Algebraic function fields


 Brice Willis
 3 years ago
 Views:
Transcription
1 Algebraic function fields
2 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 is transcendental over K. K = {z F : z is algebraic over K} K is called the field of constants of F/K. K is algebraically closed in F if K = K. We have 1. K K F 2. F/K is an algebraic function field over K (why?)
3 2 Definition F/K is called rational, if F = K(x) for some x F transcendental over K. Definition A valuation ring of the function field F/K is a ring O F with the following properties. 1. K O F 2. For any z F, z O or z 1 O Example Let p(x) K[x] be an irreducible polynomial. Then { } f(x) O p(x) = : f(x), g(x) K[x] and p(x) g(x) g(x) is a valuation ring of K(x)/K.
4 3 Proposition Let O be a valuation ring of the function field F/K. Then (a) O is a local ring, i.e., O has a unique maximal ideal P = O \ O, where O is the group of units of O. (b) For 0 x F, x P if and only if x 1 O. (c) K O and K P = {0}. Proof (a) It suffices to show that P = O \ O is an ideal of O. Let x P and let z O. Then xz O (why?), implying xz P. Let x, y P. We may assume that x/y O. Then 1 + x/y O, implying x + y = y(1 + x/y) P. (b)...
5 4 (c) Let z K and assume to the contrary that z O. Then z 1 O. Since z 1 is algebraic over K, there are elements a 1,..., a r K such that a r (z 1 ) r + + a 1 z = 0, implying z 1 (a r (z 1 ) r a 1 ) = 1. Thus, z = (a r (z 1 ) r a 1 ) K[z 1 ] O which contradicts our assumption.
6 5 Theorem Let O be a valuation ring of the function field F/K and let P be its unique maximal ideal. Then (a) P is a principal ideal. (b) If P = to, then any 0 z F has a unique representation of the form z = t n u for some n Z and u O. (c) O is a principal ideal domain, and if P = to and {0} I O is an is an ideal, then I = t n O for some n N. Definition A ring possessing the above properties is called a discrete valuation ring. Lemma Let O be a valuation ring of the algebraic function field F/K, P be its maximal ideal, and let 0 x P. Let x 1,..., x n P be such that x 1 = x and x i x i+1 P, i = 1,..., n 1. Then n [F : K(x)] <.
7 6 Proof The inequality [F : K(x)] < is immediate (why?) and for the proof of n [F : K(x)] it suffices to show that x 1,..., x n are linearly independent over K(x). Assume to the contrary that n i=1 ϕ i x i = 0, where ϕ i K(x), not all equal zero. We may assume that all ϕ i K[x] and that x does not divide all of them (why). Let j = max{i : x ϕ i }. Then for all i > j there exists a polynomial g i K[x] such that ϕ i = xg i, and it follows from x j ϕ j = i j x i ϕ i. Therefore, implying ϕ j P (why?). ϕ j = i<j x i x j ϕ i + i>j x x j x i g i, n i=1 ϕ i x i = 0 that However, since ϕ j = ϕ j (0)+xg j with g j K[x] O, ϕ j (0) = ϕ j xg j P K (why?), which is impossible, because ϕ j (0) 0 (why?).
8 7 Corollary If t P, then n=1 t n O = {0}. Proof Assume to the contrary that for some z 0, z n=1 t n O. Then for all n N, the sequence x 1 = z, x 2 = t n 1, x 3 = t n 2,..., x n = t satisfies the lemma prerequisite x i x i+1 P, i = 1..., n 1. However, this is impossible for n > [F : K(x)].
