Math 40510, Algebraic Geometry


 Fay Cole
 4 years ago
 Views:
Transcription
1 Math 40510, Algebraic Geometry Problem Set 1, due February 10, 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: Recall from MATH that for any a k (including 0), we have a p = a in k. We did a problem similar to this one in class.] f(x, y) = x p + (p 1)x + y p + (p 1)y. 2. Let k be any field. If V 1,..., V s are (finitely many) affine varieties in k n, prove that V 1 V s is again an affine variety. (FYI: It happens to be true that in this problem we can replace finite by infinite, but we need a little more background before we can prove that.) By induction. We proved in class that the intersection of two affine varieties is again an affine variety. Now assume that the result is true for any N varieties and let V 1,..., V N+1 be affine varieties. Then V 1 V N V N+1 = (V 1 V N ) V N+1. By induction the first is an affine variety, and so by the case for two varieties we get that V 1 V N V N+1 is an affine variety. Thus the result holds for any finite number of affine varieties. then Alternatively, if 3. Let k be any field. V 1 = V (f 1,1,..., f 1,t1 ) V 2 = V (f 2,1,..., f 2,t2 ). V s = V (f s,1,..., f s,ts ) V 1 V s = V (f 1,1,..., f 1,t1, f 2,1,..., f 2,t2,..., f s,1,..., f s,ts ). a) Prove that every (single) point (a 1,..., a n ) k n is an affine variety. (a 1,..., a n ) = V (x 1 a 1,..., x n a n ). b) Let V 1,..., V s be (finitely many) affine varieties in k n. Prove that V 1 V s is again an affine variety. [Hint: Using Lemma 2 of section 2 (which we proved in class) and induction, you can actually do this problem without giving explicit polynomials.] By induction. We proved in class that the union of two affine varieties is again an affine variety. Now assume that the result is true for any N varieties and let V 1,..., V N+1 be affine varieties. Then V 1 V N V N+1 = (V 1 V N ) V N+1. By induction the first is an affine variety, and so by the case for two varieties we get that V 1 V N V N+1 is an affine variety. Thus the result holds for any finite number of affine varieties. 1
2 c) Prove that any finite set of points in k n is an affine variety. [Hint: there is a reason why this is the third part of this problem. This should take less than one line.] Immediate from parts a) and b). d) Give an example of an infinite set of points in R 2 whose union is an affine variety. [Hint: this can also be done in less than one line.] Let V be the union of all the points on the line V (x), i.e. V = {(0, t) t R}. 4. Now we ll see that part d) of Problem 3 is not true for all infinite unions of affine varieties (in fact it s almost never true for infinite unions). Let k = R. Let Z be the set of points in R 2 with integer coordinates. a) Let f(x, y) be a polynomial vanishing at every point of Z. Prove that f(x, y) must be the zero polynomial. [Hint: if f(x, y) vanishes at every point of Z, what can you say about f(x, 0)?] In particular we have f(n, 0) = 0 for all n Z. But g(x) = f(x, 0) is a polynomial, and the first sentence means that g(x) has infinitely many zeros. So g(x) is the zero polynomial. This means that plugging in y = 0 into f(x, y) gives the zero polynomial, so f(x, y) contains no terms that are pure powers of x. In a similar way we can show that f(x, y) contains no terms that are pure powers of y. Now consider f(x, 1). Since each term of f(x, y) contains both powers of x and of y, f(x, 1) converts each term of f(x, y) into a term involving only x. Now the fact that f(x, 1) has infinitely many zeros means that it, too, is the zero polynomial, so all its terms are zero. This means that all terms of f(x, y) are zero, so f is the zero polynomial. b) Conclude that Z is not an affine variety. From a), if f I(Z) then f is the zero polynomial. If Z were an affine variety then we would have Z = V (f 1,..., f s ) for some polynomials f 1,..., f s that (by definition) vanish on Z. But any polynomial vanishing on Z is the zero polynomial, so the smallest variety containing Z is R 2. In particular, Z is not an affine variety. 5. Let V k n and W k m be affine varieties. Let V W = {(a 1,..., a n, b 1,..., b m ) k n+m (a 1,..., a n ) V and (b 1,..., b m ) W }. (This set is called the Cartesian product of V and W.) Prove that V W is an affine variety in k n+m. [Hint: this is problem 15 (d) of section 2 in CLO, and they give a nice hint.] We work in the polynomial ring R = k[x 1,..., x n, y 1,..., y m ]. Let V = V(f 1,..., f s ) k n and W = V(g 1,..., g t ) k m, where f 1,..., f s k[x 1,..., x n ] and g 1,..., g t k[y 1,..., y m ]. Think of the f i and the g j as being in R. We claim (1) V W = V(f 1,..., f s, g 1,..., g t ). A picture might be helpful here. Say m = n = 1 and k = R. Let and let V = {1, 2, 3} = V((x 1)(x 2)(x 3)) W = {1, 2} = V(y 1)(y 2).
