hartshorne, Exercise II In this exercise, we compare some properties of a ring homomorphism to the induced morphism of the spectra of rings. (a) Let A be a ring, X = Spec A, and f A. Show that f is nilpotent if and only if D(f) is empty. 8
9 (b) Let ϕ : A B be a homomorphism of rings, and let f : Y = Spec B X = Spec A be the induced morphism of affine schemes. Show that ϕ is injective if and only if the map of sheaves f # : O X f O Y is injective. Show furthermore in that case f is dominant, i.e., f(y ) is dense in X. (c) With the same notation, show that if ϕ is surjective, then f is a homeomorphism of Y onto a closed subset of X, and f # : O X f O Y is surjective. (d) Prove the converse to (c), namely, if f : Y X is a homeomorphism onto a closed subset, and f # : O X f O Y is surjective, then ϕ is surjective. Proof. (a) This follows from the fact that any nilpotent element is contained in all prime ideals. (b) f # : O X f O Y is an injective map of sheaves if and only if f # U : O X(U) f O Y (U) is injective for all open U X. Take U = X to obtain a map Γ(X, O X ) = A Γ(Y, O Y ) = B. This is ϕ, so it is injective. Next, suppose that ϕ is injective, so we assume A B is a subring of B. To show that f # is an injective map of sheaves, we show it is injective on all stalks. Let p X, and notice that the stalk of O X at p is the localization A p. The stalk of f O Y at p can be calculated as follows: (f O Y ) p = lim p U O Y (f 1 (U)) = lim p D(g) O Y (f 1 (D(g))) = lim B g = S 1 B g A\p where S is the multiplicative set A \ p B. The second equality above results from the fact that sets of the form D(g) form a base for the topology on X, and the third equality results from the identification of f 1 (D(g)) = D(ϕ(g)) with D(g), viewing A as a subring of B. The map induces by ϕ on stalks is then inclusion: f p # : A p S 1 B a a a a which is clearly injective. To show that f is dominant, consider any nonempty open set U X. Then O X (U) is nonzero, so by injectivity it follows that O Y (f 1 (U)) is nonzero, so that f 1 (U) is nonzero. Hence, f(y ) meets every nonempty open set U X, and thus f(y ) is dense in X, so that f is dominant. (c) We show that f is injective and closed, so that it is a homeomorphism onto its image. Suppose f(p) = f(p ) so that ϕ 1 (p) = ϕ 1 (p ). For any b p, by surjectivity choose a with ϕ(a) = b. Then we have b p if and only if ϕ(a) p if and only if a ϕ 1 (p) = ϕ 1 (p ) if and only if ϕ(a) p if and only if b p. So f is injective. Next, consider any ideal I B and the closed set it defines V (I). Since ϕ is surjective, we have the equivalence I p if and only if ϕ 1 (I) ϕ 1 (p), so that f(v (I)) = V (ϕ 1 (I)) so that f is a closed map. Now, since ϕ is surjective, we have A/ ker ϕ = B so that the prime ideals of B are precisely the prime ideals of A containing ker ϕ. It follows that f(spec B) = V (ker ϕ) is closed. 9
10 Finally, to see that the map of sheaves is surjective, again we check on stalks. Similar to part (b), here we have (f O Y ) p = lim p U O Y (f 1 (U)) = lim p D(g) O Y (f 1 (D(g))) = lim B ϕ(g) = S 1 B g A\p where now S is the multiplicative set ϕ(a\p) B because ϕ is not necessarily injective. The map on stalks is then f # p : A p S 1 B a ϕ(a) a ϕ(a ) whose surjectivity follows from that of ϕ. (d) Let A = A/ ker ϕ and let X = Spec A. We can think of ϕ as the composition of the surjective map π : A A and the injective map i : A B. It follows that f is the composition of g : Y X induced by i and h : X X induced by π. It is sufficient to show that g is an isomorphism of schemes, so that A = B from which it follows that ϕ(a) = B. Since π is surjective, it follows from part (c) that h is a homeomorphism onto a closed subset of X and that the induced map on sheaves h # : O X h O X is surjective. Similarly, since i is injective, it follows from part (b) that g is dominant and that the induced map on sheaves g # : O X g O Y is injective. Since f = h g is a homeomorphism onto a closed set, as is h, it follows that g must also be a homeomorphism onto a closed set. Since g is dominant, it follows that g is a homeomorphism. Next, we know that g # is injective on sheaves, so we are finished if we show it is also surjective. By hypothesis, f # : O X f O Y is surjective as is h #, and since f # = h g # h # it follows that h g # is also surjective. Finally, we show that the surjectivity of h g # implies that of g #. Take any p X. We have a commutative diagram on stalks h g # (h O X ) h(p) h(p) (f O Y ) h(p) (O X ) p g # h(p) (g O X ) p where the vertical maps are actually inclusions. Explicitly, the left vertical map is given by the inclusion lim O X (h 1 (V )) lim O X (U) p h 1 (V ) p U Where V ranges over open sets in X containing h(p) and U ranges over open sets in X containing p. It turns out that this map along with the right vertical map in the diagram are surjections, which, with the commutativity of the above diagram, proves that g # is surjective. To see that the above map is actually surjective, notice that it is enough to show that the directed system of the limit on the left hand side includes all basic open sets D(a ) X which contain p, some a A. Since π is surjective, we can find find 10