9 8 Proof of the theorem (a) Assume to the contrary that P is not principal and let 0 x 1 P. Since P x 1 O, there is x 2 P \ x 1 O, implying x 2 x 1 1 O. Therefore, x 1 x 1 2 P and x 1 x 2 P. We proceed by induction and obtain an infinite sequence x 1, x 2,... P such that x i x i+1 P for all i = 1, 2,.... However, this contradicts the lemma. (b) The uniqueness of the representation is immediate (why?), and for the proof of the existence we may assume that z O (why?). If z is a unit, then z = t 0 z, i.e., u = z. Otherwise, z P (why?) and, by the corollary to the lemma, there is a maximal nonnegative integer n such that z t n O. Therefore, z = t n u, where u is a unit (why?).
10 9 (c) Let {0} I O be an ideal an let A = {r N : t r I}. Then A (why?) and let n = min(a). Since t n O I, the proof of part (c) of the theorem will be complete if we show the converse inclusion. Let 0 x I. Since x = t s u for some u O and s n, t s I as well. Therefore, x = t n t s n u t n O. Definition (a) A place P of the function field F/K is the maximal ideal of some valuation ring O of F/K. Any element t P such that P = to is called a prime element for P (or local parameter or uniformizing variable). (b) P F = {P : P is a place of F/K}.
11 10 Remark If O is a valuation ring of F/K and P is its maximal ideal, then O is uniquely determined by P : O = {z F : z 1 P }. We call O p = O the valuation ring of the place P. Definition (The arithmetic of Z { }) For all n Z, + = n + = + n > n
12 11 Definition A discrete valuation of F/K is a function v : F Z { } with the following properties. (1) v(x) = if an only if x = 0. (2) v(xy) = v(x) + v(y) for all x, y F. (3) v(x + y) min{v(x), v(y)} for all x, y F. (4) There exists an element z F with v(z) = 1. (5) v(a) = 0 for all 0 a K. It follows from (2) and (4) that v : F Z { } is surjective (how?). Also it follows from the definition that v(x) = v( x) for all x F (how?).
13 12 Lemma (Strict Triangle Inequality) Let v be a discrete valuation of F/K and let x, y F be such that v(x) v(y). Then v(x + y) = min{v(x), v(y)}. Proof Since v(x) v(y), we may assume that v(x) < v(y). Now, assume to the contrary that v(x + y) min{v(x), v(y)}. Then v(x + y) > v(x) (why?), implying a contradiction v(x) = v((x + y) y) min{v(x + y), v(y)} > v(x). Definition To a place P P F we associate a function v P : F Z { } such that v P (0) = and for 0 z F v P (z) is defined as follows. Let t be a prime element for P. Then for some u O and some n Z, z = t n u and this representation of z is unique (for a given t). We put v P (z) = n. This function v P is welldefined (why?).
14 13 Theorem Let F/K be a function field. (a) For any place P P F, the function v P F/K. In addition, is a discrete valuation of O P = {z F : v p (z) 0}, O P = {z F : v p (z) = 0}, P = {z F : v p (z) > 0}. An element z of F is a prime element for P if and only if v P (z) = 1. (b) Conversely, if v is a discrete valuation of F/K, then P v = {z F : v(z) > 0} is a place of F/K, and O Pv = {z F : v(z) 0} is the corresponding valuation ring. (c) Any valuation ring O of F/K is a maximal proper subring of F.
15 14 Proof (a) We shall prove the triangle inequality only. Let x, y F, v P (x) = n, and v P (y) = m, where n m <. Then x = t n u 1 and y = t m u 2, where u 1, u 2 O P and where z O P. If z = 0, then x + y = t n (u 1 + t m n u 2 ) = t n z, v P (x + y) = > min{n, m}. Otherwise, z = t k u, where k 0 and u O P, implying v P (x + y) = v P (t n+k u) = n + k n = min{v P (x), v P (y)}. The rest of (a)... (b)...
