Math 203A - Solution Set 4
|
|
- Emily Burke
- 6 years ago
- Views:
Transcription
1 Math 203A - Solution Set 4 Problem 1. Let X and Y be prevarieties with affine open covers {U i } and {V j }, respectively. (i) Construct the product prevariety X Y by glueing the affine varieties U i V j together. (ii) Show that there are projection morphisms π X : X Y X and π Y : X Y Y satisfying the usual universal property for products: given morphisms f : Z X and g : Z Y from any prevariety Z, there is a unique morphism h : Z X Y such that f = π X h and g = π Y h. (iii) Show that if X and Y are varieties then X Y is also a variety. Answer: (i) We have already seen that products of affine algebraic sets are well defined. Thus, by restriction, it follows that products of quasi-affine sets are well defined. Let U ij = U i U j and V ij = V i V j. We glue the products of affine algebraic sets U i V k and U j V l along the products of quasiaffine sets U ij V kl. It is clear that the cocycle condition on triple overlaps for the gluing maps is satisfied. (ii) It is clear that the projections π X and π Y are morphisms. Assume f, g, Z are given. Set theoretically, we define h(z) = (f(z), g(z)) : Z X Y. We need to check that h is a morphism. Clearly h is continuous. We check that if φ is a regular function on some open set W X Y then h φ is regular on h 1 W. Let Z ij be the preimages of U i V j under the morphism h. Clearly, Z ij is open, hence it is a prevariety, hence it can be covered by affines Zij l. Since the restriction h : Wij l U i V j is a morphism of affine sets, it must correspond to a polynomial map. Thus h φ is clearly regular on h 1 (W ) Zij l e.g. it gives rise to a section of the sheaf O Z (h 1 (W ) Zij l ). These sections agree on overlaps and give a regular section of O X (h 1 (W )) which must be equal to h φ. This proves the claim. It follows from the universal property that X Y is unique up to isomorphism. (iii) If h 1, h 2 : Z X Y, we obtain morphisms We have f 1, f 2 : Z X, g 1, g 2 : Z Y. {z Z : h 1 (z) = h 2 (z)} = {z Z : f 1 (z) = f 2 (z)} {z Z : g 1 (z) = g 2 (z)}. The latter two sets are closed since X and Y are varieties, hence their intersection is also closed as well, showing that X Y is a variety. Problem 2. Let X be a prevariety. Consider pairs (U, f) where U is an open subset of X and f O X (U) a regular function on U. We call two such pairs (U, f) and (U, f ) equivalent if there is an open subset V in X with V U U such that f V = f V. 1
2 2 (i) Show that this defines an equivalence relation. (ii) Show that the set of all such pairs modulo this equivalence relation is a field. It is called the field of rational functions on X and denoted K(X). (iii) If X is an affine variety, show that K(X) is just the field of rational functions as defined in class. Answer: (i) Checking that we have defined an equivalence relation is straightforward. For instance, to verify transitivity, let (U, f), (U, f ) and (U, f ) such that f V = f V and f W = f W for some open sets V U U and W U U. We conclude f T = f T where T = V W U U. Clearly, T is nonemtpy since V, W are open dense in the irreducible set X. (ii) It is easy to see that addition and multiplication pass through the equivalence relation. We will check that non-zero elements have inverses. Let (f, U) be a non-zero element. In particular Z(f) U. Consider the function (1/f, U \ Z(f)) which is regular on the nonempty set U \ Z(f). This is the inverse of (f, U). (iii) If f is a rational function as defined in class, we take U = Domain (f) which is open in X. We obtain an equivalence class (U, f) as in the problem. Conversely, if (f, U) is an equivalence class as in the problem set, then f is regular on U hence by definition, it must be a rational function on X. This assignment (f, U) f passes through the equivalence relation. Indeed, if (f, U ) is another representative of the equivalence class, then f = f on some open set V. Write f = g/h and f = g /h. We show that g/h and g /h define the same element of the function field. Indeed, g h = g h on V = gh g h = 0 on V = gh g h = 0 on X = g h = g as elements of K(X). h For the third implication, we used that V is dense in X. Problem 3. (i) Show that every isomorphism f : A 1 A 1 is of the form f(x) = ax + b. (ii) Show that every isomorphism f : P 1 P 1 is of the form f(x) = ax+b cx+d for some a, b, c, d k, where x is an affine coordinate on A 1 P1. (iii) Given three distinct points P 1, P 2, P 3 P 1 and three distinct points Q 1, Q 2, Q 3 P 1, show that there is a unique isomorphism f : P 1 P 1 such that f(p i ) = Q i for i = 1, 2, 3. Answer: (i) Clearly, a morphism f : A 1 A 1 must be a polynomial map. If f has degree d, the equation f(x) = 0 will have d roots. Therefore, d = 1 hence f(x) = ax + b.