11 a A with π(a) = a. Then h 1 (D(a)) = D(a ). Hence the limit on the left hand side actually recovers the full stalk, and the map is surjective. Hartshorne, Exercise II.4.7. Schemes over R. For any scheme X 0 over R, let X = X 0 R C. Let α : C C be complex conjugation, and let σ : X X be the automorphism obtained by keeping X 0 fixed and applying α to C. Then X is a scheme over C, and σ is a semilinear automorphism, in the sense that we have a commutative diagram X σ X Since σ 2 = id, we call σ an involution. Spec C α Spec C (a) Now let X be a separated scheme of finite type over C, let σ be a semilinear involution on X, and assume that for any two points x 1, x 2 X, there is an open affine subset containing both of them. (This last condition is satisfied for example if X is quasiprojective.) Show that there is a unique separated scheme X 0 of finite type over R, such that X 0 R C = X, and such that this isomorphism identifies the given involution of X with the one on X 0 R C described above. For the following statements, X 0 will denote a separated scheme of finite type over R, and X, σ will denote the corresponding scheme with involution over C. (b) Show that X 0 is affine if and only if X is. (c) If X 0, Y 0 are two such schemes over R, then to give a morphism f 0 : X 0 Y 0 is equivalent to giving a morphism f : X Y which commutes with the involutions, i.e., f σ X = σ Y f. (d) If X = A 1 C, then X 0 = A 1 R. (e) If X = P 1 C, then either X 0 = P 1 R, or X 0 is isomorphic to the conic in P 2 R given by the homogeneous equation x x x 2 2 = 0. Proof. (a) The case for X = Spec A is treated in part (b), where X 0 = Spec A σ with A σ the subring of fixed elements under σ : A A. (b) If X 0 = Spec A, then X = Spec A R C is affine. If X = Spec A, then X 0 = Spec A σ, where A σ is the subring of fixed elements under σ : A A. To see this, note that A σ R C = A under the map a 1 + b i a + bi for a, b A σ. This map is surjective because for any a A we can write: ( ) ( ) a + σ (a) a σ (a) a = + i 2 2i (c) By the universal property of the fiber product, given a map of schemes over R, f 0 : X 0 Y 0, we get map of schemes over C, f : X Y, satisfying the following commutative 11
12 diagram: X 0 X Spec C f 0 Y 0 f Y id Spec C Spec R Now σ X is a map X X, where X is considered a scheme over X 0, and similarly for σ Y. It follows that the above diagram commutes when we replace f with σ Y f σ X. By uniqueness, it follows that σ Y f = f σ X. For the reverse direction, first suppose X = Spec A and Y = Spec B are affine. Then, the induced map f : B A restricts to a ring homomorphism f 0 : B σ A σ because f commutes with σ. Then f 0 : X 0 Y 0 is a morphism of schemes. (d) This follows from (b) and the observation that C[t] σ = R[t]. (e) Hartshorne, Exercise II.5.7. Let X be a noetherian scheme, and let F be a coherent sheaf. (a) If the stalk F x is a free O x module for some point x X, then there is a neighborhood U of x such that F U is free. (b) F is locally free if and only if its stalks F x are free O x modules for all x X. (c) F is invertible (i.e., locally free of rank 1) if and only if there is a coherent sheaf G such that F G = O X. Proof. (a) We can assume that X = Spec A, and that F = M for some finitely generated A module M = x 1,..., x k. Let x = p X, so by hypothesis M p = A n p. Let m 1,..., m n generate M p freely. Consider the equations n a ij x i m j b ij and let b = b ij. Then we have a surjective map of A b modules j=1 f : A n b M b (a 1,..., a n ) a i m i The kernel of this map is finitely generated because A n b is noetherian. For each generator (a 1,..., a n ), there exists some s / p such that s (a 1,..., a n ) = 0. Let g be the product of b and all such s. I claim that m 1,..., m n generate M g freely over A g. This follows from the fact that the localization of f at g is an isomorphism. (b) The reverse direction follows from (a). For the forward direction, take any x X and find an open affine neighborhood U containing x such that F U = O n X U (we know there must be finitely many direct summands because F is coherent). Now we have n n F x = lim F U (V ) = x V U lim O n X U(V ) = x V U 12 lim x V U i=1 O X (V ) = i=1 O x