16 15 (c) Let O be a valuation ring of F/K and let z F \ O. We shall prove that O[z] = F. Let y F and let P be the maximal ideal of O. Then, for sufficiently large k, v P (yz k ) 0 (why?). Consequently, w = yz k O, implying y = wz k O[z]. Since P is a maximal ideal of O P, O P /P is a field, and for z O P we define z(p ) to be the residue class of z modulo P. If z F \ O P, we put z(p ) =. 1 Sometimes we shall also write z + P instead of z(p ) for z O P. Since K O P (K O P ) (why?) and K P = {0} (K P = {0}) (why?), the residue class map O P O P /P induces a canonical embedding of K (K) into O P /P. We shall always consider K (K) as a subfield of O P /P via this embedding. 1 This is not the same as in discrete valuations.
17 16 Definition Let P P F. (a) F P = O P /P is called the residue class field of P and the map x x(p ) from F to F P { } is called the residue class map with respect to P. (b) deg P = [F P : K] is called the degree of P. What happens when deg P = 1? Proposition Let P be a place of F/K and let 0 x P. Then deg P [F : K(x)] <. Proof We have already seen that [F : K(x)] < (where?), and for the proof of deg P [F : K(x)] it suffices to show that any elements z 1,..., z n of O P, whose residue classes z 1 (P ),..., z n (P ) are linearly independent over K, are linearly independent over K(x).
18 17 So, let z 1 (P ),..., z n (P ) be linearly independent over K, and assume to the contrary that there is a nontrivial linear combination with ϕ i K(x). n i=1 ϕ i z i = 0 We may assume that the ϕ i are polynomials in x and not all of them are divisible by x (why?). That is, ϕ = a i + xg i, where a i K and g i K[x], and not all a i = 0. Applying the residue class map to n i=1 ϕ i z i = 0 we obtain 0 = 0(P ) = n ϕ i (P )z i (P ) = i=1 n a i z i (P ), i=1 which contradicts the linear independence of z 1 (P ),..., z n (P ) over K.
19 18 Definition Let z F and let P P f. We say that P is a zero (respectively, a pole) of z if and only if v P (z) > 0 (respectively, v P (z) < 0). If v P (z) = m > 0 (respectively, v P (m) = m < 0), then P is a zero (respectively, pole) of P of order m. Theorem Let F/K be a function field, K R be a subring of F, and let I {0} be a proper ideal of R. Then there is a place P P F such that I P and R O P. Proof Let F = {S : S is a subring of F with R S and IS S}. 2 F, because R F, and we shall prove that F is inductively ordered by inclusion (what does it mean?). 2 IS is the ideal of S generated by I.
20 19 Indeed, if H F is a totally ordered subset of F, then T = {S : S H} is a subring of S with R T. We have to show that IT T. For this, assume to the contrary that n i=1 a is i = 1, with a i I and s i T. Since H is totally ordered, there is an S H such that s 1,..., s n S. Therefore, 1 = n i=1 a is i S as well, in contradiction with the definition of F (which contradiction?). By Zorn s lemma, F contains a maximal element O that, as we prove below, is a valuation ring of F/K. We have K O F (why?). Assume to the contrary that there exists an element z of F with z O and z 1 O. Then IO[z] = O[z] and IO[z 1 ] = O[z 1 ] (why?). Consequently, there are a 0,..., a n, b 0,..., b m IO such that 1 = a 0 + a 1 z + + a n z n 1 = b 0 + b 1 z a n z m
21 20 It follows that m, n 1 (how?) and we may assume that m and n are chosen minimally and m n (why?). Multiplying the first equality by 1 b 0 and the second by a n z n we obtain 1 b 0 = (1 b 0 )a 0 + (1 b 0 )a 1 z + + (1 b 0 )a n z n 0 = (b 0 1)a n z n + b 1 a n z n b m a n z n m Adding these equalities yields 1 = c 0 + c 1 z + + c n z n 1 with coefficients c i in IO (why?). This contradicts the minimality of n. That is, we have shown that for any z F, z O or z 1 O.
