# Lecture 2: Abhyankar s Results on the Newton Polygon

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Lecture 2: Abhyankar s Results on the Newton Polygon Definition. Let P(X, Y ) = p ij X i Y j C[X, Y ]. The support of P, S(P), is the set {(i, j) N N f ij 0}. The Newton polygon of P, N(P), is the convex hull in R 2 of S(P) {(0, 0)}. In the next lecture, in connection with valuations, we will use a slightly different kind of Newton polygon. The ordinary degree of F(X, Y ) arises by giving each of X and Y degree one. One can also have degree in X and degree in Y. More generally Definition. Let (a, b) Z Z. The (a, b)-degree, d = d(a, b) on C[X, Y ] is defined by d(p) = max{ai+bj p ij 0} if P = p ij 0. By convention, d(0) =. P is (a, b)-homogeneous of degree r if ai + bj = r for every (i, j) in the support of P. Thus the (1, 1)-degree is the ordinary degree, the (1, 0)-degree is the X-degree, and the (0, 1)-degree is the Y -degree. A polynomial is (a, b)-homogeneous if its support lies on a line of slope b/a. Note that the zero polynomial is (a, b)- homogeneous of all degrees. In the first lecture, J(F, G) denoted a 2 2 matrix of partial derivatives. In this lecture J(F, G) will denote the determinant of this matrix. The starting point of Abhyankar s work is the observation that J(X i Y j, X r Y s ) = (is jr)x i+r 1 Y j+s 1. Thus the Jacobian of monomials X i Y j and X r Y s is, up to a scalar factor, their product divided by XY. The scalar factor is zero iff is jr = 0 iff X i Y j and X r Y s are powers of the same monomial iff X i Y j and X r Y s are algebraically dependent. Thus the above observation implies Lemma 1. Let d be the (a, b)-degree. If P and Q are d-homogeneous and algebraically independent, then J(P, Q) is d-homogeneous of degree d(p) + d(q) d(xy ). E. Noether proved a refined version of Lüroth s Theorem. Theorem 2. Let K be a subfield of C(X 1,...,X n ) which contains C and has transcendence degree one over C. (a) (J. Lüroth.) There is P C(X 1,..., X n ) such that K = C(P). (b) (E. Noether.) If K contains a nonconstant polynomial in C[X 1,..., X n ], then K = C(P) for some P C[X 1,..., X n ]. It is easy to see that if P C[X, Y ] and Q (C[X, Y ] C(P)) is non-constant and d-homogeneous, then P must be d-homogeneous. This leads to Lemma 3. If P, Q are d-homogeneous and algebraically dependent, then there is a d-homogeneous polynomial H such that P = ah r and Q = bh s for some a, b C, r, s N. Recall that P, Q C[X, Y ] are algebraically dependent iff J(P, Q) = 0. This leads to the following lemma, 1