3 (y 1)(y 2) W = 2 points V W = 6 points V = 3 points (x 1)(x 2)(x 3) Now we want to prove (1). We ll prove the two inclusions. To prove, let P V W. So P = (a 1,..., a n, b 1,..., b m ) k n+m where (a 1,..., a n ) V and (b 1,..., b m ) W. Since (a 1,..., a n ) V we have f i (a 1,..., a n ) = 0 for 1 i s. So thinking of f i as a polynomial in n+m variables, we have f i (a 1,..., a n, b 1,..., b m ) = 0 (since the b i don t do anything). Similarly we have g j (a 1,..., a n, b 1,..., b m ) = 0 for 1 j t. Hence P V(f 1,..., f s, g 1,..., g t ). Finally we ll prove in (1). Let P = (a 1,..., a n, b 1,..., b m ) V(f 1,..., f s, g 1,..., g t ). So f 1,..., f s, g 1,..., g t all vanish at P. Since the f i only involve x 1,..., x n, we have f i (a 1,..., a m ) = 0 for 1 i s, so (a 1,..., a n ) V. Similarly we see (b 1,..., b m ) W, so by definition P V W. 6. Recall we said that if f 1, f 2 R = k[x 1,..., x n ] and if f f 1, f 2 then there can be more than one way to find polynomials h 1 and h 2 so that f = h 1 f 1 + h 2 f 2. Let s explore that a bit. Let R = k[x, y] and let f = x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5. a) If f 1 = x 3 and f 2 = y 3, prove that there are unique polynomials h 1 and h 2 of the form h 1 = a 1 x 2 + a 2 xy + a 3 y 2, h 2 = a 4 x 2 + a 5 xy + a 6 y 2 (where the a i are elements of k, i.e. scalars) so that f = h 1 f 1 + h 2 f 2. In particular, find a 1, a 2, a 3, a 4, a 5, a 6. We want to solve the equation x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5 = (a 1 x 2 + a 2 xy + a 3 y 2 ) x 3 + (a 4 x 2 + a 5 xy + a 6 y 2 ) y 3 We immediately get = a 1 x 5 + a 2 x 4 y + a 3 x 3 y 2 + a 4 x 2 y 3 + a 5 xy 4 + a 6 y 5 a 1 = 1 a 2 = 2 a 3 = 3 a 4 = 4 a 5 = 5 a 6 = 6
4 Since these are uniquely determined, we are done. b) If f 1 = x 3 and f 2 = y 2, find all possible pairs (h 1, h 2 ) of the form h 1 = a 1 x 2 + a 2 xy + a 3 y 2, h 2 = a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 (again the a i are in k) so that f = h 1 f 1 + h 2 f 2. In a similar way we want to solve the equation x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5 = (a 1 x 2 + a 2 xy + a 3 y 2 ) x 3 + (a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 ) y 2 We immediately get = a 1 x 5 + a 2 x 4 y + a 3 x 3 y 2 + a 4 x 3 y 2 + a 5 x 2 y 3 + a 6 xy 4 + a 7 y 5 a 1 = 1 a 2 = 2 a 3 + a 4 = 3 a 5 = 4 a 6 = 5 a 7 = 6 So the set of all possible (h 1, h 2 ) = (a 1 x 2 + a 2 xy + a 3 y 2, a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 ) consists of all ordered pairs where the coefficients satisfy the above equations. (As we mentioned in class, this problem is related to the topic of syzygies, which hopefully we will talk about later.) 