22 21 Corollary Let F/K be a function field and let z F be transcendental over K. Then z has at least one zero and one pole. 3 Proof Let R = K[z]. Then I = zk[z] is a proper ideal of R. Thus, there is a place P P F with z P (hence P is a zero of z). To prove that z has a pole we start with... Corollary The field K of constants of F/K is a finite extension of K. Proof Let P P F. Since K embeds into F P via the residue class map O P F P, it follows that 3 In particular, P F. [K : K] [F P : K] (< ).
From now on we assume that K = K.
Divisors From now on we assume that K = K. Definition The (additively written) free abelian group generated by P F is denoted by D F and is called the divisor group of F/K. The elements of D F are called
More informationABSOLUTE VALUES AND VALUATIONS
ABSOLUTE VALUES AND VALUATIONS YIFAN WU, wuyifan@umich.edu Abstract. We introduce the basis notions, properties and results of absolute values, valuations, discrete valuation rings and higher unit groups.
More informationPlaces of Number Fields and Function Fields MATH 681, Spring 2018
Places of Number Fields and Function Fields MATH 681, Spring 2018 From now on we will denote the field Z/pZ for a prime p more compactly by F p. More generally, for q a power of a prime p, F q will denote
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 informationSome consequences of the RiemannRoch theorem
Some consequences of the RiemannRoch theorem Proposition Let g 0 Z and W 0 D F be such that for all A D F, dim A = deg A + 1 g 0 + dim(w 0 A). Then g 0 = g and W 0 is a canonical divisor. Proof We have
More informationLocal Fields. Chapter Absolute Values and Discrete Valuations Definitions and Comments
Chapter 9 Local Fields The definition of global field varies in the literature, but all definitions include our primary source of examples, number fields. The other fields that are of interest in algebraic
More informationMaterial covered: Class numbers of quadratic fields, Valuations, Completions of fields.
ALGEBRAIC NUMBER THEORY LECTURE 6 NOTES Material covered: Class numbers of quadratic fields, Valuations, Completions of fields. 1. Ideal class groups of quadratic fields These are the ideal class groups
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 informationwhere c R and the content of f is one. 1
9. Gauss Lemma Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is as follows.
More informationDefinability in fields Lecture 2: Defining valuations
Definability in fields Lecture 2: Defining valuations 6 February 2007 Model Theory and Computable Model Theory Gainesville, Florida Defining Z in Q Theorem (J. Robinson) Th(Q) is undecidable. Defining
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 informationIn Theorem 2.2.4, we generalized a result about field extensions to rings. Here is another variation.
Chapter 3 Valuation Rings The results of this chapter come into play when analyzing the behavior of a rational function defined in the neighborhood of a point on an algebraic curve. 3.1 Extension Theorems
More informationDefining Valuation Rings
East Carolina University, Greenville, North Carolina, USA June 8, 2018 Outline 1 What? Valuations and Valuation Rings Definability Questions in Number Theory 2 Why? Some Questions and Answers Becoming
More informationThe RiemannRoch Theorem
The RiemannRoch Theorem In this lecture F/K is an algebraic function field of genus g. Definition For A D F, is called the index of specialty of A. i(a) = dim A deg A + g 1 Definition An adele of F/K
More informationChapter 6: Rational Expr., Eq., and Functions Lecture notes Math 1010
Section 6.1: Rational Expressions and Functions Definition of a rational expression Let u and v be polynomials. The algebraic expression u v is a rational expression. The domain of this rational expression
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
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 informationPolynomials, Ideals, and Gröbner Bases
Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields
More informationPart 1. For any Amodule, let M[x] denote the set of all polynomials in x with coefficients in M, that is to say expressions of the form
Commutative Algebra Homework 3 David Nichols Part 1 Exercise 2.6 For any Amodule, let M[x] denote the set of all polynomials in x with coefficients in M, that is to say expressions of the form m 0 + m
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 informationx = π m (a 0 + a 1 π + a 2 π ) where a i R, a 0 = 0, m Z.