2 2 Lemma 4. Suppose P, Q C[X, Y ] are d-homogeneous and J(P, Q) = 0. Then up r = vq s for some u, v C (not both 0) and r, s N. Now let d be the (a, b)-degree, where a and b are not both zero. Any nonzero P = P(X, Y ), Q = Q(X, Y ) can be written as sums of homogeneous elements P = P i1 + + P ik, Q = Q j1 + + Q jl, where P is is a nonzero homogeneous polynomial of (a, b)-degree i s, k 1, i 1 < < i k, and similarly for Q. If P ik and Q jl are algebraically independent, then J(P ik, Q jl ) will be nonzero and homogeneous of (a, b)-degree d(p)+d(q) d(xy ), and will be the highest degree homogeneous component of J(P, Q). On the other hand, if P ik and Q jl are algebraically dependent, then P r will be a C-multiple of Q s, where r/s = d(q)/d(p). Thus Lemma 5. Let d be the (a, b)-degree, where a and b are not both zero, and let P(X, Y ), Q(X, Y ) be nonzero polynomials. (a) If the highest degree homogeneous components of P and Q are algebraically independent, then d(j(p, Q)) = d(p) + d(q) d(xy ). (b) If the highest degree homogeneous components of P and Q are algebraically dependent, then d(j(p, Q)) < d(p) + d(q) d(xy ). In this case there is u C such that deg(p r uq s ) < r d(p), where r/s = d(q)/d(p). Let us consider a polynomial P(X, Y ) = p ij X i Y j and its support S(P) and Newton polygon N(P). If we are given a nonzero degree (a, b) and an integer r, then the points {(i, j) ai + bj = r} lie on a line of slope b/a (if a = 0 we say the slope is ), and the polynomial {pij X i Y j ai + pj = r} is the homogeneous component of P of degree r. The boundary of the Newton polygon of P will consist of line segments on the positive sides of the coordinate axes (which may reduce to the point (0, 0)) whose endpoints are joined by a polygonal line which goes from the Y -axis clockwise to to the X-axis. Ignoring cases where the Newton polygon reduces to a point or a line, we can traverse its boundary in a clockwise direction by starting at the origin, going up the Y -axis, around to the X-axis, and back to the origin. With each edge E of the Newton polygon we can associate an edge polynomial P E (X, Y ) = {p ij X i Y j (i, j) E}. Each edge polynomial is the highest degree component of P with respect to some degree (a, b). The edge polynomial of the Y -axis corresponds to the ( 1, 0)-degree; proceeding clockwise, the next edge may have positive, zero, or negative slope, which will correspond, respectively, to degrees (a, b), with a < 0 < b, (a, b) = (0, 1), or 0 < a, b. Then, descending to the X-axis, edges may have infinite or positive slope, which will correspond, respectively, to (a, b) = (1, 0) and a > 0 > b. Note that (a, b)-degrees with a, b < 0 do not occur. For such degrees, the highest degree component of P (assuming p 00 0) is simply p 00, which is of no interest. The ( 1, 0)-degree and (0, 1)-degrees are also of minimal interest. For if we write P(X, Y ) and Q(X, Y ) as polynomials in X with coefficients in C[Y ] P(X, Y ) = P 0 (Y ) + P 1 (Y )X +, Q(X, Y ) = Q 0 (Y ) + Q 1 (Y )X +,

3 3 and assume (as will be the case when trying to prove the Jacobian Conjecture) that P 0 (Y ), Q 0 (Y ) 0, then the edge polynomials P 0 (Y ) and Q 0 (Y ) will be the highest degree homogeneous components with respect to the ( 1, 0)-degree. We will automatically have J(P 0 (Y ), Q 0 (Y )) = 0, but this will not in itself be of particular use. Now let us consider a Jacobian pair (F, G), which means that J(F, G) is a nonzero complex number. If the Newton polygon of F does not contain the point (1, 1) then either F(X, Y ) = ax + f(y ) or F(X, Y ) = f(x) + ay, where a C and f is a polynomial in one variable; in either case it is easy to show that (F, G) is not a counterexample to the Jacobian Conjecture. Likewise, it is easy to see that if F(X, Y ) = F 0 (X) + F 1 (X)Y +, and F 0 (X) is a constant, then (F, G) is not a counterexample to the Jacobian Conjecture. Next suppose that (F, G) is a counterexample to the Jacobian Conjecture. By the remarks above, both N(F) and N(G) contain the point (1, 1) and points on both axes other than the origin. Suppose d = (a, b) is a degree with at least one of a, b positive. Then d(f) d(xy ) because (1, 1) N(F) and d(g) > 0 because N(G) contains points (t, 0) and (0, u) with t, u > 0. This leads to Theorem 6. Suppose that (F, G) is a counterexample to the two-variable Jacobian Conjecture. (I.e. J(F, G) C, but C[F, G] C[X, Y ].) Then (a) Let d = (a, b) be a degree with at least one of a, b positive, and let P, Q be the highest degree components of F and G with respect to d. Let d(g)/d(f) = d(q)/d(p) = r/s. Then there is c C such that P r = cq s. (b) The Newton polygons of F and G are similar. More precisely, the constants c and r/s in (a) are independent of the particular degree d. Except for the edge polynomials on the coordinate axes, the edge polynomials of F s and cg r are equal. Proof. (a) We have observed that d(p) d(xy ) and d(q) > 0. If P and Q were algebraically independent, then Lemma 5(a) would give the contradiction 0 = d(j(p, Q)) = d(p) + d(q) d(xy ) > 0. Hence P and Q are algebraically dependent, and then Lemma 4 implies that there is c C such that P r = cq s, where r/s = d(q)/d(p). (b) Proceeding clockwise from the Y -axis to the X-axis on the boundary of N(F) or N(G), the endpoint of an edge is the initial point of the next edge. Thus the constant c and the ratio r/s are independent of the particular edge, which implies that P s = cq r for every corresponding pair of edge polynomials of F and G except those on the coordinate axes. In particular, N(F) and N(G) are similar. A Resolution Process for Edge Polynomials The next step in Abhyankar s program is to study the edge polynomials which arise if (F, G) is a counterexample to the Jacobian Conjecture. It turns out that they satisfy some unusual Diophantine conditions. Polynomials satisfying these conditions are rare, but they do exist. Thus, the Jacobian Conjecture cannot be established by showing that these Diophantine conditions lead to a contradiction. What Abhyankar is able to show is that if (F, G) is a counterexample to the Jacobian Conjecture and the Newton polygon of F has an edge of negative slope, then there is an automorphism ϕ of C[X, Y ] such that