13 where the last equality is justified by the following isomorphism: n n O x lim O X (V ) i=1 x V U i=1 n ( s 1, V 1,..., s n, V n ) (s 1,..., s n ), V i Notice that the rank of F x is equal to the rank of F at x. (c) Suppose F is invertible and let G = F ˇ = H om OX (F, O X ). Define a map of sheaves ϕ U : F (U) OX (U) Hom OX U (F U, O X U ) O X (U) by s, U f f U (s), U. By the sheaf axiom, it is enough to show that this induces an isomorphism on an open cover of X. Choose an open cover U i such that F Ui is free on U i of rank 1. On such a U, the above map has inverse given by s, U 1, U f s where (f s ) V : O X (V ) O X (V ) is defined by t, V s V t, V. To see that Fˇ is coherent... For the reverse direction, I first claim that Pic(Spec R) is trivial for R a commutative local ring, that is, if M and N are finitely generated Rmodules such that M R N = R, then M = N = R. To see this, let k be the residue field of R and notice that by extension of scalars we have (M R k) k (N R k) = k. It follows that both M R k and N R k are 1 dimensional vector spaces over k. We now need to invoke the following form of Nakayama s lemma: if M is a finitely generated Rmodule, and J is the Jacobson radical of R, and the images of m 1,..., m n M generate the quotient M/JM as an Rmodule, then the m 1,..., m n also generate M as an Rmodule. In our case, J = m R, and M/mM = M R k is generated as an Rmodule by one element x + mm. Nakayama s lemma then says that x M generates M as an Rmodule, so that M = R. The same follows for N. Back to our problem, if we assume that F G = O X, then taking stalks at any point x X we have F x Ox G x = Ox. By what we have just show, it follows that F x = Ox, so that F is invertible by part (b). Hartshorne, Exercise II.5.8. Again let X be a noetherian scheme, and F a coherent sheaf on X. We will consider the function ϕ(x) = dim k(x) F x Ox k(x), where k(x) = O x /m x is the residue field at the point x. Use Nakayama s lemma to prove the following results. (a) The function ϕ is uppersemicontinuous, i.e., for any n Z, the set {x X ϕ(x) n} is closed. (b) If F is locally free, and X is connected, then ϕ is a constant function. (c) Conversely, if X is reduced, and ϕ is constant, then F is locally free. Proof. (a) We will show that {x X ϕ(x) < n} is open. First, find an open affine neighborhood of x, U = Spec A, such that F U = M for some finitely generated A module M, and let m 1,..., m n be generators of M over A. Next, if x = p Spec A, 13 i=1
14 let [ x1 ] [ xk ],..., 1 1 be a k(p)basis of M p /pm p. Nakayama s lemma then states that (1) x 1 1,..., x k 1 is a minimal generating set of M p over A p. Write m i k 1 = a ij x j b ij 1 j=1 and let b = b ij and V = D(b). For and q D(b), (1) is still a generating set of M q over A q. By Nakayama s lemma again, it follows that ϕ(q) k < n so that p D(b) {x X ϕ(x) < n}, and thus {x X ϕ(x) < n} is open. (b) If F is locally free, then by part (b) of the previous exercise, its stalks F x are free O x  modules for all x X, and the rank of F x is equal to the rank of F at x. Thus for any open set U on which F is free, the value of ϕ on U is the rank of F on U, which must be finite because F is coherent. For any n, by the definition of locally free, both the set on which F has rank n and the set on which F has rank different from n are open. Since X is connected, it follows that there is some n such that ϕ(x) = n for all x X. (c) We may assume X = Spec A and F = M for some finitely generated Amodule M. Let p X, and suppose that ϕ(p) = n, so that by Nakayama s lemma there is a set m 1,..., m n M generating M p over A p (by clearing denominators we may assume the m i M). If x 1,..., x k generate M over A, write x i = n j=1 a ij b ij m j Let b = b ij. The same equation above then shows that m 1,..., m n generate M b over A b. Also, for any q D(b), the m i form a k(q)basis for M q /qm q because they are a generating set of nelements, and ϕ is constant. We now show that the generation of M b over A b is free. Suppose that we have a dependence relation n i=1 a i s i m i = 0 This gives a dependence relation in M q /qm q for each q D(b), which by linear independence, gives that a i q. It follows that a i N (A b ) = (N (A)) b = 0, because localization commutes with quotients, and A is reduced. This shows that the m i generate M b freely because coherent sheaves on an affine noetherian scheme are determined by their global sections (Corollary 5.5 in Hartshorne). 14 over A b. This is enough to show that M D(b) = O X n U
15 Hartshorne, Exercise II Vector Bundles. Let Y be a scheme. A (geometric) vector bundle of rank n over Y is a scheme X and a morphism f : X Y, together with additional data consisting of an open covering {U i } of Y, and isomorphisms ψ i : f 1 (U i ) A n U i, such that for any i, j, and for any open affine subset V = Spec A U i U j, the automorphism ψ = ψ j ψ 1 i of A n V = Spec A[x 1,..., x n ] is given by a linear automorphism θ of A[x 1,..., x n ], i.e., θ(a) = a for any a A, and θ(x i ) = a ij x j for suitable a ij A. An isomorphism g : (X, f, {U i }, {ψ i }) (X, f, {U i}, {ψ i}) of one vector bundle of rank n to another one is an isomorphism X X of underlying schemes, such that f = f g, and such that x, f together with the covering of Y consisting of all the U i and U i, and the isomorphisms ψ i and ψ i g, is also a vector bundle structure on X. (a) Let E be a locally free sheaf of rank n on a scheme Y. Let S(E ) be the symmetric algebra on E, and let X = Spec S(E ), with projection morphism f : X Y. For each open affine subset U Y for which E U is free, choose a basis of E, and let ψ : f 1 (U) A n U be the isomorphism resulting from the identification of S(E (U)) with O(U)[x 