7. Let R = k[x 1,..., x n ]. Let I be an ideal in R. We define an ideal to be radical if the following condition holds: If f m I for some positive integer m then f I. (Keep this definition in mind because we ll come back to it in class.) a) Prove that if f I then f m I for all positive integers m. If f I we certainly have f m 1 R, so f m = f m 1 f I by definition of an ideal. b) If X k n is any set, prove that I(X) is always a radical ideal. If f m I(X) then 0 = f m (P ) = f(p ) m for all P X, so f(p ) = 0 for all P X, i.e. f I(X). c) Give an example of an ideal that is not radical, and prove that it s not radical. Let I = x 2. Note that f(x) = x satisfies f 2 I. But every element of x 2 is a multiple of x 2 and clearly x is not a multiple of x 2, so f / I. Thus I is not radical. (Keep this definition in mind because we ll come back to it in class.) 8. Let I and J be ideals in k[x 1,..., x n ]. We define I J = {f R f I and f J}. We define IJ to be the set of polynomials that can be written as finite sums in the following way: { m } IJ = f i g i f i I, g i J.
5 a) Prove that I J is an ideal. We use the fact that both I and J are ideals. Since 0 I and 0 J, we have 0 I J. If f, g I J then f and g are both in I and both in J, so f + g I J. If f I J and h R then hf I and hf J so hf I J. b) Prove that IJ is an ideal. 0 I and 0 J so 0 = 0 0 IJ. Assume f = m f ig i for some f i I, g i J and g = m f i g i for some f i I, g i J. Then f + g = m m f i g j + f ig j IJ. Finally, if f = m f ig i for some f i I, g i J and h R then m m hf = h f i g i = (hf i )g i IJ since hf i I (because I is an ideal). c) Show that IJ I J (as ideals). It s enough to prove that each generator of IJ is in I J. If I = f 1,..., f s and J = g 1,..., g t then the generators of IJ have the form f i g j for 1 i s and 1 j t. But then f i g j I (since f i I) and also f i g j J (since g j J) so f i g i I J. d) Give an example to show that IJ is not necessarily equal to I J. Justify your answer! For example take I = x and J = x. Then I J is clearly equal to x, while IJ = x 2. We have already seen that these two ideals are not equal. e) If I and J are ideals in k[x 1,..., x n ], prove that V (IJ) = V (I) V (J). [Hint: this is closely related to our proof that V (I) V (J) is again an affine variety.] Let s prove the two inclusions. : Let P V (IJ). We want to show that P V (I) V (J). If P V (I) then we re done, so assume P / V (I); we want to show that then P V (J). Since P / V (I), there is some f I such that f(p ) 0. But fg IJ for all g J; hence (fg)(p ) = 0 for all g J. Thus P V (J) as desired. : Let P V (I) V (J). So either P V (I) or P V (J) (or both). Assume without loss of generality that P V (I). Then f(p ) = 0 for all f I. Let g IJ, so m g = f i g i f i I and g i J. Then we get g(p ) = m f i(p )g i (P ) = 0. Hence g(p ) = 0 for all g IJ, and so P V (IJ).