ALGEBRAIC NUMBER THEORY LECTURE 7 NOTES Material covered: Local fields, Hensel s lemma. Remark. The nonarchimedean topology: Recall that if K is a field with a valuation, then it also is a metric space
More informationIntegral Extensions. Chapter Integral Elements Definitions and Comments Lemma
Chapter 2 Integral Extensions 2.1 Integral Elements 2.1.1 Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients
More informationChapter 1 : The language of mathematics.
MAT 200, Logic, Language and Proof, Fall 2015 Summary Chapter 1 : The language of mathematics. Definition. A proposition is a sentence which is either true or false. Truth table for the connective or :
More informationMA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES
MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2018 57 5. padic Numbers 5.1. Motivating examples. We all know that 2 is irrational, so that 2 is not a square in the rational field Q, but that we can
More information4. Noether normalisation
4. Noether normalisation We shall say that a ring R is an affine ring (or affine kalgebra) if R is isomorphic to a polynomial ring over a field k with finitely many indeterminates modulo an ideal, i.e.,
More informationExtension theorems for homomorphisms
Algebraic Geometry Fall 2009 Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields. The prototype is the following
More informationRINGS: SUMMARY OF MATERIAL
RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 1113 of Artin. Definitions not included here may be considered
More informationPolynomial Rings. i=0. i=0. n+m. i=0. k=0
Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients
More informationCHAPTER 14. Ideals and Factor Rings
CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (twosided) ideal of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A. Note. (1) A absorbs elements
More informationNotes on QE for ACVF
Notes on QE for ACVF Fall 2016 1 Preliminaries on Valuations We begin with some very basic definitions and examples. An excellent survey of this material with can be found in van den Dries [3]. Engler
More informationpadic Analysis in Arithmetic Geometry
padic Analysis in Arithmetic Geometry Winter Semester 2015/2016 University of Bayreuth Michael Stoll Contents 1. Introduction 2 2. padic numbers 3 3. Newton Polygons 14 4. Multiplicative seminorms and
More information1 Absolute values and discrete valuations
18.785 Number theory I Lecture #1 Fall 2015 09/10/2015 1 Absolute values and discrete valuations 1.1 Introduction At its core, number theory is the study of the ring Z and its fraction field Q. Many questions
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 information55 Separable Extensions
55 Separable Extensions In 54, we established the foundations of Galois theory, but we have no handy criterion for determining whether a given field extension is Galois or not. Even in the quite simple
More informationChapter 8. Padic numbers. 8.1 Absolute values
Chapter 8 Padic numbers Literature: N. Koblitz, padic Numbers, padic Analysis, and ZetaFunctions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap.
More informationGauss s Theorem. Theorem: Suppose R is a U.F.D.. Then R[x] is a U.F.D. To show this we need to constuct some discrete valuations of R.
Gauss s Theorem Theorem: Suppose R is a U.F.D.. Then R[x] is a U.F.D. To show this we need to constuct some discrete valuations of R. Proposition: Suppose R is a U.F.D. and that π is an irreducible element
More informationpadic fields Chapter 7
Chapter 7 padic fields In this chapter, we study completions of number fields, and their ramification (in particular in the Galois case). We then look at extensions of the padic numbers Q p and classify
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 informationWEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS
WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS YIFEI ZHAO Contents. The Weierstrass factorization theorem 2. The Weierstrass preparation theorem 6 3. The Weierstrass division theorem 8 References
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 FINAL EXAM REVIEW HINTS
MATH 109  FINAL EXAM REVIEW HINTS Answer: Answer: 1. Cardinality (1) Let a < b be two real numbers and define f : (0, 1) (a, b) by f(t) = (1 t)a + tb. (a) Prove that f is a bijection. (b) Prove that any
More information2. THE EUCLIDEAN ALGORITHM More ring essentials
2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there
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 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 informationFields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory.
