Math 203A - Solution Set 1
|
|
- Anthony Lawrence
- 5 years ago
- Views:
Transcription
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 the Zariski topology of A 2. We claim this Z is not closed in the product topology of A 1 A 1. Assume otherwise. Then, the complement A 2 \ Z is open in the product topology. Therefore, there are nonempty open sets U, V in A 1 such that U V A 2 \ Z. Now, the sets U and V have finite complements: We obtain a contradiction picking In this case, we clearly have U = A 1 \ {p 1,..., p n }, V = A 1 \ {q 1,..., q m }. z A 1 \ {p 1,..., p n, q 1,..., q m }. (z, z) U V, while (z, z) A 2 \ Z. Problem 2. A topological space X is said to be Noetherian if it satisfies the ascending chain condition on open sets, i.e. any ascending chain of open sets eventually stabilizes. (i) Check that any subset Y X of a Noetherian space is also Noetherian in the subspace topology. (ii) Show that a Noetherian space is quasi-compact i.e., show that any open cover of a Noetherian space has a finite subcover. (iii) Check that a Noetherian space which is also Hausdorff must be a finite set of points. (iv) Show that A n is Noetherian in the Zariski topology. Conclude that any affine algebraic set is Noetherian. Answer: (i) If {U i } i 1 is an ascending chain of open subsets of Y, there exists {V i } i 1 open subsets of X such that V i Y = U i for all i. Note that {V i } i 1 may not be ascending, but let W i = i k=1 V k. Then {W i } i 1 is clearly ascending and W i Y = ( i k=1 V k) Y = i k=1 (V k Y ) = i k=1 U k = U i. Therefore {W i } i 1 is an ascending chain of open subsets of X. Since X is Noetherian, there exists n 0 such that W n = W n0 for all n n 0. Therefore, U n = W n Y = W n0 Y = U n0 for all n n 0, i.e. {U i } i 1 eventually stabilizes. This implies Y is Noetherian. 1
2 2 (ii) Order the open sets in X by inclusion. It is clear that any collection U of open sets in X must have a maximal element. Indeed, if this was false, we could inductively produce a strictly increasing chain of open sets, violating the fact that X is Noetherian. Now, consider an open cover X = α U α, and define the collection U consisting in finite union of the open sets U α. The collection U must have a maximal element U α1... U αn. We claim Indeed, otherwise we could find X = U α1... U αn. x X \ (U α1... U αn ). Now, x U β for some index β. Then, U β U α1... U αn belongs to the collection U, and strictly contains the maximal element of U, namely U α1... U αn. This is absurd, thus proving our claim. (iii) Since X is Noetherian and Hausdorff, every subset Y X is also Noetherian and Hausdorff. By part (ii), Y is quasi-compact and Hausdorff, hence compact. In particular, X is compact. Since Y is compact, Y is closed. Since Y was arbitrary, X must have the discrete topology. In particular, X can be covered by point-sets {x}, for x X. Since X is compact, X must be finite. (iv) It suffices to show that A n satisfies the ascending chain condition on open sets. If {U i } be an ascending chain of open sets, let Y i be the complement of U i in A n. Then {Y i } is a descending chain of closed sets. Since taking ideal reverses inclusions, we conclude that I(Y i ) is an ascending chain of ideals in k[x 1,..., X n ]. This chain should eventually stabilize, so I(Y n ) = I(Y n+1 ) =..., for n large enough. Applying Z to these equalities, we obtain Y n = Y n+1 =.... This clearly implies that the chain {U i } stabilizes. Problem 3. Let A 3 be the 3-dimensional affine space with coordinates x, y, z. Find the ideals of the following algebraic sets: (i) The union of the three coordinate axes. (ii) The union of the (x, y)-plane with the z-axis. Answer: (i) The x-axis has the ideal (y, z), the y-axis has the ideal (x, z), and the z-axis has the ideal (x, y). Thus the union of the three axes has ideal (y, z) (x, z) (x, y) = (xy, yz, zx). (ii) The ideal of the (x, y)-plane is (z). The ideal of the z-axis is (x, y). By Problem 6, the ideal we re looking for is (z) (x, y) = (xz, yz).