6 f) If I and J are ideals in k[x 1,..., x n ], prove that V (I J) = V (I) V (J). Combined with the previous part, conclude that V (IJ) = V (I J). Again we prove the two inclusions. : Let P V (I J), so h(p ) = 0 for all h I J. Suppose that P / V (I). We want to show P V (J), i.e. we want to show that g(p ) = 0 for all g J. Since P / V (I), there is some f I such that f(p ) 0. Then for any g J, we know that fg I J so (fg)(p ) = 0. Sincef(P ) 0, this forces g(p ) = 0 for all g J, so P V (J) as desired. : Let P V (I) V (J), so either P V (I) or P V (J) or both. Let f I J. Since f is in both I and J, we must have f(p ) = 0. So P V (I J). g) It happens to be true that if P = (0, 1) k 2 then I(P ) = x, y 1. (You can accept this without proof.) Let Q = (0, 0). We saw in class that I(Q) = x, y. You can also accept without proof the fact that I(P ) I(Q) = x, y(y 1). Show that I(P ) I(Q) = I(P ) I(Q) = I(P Q). We ll show I(P ) I(Q) I(P ) I(Q) I(P Q) I(P ) I(Q). (1) (2) (3) (1) It s always true for two ideals I and J that IJ I J (we showed this in part c) above), so the first inclusion is immediate. (2) If f I(P ) I(Q) then f vanishes on both P and Q, so f I(P Q). In fact it s convenient to observe here that the reverse is true: if f I(P Q) then f vanishes on each of P and Q so f I(P ) I(Q). (I.e. I(P ) I(Q) = I(P Q).) (3) We know from the statement of the problem that I(P ) = x, y 1, I(Q) = x, y and I(P Q) = x, y(y 1) = x, y 2 y. (The last one also uses what we observed in (2) above.) To show the inclusion (3), it s enough to show that each generator of I(P Q) is in I(P ) I(Q). So we have to show that x I(P ) I(Q) and that y(y 1) I(P ) I(Q). The second one is obvious since y 1 I(P ) and y I(Q). So let s look at the first one. We have to show that x x, y 1 x, y. But this is true because x = (y 1)( x) + (x)(y). (The point of d) and g) is that sometimes IJ = I J and sometimes IJ I J.)
Math 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 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 informationIntegerValued Polynomials
IntegerValued Polynomials LA Math Circle High School II Dillon Zhi October 11, 2015 1 Introduction Some polynomials take integer values p(x) for all integers x. The obvious examples are the ones where
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 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 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 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 informationChapter2 Relations and Functions. Miscellaneous
1 Chapter2 Relations and Functions Miscellaneous Question 1: The relation f is defined by The relation g is defined by Show that f is a function and g is not a function. The relation f is defined as It
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 informationCommutative Algebra. Andreas Gathmann. Class Notes TU Kaiserslautern 2013/14
Commutative Algebra Andreas Gathmann Class Notes TU Kaiserslautern 2013/14 Contents 0. Introduction......................... 3 1. Ideals........................... 9 2. Prime and Maximal Ideals.....................
More informationRings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.
Rings 10262008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over
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 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 informationResultants. Chapter Elimination Theory. Resultants
Chapter 9 Resultants 9.1 Elimination Theory We know that a line and a curve of degree n intersect in exactly n points if we work in the projective plane over some algebraically closed field K. Using the
More informationWriting proofs for MATH 51H Section 2: Set theory, proofs of existential statements, proofs of uniqueness statements, proof by cases
Writing proofs for MATH 51H Section 2: Set theory, proofs of existential statements, proofs of uniqueness statements, proof by cases September 22, 2018 Recall from last week that the purpose of a proof
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 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 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 (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 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 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 4370 Exam 1. Handed out March 9th 2010 Due March 18th 2010
Math 4370 Exam 1 Handed out March 9th 2010 Due March 18th 2010 Problem 1. Recall from problem 1.4.6.e in the book, that a generating set {f 1,..., f s } of I is minimal if I is not the ideal generated
More informationA Harvard Sampler. Evan Chen. February 23, I crashed a few math classes at Harvard on February 21, Here are notes from the classes.
A Harvard Sampler Evan Chen February 23, 2014 I crashed a few math classes at Harvard on February 21, 2014. Here are notes from the classes. 1 MATH 123: Algebra II In this lecture we will make two assumptions.