4 4 D(F(X, Y )) < D(ϕ(F(X, Y )) = D(F(ϕ(X), ϕ(y )), where D(P) = d X (P) + d Y (P), the sum of the X-degree and the Y -degree. This leads to Abhyankar s main result, which says that if the two-variable Jacobian Conjecture is false, then there is a counterexample (F, G) for which the Newton polygons of F and G have no edges with negative slope. Suppose that P, Q are nonzero polynomials in C, and d is the (a, b)-degree, where at least one of a, b is positive. By Lemma 5(b), d(p) + d(q) d(xy ) d(j(p, Q)) 0, with equality iff the highest degree homogeneous components of P and Q are algebraically independent. We now define a nonnegative integer (P, Q) by (P, Q) = d(p) + d(q) d(xy ) d(j(p, Q)). Now suppose further that (F, G) is a Jacobian pair which is a counterexample to the Jacobian Conjecture. Then we know that (F, G) > 0. Set F 1 = F, f 1 = d(f 1 ), P 1 = highest degree d-homogeneous component of F 1, g = d(g), Q = highest degree d-homogeneous component of G, and define polynomials F 2, F 3,... with highest degree d-homogeneous components P 2, P 3,... and degrees f 2, f 3,... inductively as follows: By Theorem 6, there is c 1 C such that P g 1 = c 1Q f1. This implies that d(f g 1 c 1G f1 ) < d(f g 1 ) = gd(f 1) = gf 1, and we set F 2 = F g 1 c 1G f1, δ 1 = d(f g 1 ) d(f 2) > 0. Then J(F 2, G) = J(F g 1 c 1G f1, G) = gf g 1 1, and (F 2, G) = d(f 2 ) + d(g) d(xy ) d(j(f 2, G)) = (gf 1 δ 1 ) + g d(xy ) (g 1)f 1 = f 1 + g 1 d(xy ) δ 1 = (F 1, G) δ 1. It may happen that (F 2, G) = 0, in which case the procedure terminates. On the other hand, if (F 2, G) > 0, then there is a c 2 C such that F 3 = F g 2 c 2G f2 satisfies d(f 3 ) = gd(f 2 ) δ 2, J(F 3, G) = (F 1 F 2 ) g 1, and (F 3, G) = (F 1, G) δ 1 δ 2. As long as (F n, G) > 0 we continue, letting F n+1 = F g n c n G fn, where J(F n+1, G) = (F 1 F n ) g 1 and (F 1, G) > (F 2, G) > > (F n+1, G). Since (F i+1, G) = (F i, G) δ i, where δ i is a positive integer, the process must terminate with (F n+1, G) = 0 for some n. Consider the highest degree d-homogeneous components of G, F 1, F 2,..., F n. Since (F 1, G),..., (F n, G) are all positive, they are all C-multiples of powers of some d-homogeneous polynomial H, which we may assume is not a proper power (using the fact that C[X, Y ] is a unique factorization domain). Letting K be the highest degree d-homogeneous component of F n+1, the equation J(F n+1, G) = (F 1 F n ) g 1 implies that J(K, H s ) = ch t for some positive integers s and t and c C. An easy calculation shows that s t and that J(K, H) = c H u for u = t s + 1 and some c C. This yields