3 Problem 4. Let f : A n A m be a polynomial map i.e. f(p) = (f 1 (p),..., f m (p)) for p A n, where f 1,..., f m are polynomials in n variables. Are the following true or false: (1) The image f(x) A m of an affine algebraic set X A n is an affine algebraic set. (2) The inverse image f 1 (X) A n of an affine algebraic set X A m is an affine algebraic set. (3) If X A n is an affine algebraic set, then the graph Γ = {(x, f(x)) : x X} A n+m is an affine algebraic set. Answer: (i) False. Consider the following counterexample which works over infinite ground fields. Consider the hyperbola Clearly, X is an algebraic set. Let X = {(x, y) : xy 1 = 0} A 2. f : A 2 A 1, f(x, y) = x be the projection onto the first axis. The image of f in A 1 is {x : x 0}. This is not an algebraic set because the only algebraic subsets of A 1 are empty set, finitely many points or A 1. (ii) True. Suppose X = Z(F 1,..., F s ) A m, where F 1,..., F s are polynomials in m variables. Define G i = F i (f 1 (X 1,..., X n ),..., f m (X 1,..., X n )), 1 i s. These are clearly polynomials in n variables. In short, we set G i = F i f, 1 i s We claim f 1 (X) = Z(G 1,..., G s ) A n, which clearly exihibits f 1 (X) as an algebraic set. This claim follows from the remarks p f 1 (X) f(p) Z(F 1,..., F s ) F 1 (f(p)) = F 2 (f(p)) =... = F s (f(p)) = 0 G 1 (p) =... = G s (p) p Z(G 1,..., G s ). (iii) True. Assume that X is defined by the vanishing of the polynomials F 1,..., F s in the variables X 1,..., X n. Set G 1 = Y 1 f 1 (X 1,..., X n ),..., G m = Y m f m (X 1,..., X n ). We regard the F and G s as polynomials in the n+m variables X 1,..., X n, Y 1,..., Y m. We claim that the graph Γ is defined by the equations Indeed, Γ = Z(F 1,..., F s, G 1,..., G s ) A n+m. (x, y) Z(F 1,..., F s, G 1,..., G s ) F 1 (x) =... = F s (x) = 0, y 1 = f 1 (x),..., y m = f m (x) x X, y = f(x) (x, y) Γ. 3
4 4 Problem 5. Let X 1, X 2 be affine algebraic sets in A n. Show that (i) I(X 1 X 2 ) = I(X 1 ) I(X 2 ), (ii) I(X 1 X 2 ) = I(X 1 ) + I(X 2 ). Show by an example that taking radicals in (ii) is necessary. Answer: (i) If f I(X 1 X 2 ), f vanishes on X 1 X 2. In particular, f vanishes on X 1. Thus, f I(X 1 ). Similarly, we have f I(X 2 ). Then f I(X 1 ) I(X 2 ). We proved that I(X 1 X 2 ) I(X 1 ) I(X 2 ). On the other hand, if f I(X 1 ) I(X 2 ), then f I(X 1 ) and f I(X 2 ). Therefore, f vanishes on both X 1 and X 2, and then also on X 1 X 2. This implies that f I(X 1 X 2 ). Hence I(X 1 ) I(X 2 ) I(X 1 X 2 ). (ii) We assume the base field is algebraically closed. Because X 1, X 2 be affine algebraic sets, we can assume X 1 = Z(a), X 2 = Z(b). Furthermore, we may assume a and b are radical. Otherwise, if for instance a is not radical, we can replace a by a. This doesn t change the algebraic sets in question as X 1 = Z(a) = Z( a). By Hilbert s Nullstellensatz, I(X1 ) + I(X 2 ) = a I(Z(a)) + I(Z(b)) = + b = a + b and I(X 1 X 2 ) = I(Z(a) Z(b)) = I(Z(a + b)) = a + b. Therefore, I(X 1 X 2 ) = I(X 1 ) + I(X 2 ). Note: In the above, we made use of the equality Z(a) Z(b) = Z(a + b). This can be argued as follows. If x Z(a) Z(b), then f(x) = g(x) = 0 for all f a and g b. Thus x Z(a + b) proving that Z(a) Z(b) Z(a + b). Conversely, if x Z(a + b), then (f + g)(x) = 0 for every f a and g b. In particular, picking g = 0, we get f(x) = 0 for every f a, so x Z(a). Similarly x Z(b). Therefore Z(a + b) Z(a) Z(b). Counterexample: Consider the following two algebraic subsets of A 2 : X 1 = (y x 2 = 0), X 2 = (y = 0). Graphically, these can be represented by a parabola and a line tangent to it. It is clear that X 1 X 2 = {0}, so I(X 1 X 2 ) = (x, y).