More informationOn Invariants of Complex Filiform Leibniz Algebras ABSTRACT INTRODUCTION
Malaysian Journal of Mathematical Sciences (): 4759 (009) Isamiddin S.Rakhimov Department of Mathematics Faculty of Science and Institute for Mathematical Research Universiti Putra Malaysia 4400 UPM Serdang
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 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 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 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 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 informationMATH 115, SUMMER 2012 LECTURE 12
MATH 115, SUMMER 2012 LECTURE 12 JAMES MCIVOR  last time  we used hensel s lemma to go from roots of polynomial equations mod p to roots mod p 2, mod p 3, etc.  from there we can use CRT to construct
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 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 informationMATH 243E Test #3 Solutions
MATH 4E Test # Solutions () Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s. You do not need to solve this recurrence relation. (Hint: Consider
More informationAlgebraic function fields
Algebraic function fields 1 Places Definition An algebraic function field F/K of one variable over K is an extension field F K such that F is a finite algebraic extension of K(x) for some element x F which
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013 As usual, k is a perfect field and k is a fixed algebraic closure of k. Recall that an affine (resp. projective) variety is an
More informationSets and Functions. (As we will see, in describing a set the order in which elements are listed is irrelevant).
Sets and Functions 1. The language of sets Informally, a set is any collection of objects. The objects may be mathematical objects such as numbers, functions and even sets, or letters or symbols of any
More informationINTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 1. Contents 1. Commutative algebra 2 2. Algebraic sets 2 3. Nullstellensatz (theorem of zeroes) 4
INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 1 RAVI VAKIL Contents 1. Commutative algebra 2 2. Algebraic sets 2 3. Nullstellensatz (theorem of zeroes) 4 I m going to start by telling you about this course,
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 informationDMATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 6. Unique Factorization Domains
DMATH Algebra I HS18 Prof. Rahul Pandharipande Solution 6 Unique Factorization Domains 1. Let R be a UFD. Let that a, b R be coprime elements (that is, gcd(a, b) R ) and c R. Suppose that a c and b c.
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 socalled prime and maximal ideals. Let
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 informationOn Inflection Points of Plane Curves
On Inflection Points of Plane Curves Lois van der Meijden 3931919 Supervisor: Prof. Dr. C. Faber 2015 Utrecht University Introduction The field of algebraic geometry has a long history. Its origin lies
More information8 Appendix: Polynomial Rings
8 Appendix: Polynomial Rings Throughout we suppose, unless otherwise specified, that R is a commutative ring. 8.1 (Largely) a reminder about polynomials A polynomial in the indeterminate X with coefficients
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going
More informationLECTURE 5, FRIDAY
LECTURE 5, FRIDAY 20.02.04 FRANZ LEMMERMEYER Before we start with the arithmetic of elliptic curves, let us talk a little bit about multiplicities, tangents, and singular points. 1. Tangents How do we
More informationElliptic Curves and Public Key Cryptography
Elliptic Curves and Public Key Cryptography Jeff Achter January 7, 2011 1 Introduction to Elliptic Curves 1.1 Diophantine equations Many classical problems in number theory have the following form: Let
More informationMATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA
MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA These are notes for our first unit on the algebraic side of homological algebra. While this is the last topic (Chap XX) in the book, it makes sense to
More informationbe any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore
More informationWEAK NULLSTELLENSATZ
WEAK NULLSTELLENSATZ YIFAN WU, wuyifan@umich.edu Abstract. We prove weak Nullstellensatz which states if a finitely generated k algebra is a field, then it is a finite algebraic field extension of k. We
More informationTheorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.