5 5 Theorem 7. Suppose that (F, G) is a counterexample to the two-variable Jacobian Conjecture, and d = (a, b) is a degree with a, b 0 such that N(F) has an edge of slope b/a, and let H = H(X, Y ) be a d-homogeneous polynomial such that H is not a proper power and the d-homogeneous component of F is a power of H. Then there is a d-homogeneous polynomial K = K(X, Y ) and a positive integer s such that J(K, H) = H s. Homogeneous polynomial solutions of the equation J(K, H) = H s do exist, although in a sense they are rare. They are of four essentially different types: (1) Those corresponding to edges of negative slope, or degree functions d = (a, b) with both a and b positive. (2) Lower edges of slope > 1 (d = (a, b), with a > 0 > b, a > b) or upper edges of slope < 1 (d = (a, b), with a < 0 < b, b > a). (3) Lower edges of slope < 1 (d = (a, b), with a > 0 > b, a < b), or upper edges of slope > 1(d = (a,b), with a < 0 < b, b < a. (4) Lower edges of slope 1 (d = (1, 1)), or upper edges of slope 1 (d = ( 1, 1)). Although all four types of solutions exist, it is the first type which has so far proved the most useful. This is because of the following result. Lemma 8. Let d = (a, b) be a degree with a, b positive, and let H be a d- homogeneous polynomial in C[X, Y ] which is not a monomial. Suppose that there is a d-homogeneous polynomial K 0 and a positive integer s such that J(K 0, H) = H s. Then (a) There is a d-homogeneous polynomial K such that J(K, H) = H. (b) The support of K either consists of the point (1, 1) and one other point on either the X-axis or the Y -axis (but not (0, 0)), or is contained in the set {(1, 0), (1, 1), (0, 1)}. Thus, up to multiplication by an element of C, K is equal to XY + ry n+1 = Y (X+rY n ), XY +rx n+1 = X(Y +rx n ), or r(x+sy )(X+tY ), where r, s, t C,. Moreover, the slope, a/b, of H and K is either a negative integer or the reciprocal of a negative integer. Proof. (a) is proved by a direct calculation which we omit. (b) is simply the observation that if a line segment lies in the first quadrant, has negative slope, contains the point (1, 1), and has endpoints with integer coordinates, then it has the indicated form. Now suppose that (F, G) is a counterexample to the two-variable Jacobian Conjecture, and the Newton polygon of F has an edge of negative slope, and let H and K be the polynomials given by Theorem 7. Let d X = (1, 0) and d Y = (0, 1) denote the ordinary and X and Y -degrees, respectively. We now apply Lemma 8, which says that one of four possibilities occurs. (a) K = X(Y + rx n ), where r C and n N, n 1. n 1. The equation J(K, H) = H implies that H = qx m (Y +rx n ) for some q, r C. Let ϕ = (X, Y rx n ), an elementary automorphism. d Y (ϕ(f)) = d Y (F) and d X (ϕ(f)) < d X (F), so d X (ϕ(f)) + d Y (ϕ(f) < d X (F) + d Y (F). (b) K = Y (X + ry n ). Proceed as in (a) to reduce d X (F) + d Y (F).