5 On the other hand, Therefore I(X 1 ) + I(X 2 ) = (y x 2 ) + (y) = (y x 2, y) = (y, x 2 ). I(X 1 X 2 ) I(X 1 ) + I(X 2 ). Problem 6. Find the irreducible components of the affine algebraic set x 2 yz = xz x = 0 in A 3. What is the dimension of this algebraic set? Answer: Let X be the affine algebraic set given by the two equations above. The second equation can be written as Case 1. If x = 0, the first equation becomes xz x = x(z 1) = 0 = x = 0 or z = 1. x 2 yz = yz = 0 = y = 0 or z = 0. Therefore, there are two possibilities: x = y = 0 or x = z = 0. These correspond to two algebraic sets X 1 = the z-axis and X 2 = the y-axis, defined by the equations Case 2. If z = 1, the first equation becomes X 1 = Z(y, z) and X 2 = Z(x, z). x 2 yz = x 2 y = 0. This zero set is a parabola which can be rewritten as According to discussion above, X 3 = Z(z 1, x 2 y). X = X 1 X 2 X 3. We claim that the X i s are irreducible. This can be seen at once since all these sets are homeomorphic to A 1. This shows X 1, X 2, X 3 are irreducible components of the affine algebraic set X. These three sets are homeomorphic to A 1, hence their dimension is 1. Alternatively one may argue as follows. X 1 is irreducible if and only if (y, z) is a prime ideal in C[x, y, z], which is equivalent to C[x, y, z]/(y, z) = C[x] is an integral domain. By the same argument, X 2 is also irreducible. Finally, X 3 = Z(z 1, x 2 y) is irreducible. Indeed (z 1, x 2 y) is a prime ideal in C[x, y, z]. This is the case since is an integral domain. C[x, y, z]/(z 1, x 2 y) = C[x, y]/(x 2 y) = C[x], Problem 7. Find the irreducible components of the affine algebraic set xz y 2 = z 3 x 5 = 0 in A 3. 5
6 6 Answer: If x = 0, then the two equations imply that y = z = 0. Similarly, if y = 0, then x = z = 0. Let us assume that x 0, y 0. Write the first equation as y x = z := t. y Thus, The second equation becomes This implies y = tx, z = t 2 x. z 3 = x 5 t 6 x 3 = x 5 x 2 = t 6 x = ±t 3. y = ±t 4, z = ±t 5. Thus (x, y, z) = (t 3, t 4, t 5 ) or ( t 3, t 4, t 5 ) for some t C. Letting we obtain X 1 = {(t 3, t 4, t 5 ) t C } and X 2 = {( t 3, t 4, t 5 ) t C } X = X 1 X 2. We claim that X 1, X 2 are the irreducible components of X. First, there are polynomial maps f 1 : A 1 A 3 t (t 3, t 4, t 5 ) and f 2 : A 1 A 3 t ( t 3, t 4, t 5 ). Since X 1 = f 1 (A 1 ), X 2 = f 2 (A 1 ) and A 1 is irreducible, the images X 1 and X 2 are irreducible. It remains to explain that X 1 and X 2 are algebraic sets. We claim X 1 = Z(y 3 x 4, z 3 x 5, z 4 y 5 ), X 2 = Z(y 3 + x 4, z 3 x 5, z 4 + y 5 ). Let us check this claim for X 1 only. This follows using the same strategy as before. Indeed, if x = 0 then y = z = 0, so we may assume x 0. Let t = y x. From the first equation we have ( y ) 3 x = = t 3 = y = t 4. x Dividing the last two equations we get z = y5 x 5 = t5 = (x, y, z) = (t 3, t 4, t 5 ), as claimed. The verification for X 2 is entirely similar. Problem 8. Let X be the image of the map A 1 A 3, t (t, t 3, t 5 ). Show that I(X) cannot be generated by fewer than 3 elements.