1,..., x n ]. Then (X, f, {U}, {ψ}) is a vector bundle of rank n over Y, which (up to isomorphism) does not depend on the bases of E U chosen. We call it the geometric vector bundle associated to E, and denote it by V(E ). (b) For any morphisms f : X Y, a section of f over an open set U Y is a morphism s : U X such that f s = id U. It is clear how to restrict sections to smaller open sets, or how to glue them together, so we see that the presheaf U {set of sections of f overu} is a sheaf of sets on Y, which we denote by S (X/Y ). Show that if f : X Y is a vector bundle of rank n, then the sheaf of sections S (X/Y ) has a natural structure of O Y module, which makes it a locally free O Y module of rank n. (c) Again let E be a locally free sheaf of rank n on Y, let X = V(E ), and let S = S (X/Y ) be the sheaf of sections of X over Y. Show that S = E, as follows. Given a section s Γ(V, E ) over any open set V, we think of s as an element of Hom(E V, O V ). So s determines an O V algebra homomorphism S(E V ) O V. This determines a morphism of spectra V = Spec O V Spec S(E V ) = f 1 (V ), which is a section of X/Y. Show that this construction gives an isomorphism of E to S. (d) Summing up, show that we have established a onetoone correspondence between isomorphism classes of locally free sheaves of rank n on Y, and isomorphism classes of vector bundles of rank n over Y. 15
MATH 8253 ALGEBRAIC GEOMETRY WEEK 12
MATH 8253 ALGEBRAIC GEOMETRY WEEK 2 CİHAN BAHRAN 3.2.. Let Y be a Noetherian scheme. Show that any Y scheme X of finite type is Noetherian. Moreover, if Y is of finite dimension, then so is X. Write f
More informationCHEVALLEY S THEOREM AND COMPLETE VARIETIES
CHEVALLEY S THEOREM AND COMPLETE VARIETIES BRIAN OSSERMAN In this note, we introduce the concept which plays the role of compactness for varieties completeness. We prove that completeness can be characterized
More informationAN INTRODUCTION TO AFFINE SCHEMES
AN INTRODUCTION TO AFFINE SCHEMES BROOKE ULLERY Abstract. This paper gives a basic introduction to modern algebraic geometry. The goal of this paper is to present the basic concepts of algebraic geometry,
More informationExercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo. Alex Massarenti
Exercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo Alex Massarenti SISSA, VIA BONOMEA 265, 34136 TRIESTE, ITALY Email address: alex.massarenti@sissa.it These notes collect a series of
More informationAlgebraic varieties and schemes over any scheme. Non singular varieties
Algebraic varieties and schemes over any scheme. Non singular varieties Trang June 16, 2010 1 Lecture 1 Let k be a field and k[x 1,..., x n ] the polynomial ring with coefficients in k. Then we have two
More information1. Algebraic vector bundles. Affine Varieties
0. Brief overview Cycles and bundles are intrinsic invariants of algebraic varieties Close connections going back to Grothendieck Work with quasiprojective varieties over a field k Affine Varieties 1.
More informationALGEBRAIC GROUPS. Disclaimer: There are millions of errors in these notes!
ALGEBRAIC GROUPS Disclaimer: There are millions of errors in these notes! 1. Some algebraic geometry The subject of algebraic groups depends on the interaction between algebraic geometry and group theory.
More information10. Smooth Varieties. 82 Andreas Gathmann
82 Andreas Gathmann 10. Smooth Varieties Let a be a point on a variety X. In the last chapter we have introduced the tangent cone C a X as a way to study X locally around a (see Construction 9.20). It
More informationABSTRACT NONSINGULAR CURVES
ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine nspace the collection A n k of points P = a 1, a,..., a
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43 RAVI VAKIL CONTENTS 1. Facts we ll soon know about curves 1 1. FACTS WE LL SOON KNOW ABOUT CURVES We almost know enough to say a lot of interesting things about
More informationLecture 6: Etale Fundamental Group
Lecture 6: Etale Fundamental Group October 5, 2014 1 Review of the topological fundamental group and covering spaces 1.1 Topological fundamental group Suppose X is a pathconnected topological space, and
More informationSystems of linear equations. We start with some linear algebra. Let K be a field. We consider a system of linear homogeneous equations over K,
Systems of linear equations We start with some linear algebra. Let K be a field. We consider a system of linear homogeneous equations over K, f 11 t 1 +... + f 1n t n = 0, f 21 t 1 +... + f 2n t n = 0,.
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. ANDREW SALCH 1. Affine schemes. About notation: I am in the habit of writing f (U) instead of f 1 (U) for the preimage of a subset
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More informationSCHEMES. David Harari. Tsinghua, FebruaryMarch 2005
SCHEMES David Harari Tsinghua, FebruaryMarch 2005 Contents 1. Basic notions on schemes 2 1.1. First definitions and examples.................. 2 1.2. Morphisms of schemes : first properties.............