6 6 (c) K = r(x + sy ) 2 for some r, s C. Then ϕ = (X sy, Y ) reduces d X (F) + d Y (F) as in (b). (d) K = r(x + sy )(X + ty ) for some r, s, t C, where s t. The equation J(K, H) = H shows that H = u(x + sy ) i (X + ty ) j for some positive integers i and j, and u C. Consider the elementary automorphism ϕ, where ϕ 1 = (X + sy, X + ty ). Since ϕ(h) = ux I Y j, we see that d X (ϕ(f)) < d X (F) and d Y (ϕ(f) < d Y (F). Note that none of the above transformations increase d(f), where d = (1, 1) is the ordinary degree. The following theorem sums up most of Abhyankar s results on the Newton polygon. Theorem 9 (S. S. Abhyankar). Suppose that the two-variable Jacobian Conjecture is false. Then there is a counterexample (F, G) with the following properties. (a) The Newton polygons of F and G are similar, and have no edges of negative slope. (I.e. There exist positive integers a, b such that (a, b) N(F) and N(F) [0, a] [0, b].) (b) d(f) and d(f) + d(g) are minimal among all counterexamples, where d is the ordinary degree. (c) Neither of d(f), d(g) divides the other. Proof. (a) follows from the discussion above. (b) follows from the fact that the automorphisms used to achieve (a) do not increase d(f) or d(g). (c) Assume for definiteness that d(g) = rd(f) for some positive integer r. Then, by Theorem 6(b), there would be c C such that d(g cf r ) < d(g), and the pair (F, G cf r ) would reduce d(f) + d(g) while leaving d(f) unchanged. R. Heitmann, J. Lang and M. Nagata have independently refined Theorem 9(a) as follows: Assume that a b. Then N(F) is contained in the quadrilateral with vertices (0, 0), (0, b a), (a, b), (a, 0). In other words, the upper edges of N(F) have slope 1.

### it is easy to see that α = a

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 UF. Therefore

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Interpolating Polynomials Handout March 7, 2012

Interpolating Polynomials Handout March 7, 212 Again we work over our favorite field F (such as R, Q, C or F p ) We wish to find a polynomial y = f(x) passing through n specified data points (x 1,y 1 ),

### 2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

### Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

### MATHEMATICAL BACKGROUND

Chapter 1 MATHEMATICAL BACKGROUND This chapter discusses the mathematics that is necessary for the development of the theory of linear programming. We are particularly interested in the solutions of a

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### Limits at Infinity Limits at Infinity for Polynomials Limits at Infinity for the Exponential Function Function Dominance More on Asymptotes

Lecture 5 Limits at Infinity and Asymptotes Limits at Infinity Horizontal Asymptotes Limits at Infinity for Polynomials Limit of a Reciprocal Power The End Behavior of a Polynomial Evaluating the Limit

### Summary of week 8 (Lectures 22, 23 and 24)

WEEK 8 Summary of week 8 (Lectures 22, 23 and 24) This week we completed our discussion of Chapter 5 of [VST] Recall that if V and W are inner product spaces then a linear map T : V W is called an isometry

### MATH 110 Spring 2015 Homework 6 Solutions

MATH 110 Spring 2015 Homework 6 Solutions Section 2.6 2.6.4 Let α denote the standard basis for V = R 3. Let α = {e 1, e 2, e 3 } denote the dual basis of α for V. We would first like to show that β =

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 20

CS 70 Discrete Mathematics and Probability Theory Fall 009 Satish Rao, David Tse Note 0 Infinity and Countability Consider a function (or mapping) f that maps elements of a set A (called the domain of

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

### 1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### Diagonal, Symmetric and Triangular Matrices

Contents 1 Diagonal, Symmetric Triangular Matrices 2 Diagonal Matrices 2.1 Products, Powers Inverses of Diagonal Matrices 2.1.1 Theorem (Powers of Matrices) 2.2 Multiplying Matrices on the Left Right by

### 1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]

1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not

### Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.

Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.

MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar

### 1 VECTOR SPACES AND SUBSPACES

1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such

### UNIT 2 MATRICES - I 2.0 INTRODUCTION. Structure

UNIT 2 MATRICES - I Matrices - I Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices 2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress

### NOTES ON LINEAR TRANSFORMATIONS

NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all

### A matrix over a field F is a rectangular array of elements from F. The symbol

Chapter MATRICES Matrix arithmetic A matrix over a field F is a rectangular array of elements from F The symbol M m n (F) denotes the collection of all m n matrices over F Matrices will usually be denoted

### k=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =

Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem

### THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. Saturday, March 5, 2016

THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION In Memory of Robert Barrington Leigh Saturday, March 5, 2016 Time: 3 1 2 hours No aids or calculators permitted. The grading is designed

### Functions and Equations

Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c

### Class 11 Maths Chapter 10. Straight Lines

1 P a g e Class 11 Maths Chapter 10. Straight Lines Coordinate Geometry The branch of Mathematics in which geometrical problem are solved through algebra by using the coordinate system, is known as coordinate

### Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if

1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c

### x n = 1 x n In other words, taking a negative expoenent is the same is taking the reciprocal of the positive expoenent.