7 Answer: Let I be the ideal of the set (t 3, t 4, t 5 ) in A 3, as t A 1. Observe that (xz y 2, yz x 3, x 2 y z 2 ) I. We assume that I can be generated by at most two elements f and g. Since I contains no polynomials of degree 0 or 1, it follows that f and g contain only quadratic terms or higher. Writing I (2) for the quadratic terms of the polynomials in I, it follows that f (2) and g (2) generate I (2). However, observe that the part I (2) is a k-vector space of dimension at least 3 since it contains (yz, y 2, z 2 ). hence I (2) cannot be generated by 2 polynomials. This contradication shows that I needs at least 3 generators. 7
Math 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 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 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 informationLECTURE Affine Space & the Zariski Topology. It is easy to check that Z(S)=Z((S)) with (S) denoting the ideal generated by elements of S.
LECTURE 10 1. Affine Space & the Zariski Topology Definition 1.1. Let k a field. Take S a set of polynomials in k[t 1,..., T n ]. Then Z(S) ={x k n f(x) =0, f S}. It is easy to check that Z(S)=Z((S)) with
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 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 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 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 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 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 informationMAT4210 Algebraic geometry I: Notes 2
MAT4210 Algebraic geometry I: Notes 2 The Zariski topology and irreducible sets 26th January 2018 Hot themes in notes 2: The Zariski topology on closed algebraic subsets irreducible topological spaces
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 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 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 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 informationAlgebraic geometry of the ring of continuous functions
Algebraic geometry of the ring of continuous functions Nicolas Addington October 27 Abstract Maximal ideals of the ring of continuous functions on a compact space correspond to points of the space. For
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 informationTHE ALGEBRAIC GEOMETRY DICTIONARY FOR BEGINNERS. Contents
THE ALGEBRAIC GEOMETRY DICTIONARY FOR BEGINNERS ALICE MARK Abstract. This paper is a simple summary of the first most basic definitions in Algebraic Geometry as they are presented in Dummit and Foote ([1]),
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 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 informationPure Math 764, Winter 2014
Compact course notes Pure Math 764, Winter 2014 Introduction to Algebraic Geometry Lecturer: R. Moraru transcribed by: J. Lazovskis University of Waterloo April 20, 2014 Contents 1 Basic geometric objects
More informationAlgebraic Varieties. Brian Osserman
Algebraic Varieties Brian Osserman Preface This book is largely intended as a substitute for Chapter I (and an invitation to Chapter IV) of Hartshorne [Har77], to be taught as an introduction to varieties
More informationAlgebraic Geometry. Instructor: Stephen Diaz & Typist: Caleb McWhorter. Spring 2015
Algebraic Geometry Instructor: Stephen Diaz & Typist: Caleb McWhorter Spring 2015 Contents 1 Varieties 2 1.1 Affine Varieties....................................... 2 1.50 Projective Varieties.....................................
More informationExploring the Exotic Setting for Algebraic Geometry
Exploring the Exotic Setting for Algebraic Geometry Victor I. Piercey University of Arizona Integration Workshop Project August 6-10, 2010 1 Introduction In this project, we will describe the basic topology
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. Andreas Gathmann. Notes for a class. taught at the University of Kaiserslautern 2002/2003
Algebraic Geometry Andreas Gathmann Notes for a class taught at the University of Kaiserslautern 2002/2003 CONTENTS 0. Introduction 1 0.1. What is algebraic geometry? 1 0.2. Exercises 6 1. Affine varieties
More informationMATH 631: ALGEBRAIC GEOMETRY: HOMEWORK 1 SOLUTIONS
MATH 63: ALGEBRAIC GEOMETRY: HOMEWORK SOLUTIONS Problem. (a.) The (t + ) (t + ) minors m (A),..., m k (A) of an n m matrix A are polynomials in the entries of A, and m i (A) = 0 for all i =,..., k if and
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 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 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 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 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 information(dim Z j dim Z j 1 ) 1 j i
Math 210B. Codimension 1. Main result and some interesting examples Let k be a field, and A a domain finitely generated k-algebra. In class we have seen that the dimension theory of A is linked to the
More informationMath General Topology Fall 2012 Homework 6 Solutions
Math 535 - General Topology Fall 202 Homework 6 Solutions Problem. Let F be the field R or C of real or complex numbers. Let n and denote by F[x, x 2,..., x n ] the set of all polynomials in n variables
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 information10. Noether Normalization and Hilbert s Nullstellensatz
10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.