5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field
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 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 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS. Contents 1. Polynomial Functions 1 2. Rational Functions 6
MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS PETE L. CLARK Contents 1. Polynomial Functions 1 2. Rational Functions 6 1. Polynomial Functions Using the basic operations of addition, subtraction,
More informationMath 762 Spring h Y (Z 1 ) (1) h X (Z 2 ) h X (Z 1 ) Φ Z 1. h Y (Z 2 )
Math 762 Spring 2016 Homework 3 Drew Armstrong Problem 1. Yoneda s Lemma. We have seen that the bifunctor Hom C (, ) : C C Set is analogous to a bilinear form on a Kvector space, : V V K. Recall that
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 informationMATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017
MATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017 Definition: A set A is finite if there exists a nonnegative integer c such that there exists a bijection from A
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 informationCoordinates for Projective Planes
Chapter 8 Coordinates for Projective Planes Math 4520, Fall 2017 8.1 The Affine plane We now have several eamples of fields, the reals, the comple numbers, the quaternions, and the finite fields. Given
More informationSpring 2016, lecture notes by Maksym Fedorchuk 51
Spring 2016, lecture notes by Maksym Fedorchuk 51 10.2. Problem Set 2 Solution Problem. Prove the following statements. (1) The nilradical of a ring R is the intersection of all prime ideals of R. (2)
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 informationCPSC 536N: Randomized Algorithms Term 2. Lecture 9
CPSC 536N: Randomized Algorithms 201112 Term 2 Prof. Nick Harvey Lecture 9 University of British Columbia 1 Polynomial Identity Testing In the first lecture we discussed the problem of testing equality
More informationMIT Algebraic techniques and semidefinite optimization February 16, Lecture 4
MIT 6.972 Algebraic techniques and semidefinite optimization February 16, 2006 Lecture 4 Lecturer: Pablo A. Parrilo Scribe: Pablo A. Parrilo In this lecture we will review some basic elements of abstract
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include
PUTNAM TRAINING POLYNOMIALS (Last updated: December 11, 2017) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More information15. Polynomial rings DefinitionLemma Let R be a ring and let x be an indeterminate.
15. Polynomial rings DefinitionLemma 15.1. Let R be a ring and let x be an indeterminate. The polynomial ring R[x] is defined to be the set of all formal sums a n x n + a n 1 x n +... a 1 x + a 0 = a
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 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 informationLecture I: Introduction to Tropical Geometry David Speyer
Lecture I: Introduction to Tropical Geometry David Speyer The field of Puiseux series C[[t]] is the ring of formal power series a 0 + a 1 t + and C((t)) is the field of formal Laurent series: a N t N +
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 informationCHAPTER 10: POLYNOMIALS (DRAFT)
CHAPTER 10: POLYNOMIALS (DRAFT) LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN The material in this chapter is fairly informal. Unlike earlier chapters, no attempt is made to rigorously
More informationHomework 2  Math 603 Fall 05 Solutions
Homework 2  Math 603 Fall 05 Solutions 1. (a): In the notation of AtiyahMacdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an Amodule. (b): By Noether
More informationMath 113 Homework 5. Bowei Liu, Chao Li. Fall 2013
Math 113 Homework 5 Bowei Liu, Chao Li Fall 2013 This homework is due Thursday November 7th at the start of class. Remember to write clearly, and justify your solutions. Please make sure to put your name
More informationMath 115 Spring 11 Written Homework 10 Solutions
Math 5 Spring Written Homework 0 Solutions. For following its, state what indeterminate form the its are in and evaluate the its. (a) 3x 4x 4 x x 8 Solution: This is in indeterminate form 0. Algebraically,
More information(x + 3)(x 1) lim(x + 3) = 4. lim. (x 2)( x ) = (x 2)(x + 2) x + 2 x = 4. dt (t2 + 1) = 1 2 (t2 + 1) 1 t. f(x) = lim 3x = 6,
Math 140 MT1 Sample C Solutions Tyrone Crisp 1 (B): First try direct substitution: you get 0. So try to cancel common factors. We have 0 x 2 + 2x 3 = x 1 and so the it as x 1 is equal to (x + 3)(x 1),
More informationV (v i + W i ) (v i + W i ) is pathconnected and hence is connected.
Math 396. Connectedness of hyperplane complements Note that the complement of a point in R is disconnected and the complement of a (translated) line in R 2 is disconnected. Quite generally, we claim that
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 quasiprojective varieties are Hausdorff and that projective varieties are the only compact varieties.