3 (ii) Let f : P 1 P 1. If f( ) =, then f maps A 1 A 1 and must have the form f(x) = ax + b. Otherwise, if f( ) = λ, let g(x) = 1 x λ. Then g f is an isomorphism which maps to itself hence it must have the form g f = cx + d. This shows f(x) = λ + 1 cx + d = ax + b cx + d for a = cλ, b = dλ + 1. (iii) By transitivity, it suffices to assume P 1 = 0, P 2 = 1, P 3 =. Also, let Q 0 = λ, Q 1 = µ, Q 2 = ν. Assume that none of the greek letters is. We require We may take b d = λ, a + b c + d = µ, a c = ν. a = ν(µ λ), b = λ(ν µ), c = µ λ, d = ν µ. If one of the greek letters is, say ν =, and µ, λ 0, we may apply the transformation x 1 x to reduce to the case we have already studied. If by contrast, λ = 0, then the automorphism f(x) = µx sends the P s to the Q s. If µ = 0, then we may take f(x) = λ(x 1). To prove uniqueness, assume that two morphisms f 1, f 2 have been constructed. The inverse f = f 1 f2 1 has 3 fixed points P 1, P 2, P 3. We may assume that these fixed points are 0, 1,. Since and we conclude f(x) = ax + b cx + d f(0) = 0, f(1) = 1, f( ) = b = 0, a + b = c + d, c = 0 = f = id = f 1 = f 2. Problem 4. Let X P 2 be a projective variety. A morphism f : P 1 X is a polynomial map f([x : y]) = (f 0 ([x : y]), f 1 ([x : y]), f 2 ([x : y])), where f 0, f 1, f 2 are homogeneous polynomials of the same degree, such that f(p 1 ) X. Prove the following facts about lines and conics in projective plane: (i) For any line L P 2, there is a bijective morphism f : P 1 L. (ii) For any irreducible conic C P 2, there is a bijective morphism f : P 1 C. You may wish to change coordinates so that your conic has a convenient expression. 3
4 4 (iii) Consider the elliptic curve E λ P 2 : y 2 z = x(x z)(x λz). Show that there are no nonconstant morphisms P 1 E λ P 2. Therefore, elliptic curves are not rational curves. Answer: (i) Let L be the line αx+βy+γz = 0. We may assume that γ 0, eventually relabeling the coordinates if necessary. Set f : P 1 L, [x : y] [x : y : αγ x βγ ] y, and g : L P 1 [x : y : z] [x : y]. It is easy to see that both f and g are well defined inverse isomorphisms. (ii) Changing coordinates, we may assume the conic C is given by the equation Set It is easy to check that xz = y 2. f : P 1 C, [s : t] [s 2 : st : t 2 ]. g : C P 1, g([x : y : z]) = is an inverse morphism of f. (iii) Assume that F : P 1 E λ P 2 is a morphism with where { [x : y] if (x, y) (0, 0) [y : z] if (y, z) (0, 0) F ([x : y]) = [f 0 ([x : y]) : f 1 ([x : y]) : f 2 ([x : y])], f 2 1 f 2 = f 0 (f 0 f 1 )(f 0 λf 1 ). If f 2 0, it follows from the equation above that f 0 = 0, hence F is the constant [0 : 1 : 0]. Let us assume f 2 0. Define ( f0 (t, 1) f(t) = f 2 (t, 1), f ) 1(t, 1). f 2 (t, 1) It is clear that f defines a rational map f : A 1 which must be constant by the previous pset. This implies that f 0 (t, 1)/f 2 (t, 1) and f 1 (t, 1)/f 2 (t, 1) are constants. It follows that all points [t : 1] take the same value under F. Similarly, F sends all [1 : t] to the same value, and therefore F must be constant. E λ
5 Problem 5. Part A. Show that a line and an irreducible conic in P 2 cannot intersect in 3 points. Part B. (i) Four points in P 2 are said to be in general position if no three are collinear (i.e. lie on a projective line in the projective plane). Show that if p 1,..., p 4 are points in general position, there exists a linear change of coordinates with T : P 2 P 2 T ([1 : 0 : 0]) = p 1, T ([0 : 1 : 0]) = p 2, T ([0 : 0 : 1]) = p 3, T ([1 : 1 : 1]) = p 4. (ii) Given five distinct points in P 2, no three of which are collinear, show that there is an unique irreducible projective conic passing though all five points. You may want to use part (i) to assume that four of the points are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1]. (iii) Deduce that two distinct irreducible conics in P 2 cannot intersect in 5 points. (We will see later that they intersect in exactly 4 points counted with multiplicity.) Answer: Part A. Changing coordinates, we may assume that the conic C is xz = y 2. Let L be the line αx + βy + γz = 0. If β = 0, it is clear that C and L intersect at the points If β 0, then which gives [ γ : ± αγ : α]. y = α β x γ β z xz = 1 β 2 (αx + γz)2. If x = 0, then y = z = 0 which is not allowed. Therefore, we may assume x 0. Dividing by x 2 we obtain the quadratic equation in z x : γ 2 ( z ) ( ) 2 2αγ z + β 2 x β 2 1 x + α2 β 2 = 0. Letting λ 1, λ 2 be the solutions of this equation, we see that the intersection points are [1 : αβ γβ λ i : λ i ]. Part B. (i) Let p i = [a i : b i : c i ] for 1 i 4. Define A = αa 1 αb 1 αc 1 βa 2 βb 2 βc 2, γa 3 γb 3 γc 3 where α, β, γ will be specified later. In fact, we will require that α, β, γ solve the system αa 1 + βb 1 + γc 1 = a 4, 5