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 informationINTRODUCTION TO ALGEBRAIC GEOMETRY. Throughout these notes all rings will be commutative with identity. k will be an algebraically
INTRODUCTION TO ALGEBRAIC GEOMETRY STEVEN DALE CUTKOSKY Throughout these notes all rings will be commutative with identity. k will be an algebraically closed field. 1. Preliminaries on Ring Homomorphisms
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 informationB 1 = {B(x, r) x = (x 1, x 2 ) H, 0 < r < x 2 }. (a) Show that B = B 1 B 2 is a basis for a topology on X.
Math 6342/7350: Topology and Geometry Sample Preliminary Exam Questions 1. For each of the following topological spaces X i, determine whether X i and X i X i are homeomorphic. (a) X 1 = [0, 1] (b) X 2
More informationChapter 1. Affine algebraic geometry. 1.1 The Zariski topology on A n
Chapter 1 Affine algebraic geometry We shall restrict our attention to affine algebraic geometry, meaning that the algebraic varieties we consider are precisely the closed subvarieties of affine n- space
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 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 informationAlgebraic Varieties. Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra
Algebraic Varieties Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra Algebraic varieties represent solutions of a system of polynomial
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 informationMaths 212: Homework Solutions
Maths 212: Homework Solutions 1. The definition of A ensures that x π for all x A, so π is an upper bound of A. To show it is the least upper bound, suppose x < π and consider two cases. If x < 1, then
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 informationNAME: Mathematics 205A, Fall 2008, Final Examination. Answer Key
NAME: Mathematics 205A, Fall 2008, Final Examination Answer Key 1 1. [25 points] Let X be a set with 2 or more elements. Show that there are topologies U and V on X such that the identity map J : (X, U)
More informationTopology Exercise Sheet 2 Prof. Dr. Alessandro Sisto Due to March 7
Topology Exercise Sheet 2 Prof. Dr. Alessandro Sisto Due to March 7 Question 1: The goal of this exercise is to give some equivalent characterizations for the interior of a set. Let X be a topological
More informationMath 214A. 1 Affine Varieties. Yuyu Zhu. August 31, Closed Sets. These notes are based on lectures given by Prof. Merkurjev in Winter 2015.
Math 214A Yuyu Zhu August 31, 2016 These notes are based on lectures given by Prof. Merkurjev in Winter 2015. 1 Affine Varieties 1.1 Closed Sets Let k be an algebraically closed field. Denote A n = A n
More informationMath 203A - Solution Set 4
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
More informationThe Topology and Algebraic Functions on Affine Algebraic Sets Over an Arbitrary Field
Georgia State University ScholarWorks @ Georgia State University Mathematics Theses Department of Mathematics and Statistics Fall 11-15-2012 The Topology and Algebraic Functions on Affine Algebraic Sets
More informationAlgebraic Geometry. Andreas Gathmann. Class Notes TU Kaiserslautern 2014
Algebraic Geometry Andreas Gathmann Class Notes TU Kaiserslautern 2014 Contents 0. Introduction......................... 3 1. Affine Varieties........................ 9 2. The Zariski Topology......................