Fields and Galois Theory Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. This should be a reasonably logical ordering, so that a result here should
More informationSchool of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information
MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon
More informationPolynomial Rings. (Last Updated: December 8, 2017)
Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters
More informationPolynomials. Chapter 4
Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation
More informationZEROS OF SPARSE POLYNOMIALS OVER LOCAL FIELDS OF CHARACTERISTIC p
ZEROS OF SPARSE POLYNOMIALS OVER LOCAL FIELDS OF CHARACTERISTIC p BJORN POONEN 1. Statement of results Let K be a field of characteristic p > 0 equipped with a valuation v : K G taking values in an ordered
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
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 informationModule MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013
Module MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents 1 Basic Principles of Group Theory 1 1.1 Groups...............................
More informationAlgebraic Cryptography Exam 2 Review
Algebraic Cryptography Exam 2 Review You should be able to do the problems assigned as homework, as well as problems from Chapter 3 2 and 3. You should also be able to complete the following exercises:
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 informationCRing Project, Chapter 7
Contents 7 Integrality and valuation rings 3 1 Integrality......................................... 3 1.1 Fundamentals................................... 3 1.2 Le sorite for integral extensions.........................
More informationAlgebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...
Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2
More informationLINEAR EQUATIONS WITH UNKNOWNS FROM A MULTIPLICATIVE GROUP IN A FUNCTION FIELD. To Professor Wolfgang Schmidt on his 75th birthday
LINEAR EQUATIONS WITH UNKNOWNS FROM A MULTIPLICATIVE GROUP IN A FUNCTION FIELD JANHENDRIK EVERTSE AND UMBERTO ZANNIER To Professor Wolfgang Schmidt on his 75th birthday 1. Introduction Let K be a field
More informationTHROUGH THE FIELDS AND FAR AWAY
THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents
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 informationLecture 6. s S} is a ring.
Lecture 6 1 Localization Definition 1.1. Let A be a ring. A set S A is called multiplicative if x, y S implies xy S. We will assume that 1 S and 0 / S. (If 1 / S, then one can use Ŝ = {1} S instead of
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 informationAlgorithmic Number Theory in Function Fields
Algorithmic Number Theory in Function Fields Renate Scheidler UNCG Summer School in Computational Number Theory 2016: Function Fields May 30 June 3, 2016 References Henning Stichtenoth, Algebraic Function
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local
More information38 Irreducibility criteria in rings of polynomials
38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More informationLecture 7: Polynomial rings
Lecture 7: Polynomial rings Rajat Mittal IIT Kanpur You have seen polynomials many a times till now. The purpose of this lecture is to give a formal treatment to constructing polynomials and the rules
More information1 Adeles over Q. 1.1 Absolute values
1 Adeles over Q 1.1 Absolute values Definition 1.1.1 (Absolute value) An absolute value on a field F is a nonnegative real valued function on F which satisfies the conditions: (i) x = 0 if and only if
More information54.1 Definition: Let E/K and F/K be field extensions. A mapping : E
54 Galois Theory This paragraph gives an exposition of Galois theory. Given any field extension E/K we associate intermediate fields of E/K with subgroups of a group called the Galois group of the extension.
More informationAbstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications
1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the
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 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 informationFormal groups. Peter Bruin 2 March 2006
Formal groups Peter Bruin 2 March 2006 0. Introduction The topic of formal groups becomes important when we want to deal with reduction of elliptic curves. Let R be a discrete valuation ring with field
More informationPROBLEMS, MATH 214A. Affine and quasiaffine varieties
PROBLEMS, MATH 214A k is an algebraically closed field Basic notions Affine and quasiaffine varieties 1. Let X A 2 be defined by x 2 + y 2 = 1 and x = 1. Find the ideal I(X). 2. Prove that the subset
More information= 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2
8. padic numbers 8.1. Motivation: Solving x 2 a (mod p n ). Take an odd prime p, and ( an) integer a coprime to p. Then, as we know, x 2 a (mod p) has a solution x Z iff = 1. In this case we can suppose
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 informationPolynomial Rings. i=0
Polynomial Rings 4152018 If R is a ring, the ring of polynomials in x with coefficients in R is denoted R[x]. It consists of all formal sums a i x i. Here a i = 0 for all but finitely many values of
More informationFormal power series with cyclically ordered exponents
Formal power series with cyclically ordered exponents M. Giraudet, F.V. Kuhlmann, G. Leloup January 26, 2003 Abstract. We define and study a notion of ring of formal power series with exponents in a cyclically
More informationAN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS
AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS SAMUEL MOY Abstract. Assuming some basic knowledge of groups, rings, and fields, the following investigation will introduce the reader to the theory of
More informationSection 31 Algebraic extensions
Section 31 Algebraic extensions Instructor: Yifan Yang Spring 2007 Vector spaces over a field Definition Let F be a field. A vector space over F is an additive group V, together with a scalar multiplication
More informationDedekind Domains. Mathematics 601
Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the RiemannRoch theorem. The main theorem shows that if K/F is a finite
More informationAlgebra Homework, Edition 2 9 September 2010
Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.
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 informationA finite universal SAGBI basis for the kernel of a derivation. Osaka Journal of Mathematics. 41(4) P.759P.792
Title Author(s) A finite universal SAGBI basis for the kernel of a derivation Kuroda, Shigeru Citation Osaka Journal of Mathematics. 4(4) P.759P.792 Issue Date 20042 Text Version publisher URL https://doi.org/0.890/838
More informationUNDERSTANDING RULER AND COMPASS CONSTRUCTIONS WITH FIELD THEORY
UNDERSTANDING RULER AND COMPASS CONSTRUCTIONS WITH FIELD THEORY ISAAC M. DAVIS Abstract. By associating a subfield of R to a set of points P 0 R 2, geometric properties of ruler and compass constructions
More informationMathematical Olympiad Training Polynomials
Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, a i R, for 0 i n. If a n 0,
More informationDMATH Algebra I HS17 Prof. Emmanuel Kowalski. Solution 12. Algebraic closure, splitting field
DMATH Algebra I HS17 Prof. Emmanuel Kowalski Solution 1 Algebraic closure, splitting field 1. Let K be a field of characteristic and L/K a field extension of degree. Show that there exists α L such that
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field. 17.1 Tangent spaces and hypersurfaces For any polynomial f k[x
More 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 informationCYCLOTOMIC POLYNOMIALS
CYCLOTOMIC POLYNOMIALS 1. The Derivative and Repeated Factors The usual definition of derivative in calculus involves the nonalgebraic notion of limit that requires a field such as R or C (or others) where
More informationModel theory of valued fields Lecture Notes
Model theory of valued fields Lecture Notes Lou van den Dries Fall Semester 2004 Contents 1 Introduction 2 2 Henselian local rings 5 2.2 Hensel s Lemma........................... 6 2.3 Completion..............................
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 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 informationABSTRACT NONSINGULAR CURVES
ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine nspace the collection A n k of points P = a 1, a,..., a
More 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 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 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 Email address: ford@fau.edu
More information1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism
1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials
More informationCHAPTER 0 PRELIMINARY MATERIAL. Paul Vojta. University of California, Berkeley. 18 February 1998
CHAPTER 0 PRELIMINARY MATERIAL Paul Vojta University of California, Berkeley 18 February 1998 This chapter gives some preliminary material on number theory and algebraic geometry. Section 1 gives basic
More informationR S. with the property that for every s S, φ(s) is a unit in R S, which is universal amongst all such rings. That is given any morphism
8. Nullstellensatz We will need the notion of localisation, which is a straightforward generalisation of the notion of the field of fractions. Definition 8.1. Let R be a ring. We say that a subset S of
More information