More informationCHAPTER 1. AFFINE ALGEBRAIC VARIETIES
CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the
More informationMATH 221 NOTES BRENT HO. Date: January 3, 2009.
MATH 22 NOTES BRENT HO Date: January 3, 2009. 0 Table of Contents. Localizations......................................................................... 2 2. Zariski Topology......................................................................
More informationNONSINGULAR CURVES BRIAN OSSERMAN
NONSINGULAR CURVES BRIAN OSSERMAN The primary goal of this note is to prove that every abstract nonsingular curve can be realized as an open subset of a (unique) nonsingular projective curve. Note that
More informationReid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed.
Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed. Answer: Note that the first generator factors as (y
More information3. The Sheaf of Regular Functions
24 Andreas Gathmann 3. The Sheaf of Regular Functions After having defined affine varieties, our next goal must be to say what kind of maps between them we want to consider as morphisms, i. e. as nice
More information1 Flat, Smooth, Unramified, and Étale Morphisms
1 Flat, Smooth, Unramified, and Étale Morphisms 1.1 Flat morphisms Definition 1.1. An Amodule M is flat if the (rightexact) functor A M is exact. It is faithfully flat if a complex of Amodules P N Q
More informationMath 418 Algebraic Geometry Notes
Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = {P R
More informationSection Projective Morphisms
Section 2.7  Projective Morphisms Daniel Murfet October 5, 2006 In this section we gather together several topics concerned with morphisms of a given scheme to projective space. We will show how a morphism
More informationA course in. Algebraic Geometry. Taught by Prof. Xinwen Zhu. Fall 2011
A course in Algebraic Geometry Taught by Prof. Xinwen Zhu Fall 2011 1 Contents 1. September 1 3 2. September 6 6 3. September 8 11 4. September 20 16 5. September 22 21 6. September 27 25 7. September
More information12. Linear systems Theorem Let X be a scheme over a ring A. (1) If φ: X P n A is an Amorphism then L = φ O P n
12. Linear systems Theorem 12.1. Let X be a scheme over a ring A. (1) If φ: X P n A is an Amorphism then L = φ O P n A (1) is an invertible sheaf on X, which is generated by the global sections s 0, s
More informationHere is another way to understand what a scheme is 1.GivenaschemeX, and a commutative ring R, the set of Rvalued points
Chapter 7 Schemes III 7.1 Functor of points Here is another way to understand what a scheme is 1.GivenaschemeX, and a commutative ring R, the set of Rvalued points X(R) =Hom Schemes (Spec R, X) This is
More informationArithmetic Algebraic Geometry
Arithmetic Algebraic Geometry 2 Arithmetic Algebraic Geometry Travis Dirle December 4, 2016 2 Contents 1 Preliminaries 1 1.1 Affine Varieties.......................... 1 1.2 Projective Varieties........................
More informationDirect Limits. Mathematics 683, Fall 2013
Direct Limits Mathematics 683, Fall 2013 In this note we define direct limits and prove their basic properties. This notion is important in various places in algebra. In particular, in algebraic geometry
More informationResolution of Singularities in Algebraic Varieties
Resolution of Singularities in Algebraic Varieties Emma Whitten Summer 28 Introduction Recall that algebraic geometry is the study of objects which are or locally resemble solution sets of polynomial equations.
More information214A HOMEWORK KIM, SUNGJIN
214A HOMEWORK KIM, SUNGJIN 1.1 Let A = k[[t ]] be the ring of formal power series with coefficients in a field k. Determine SpecA. Proof. We begin with a claim that A = { a i T i A : a i k, and a 0 k }.
More informationAlgebraic Geometry Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry
More informationCHAPTER 0 PRELIMINARY MATERIAL. Paul Vojta. University of California, Berkeley. 18 February 1998
CHAPTER 0 PRELIMINARY MATERIAL Paul Vojta University of California, Berkeley 18 February 1998 This chapter gives some preliminary material on number theory and algebraic geometry. Section 1 gives basic
More informationwhich is a group homomorphism, such that if W V U, then
4. Sheaves Definition 4.1. Let X be a topological space. A presheaf of groups F on X is a a function which assigns to every open set U X a group F(U) and to every inclusion V U a restriction map, ρ UV
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local
More informationAlgebraic Geometry I Lectures 14 and 15
Algebraic Geometry I Lectures 14 and 15 October 22, 2008 Recall from the last lecture the following correspondences {points on an affine variety Y } {maximal ideals of A(Y )} SpecA A P Z(a) maximal ideal
More informationA course in. Algebraic Geometry. Taught by C. Birkar. Michaelmas 2012
A course in Algebraic Geometry Taught by C. Birkar Michaelmas 2012 Last updated: May 30, 2013 1 Disclaimer These are my notes from Caucher Birkar s Part III course on algebraic geometry, given at Cambridge
More informationne varieties (continued)
Chapter 2 A ne varieties (continued) 2.1 Products For some problems its not very natural to restrict to irreducible varieties. So we broaden the previous story. Given an a ne algebraic set X A n k, we
More informationMATH 233B, FLATNESS AND SMOOTHNESS.