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 informationProblems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9;
Math 553 - Topology Todd Riggs Assignment 2 Sept 17, 2014 Problems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9; 17.2) Show that if A is closed in Y and
More information2. Prime and Maximal Ideals
18 Andreas Gathmann 2. Prime and Maximal Ideals There are two special kinds of ideals that are of particular importance, both algebraically and geometrically: the so-called prime and maximal ideals. Let
More informationMath 6140 Notes. Spring Codimension One Phenomena. Definition: Examples: Properties:
Math 6140 Notes. Spring 2003. 11. Codimension One Phenomena. A property of the points of a variety X holds in codimension one if the locus of points for which the property fails to hold is contained in
More informationCRing Project, Chapter 5
Contents 5 Noetherian rings and modules 3 1 Basics........................................... 3 1.1 The noetherian condition............................ 3 1.2 Stability properties................................
More information8. Prime Factorization and Primary Decompositions
70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings
More information2. Intersection Multiplicities
2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.
More informationLecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).
Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an A-linear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is
More informationReading Mathematical Expressions & Arithmetic Operations Expression Reads Note
Math 001 - Term 171 Reading Mathematical Expressions & Arithmetic Operations Expression Reads Note x A x belongs to A,x is in A Between an element and a set. A B A is a subset of B Between two sets. φ
More informationALGEBRAIC GEOMETRY HOMEWORK 3
ALGEBRAIC GEOMETRY HOMEWORK 3 (1) Consider the curve Y 2 = X 2 (X + 1). (a) Sketch the curve. (b) Determine the singular point P on C. (c) For all lines through P, determine the intersection multiplicity
More informationz -FILTERS AND RELATED IDEALS IN C(X) Communicated by B. Davvaz
Algebraic Structures and Their Applications Vol. 2 No. 2 ( 2015 ), pp 57-66. z -FILTERS AND RELATED IDEALS IN C(X) R. MOHAMADIAN Communicated by B. Davvaz Abstract. In this article we introduce the concept
More informationMath 40510, Algebraic Geometry
Math 40510, Algebraic Geometry Problem Set 1, due February 10, 2016 1. Let k = Z p, the field with p elements, where p is a prime. Find a polynomial f k[x, y] that vanishes at every point of k 2. [Hint:
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More information4.4 Noetherian Rings
4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)
More informationTopological properties
CHAPTER 4 Topological properties 1. Connectedness Definitions and examples Basic properties Connected components Connected versus path connected, again 2. Compactness Definition and first examples Topological
More informationTHE CLOSED-POINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS. K. R. Goodearl and E. S. Letzter
THE CLOSED-POINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS K. R. Goodearl and E. S. Letzter Abstract. In previous work, the second author introduced a topology, for spaces of irreducible representations,
More informationIntroduction to Algebraic Geometry. Jilong Tong
Introduction to Algebraic Geometry Jilong Tong December 6, 2012 2 Contents 1 Algebraic sets and morphisms 11 1.1 Affine algebraic sets.................................. 11 1.1.1 Some definitions................................
More informationLocal properties of plane algebraic curves
Chapter 7 Local properties of plane algebraic curves Throughout this chapter let K be an algebraically closed field of characteristic zero, and as usual let A (K) be embedded into P (K) by identifying
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 information3 Hausdorff and Connected Spaces
3 Hausdorff and Connected Spaces In this chapter we address the question of when two spaces are homeomorphic. This is done by examining two properties that are shared by any pair of homeomorphic spaces.
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 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 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 informationOn z -ideals in C(X) F. A z a r p a n a h, O. A. S. K a r a m z a d e h and A. R e z a i A l i a b a d (Ahvaz)
F U N D A M E N T A MATHEMATICAE 160 (1999) On z -ideals in C(X) by F. A z a r p a n a h, O. A. S. K a r a m z a d e h and A. R e z a i A l i a b a d (Ahvaz) Abstract. An ideal I in a commutative ring
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 informationFinal Exam Solutions June 10, 2004
Math 0400: Analysis in R n II Spring 004 Section 55 P. Achar Final Exam Solutions June 10, 004 Total points: 00 There are three blank pages for scratch work at the end of the exam. Time it: hours 1. True
More informationCartan s Criteria. Math 649, Dan Barbasch. February 26
Cartan s Criteria Math 649, 2013 Dan Barbasch February 26 Cartan s Criteria REFERENCES: Humphreys, I.2 and I.3. Definition The Cartan-Killing form of a Lie algebra is the bilinear form B(x, y) := Tr(ad
More informationDimension Theory. Mathematics 683, Fall 2013
Dimension Theory Mathematics 683, Fall 2013 In this note we prove some of the standard results of commutative ring theory that lead up to proofs of the main theorem of dimension theory and of the Nullstellensatz.