Rules of Exponents: If n > 0, m > 0 are positive integers and x, y are any real numbers, then: x m x n = x m+n x m x n = xm n, if m n (x m ) n = x mn (xy) n = x n y n ( x y ) n = xn y n 1 Can we make sense

### Problem Set 1 Solutions Math 109

Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twenty-four symmetries if reflections and products of reflections are allowed. Identify a symmetry which is

### MTH 06 LECTURE NOTES (Ojakian) Topic 2: Functions

MTH 06 LECTURE NOTES (Ojakian) Topic 2: Functions OUTLINE (References: Iyer Textbook - pages 4,6,17,18,40,79,80,81,82,103,104,109,110) 1. Definition of Function and Function Notation 2. Domain and Range

### Computer Algebra for Computer Engineers

p.1/14 Computer Algebra for Computer Engineers Preliminaries Priyank Kalla Department of Electrical and Computer Engineering University of Utah, Salt Lake City p.2/14 Notation R: Real Numbers Q: Fractions

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### by the matrix A results in a vector which is a reflection of the given

Eigenvalues & Eigenvectors Example Suppose Then So, geometrically, multiplying a vector in by the matrix A results in a vector which is a reflection of the given vector about the y-axis We observe that

### GRA6035 Mathematics. Eivind Eriksen and Trond S. Gustavsen. Department of Economics

GRA635 Mathematics Eivind Eriksen and Trond S. Gustavsen Department of Economics c Eivind Eriksen, Trond S. Gustavsen. Edition. Edition Students enrolled in the course GRA635 Mathematics for the academic

### Math 54. Selected Solutions for Week Is u in the plane in R 3 spanned by the columns

Math 5. Selected Solutions for Week 2 Section. (Page 2). Let u = and A = 5 2 6. Is u in the plane in R spanned by the columns of A? (See the figure omitted].) Why or why not? First of all, the plane in

### DIFFERENTIABILITY OF COMPLEX FUNCTIONS. Contents

DIFFERENTIABILITY OF COMPLEX FUNCTIONS Contents 1. Limit definition of a derivative 1 2. Holomorphic functions, the Cauchy-Riemann equations 3 3. Differentiability of real functions 5 4. A sufficient condition

### Introduction to polynomials

Worksheet 4.5 Polynomials Section 1 Introduction to polynomials A polynomial is an expression of the form p(x) = p 0 + p 1 x + p 2 x 2 + + p n x n, (n N) where p 0, p 1,..., p n are constants and x os

### MATH : HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS

MATH 16300-33: HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS 25-1 Find the absolute value and argument(s) of each of the following. (ii) (3 + 4i) 1 (iv) 7 3 + 4i (ii) Put z = 3 + 4i. From z 1 z = 1, we have

### HTP in Positive Characteristic

HTP in Positive Characteristic Alexandra Shlapentokh East Carolina University November 2007 Table of Contents 1 A Brief History of Diophantine Undecidability over Number Fields The Original Problem Extensions

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### (Refer Slide Time 1.50)

Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Module -2 Lecture #11 Induction Today we shall consider proof

### Solutions to Practice Problems

Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### 1.3 Induction and Other Proof Techniques

4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.

### Chapter 1 Basic Number Concepts

Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section. 1.2. Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11:

### Section 3.2 Polynomial Functions and Their Graphs

Section 3.2 Polynomial Functions and Their Graphs EXAMPLES: P(x) = 3, Q(x) = 4x 7, R(x) = x 2 +x, S(x) = 2x 3 6x 2 10 QUESTION: Which of the following are polynomial functions? (a) f(x) = x 3 +2x+4 (b)

### 3 Congruence arithmetic

3 Congruence arithmetic 3.1 Congruence mod n As we said before, one of the most basic tasks in number theory is to factor a number a. How do we do this? We start with smaller numbers and see if they divide