MATH 233B, FLATNESS AND SMOOTHNESS. The discussion of smooth morphisms is one place were Hartshorne doesn t do a very good job. Here s a summary of this week s material. I ll also insert some (optional)
More informationCOMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY
COMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY BRIAN OSSERMAN Classical algebraic geometers studied algebraic varieties over the complex numbers. In this setting, they didn t have to worry about the Zariski
More informationSynopsis of material from EGA Chapter II, 4. Proposition (4.1.6). The canonical homomorphism ( ) is surjective [(3.2.4)].
Synopsis of material from EGA Chapter II, 4 4.1. Definition of projective bundles. 4. Projective bundles. Ample sheaves Definition (4.1.1). Let S(E) be the symmetric algebra of a quasicoherent O Y module.
More informationAlgebraic Geometry Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry
More informationVector Bundles vs. Jesko Hüttenhain. Spring Abstract
Vector Bundles vs. Locally Free Sheaves Jesko Hüttenhain Spring 2013 Abstract Algebraic geometers usually switch effortlessly between the notion of a vector bundle and a locally free sheaf. I will define
More information14 Lecture 14: Basic generallities on adic spaces
14 Lecture 14: Basic generallities on adic spaces 14.1 Introduction The aim of this lecture and the next two is to address general adic spaces and their connection to rigid geometry. 14.2 Two open questions
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More informationSMA. Grothendieck topologies and schemes
SMA Grothendieck topologies and schemes Rafael GUGLIELMETTI Semester project Supervised by Prof. Eva BAYER FLUCKIGER Assistant: Valéry MAHÉ April 27, 2012 2 CONTENTS 3 Contents 1 Prerequisites 5 1.1 Fibred
More informationMATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1
MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 CİHAN BAHRAN I discussed several of the problems here with Cheuk Yu Mak and Chen Wan. 4.1.12. Let X be a normal and proper algebraic variety over a field k. Show
More informationLecture 3: Flat Morphisms
Lecture 3: Flat Morphisms September 29, 2014 1 A crash course on Properties of Schemes For more details on these properties, see [Hartshorne, II, 15]. 1.1 Open and Closed Subschemes If (X, O X ) is a
More informationPROBLEMS, MATH 214A. Affine and quasiaffine varieties
PROBLEMS, MATH 214A k is an algebraically closed field Basic notions Affine and quasiaffine varieties 1. Let X A 2 be defined by x 2 + y 2 = 1 and x = 1. Find the ideal I(X). 2. Prove that the subset
More informationALGEBRAIC GEOMETRY: GLOSSARY AND EXAMPLES
ALGEBRAIC GEOMETRY: GLOSSARY AND EXAMPLES HONGHAO GAO FEBRUARY 7, 2014 Quasicoherent and coherent sheaves Let X Spec k be a scheme. A presheaf over X is a contravariant functor from the category of open
More informationAlgebraic Geometry Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry
More informationLocally Free Sheaves
Locally Free Sheaves Patrick Morandi Algebra Seminar, Spring 2002 In these talks we will discuss several important examples of locally free sheaves and see the connection between locally free sheaves and
More informationDIVISORS ON NONSINGULAR CURVES
DIVISORS ON NONSINGULAR CURVES BRIAN OSSERMAN We now begin a closer study of the behavior of projective nonsingular curves, and morphisms between them, as well as to projective space. To this end, we introduce
More informationNOTES ON ALGEBRAIC GEOMETRY MATH 202A. Contents Introduction Affine varieties 22
NOTES ON ALGEBRAIC GEOMETRY MATH 202A KIYOSHI IGUSA BRANDEIS UNIVERSITY Contents Introduction 1 1. Affine varieties 2 1.1. Weak Nullstellensatz 2 1.2. Noether s normalization theorem 2 1.3. Nullstellensatz
More informationSynopsis of material from EGA Chapter II, 3
Synopsis of material from EGA Chapter II, 3 3. Homogeneous spectrum of a sheaf of graded algebras 3.1. Homogeneous spectrum of a graded quasicoherent O Y algebra. (3.1.1). Let Y be a prescheme. A sheaf
More informationwhere m is the maximal ideal of O X,p. Note that m/m 2 is a vector space. Suppose that we are given a morphism
8. Smoothness and the Zariski tangent space We want to give an algebraic notion of the tangent space. In differential geometry, tangent vectors are equivalence classes of maps of intervals in R into the
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 27
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 27 RAVI VAKIL CONTENTS 1. Proper morphisms 1 2. Schemetheoretic closure, and schemetheoretic image 2 3. Rational maps 3 4. Examples of rational maps 5 Last day:
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37 RAVI VAKIL CONTENTS 1. Motivation and game plan 1 2. The affine case: three definitions 2 Welcome back to the third quarter! The theme for this quarter, insofar
More informationPreliminary Exam Topics Sarah Mayes
Preliminary Exam Topics Sarah Mayes 1. Sheaves Definition of a sheaf Definition of stalks of a sheaf Definition and universal property of sheaf associated to a presheaf [Hartshorne, II.1.2] Definition
More informationBEZOUT S THEOREM CHRISTIAN KLEVDAL
BEZOUT S THEOREM CHRISTIAN KLEVDAL A weaker version of Bézout s theorem states that if C, D are projective plane curves of degrees c and d that intersect transversally, then C D = cd. The goal of this
More informationMath 632 Notes Algebraic Geometry 2
Math 632 Notes Algebraic Geometry 2 Lectures by Karen Smith Notes by Daniel Hast Winter 2013 Contents 1 Affine schemes 4 1.1 Motivation and review of varieties........................ 4 1.2 First attempt
More informationThe most important result in this section is undoubtedly the following theorem.
28 COMMUTATIVE ALGEBRA 6.4. Examples of Noetherian rings. So far the only rings we can easily prove are Noetherian are principal ideal domains, like Z and k[x], or finite. Our goal now is to develop theorems
More informationLecture 2 Sheaves and Functors
Lecture 2 Sheaves and Functors In this lecture we will introduce the basic concept of sheaf and we also will recall some of category theory. 1 Sheaves and locally ringed spaces The definition of sheaf
More informationThis is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:
Math 6130 Notes. Fall 2002. 7. Basic Maps. Recall from 3 that a regular map of affine varieties is the same as a homomorphism of coordinate rings (going the other way). Here, we look at how algebraic properties
More informationSUMMER COURSE IN MOTIVIC HOMOTOPY THEORY
SUMMER COURSE IN MOTIVIC HOMOTOPY THEORY MARC LEVINE Contents 0. Introduction 1 1. The category of schemes 2 1.1. The spectrum of a commutative ring 2 1.2. Ringed spaces 5 1.3. Schemes 10 1.4. Schemes
More information9. Integral Ring Extensions
80 Andreas Gathmann 9. Integral ing Extensions In this chapter we want to discuss a concept in commutative algebra that has its original motivation in algebra, but turns out to have surprisingly many applications
More informationChern classes à la Grothendieck
Chern classes à la Grothendieck Theo Raedschelders October 16, 2014 Abstract In this note we introduce Chern classes based on Grothendieck s 1958 paper [4]. His approach is completely formal and he deduces
More informationRepresentations and Linear Actions
Representations and Linear Actions Definition 0.1. Let G be an Sgroup. A representation of G is a morphism of Sgroups φ G GL(n, S) for some n. We say φ is faithful if it is a monomorphism (in the category
More informationMatsumura: Commutative Algebra Part 2
Matsumura: Commutative Algebra Part 2 Daniel Murfet October 5, 2006 This note closely follows Matsumura s book [Mat80] on commutative algebra. Proofs are the ones given there, sometimes with slightly more
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationBasic Arithmetic Geometry. Lucien Szpiro. Based on notes by Florian Lengyel
Basic Arithmetic Geometry Lucien Szpiro Based on notes by Florian Lengyel Contents Notation and conventions 1 Chapter 1. The Picard Group 3 1. Tensor products and localization 3 2. Universal algebras
More informationALGEBRAIC GEOMETRY CAUCHER BIRKAR
ALGEBRAIC GEOMETRY CAUCHER BIRKAR Contents 1. Introduction 1 2. Affine varieties 3 Exercises 10 3. Quasiprojective varieties. 12 Exercises 20 4. Dimension 21 5. Exercises 24 References 25 1. Introduction
More informationYuriy Drozd. Intriduction to Algebraic Geometry. Kaiserslautern 1998/99
Yuriy Drozd Intriduction to Algebraic Geometry Kaiserslautern 1998/99 CHAPTER 1 Affine Varieties 1.1. Ideals and varieties. Hilbert s Basis Theorem Let K be an algebraically closed field. We denote by
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More informationSection Higher Direct Images of Sheaves
Section 3.8  Higher Direct Images of Sheaves Daniel Murfet October 5, 2006 In this note we study the higher direct image functors R i f ( ) and the higher coinverse image functors R i f! ( ) which will
More informationSmooth morphisms. Peter Bruin 21 February 2007
Smooth morphisms Peter Bruin 21 February 2007 Introduction The goal of this talk is to define smooth morphisms of schemes, which are one of the main ingredients in Néron s fundamental theorem [BLR, 1.3,
More informationCHEAT SHEET: PROPERTIES OF MORPHISMS OF SCHEMES
CHEAT SHEET: PROPERTIES OF MORPHISMS OF SCHEMES BRIAN OSSERMAN The purpose of this cheat sheet is to provide an easy reference for definitions of various properties of morphisms of schemes, and basic results
More informationThe Proj Construction
The Proj Construction Daniel Murfet May 16, 2006 Contents 1 Basic Properties 1 2 Functorial Properties 2 3 Products 6 4 Linear Morphisms 9 5 Projective Morphisms 9 6 Dimensions of Schemes 11 7 Points of
More informationINTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 14
INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 14 RAVI VAKIL Contents 1. Dimension 1 1.1. Last time 1 1.2. An algebraic definition of dimension. 3 1.3. Other facts that are not hard to prove 4 2. Nonsingularity:
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 5
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 5 RAVI VAKIL CONTENTS 1. The inverse image sheaf 1 2. Recovering sheaves from a sheaf on a base 3 3. Toward schemes 5 4. The underlying set of affine schemes 6 Last
More informationALGEBRA QUALIFYING EXAM SPRING 2012
ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.
More informationn P say, then (X A Y ) P
COMMUTATIVE ALGEBRA 35 7.2. The Picard group of a ring. Definition. A line bundle over a ring A is a finitely generated projective Amodule such that the rank function Spec A N is constant with value 1.
More informationLINE BUNDLES ON PROJECTIVE SPACE
LINE BUNDLES ON PROJECTIVE SPACE DANIEL LITT We wish to show that any line bundle over P n k is isomorphic to O(m) for some m; we give two proofs below, one following Hartshorne, and the other assuming
More information6.1 Finite generation, finite presentation and coherence
! Chapter 6 Coherent Sheaves 6.1 Finite generation, finite presentation and coherence Before getting to sheaves, let us discuss some finiteness properties for modules over a commutative ring R. Recall
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 18
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 18 CONTENTS 1. Invertible sheaves and divisors 1 2. Morphisms of schemes 6 3. Ringed spaces and their morphisms 6 4. Definition of morphisms of schemes 7 Last day:
More informationModules over a Scheme
Modules over a Scheme Daniel Murfet October 5, 2006 In these notes we collect various facts about quasicoherent sheaves on a scheme. Nearly all of the material is trivial or can be found in [Gro60]. These
More informationAlgebraic varieties. Chapter A ne varieties
Chapter 4 Algebraic varieties 4.1 A ne varieties Let k be a field. A ne nspace A n = A n k = kn. It s coordinate ring is simply the ring R = k[x 1,...,x n ]. Any polynomial can be evaluated at a point
More informationMATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties
MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently,
More informationSummer Algebraic Geometry Seminar
Summer Algebraic Geometry Seminar Lectures by Bart Snapp About This Document These lectures are based on Chapters 1 and 2 of An Invitation to Algebraic Geometry by Karen Smith et al. 1 Affine Varieties
More informationSolutions to some of the exercises from Tennison s Sheaf Theory
Solutions to some of the exercises from Tennison s Sheaf Theory Pieter Belmans June 19, 2011 Contents 1 Exercises at the end of Chapter 1 1 2 Exercises in Chapter 2 6 3 Exercises at the end of Chapter
More informationMath 145. Codimension
Math 145. Codimension 1. Main result and some interesting examples In class we have seen that the dimension theory of an affine variety (irreducible!) is linked to the structure of the function field in
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More informationAlgebraic v.s. Analytic Point of View
Algebraic v.s. Analytic Point of View Ziwen Zhu September 19, 2015 In this talk, we will compare 3 different yet similar objects of interest in algebraic and complex geometry, namely algebraic variety,
More informationSheaves. S. Encinas. January 22, 2005 U V. F(U) F(V ) s s V. = s j Ui Uj there exists a unique section s F(U) such that s Ui = s i.
Sheaves. S. Encinas January 22, 2005 Definition 1. Let X be a topological space. A presheaf over X is a functor F : Op(X) op Sets, such that F( ) = { }. Where Sets is the category of sets, { } denotes
More informationMASTER S THESIS MAT CHOW VARIETIES
MASTER S THESIS MAT200306 CHOW VARIETIES David Rydh DEPARTMENT OF MATHEMATICS ROYAL INSTITUTE OF TECHNOLOGY SE100 44 STOCKHOLM, SWEDEN Chow Varieties June, 2003 David Rydh Master s Thesis Department
More informationExtended Index. 89f depth (of a prime ideal) 121f ArtinRees Lemma. 107f descending chain condition 74f Artinian module
Extended Index cokernel 19f for Atiyah and MacDonald's Introduction to Commutative Algebra colon operator 8f Key: comaximal ideals 7f  listings ending in f give the page where the term is defined commutative
More informationConstruction of M B, M Dol, M DR
Construction of M B, M Dol, M DR Hendrik Orem Talbot Workshop, Spring 2011 Contents 1 Some Moduli Space Theory 1 1.1 Moduli of Sheaves: Semistability and Boundedness.............. 1 1.2 Geometric Invariant
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE ABOUT VARIETIES AND REGULAR FUNCTIONS.
ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS. ANDREW SALCH. More about some claims from the last lecture. Perhaps you have noticed by now that the Zariski topology
More informationRings and groups. Ya. Sysak
Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...
More informationMath 248B. Applications of base change for coherent cohomology
Math 248B. Applications of base change for coherent cohomology 1. Motivation Recall the following fundamental general theorem, the socalled cohomology and base change theorem: Theorem 1.1 (Grothendieck).
More informationProjective Varieties. Chapter Projective Space and Algebraic Sets
Chapter 1 Projective Varieties 1.1 Projective Space and Algebraic Sets 1.1.1 Definition. Consider A n+1 = A n+1 (k). The set of all lines in A n+1 passing through the origin 0 = (0,..., 0) is called the
More information