More informationP1 Chapter 3 :: Equations and Inequalities
P1 Chapter 3 :: Equations and Inequalities jfrost@tiffin.kingston.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 26 th August 2017 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework
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 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 informationWhen a function is defined by a fraction, the denominator of that fraction cannot be equal to zero
As stated in the previous lesson, when changing from a function to its inverse the inputs and outputs of the original function are switched, because we take the original function and solve for x. This
More informationIntroduction to Real Analysis Alternative Chapter 1
Christopher Heil Introduction to Real Analysis Alternative Chapter 1 A Primer on Norms and Banach Spaces Last Updated: March 10, 2018 c 2018 by Christopher Heil Chapter 1 A Primer on Norms and Banach Spaces
More information88 CHAPTER 3. SYMMETRIES
88 CHAPTER 3 SYMMETRIES 31 Linear Algebra Start with a field F (this will be the field of scalars) Definition: A vector space over F is a set V with a vector addition and scalar multiplication ( scalars
More informationADVANCED TOPICS IN ALGEBRAIC GEOMETRY
ADVANCED TOPICS IN ALGEBRAIC GEOMETRY DAVID WHITE Outline of talk: My goal is to introduce a few more advanced topics in algebraic geometry but not to go into too much detail. This will be a survey of
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 informationNumber Theory Math 420 Silverman Exam #1 February 27, 2018
Name: Number Theory Math 420 Silverman Exam #1 February 27, 2018 INSTRUCTIONS Read Carefully Time: 50 minutes There are 5 problems. Write your name neatly at the top of this page. Write your final answer
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 informationSupplementary Material for MTH 299 Online Edition
Supplementary Material for MTH 299 Online Edition Abstract This document contains supplementary material, such as definitions, explanations, examples, etc., to complement that of the text, How to Think
More informationAbsolute Values and Completions
Absolute Values and Completions B.Sury This article is in the nature of a survey of the theory of complete fields. It is not exhaustive but serves the purpose of familiarising the readers with the basic
More informationMATH 115, SUMMER 2012 LECTURE 4 THURSDAY, JUNE 21ST
MATH 115, SUMMER 2012 LECTURE 4 THURSDAY, JUNE 21ST JAMES MCIVOR Today we enter Chapter 2, which is the heart of this subject. Before starting, recall that last time we saw the integers have unique factorization
More informationFactorization of integervalued polynomials with squarefree denominator
accepted by Comm. Algebra (2013) Factorization of integervalued polynomials with squarefree denominator Giulio Peruginelli September 9, 2013 Dedicated to Marco Fontana on the occasion of his 65th birthday
More informationMath 480 The Vector Space of Differentiable Functions
Math 480 The Vector Space of Differentiable Functions The vector space of differentiable functions. Let C (R) denote the set of all infinitely differentiable functions f : R R. Then C (R) is a vector space,
More informationGeneralized eigenspaces
Generalized eigenspaces November 30, 2012 Contents 1 Introduction 1 2 Polynomials 2 3 Calculating the characteristic polynomial 5 4 Projections 7 5 Generalized eigenvalues 10 6 Eigenpolynomials 15 1 Introduction
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 informationDMATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski. Solutions Sheet 1. Classical Varieties
DMATH 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 informationDescribing the Real Numbers
Describing the Real Numbers Anthony Várilly Math 25a, Fall 2001 1 Introduction The goal of these notes is to uniquely describe the real numbers by taking certain statements as axioms. This exercise might
More informationADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS
ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material.
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 1060 Linear Algebra Homework Exercises 1 1. Find the complete solutions (if any!) to each of the following systems of simultaneous equations:
Homework Exercises 1 1 Find the complete solutions (if any!) to each of the following systems of simultaneous equations: (i) x 4y + 3z = 2 3x 11y + 13z = 3 2x 9y + 2z = 7 x 2y + 6z = 2 (ii) x 4y + 3z =
More informationMath 210B. Artin Rees and completions
Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an Amodule. In class we defined the Iadic completion of M to be M = lim M/I n M. We will soon show
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 informationTangent spaces, normals and extrema
Chapter 3 Tangent spaces, normals and extrema If S is a surface in 3space, with a point a S where S looks smooth, i.e., without any fold or cusp or selfcrossing, we can intuitively define the tangent
More information