### Homework: 2.1 (page 56): 7, 9, 13, 15, 17, 25, 27, 35, 37, 41, 46, 49, 67

Chapter Matrices Operations with Matrices Homework: (page 56):, 9, 3, 5,, 5,, 35, 3, 4, 46, 49, 6 Main points in this section: We define a few concept regarding matrices This would include addition of

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### since by using a computer we are limited to the use of elementary arithmetic operations

> 4. Interpolation and Approximation Most functions cannot be evaluated exactly: x, e x, ln x, trigonometric functions since by using a computer we are limited to the use of elementary arithmetic operations

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### STRAIGHT LINES. , y 1. tan. and m 2. 1 mm. If we take the acute angle between two lines, then tan θ = = 1. x h x x. x 1. ) (x 2

STRAIGHT LINES Chapter 10 10.1 Overview 10.1.1 Slope of a line If θ is the angle made by a line with positive direction of x-axis in anticlockwise direction, then the value of tan θ is called the slope

### 13 Groups of Isometries

13 Groups of Isometries For any nonempty set X, the set S X of all one-to-one mappings from X onto X is a group under the composition of mappings (Example 71(d)) In particular, if X happens to be the Euclidean

### 4. How many integers between 2004 and 4002 are perfect squares?

5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started

### Cylinder Maps and the Schwarzian

Cylinder Maps and the Schwarzian John Milnor Stony Brook University (www.math.sunysb.edu) Bremen June 16, 2008 Cylinder Maps 1 work with Araceli Bonifant Let C denote the cylinder (R/Z) I. (.5, 1) fixed

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### Lecture 5 : Continuous Functions Definition 1 We say the function f is continuous at a number a if

Lecture 5 : Continuous Functions Definition We say the function f is continuous at a number a if f(x) = f(a). (i.e. we can make the value of f(x) as close as we like to f(a) by taking x sufficiently close

### Practice with Proofs

Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using

### Chapter 12. The Straight Line

302 Chapter 12 (Plane Analytic Geometry) 12.1 Introduction: Analytic- geometry was introduced by Rene Descartes (1596 1650) in his La Geometric published in 1637. Accordingly, after the name of its founder,

### Integral Domains. As always in this course, a ring R is understood to be a commutative ring with unity.

Integral Domains As always in this course, a ring R is understood to be a commutative ring with unity. 1 First definitions and properties Definition 1.1. Let R be a ring. A divisor of zero or zero divisor

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Appendix A. Appendix. A.1 Algebra. Fields and Rings

Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.

### 3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.

Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible. 3.1.1 Definition A Dedekind

### Rational Polynomial Functions

Rational Polynomial Functions Rational Polynomial Functions and Their Domains Today we discuss rational polynomial functions. A function f(x) is a rational polynomial function if it is the quotient of

### Linear Algebra I. Ronald van Luijk, 2012

Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.

### A COUNTEREXAMPLE FOR SUBADDITIVITY OF MULTIPLIER IDEALS ON TORIC VARIETIES

Communications in Algebra, 40: 1618 1624, 2012 Copyright Taylor & Francis Group, LLC ISSN: 0092-7872 print/1532-4125 online DOI: 10.1080/00927872.2011.552084 A COUNTEREXAMPLE FOR SUBADDITIVITY OF MULTIPLIER

### Lecture 3. Mathematical Induction

Lecture 3 Mathematical Induction Induction is a fundamental reasoning process in which general conclusion is based on particular cases It contrasts with deduction, the reasoning process in which conclusion

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### 3. Some commutative algebra Definition 3.1. Let R be a ring. We say that R is graded, if there is a direct sum decomposition, M n, R n,

3. Some commutative algebra Definition 3.1. Let be a ring. We say that is graded, if there is a direct sum decomposition, = n N n, where each n is an additive subgroup of, such that d e d+e. The elements

### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

### Solutions to Homework Problems from Chapter 3

Solutions to Homework Problems from Chapter 3 31 311 The following subsets of Z (with ordinary addition and multiplication satisfy all but one of the axioms for a ring In each case, which axiom fails (a