6 6 αa 2 + βb 2 + γc 2 = b 4, αa 3 + βb 3 + γc 3 = c 4. A solution exists since the matrix of coefficients B = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 is invertible. Indeed, the rows of B are independent. Otherwise, a nontrivial linear relation between the rows would give a line on which the points p 1, p 2, p 3 lie. Thus B is invertible. Now, the system above has the solution α β γ = B 1 a 4 b 4 c 4 Note that the same argument shows that A is invertible. Let S : P 2 P 2 be the linear transformation defined by A. Then S is invertible. A direct computation shows S([1 : 0 : 0]) = p 1, S([0 : 1 : 0]) = p 2, S([0 : 0 : 1]) = p 3, S([1 : 1 : 1]) = p 4. The proof is completed letting T be the inverse of S. (ii) After a linear change of coordinates, we may assume that the five points are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1] and [u : v : w]. The equation f(x, y, z) of any conic passing through the first three points can t contain x 2, y 2, z 2, so. f(x, y, z) = ayz + bxz + cxy. The remaining two points impose the conditions Letting a + b + c = avw + buw + cuv = 0. A = ( vw uw uv we see that (a, b, c) must be in the null space of A. The rank of A is 2 (the rank cannot be 1 since then the rows would be proportional, hence u = v = w which is not allowed). Therefore, the null space of this matrix is one dimensional, hence the conic passing through the 5 points is unique. The conic cannot be reducible since then it would be union of two lines. One of the lines would have to contain 3 points but that contradicts the general position assumption. (iii) By Part A, a conic and a line cannot intersect in 3 points. Therefore, any 5 points on an irreducible conic are in general position. Now, the claim is evident by (ii). Problem 6. We will make the space of all lines in P n into a projective variety. We define a set-theoretic map φ : {lines in P n } P N ),
7 7 with ( ) n + 1 N = 1 2 as follows. For every line L P n, choose two distinct points P = (a 0... a n ) and Q = (b 0... b n ) on L and define φ(l) to be the point in P N whose homogeneous coordinates are the maximal minors of the matrix ( ) a0 :... : a n in any fixed order. Show that: b 0 :... : b n (i) The map φ is well-defined and injective. The map φ is called the Plucker embedding. (ii) The image of φ is a projective variety that has a finite cover by affine spaces A 2(n 1). You may want to recall the Gaussian algorithm which brings almost any matrix as above into the form ( 1 0 a 2... a ) n 0 1 b b n (iii) Show that G(1, 1) is a point, G(1, 2) = P 2, and G(1, 3) is the zero locus of a quadratic equation in P 5. Answer: (i) Let e 0,..., e n be the standard basis of the vector space V = k n+1. A line L corresponds to a 2-dimensional subspace in k n+1, also denoted by L. We claim that the map φ can be described as follows. Picking a basis v, w for the subspace, φ(l) = [v w] P(Λ 2 V ) = P N. Indeed, if P = (a 0 :... : a n ) and Q = (b 0 :... : b n ) are two distinct points, then we may take Thus v w = i,j v = a i e i, w = b i e i. a i b j e i e j = i<j (a i b j a j b i )e i e j. Therefore, in the basis e i e j, the coordinates are the 2 2-minors of the matrix in (i). This shows that φ is well-defined. To check injectivity, let L be another line corresponding to a 2-dimensional subspace. If L L = 0, then pick a basis v 0, v 1, v 2, v 3 for L L with v 0, v 1 L, v 2, v 3 L, and extend it to a basis v 0,..., v n of V = k n+1. Thus v i v j for i < j is a basis for Λ 2 V. In particular, v 0 v 1 and v 2 v 3 are not multiples, or equivalently φ(l) φ(l ). The same argument applies if L and L have a 1 dimensional intersection.
8 8 (ii) To show projectivity, we will prove that then ω ω = In particular, if i<j,k<l ω Λ 2 V splits as ω = v w if and only if ω ω = 0. ω ij ω kl e i e j e k e l = ω = ω ij e i e j, i<j<k<l so the image of φ is cut by the quadrics ω ij ω kl ω ik ω jl + ω il ω jk = 0. (ω ij ω kl ω ik ω jl + ω il ω jk )e i e j e k e l We now prove the claim. It is clear that if ω = v w then ω ω = 0. Conversely, we will induct on n, the base case n = 2 being clear. Let us write ω = e 0 η + ω where ω, η do not contain the vector e 0. Thus This implies that hence by induction 0 = ω ω = 2e 0 η ω + ω ω. ω ω = 0 ω = v w, with v, w being in the subspace spanned by e 1,..., e n. Also, we know e 0 η ω = 0 = η v w = 0. This shows that η cannot be independent of v, w hence Collecting terms we find η = av + bw. ω = e 0 (av + bw) + v w = (v + be 0 ) (w + ae 0 ) as claimed. For the last part, note that one of the coordinates of φ(l) must be non-zero. Without loss of generality let us assume it is the coordinate corresponding to e 0 e 1. This means that the first 2 2 minor of the matrix ( ) a0 :... : a n b 0 :... : b n is non-zero. The Gaussian algorithm brings this matrix into the form ( 1 0 a 2... a ) n 0 1 b b n The association L (a 2,..., a n, b 2,..., b n) A 2(n 1) shows how to cover the image of φ by affine opens isomorphic to A 2(n 1).
9 (iii) All these statements are particular cases of what we proved in part (i). For instance, to see that G(1, 3) is a quadric in P 5 given by x 0 x 5 x 1 x 4 + x 2 x 3 = 0. 9
Math Midterm Solutions
Math 145 - Midterm Solutions Problem 1. (10 points.) Let n 2, and let S = {a 1,..., a n } be a finite set with n elements in A 1. (i) Show that the quasi-affine set A 1 \ S is isomorphic to an affine set.
More informationMath 203, Solution Set 4.
Math 203, Solution Set 4. Problem 1. Let V be a finite dimensional vector space and let ω Λ 2 V be such that ω ω = 0. Show that ω = v w for some vectors v, w V. Answer: It is clear that if ω = v w then
More informationMath 203A - Solution Set 1
Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationMODEL ANSWERS TO HWK #3
MODEL ANSWERS TO HWK #3 1. Suppose that the point p = [v] and that the plane H corresponds to W V. Then a line l containing p, contained in H is spanned by the vector v and a vector w W, so that as a point
More informationExercise Sheet 3 - Solutions
Algebraic Geometry D-MATH, FS 2016 Prof. Pandharipande Exercise Sheet 3 - Solutions 1. Prove the following basic facts about algebraic maps. a) For f : X Y and g : Y Z algebraic morphisms of quasi-projective
More informationExercise Sheet 7 - Solutions
Algebraic Geometry D-MATH, FS 2016 Prof. Pandharipande Exercise Sheet 7 - Solutions 1. Prove that the Zariski tangent space at the point [S] Gr(r, V ) is canonically isomorphic to S V/S (or equivalently
More informationD-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski. Solutions Sheet 1. Classical Varieties
D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski Solutions Sheet 1 Classical Varieties Let K be an algebraically closed field. All algebraic sets below are defined over K, unless specified otherwise.
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 informationMath 203A - Solution Set 3
Math 03A - Solution Set 3 Problem 1 Which of the following algebraic sets are isomorphic: (i) A 1 (ii) Z(xy) A (iii) Z(x + y ) A (iv) Z(x y 5 ) A (v) Z(y x, z x 3 ) A Answer: We claim that (i) and (v)
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 informationLECTURE 10, MONDAY MARCH 15, 2004
LECTURE 10, MONDAY MARCH 15, 2004 FRANZ LEMMERMEYER 1. Minimal Polynomials Let α and β be algebraic numbers, and let f and g denote their minimal polynomials. Consider the resultant R(X) of the polynomials
More informationMath 203A - Solution Set 1
Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationPROBLEMS, MATH 214A. Affine and quasi-affine varieties
PROBLEMS, MATH 214A k is an algebraically closed field Basic notions Affine and quasi-affine 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 informationSPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS
SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly
More informationVector bundles in Algebraic Geometry Enrique Arrondo. 1. The notion of vector bundle
Vector bundles in Algebraic Geometry Enrique Arrondo Notes(* prepared for the First Summer School on Complex Geometry (Villarrica, Chile 7-9 December 2010 1 The notion of vector bundle In affine geometry,
More information2.3. VECTOR SPACES 25
2.3. VECTOR SPACES 25 2.3 Vector Spaces MATH 294 FALL 982 PRELIM # 3a 2.3. Let C[, ] denote the space of continuous functions defined on the interval [,] (i.e. f(x) is a member of C[, ] if f(x) is continuous
More information12. Linear systems Theorem Let X be a scheme over a ring A. (1) If φ: X P n A is an A-morphism 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 A-morphism 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 informationMath 203A, Solution Set 6.
Math 203A, Solution Set 6. Problem 1. (Finite maps.) Let f 0,..., f m be homogeneous polynomials of degree d > 0 without common zeros on X P n. Show that gives a finite morphism onto its image. f : X P
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 informationSpring, 2006 CIS 610. Advanced Geometric Methods in Computer Science Jean Gallier Homework 1
Spring, 2006 CIS 610 Advanced Geometric Methods in Computer Science Jean Gallier Homework 1 January 23, 2006; Due February 8, 2006 A problems are for practice only, and should not be turned in. Problem
More informationELEMENTARY SUBALGEBRAS OF RESTRICTED LIE ALGEBRAS
ELEMENTARY SUBALGEBRAS OF RESTRICTED LIE ALGEBRAS J. WARNER SUMMARY OF A PAPER BY J. CARLSON, E. FRIEDLANDER, AND J. PEVTSOVA, AND FURTHER OBSERVATIONS 1. The Nullcone and Restricted Nullcone We will need
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 informationInstitutionen för matematik, KTH.
Institutionen för matematik, KTH. Contents 7 Affine Varieties 1 7.1 The polynomial ring....................... 1 7.2 Hypersurfaces........................... 1 7.3 Ideals...............................
More informationMath 203A - Solution Set 1
Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
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 informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013 As usual, a curve is a smooth projective (geometrically irreducible) variety of dimension one and k is a perfect field. 23.1
More informationTopics In Algebra Elementary Algebraic Geometry
Topics In Algebra Elementary Algebraic Geometry David Marker Spring 2003 Contents 1 Algebraically Closed Fields 2 2 Affine Lines and Conics 14 3 Projective Space 23 4 Irreducible Components 40 5 Bézout
More informationFOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 48
FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 48 RAVI VAKIL CONTENTS 1. A little more about cubic plane curves 1 2. Line bundles of degree 4, and Poncelet s Porism 1 3. Fun counterexamples using elliptic curves
More informationSection Blowing Up
Section 2.7.1 - Blowing Up Daniel Murfet October 5, 2006 Now we come to the generalised notion of blowing up. In (I, 4) we defined the blowing up of a variety with respect to a point. Now we will define
More informationLinear Algebra. Paul Yiu. Department of Mathematics Florida Atlantic University. Fall 2011
Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1A: Vector spaces Fields
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 informationMATH 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 informationAlgebraic Geometry. Question: What regular polygons can be inscribed in an ellipse?
Algebraic Geometry Question: What regular polygons can be inscribed in an ellipse? 1. Varieties, Ideals, Nullstellensatz Let K be a field. We shall work over K, meaning, our coefficients of polynomials
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 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 informationOn Maps Taking Lines to Plane Curves
Arnold Math J. (2016) 2:1 20 DOI 10.1007/s40598-015-0027-1 RESEARCH CONTRIBUTION On Maps Taking Lines to Plane Curves Vsevolod Petrushchenko 1 Vladlen Timorin 1 Received: 24 March 2015 / Accepted: 16 October
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 informationExercises for algebraic curves
Exercises for algebraic curves Christophe Ritzenthaler February 18, 2019 1 Exercise Lecture 1 1.1 Exercise Show that V = {(x, y) C 2 s.t. y = sin x} is not an algebraic set. Solutions. Let us assume that
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 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 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 informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field. 17.1 Tangent spaces and hypersurfaces For any polynomial f k[x
More informationBasic facts and definitions
Synopsis Thursday, September 27 Basic facts and definitions We have one one hand ideals I in the polynomial ring k[x 1,... x n ] and subsets V of k n. There is a natural correspondence. I V (I) = {(k 1,
More informationLinear Vector Spaces
CHAPTER 1 Linear Vector Spaces Definition 1.0.1. A linear vector space over a field F is a triple (V, +, ), where V is a set, + : V V V and : F V V are maps with the properties : (i) ( x, y V ), x + y
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 informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #24 12/03/2013
18.78 Introduction to Arithmetic Geometry Fall 013 Lecture #4 1/03/013 4.1 Isogenies of elliptic curves Definition 4.1. Let E 1 /k and E /k be elliptic curves with distinguished rational points O 1 and
More informationbut no smaller power is equal to one. polynomial is defined to be
13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said
More informationFinite affine planes in projective spaces
Finite affine planes in projective spaces J. A.Thas H. Van Maldeghem Ghent University, Belgium {jat,hvm}@cage.ugent.be Abstract We classify all representations of an arbitrary affine plane A of order q
More informationMATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11
MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11 3. Examples I did some examples and explained the theory at the same time. 3.1. roots of unity. Let L = Q(ζ) where ζ = e 2πi/5 is a primitive 5th root of
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 information9. Birational Maps and Blowing Up
72 Andreas Gathmann 9. Birational Maps and Blowing Up In the course of this class we have already seen many examples of varieties that are almost the same in the sense that they contain isomorphic dense
More informationNoetherian property of infinite EI categories
Noetherian property of infinite EI categories Wee Liang Gan and Liping Li Abstract. It is known that finitely generated FI-modules over a field of characteristic 0 are Noetherian. We generalize this result
More informationAlgebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.
More informationNOTES (1) FOR MATH 375, FALL 2012
NOTES 1) FOR MATH 375, FALL 2012 1 Vector Spaces 11 Axioms Linear algebra grows out of the problem of solving simultaneous systems of linear equations such as 3x + 2y = 5, 111) x 3y = 9, or 2x + 3y z =
More informationHARTSHORNE EXERCISES
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
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 quasi-projective varieties over a field k Affine Varieties 1.
More informationIRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.
IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. IAN KIMING 1. Non-singular points and tangents. Suppose that k is a field and that F (x 1,..., x n ) is a homogeneous polynomial in n variables
More informationAlgebraic Varieties. Chapter Algebraic Varieties
Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) :
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 n-space the collection A n k of points P = a 1, a,..., a
More information(1) is an invertible sheaf on X, which is generated by the global sections
7. Linear systems First a word about the base scheme. We would lie to wor in enough generality to cover the general case. On the other hand, it taes some wor to state properly the general results if one
More informationSPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS
SPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS ROYA BEHESHTI AND N. MOHAN KUMAR Abstract. We prove that the space of smooth rational curves of degree e in a general complete intersection of multidegree
More information12. Hilbert Polynomials and Bézout s Theorem
12. Hilbert Polynomials and Bézout s Theorem 95 12. Hilbert Polynomials and Bézout s Theorem After our study of smooth cubic surfaces in the last chapter, let us now come back to the general theory of
More informationπ X : X Y X and π Y : X Y Y
Math 6130 Notes. Fall 2002. 6. Hausdorffness and Compactness. We would like to be able to say that all quasi-projective varieties are Hausdorff and that projective varieties are the only compact varieties.
More informationALGEBRAIC GROUPS J. WARNER
ALGEBRAIC GROUPS J. WARNER Let k be an algebraically closed field. varieties unless otherwise stated. 1. Definitions and Examples For simplicity we will work strictly with affine Definition 1.1. An algebraic
More informationdiv(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let:
Algebraic Curves/Fall 015 Aaron Bertram 4. Projective Plane Curves are hypersurfaces in the plane CP. When nonsingular, they are Riemann surfaces, but we will also consider plane curves with singularities.
More informationALGEBRAIC GEOMETRY (NMAG401) Contents. 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30
ALGEBRAIC GEOMETRY (NMAG401) JAN ŠŤOVÍČEK Contents 1. Affine varieties 1 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30 1. Affine varieties The basic objects
More informationChapter 2: Linear Independence and Bases
MATH20300: Linear Algebra 2 (2016 Chapter 2: Linear Independence and Bases 1 Linear Combinations and Spans Example 11 Consider the vector v (1, 1 R 2 What is the smallest subspace of (the real vector space
More information1 Structures 2. 2 Framework of Riemann surfaces Basic configuration Holomorphic functions... 3
Compact course notes Riemann surfaces Fall 2011 Professor: S. Lvovski transcribed by: J. Lazovskis Independent University of Moscow December 23, 2011 Contents 1 Structures 2 2 Framework of Riemann surfaces
More informationAlgebraic Geometry of Matrices II
Algebraic Geometry of Matrices II Lek-Heng Lim University of Chicago July 3, 2013 today Zariski topology irreducibility maps between varieties answer our last question from yesterday again, relate to linear
More informationRepresentations and Linear Actions
Representations and Linear Actions Definition 0.1. Let G be an S-group. A representation of G is a morphism of S-groups φ G GL(n, S) for some n. We say φ is faithful if it is a monomorphism (in the category
More informationSolutions of exercise sheet 6
D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 6 1. (Irreducibility of the cyclotomic polynomial) Let n be a positive integer, and P Z[X] a monic irreducible factor of X n 1
More informationPolynomials, Ideals, and Gröbner Bases
Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields
More informationReview of Linear Algebra
Review of Linear Algebra Throughout these notes, F denotes a field (often called the scalars in this context). 1 Definition of a vector space Definition 1.1. A F -vector space or simply a vector space
More informationReview of Some Concepts from Linear Algebra: Part 2
Review of Some Concepts from Linear Algebra: Part 2 Department of Mathematics Boise State University January 16, 2019 Math 566 Linear Algebra Review: Part 2 January 16, 2019 1 / 22 Vector spaces A set
More informationAlgebraic Geometry. 1 4 April David Philipson. 1.1 Basic Definitions. 1.2 Noetherian Rings
Algebraic Geometry David Philipson 1 4 April 2008 Notes for this day courtesy of Yakov Shlapentokh-Rothman. 1.1 Basic Definitions Throughout this course, we let k be an algebraically closed field with
More informationMath 113 Winter 2013 Prof. Church Midterm Solutions
Math 113 Winter 2013 Prof. Church Midterm Solutions Name: Student ID: Signature: Question 1 (20 points). Let V be a finite-dimensional vector space, and let T L(V, W ). Assume that v 1,..., v n is a basis
More informationMath 121 Homework 2 Solutions
Math 121 Homework 2 Solutions Problem 13.2 #16. Let K/F be an algebraic extension and let R be a ring contained in K that contains F. Prove that R is a subfield of K containing F. We will give two proofs.
More information4. Images of Varieties Given a morphism f : X Y of quasi-projective varieties, a basic question might be to ask what is the image of a closed subset
4. Images of Varieties Given a morphism f : X Y of quasi-projective varieties, a basic question might be to ask what is the image of a closed subset Z X. Replacing X by Z we might as well assume that Z
More informationFILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.
FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0
More informationWell Ordered Sets (continued)
Well Ordered Sets (continued) Theorem 8 Given any two well-ordered sets, either they are isomorphic, or one is isomorphic to an initial segment of the other. Proof Let a,< and b, be well-ordered sets.
More informationOHSX XM511 Linear Algebra: Multiple Choice Exercises for Chapter 2
OHSX XM5 Linear Algebra: Multiple Choice Exercises for Chapter. In the following, a set is given together with operations of addition and scalar multiplication. Which is not a vector space under the given
More informationALGEBRAIC GEOMETRY CAUCHER BIRKAR
ALGEBRAIC GEOMETRY CAUCHER BIRKAR Contents 1. Introduction 1 2. Affine varieties 3 Exercises 10 3. Quasi-projective varieties. 12 Exercises 20 4. Dimension 21 5. Exercises 24 References 25 1. Introduction
More informationChapter 2 Linear Transformations
Chapter 2 Linear Transformations Linear Transformations Loosely speaking, a linear transformation is a function from one vector space to another that preserves the vector space operations. Let us be more
More informationGALOIS THEORY BRIAN OSSERMAN
GALOIS THEORY BRIAN OSSERMAN Galois theory relates the theory of field extensions to the theory of groups. It provides a powerful tool for studying field extensions, and consequently, solutions to polynomial
More informationCLASSIFYING QUADRATIC QUANTUM P 2 S BY USING GRADED SKEW CLIFFORD ALGEBRAS
CLASSIFYING QUADRATIC QUANTUM P 2 S BY USING GRADED SKEW CLIFFORD ALGEBRAS Manizheh Nafari 1 manizheh@uta.edu Michaela Vancliff 2 vancliff@uta.edu uta.edu/math/vancliff Jun Zhang zhangjun19@gmail.com Department
More informationmult V f, where the sum ranges over prime divisor V X. We say that two divisors D 1 and D 2 are linearly equivalent, denoted by sending
2. The canonical divisor In this section we will introduce one of the most important invariants in the birational classification of varieties. Definition 2.1. Let X be a normal quasi-projective variety
More informationPolynomials with nontrivial relations between their roots
ACTA ARITHMETICA LXXXII.3 (1997) Polynomials with nontrivial relations between their roots by John D. Dixon (Ottawa, Ont.) 1. Introduction. Consider an irreducible polynomial f(x) over a field K. We are
More informationElliptic curves over Q p
Rosa Winter rwinter@math.leidenuniv.nl Elliptic curves over Q p Bachelor thesis, August 23, 2011 Supervisor: Drs. R. Pannekoek Mathematisch Instituut, Universiteit Leiden Contents Introduction 3 1 The
More informationMAS345 Algebraic Geometry of Curves: Exercises
MAS345 Algebraic Geometry of Curves: Exercises AJ Duncan, September 22, 2003 1 Drawing Curves In this section of the exercises some methods of visualising curves in R 2 are investigated. All the curves
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 informationLecture Notes for the Algebraic Geometry course held by Rahul Pandharipande
Lecture Notes for the Algebraic Geometry course held by Rahul Pandharipande Endrit Fejzullahu, Nikolas Kuhn, Vlad Margarint, Nicolas Müller, Samuel Stark, Lazar Todorovic July 28, 2014 Contents 0 References
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 informationMath 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d
Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v).
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS
ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS Your Name: Conventions: all rings and algebras are assumed to be unital. Part I. True or false? If true provide a brief explanation, if false provide a counterexample
More informationUnless otherwise specified, V denotes an arbitrary finite-dimensional vector space.
MAT 90 // 0 points Exam Solutions Unless otherwise specified, V denotes an arbitrary finite-dimensional vector space..(0) Prove: a central arrangement A in V is essential if and only if the dual projective
More informationMATH 225 Summer 2005 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 2005
MATH 225 Summer 25 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 25 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 224. #2] The set of all
More informationA ne Algebraic Varieties Undergraduate Seminars: Toric Varieties
A ne Algebraic Varieties Undergraduate Seminars: Toric Varieties Lena Ji February 3, 2016 Contents 1. Algebraic Sets 1 2. The Zariski Topology 3 3. Morphisms of A ne Algebraic Sets 5 4. Dimension 6 References
More informationALGEBRAIC GEOMETRY I, FALL 2016.
ALGEBRAIC GEOMETRY I, FALL 2016. DIVISORS. 1. Weil and Cartier divisors Let X be an algebraic variety. Define a Weil divisor on X as a formal (finite) linear combination of irreducible subvarieties of
More informationGLUING SCHEMES AND A SCHEME WITHOUT CLOSED POINTS
GLUING SCHEMES AND A SCHEME WITHOUT CLOSED POINTS KARL SCHWEDE Abstract. We first construct and give basic properties of the fibered coproduct in the category of ringed spaces. We then look at some special
More information