More informationCHAPTER 7. Connectedness
CHAPTER 7 Connectedness 7.1. Connected topological spaces Definition 7.1. A topological space (X, T X ) is said to be connected if there is no continuous surjection f : X {0, 1} where the two point set
More informationV (f) :={[x] 2 P n ( ) f(x) =0}. If (x) ( x) thenf( x) =
20 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2. Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F -vector subspaces of F n+.
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 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 information12 Hilbert polynomials
12 Hilbert polynomials 12.1 Calibration Let X P n be a (not necessarily irreducible) closed algebraic subset. In this section, we ll look at a device which measures the way X sits inside P n. Throughout
More informationMAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9
MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended
More informationSOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 1. I. Foundational material
SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 1 Fall 2014 I. Foundational material I.1 : Basic set theory Problems from Munkres, 9, p. 64 2. (a (c For each of the first three parts, choose a 1 1 correspondence
More informationMAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton. Timothy J. Ford April 4, 2016
MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton Timothy J. Ford April 4, 2016 FLORIDA ATLANTIC UNIVERSITY, BOCA RATON, FLORIDA 33431 E-mail address: ford@fau.edu
More information14. Rational maps It is often the case that we are given a variety X and a morphism defined on an open subset U of X. As open sets in the Zariski
14. Rational maps It is often the case that we are given a variety X and a morphism defined on an open subset U of X. As open sets in the Zariski topology are very large, it is natural to view this as
More informationKrull Dimension and Going-Down in Fixed Rings
David Dobbs Jay Shapiro April 19, 2006 Basics R will always be a commutative ring and G a group of (ring) automorphisms of R. We let R G denote the fixed ring, that is, Thus R G is a subring of R R G =
More informationSome Basic Logic. Henry Liu, 25 October 2010
Some Basic Logic Henry Liu, 25 October 2010 In the solution to almost every olympiad style mathematical problem, a very important part is existence of accurate proofs. Therefore, the student should be
More informationCommutative Algebra and Algebraic Geometry. Robert Friedman
Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions
More informationHomework 5. Solutions
Homework 5. Solutions 1. Let (X,T) be a topological space and let A,B be subsets of X. Show that the closure of their union is given by A B = A B. Since A B is a closed set that contains A B and A B is
More informationOn Invariants of Complex Filiform Leibniz Algebras ABSTRACT INTRODUCTION
Malaysian Journal of Mathematical Sciences (): 47-59 (009) Isamiddin S.Rakhimov Department of Mathematics Faculty of Science and Institute for Mathematical Research Universiti Putra Malaysia 4400 UPM Serdang
More informationFoundations of Mathematics MATH 220 FALL 2017 Lecture Notes
Foundations of Mathematics MATH 220 FALL 2017 Lecture Notes These notes form a brief summary of what has been covered during the lectures. All the definitions must be memorized and understood. Statements
More informationMATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION. Problem 1
MATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION ELEMENTS OF SOLUTION Problem 1 1. Let X be a Hausdorff space and K 1, K 2 disjoint compact subsets of X. Prove that there exist disjoint open sets U 1 and
More informationD-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 1. Arithmetic, Zorn s Lemma.
D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 1 Arithmetic, Zorn s Lemma. 1. (a) Using the Euclidean division, determine gcd(160, 399). (b) Find m 0, n 0 Z such that gcd(160, 399) = 160m 0 +
More information4.2 Chain Conditions
4.2 Chain Conditions Imposing chain conditions on the or on the poset of submodules of a module, poset of ideals of a ring, makes a module or ring more tractable and facilitates the proofs of deep theorems.
More information11. Dimension. 96 Andreas Gathmann
96 Andreas Gathmann 11. Dimension We have already met several situations in this course in which it seemed to be desirable to have a notion of dimension (of a variety, or more generally of a ring): for
More information