### 1.5 Elementary Matrices and a Method for Finding the Inverse

.5 Elementary Matrices and a Method for Finding the Inverse Definition A n n matrix is called an elementary matrix if it can be obtained from I n by performing a single elementary row operation Reminder:

### (x + a) n = x n + a Z n [x]. Proof. If n is prime then the map

22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find

### TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

### Homework Set 2 (Due in class on Thursday, Sept. 25) (late papers accepted until 1:00 Friday)

Math 202 Homework Set 2 (Due in class on Thursday, Sept. 25) (late papers accepted until 1:00 Friday) Jerry L. Kazdan The problem numbers refer to the D Angelo - West text. 1. [# 1.8] In the morning section

### Some Basic Properties of Vectors in n

These notes closely follow the presentation of the material given in David C. Lay s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation

### Math 315: Linear Algebra Solutions to Midterm Exam I

Math 35: Linear Algebra s to Midterm Exam I # Consider the following two systems of linear equations (I) ax + by = k cx + dy = l (II) ax + by = 0 cx + dy = 0 (a) Prove: If x = x, y = y and x = x 2, y =

### Polynomial Invariants

Polynomial Invariants Dylan Wilson October 9, 2014 (1) Today we will be interested in the following Question 1.1. What are all the possible polynomials in two variables f(x, y) such that f(x, y) = f(y,

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### 14. Dirichlet polygons: the construction

14. Dirichlet polygons: the construction 14.1 Recap Let Γ be a Fuchsian group. Recall that a Fuchsian group is a discrete subgroup of the group Möb(H) of all Möbius transformations. In the last lecture,

### Lecture 14: Section 3.3

Lecture 14: Section 3.3 Shuanglin Shao October 23, 2013 Definition. Two nonzero vectors u and v in R n are said to be orthogonal (or perpendicular) if u v = 0. We will also agree that the zero vector in

### 1 Local Brouwer degree

1 Local Brouwer degree Let D R n be an open set and f : S R n be continuous, D S and c R n. Suppose that the set f 1 (c) D is compact. (1) Then the local Brouwer degree of f at c in the set D is defined.

### CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS

CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence

### Markov Chains, part I

Markov Chains, part I December 8, 2010 1 Introduction A Markov Chain is a sequence of random variables X 0, X 1,, where each X i S, such that P(X i+1 = s i+1 X i = s i, X i 1 = s i 1,, X 0 = s 0 ) = P(X

### CHAPTER 3. Mapping Concepts and Mapping Problems for. Scalar Valued Functions of a Scalar Variable

A SERIES OF CLASS NOTES TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS REMEDIAL CLASS NOTES A COLLECTION OF HANDOUTS FOR REMEDIATION IN FUNDAMENTAL CONCEPTS

### Chapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and Exponential Functions Quiz 5 Exam 4

Chapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and QUIZ AND TEST INFORMATION: The material in this chapter is on Quiz 5 and Exam 4. You should complete at least one attempt of

### 7 - Linear Transformations

7 - Linear Transformations Mathematics has as its objects of study sets with various structures. These sets include sets of numbers (such as the integers, rationals, reals, and complexes) whose structure

### ZERO-DIVISOR GRAPHS OF POLYNOMIALS AND POWER SERIES OVER COMMUTATIVE RINGS

ZERO-DIVISOR GRAPHS OF POLYNOMIALS AND POWER SERIES OVER COMMUTATIVE RINGS M. AXTELL, J. COYKENDALL, AND J. STICKLES Abstract. We recall several results of zero divisor graphs of commutative rings. We

### 18.312: Algebraic Combinatorics Lionel Levine. Lecture 8

18.312: Algebraic Combinatorics Lionel Levine Lecture date: March 1, 2011 Lecture 8 Notes by: Christopher Policastro Remark: In the last lecture, someone asked whether all posets could be constructed from

### MATH 304 Linear Algebra Lecture 4: Matrix multiplication. Diagonal matrices. Inverse matrix.

MATH 304 Linear Algebra Lecture 4: Matrix multiplication. Diagonal matrices. Inverse matrix. Matrices Definition. An m-by-n matrix is a rectangular array of numbers that has m rows and n columns: a 11

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization