MATH 361: NUMBER THEORY TENTH LECTURE


 Annabella Rose
 4 years ago
 Views:
Transcription
1 MATH 361: NUMBER THEORY TENTH LECTURE The subject of this lecture is finite fields. 1. Root Fields Let k be any field, and let f(x) k[x] be irreducible and have positive degree. We want to construct a superfield K of k in which f has a root. To do so, consider the quotient ring R = k[x]/ f, where f is the principal ideal f(x)k[x] of k[x]. That is, R is the usual ring of polynomials over k subject to the additional rule f(x) = 0. Specifically, the ringelements are cosets and the operations are (g + f ) + (h + f ) = (g + h) + f, (g + f )(h + f ) = gh + f. The fact that f is irreducible gives R the structure of a field, not only a ring. The only matter in question is multiplicative inverses. To see that they exist, suppose that g + f f in R. The condition is that g / f, i.e., f g. so (f, g) = 1 (because f is irreducible), and so there exist F, G k[x] such that F f + Gg = 1. That is, f Gg 1, i.e., Gg 1 f, so that Gg + f = 1 + f in R. That is, (G + f )(g + f ) = 1 + f in R, showing that G + f inverts g + f in R. Now use the field R to create a set K of symbols that is a superset of k and is in bijective correspondence with R. That is, there is a bijection σ : R K, σ(a + f ) = a for all a k. Endow K with addition and multiplication operations that turn the set bijection into a field isomorphism. The operations on K thus extend the operations on k. Name a particular element of K, r = σ(x + f ). 1
2 2 MATH 361: NUMBER THEORY TENTH LECTURE Then f(r) = f(σ(x + f )) by definition of r = σ(f(x + f )) since algebra passes through σ = σ(f(x) + f ) by the nature of algebra in R = σ( f ) by the nature of algebra in R = 0 by construction of σ. Thus K is a superfield of k containing an element r such that f(r) = 0. For example, since the polynomial f(x) = X 3 2 is irreducible over Q, the corresponding quotient ring R = Q[X]/ X 3 2 = {a + bx + cx 2 + X 3 2 : a, b, c Q} is a field. And from R we construct a field (denoted Q(r) or Q[r]) such that r 3 = 2. Yes, we know that there exist cube roots of 2 in the superfield C of Q, but the construction given here is purely algebraic and makes no assumptions about the nature of the starting field k to which we want to adjoin a root of a polynomial. 2. Splitting Fields Again let k be a field and consider a nonunit polynomial f(x) k[x]. We can construct an extension field k 1 = k(r 1 ), where r 1 satisfies some irreducible factor of f. Let We can construct an extension field f 2 (X) = f(x)/(x r 1 ) k 1 [X]. k 2 = k 1 (r 2 ) = k(r 1, r 2 ), where r 2 satisfies some irreducible factor of f 2. Continue in this fashion until reaching a field where the original polynomial f factors down to linear terms. The resulting field is the splitting field of f over k, denoted spl k (f). Continuing the example of the previous section, compute that X 3 2 X r = X2 + rx + r 2 in Q(r)[X]. Let s = rt where t 3 = 1 but t 1. Then, working in Q(r, t) we have s 2 + rs + r 2 = r 2 t 2 + r 2 t + r 2 = r 2 (t 2 + t + 1) = r 2 0 = 0, Thus s = rt satisfies the polynomial X 2 + rx + r 2, and now compute that That is, showing that X 2 + rx + r 2 = X rt 2 in Q(r, t)[x]. X rt X 3 2 = (X r)(x rt)(x rt 2 ) Q(r, t)[x], spl Q (X 3 2) = Q(r, t).
3 MATH 361: NUMBER THEORY TENTH LECTURE 3 We already know the finite fields 3. Examples of Finite Fields F p = Z/pZ, p prime. For any positive integer n we can construct the extension field K = spl Fp (X pn X). Furthermore, the roots of X pn X in K form a subfield of K. To see this, check that if a pn = a and b pn = b then (ab) pn = a pn b pn = ab, (a + b) pn = a pn + b pn = a + b, (a 1 ) pn = (a pn ) 1 = a 1 if a 0, ( a) pn = ( 1) pn a pn = a (even if p = 2). (The modulo p result that (a + b) p = a p + b p is sometimes called the freshman s dream.) Thus the splitting field K is exactly the roots of X pn X. The roots are distinct because the derivative (X pn X) = 1 is nonzero, precluding multiple roots. Thus K contains p n elements. We give it a name, F q = spl Fp (X pn X) where q = p n. In fact the fields 4. Exhaustiveness of the Examples F q, q = p n, n 1 are the only finite fields, up to isomorphism. To see this, let K be a finite field. The natural homomorphism Z K, n n 1 K has for its kernel an ideal I = nz of Z such that Z/I K, and so Z/I is a finite integral domain. This forces I = pz for some prime p, and so F p K. Identify F p with its image in K. Then K is a finitedimensional vector space over F p, so that K = p n for some n. Every nonzero element x K satisfies the condition x pn 1 = 1, and so every element x K satisfies the condition x pn = x. In sum, K = F q up to isomorphism where again q = p n.
4 4 MATH 361: NUMBER THEORY TENTH LECTURE 5. Containments of Finite Fields A natural question is: For which m, n Z + do we have F p m F p n? Assuming that the containment holds, the larger field is a vector space over the smaller one, and so p n = (p m ) d = p md for some d, showing that m n. Conversely, if m n then X pn 1 1 X pm 1 1 = p n 1 p m 1 1 i=0 X (pm 1)i. (Note that p m 1 p n 1 by the finite geometric sum formula; specifically, their quotient is n/m 1 j=0 p mj.) So and so In sum, X pm 1 1 X pn 1 1, F p m F p n. F p m F p n m n. For example, F 27 is not a subfield of F Cyclic Structure For any prime power q = p n, the unit group F q is cyclic. The proof is exactly the same as for F p = (Z/pZ). Either we quote the structure theorem for finitelygenerated abelian groups, or we note that for any divisor d of q 1, (q 1)/d 1 X q 1 1 = (X d 1) X di, and the first factor on the right side has at most d roots while the second factor has at most q 1 d roots. But the left side has a full contingent of q 1 roots in F q. Now factor q 1, q 1 = r er. For each prime factor r, X re 1 has r e roots and X re 1 1 has r e 1 roots, showing that there are φ(r e ) roots of order r e. Thus there are φ(q 1) elements of order q 1 in F q. That is, F q has φ(q 1) generators. 7. Examples To construct the field of 9 = 3 2 elements we need an irreducible polynomial of degree 2 over F 3. The polynomial X will do. Thus up to isomorphism, i=0 F 9 = F 3 [X]/ X Here we have created F 9 by adjoining a square root of 1 to F 3. To construct the field of 16 = 2 4 elements we need an irreducible polynomial of degree 4 over F 2. (Note that the principle no roots implies irreducible is valid only for polynomials of degree 2 and 3. For example, a quartic polynomial can have two quadratic factors.)
5 MATH 361: NUMBER THEORY TENTH LECTURE 5 The polynomial X 4 + X + 1 works: it has no roots, and it doesn t factor into two quadratic terms because (X 2 + ax + 1)(X 2 + bx + 1) = X 4 + (a + b)x 3 + abx 2 + (a + b)x + 1. Thus, again up to isomorphism, F 16 = F 2 [X]/ X 4 + X + 1 = F 2 (r) where r 4 + r + 1 = 0 = {a + br + cr 2 + dr 3 : a, b, c, d F 2 }. 8. A Remarkable Polynomial Factorization Fix a prime p and a positive integer n. To construct the field F p n, we need an irreducible monic polynomial of degree n over F p. Are there such? How do we find them? For any d 1, let MI p (d) denote the set of monic irreducible polynomials over F p of degree d. We will show that X pn X = f(x) in F p [X]. d n f MI p(d) To establish the identity, consider a monic irreducible polynomial f of degree d 1. Working in F p n, whose elements form a full contingent of roots of X pn X, we show that f(x) X pn X in F p [X] f(x) has a root in F p n. For the = direction, we have f(x)g(x) = X pn X for some g(x), and so f(α)g(α) = 0 for every α F p n. But deg(g) = p n deg(f) < p n, and so g(α) 0 for at least one such α, for which necessarily f(α) = 0. For the = direction, let α be a root of f(x) in F p n. Substitute α for X in the relation X pn X = q(x)f(x) + r(x) (where r = 0 or deg(r) < deg f) to get r(α) = 0. Thus α is a root of gcd(f(x), r(x)), but since f(x) is irreducible, this greatest common divisor is 1 unless r = 0. So r = 0 and thus f(x) X pn X as desired. With the displayed equivalence established, again let f be a monic irreducible polynomial of degree d 1. Adjoining any root of f to F p gives a field of order p d. Thus f(x) X pn X in F p [X] a field of order p d lies in F p n. By the nature of finite field containments, it follows that for any monic irreducible polynomial f over F p of degree d 1, f(x) X pn X in F p [X] d n. Also, no monic irreducible f(x) of degree d 1 can divide X pn X twice, because the derivative of the latter polynomial never vanishes. The desired product formula in the box follows. (This argument can be simplified if one works in a given algebraic closure of F p, but constructing the algebraic closure requires Zorn s Lemma.) To count the monic irreducible polynomials over F p of a given degree, take the degrees of both sides of the identity X pn X = f(x) in F p [X] d n f MI p(d)
6 6 MATH 361: NUMBER THEORY TENTH LECTURE to get p n = d n d MI p (d). By Möbius inversion, MI p (n) = 1 µ(n/d)p d n d n (where µ is the Möbius function). The sum on the right side is positive because it is a basep expansion with top term p n and the coefficients of the lower powers of p all in {0, ±1}. So MI p (n) > 0 for all n > 0. That is, there do exist monic irreducible polynomials of every degree over every field F p. For example, taking p = 2 and n = 3, X 8 X = X(X 7 1) = X(X 1)(X 6 + X 5 + X 4 + X 3 + X 2 + X + 1) = X(X 1)(X 3 + X 2 + 1)(X 3 + X + 1) mod 2, a product of two linear factors and two cubic factors. And the counting formula from the previous section gives the results that it must, MI 2 (1) = 1 1 µ(1)21 = 2, MI 2 (3) = 1 3 (µ(1)23 + µ(3)2 1 ) = 1 (8 2) = 2. 3 Similarly, taking p = 3 and n = 2, X 9 X = X(X 8 1) = X(X 4 + 1)(X 2 + 1)(X + 1)(X 1) = X(X 1)(X + 1)(X 2 + 1)(X 2 + X 1)(X 2 X 1) mod 3, and MI 3 (1) = 1 1 µ(1)31 = 3, MI 3 (2) = 1 2 (µ(1)32 + µ(3)3 1 ) = 1 (9 3) = Common Errors The finite field F q where q = p n is neither of the algebraic structures Z/qZ and (Z/pZ) n as a ring. As a vector space, F q = F n p, but the ring (multiplicative) structure of F q is not that of Z/qZ or of (Z/pZ) n. The finite field F p m is not a subfield of F p n unless m n, in which case it is. 10. Primes in Extensions We return to a question from the very first lecture: Does a given odd prime p factor or remain prime in the Gaussian integer ring Z[i]? Equivalently, is the quotient ring Z[i]/pZ[i] = Z[i]/ p an integral domain? Compute, Z[i]/ p Z[X]/ p, X F p [X]/ X Thus the question is whether X is reducible or irreducible in F p [X], which is to say whether the Legendre symbol ( 1/p) is 1 or 1. By Euler s Criterion,
7 MATH 361: NUMBER THEORY TENTH LECTURE 7 ( 1/p) = ( 1) (p 1)/2, so the 1 mod 4 primes p factor in Z[i] while the 3 mod 4 primes p don t. Returning to quotient ring structure, we have shown that Z[i]/ p F p [X]/ X r F p [X]/ X + r, p = 1 mod 4, where r 2 = 1 mod p. Take, for example, p = 5, and create two ideals of Z[i] modeled on the two polynomial ideals in the previous display with r = 2. I 1 = 5, i 2, I 2 = 5, i + 2. Their product is generated by the pairwise products of their generators, I 1 I 2 = 5 2, 5(i 2), 5(i + 2), 5. Each generator of I 1 I 2 is a multiple of 5 in Z[i], and 5 is a Z[i]linear combination of the generators. That is, the methods being illustrated here have factored 5 as an ideal of Z[i], I 1 I 2 = 5Z[i]. The previous example overlooks the elementwise factorization 5 = (2 i)(2 + i) willfully. Here is another example. For any odd prime p 19, Z[ 19]/ p Z[X]/ p, X 2 19 F p [X]/ X 2 19, and this this ring decomposes if (19/p) = 1. For example, (19/5) = 1 because 2 2 = 4 = 19 mod 5, and so we compute that in Z[ 19], 5, , = 5 2, 5( 19 2), 5( ), 15 = 5Z[ 19]. Here we have factored the ideal 5Z[ 19] without factoring the element 5 itself. Similarly, letting q be an odd prime and Φ q (X) the qth cyclotomic polynomial (the smallest monic polynomial over Z satisfied by ζ q ), compute for any odd prime p q, Z[ζ q ]/ p Z[X]/ p, Φ q (X) F p [X]/ Φ q (X). Thus if g Φ q (X) = ϕ i (X) ei in F p [X] then the Sun Ze Theorem gives the corresponding ring decomposition g Z[ζ q ]/ p F p [X]/ ϕ i (X) ei, i=1 i=1 and plausibly this decomposition somehow indicates the decomposition of p in Z[ζ q ]. Factoring Φ q (X) in F p [X] is a finite problem, so these methods finiteize the determination of how p decomposes in Z[ζ q ].
RINGS: 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 informationPublickey Cryptography: Theory and Practice
Publickey Cryptography Theory and Practice Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Chapter 2: Mathematical Concepts Divisibility Congruence Quadratic Residues
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 informationFinite Fields. Sophie Huczynska. Semester 2, Academic Year
Finite Fields Sophie Huczynska Semester 2, Academic Year 200506 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,
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 informationFinite Fields. Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay
1 / 25 Finite Fields Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay September 25, 2014 2 / 25 Fields Definition A set F together
More informationMath 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d
Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v).
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 31, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Symbolic Adjunction of Roots When dealing with subfields of C it is easy to
More informationFinite Fields. Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13
Finite Fields Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13 Contents 1 Introduction 3 1 Group theory: a brief summary............................ 3 2 Rings and fields....................................
More informationg(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.
6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral
More informationIUPUI Qualifying Exam Abstract Algebra
IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order
More informationMath 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours
Math 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours Name: Please read the questions carefully. You will not be given partial credit on the basis of having misunderstood a question, and please
More informationMATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION
MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0
More informationAN INTRODUCTION TO GALOIS THEORY
AN INTRODUCTION TO GALOIS THEORY STEVEN DALE CUTKOSKY In these notes we consider the problem of constructing the roots of a polynomial. Suppose that F is a subfield of the complex numbers, and f(x) is
More information1. Algebra 1.5. Polynomial Rings
1. ALGEBRA 19 1. Algebra 1.5. Polynomial Rings Lemma 1.5.1 Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are
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 informationCHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and
CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)
More informationSection III.6. Factorization in Polynomial Rings
III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)
More informationPractice problems for first midterm, Spring 98
Practice problems for first midterm, Spring 98 midterm to be held Wednesday, February 25, 1998, in class Dave Bayer, Modern Algebra All rings are assumed to be commutative with identity, as in our text.
More informationRUDIMENTARY GALOIS THEORY
RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the
More informationMT5836 Galois Theory MRQ
MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and
More informationAlgebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.
More informationCongruences and Residue Class Rings
Congruences and Residue Class Rings (Chapter 2 of J. A. Buchmann, Introduction to Cryptography, 2nd Ed., 2004) Shoichi Hirose Faculty of Engineering, University of Fukui S. Hirose (U. Fukui) Congruences
More informationCSIR  Algebra Problems
CSIR  Algebra Problems N. Annamalai DST  INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli 620024 Email: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com
More informationMath 547, Exam 2 Information.
Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:1011:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/33/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)
More informationHomework 10 M 373K by Mark Lindberg (mal4549)
Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More informationGalois Theory, summary
Galois Theory, summary Chapter 11 11.1. UFD, definition. Any two elements have gcd 11.2 PID. Every PID is a UFD. There are UFD s which are not PID s (example F [x, y]). 11.3 ED. Every ED is a PID (and
More informationMath 553 Qualifying Exam. In this test, you may assume all theorems proved in the lectures. All other claims must be proved.
Math 553 Qualifying Exam January, 2019 Ron Ji In this test, you may assume all theorems proved in the lectures. All other claims must be proved. 1. Let G be a group of order 3825 = 5 2 3 2 17. Show that
More informationIRREDUCIBILITY TESTS IN Q[T ]
IRREDUCIBILITY TESTS IN Q[T ] KEITH CONRAD 1. Introduction For a general field F there is no simple way to determine if an arbitrary polynomial in F [T ] is irreducible. Here we will focus on the case
More informationCoding Theory ( Mathematical Background I)
N.L.Manev, Lectures on Coding Theory (Maths I) p. 1/18 Coding Theory ( Mathematical Background I) Lector: Nikolai L. Manev Institute of Mathematics and Informatics, Sofia, Bulgaria N.L.Manev, Lectures
More informationHomework 8 Solutions to Selected Problems
Homework 8 Solutions to Selected Problems June 7, 01 1 Chapter 17, Problem Let f(x D[x] and suppose f(x is reducible in D[x]. That is, there exist polynomials g(x and h(x in D[x] such that g(x and h(x
More information6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree
Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S
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 information2a 2 4ac), provided there is an element r in our
MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built
More informationSection VI.33. Finite Fields
VI.33 Finite Fields 1 Section VI.33. Finite Fields Note. In this section, finite fields are completely classified. For every prime p and n N, there is exactly one (up to isomorphism) field of order p n,
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 informationQuasireducible Polynomials
Quasireducible Polynomials Jacques Willekens 06Dec2008 Abstract In this article, we investigate polynomials that are irreducible over Q, but are reducible modulo any prime number. 1 Introduction Let
More informationTC10 / 3. Finite fields S. Xambó
TC10 / 3. Finite fields S. Xambó The ring Construction of finite fields The Frobenius automorphism Splitting field of a polynomial Structure of the multiplicative group of a finite field Structure of the
More informationFACTORIZATION OF IDEALS
FACTORIZATION OF IDEALS 1. General strategy Recall the statement of unique factorization of ideals in Dedekind domains: Theorem 1.1. Let A be a Dedekind domain and I a nonzero ideal of A. Then there are
More informationAlgebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001
Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,
More information(January 14, 2009) q n 1 q d 1. D = q n = q + d
(January 14, 2009) [10.1] Prove that a finite division ring D (a notnecessarily commutative ring with 1 in which any nonzero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.)
More informationSolutions of exercise sheet 11
DMATH Algebra I HS 14 Prof Emmanuel Kowalski Solutions of exercise sheet 11 The content of the marked exercises (*) should be known for the exam 1 For the following values of α C, find the minimal polynomial
More informationGalois theory (Part II)( ) Example Sheet 1
Galois theory (Part II)(2015 2016) Example Sheet 1 c.birkar@dpmms.cam.ac.uk (1) Find the minimal polynomial of 2 + 3 over Q. (2) Let K L be a finite field extension such that [L : K] is prime. Show that
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationEighth Homework Solutions
Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely
More informationFIELD THEORY. Contents
FIELD THEORY MATH 552 Contents 1. Algebraic Extensions 1 1.1. Finite and Algebraic Extensions 1 1.2. Algebraic Closure 5 1.3. Splitting Fields 7 1.4. Separable Extensions 8 1.5. Inseparable Extensions
More informationOutline. MSRIUP 2009 Coding Theory Seminar, Week 2. The definition. Link to polynomials
Outline MSRIUP 2009 Coding Theory Seminar, Week 2 John B. Little Department of Mathematics and Computer Science College of the Holy Cross Cyclic Codes Polynomial Algebra More on cyclic codes Finite fields
More informationFactorization in Integral Domains II
Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More informationMath 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille
Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is
More informationUniversity of Ottawa
University of Ottawa Department of Mathematics and Statistics MAT3143: Ring Theory Professor: Hadi Salmasian Final Exam April 21, 2015 Surname First Name Instructions: (a) You have 3 hours to complete
More informationMTH310 EXAM 2 REVIEW
MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not
More informationHandout  Algebra Review
Algebraic Geometry Instructor: Mohamed Omar Handout  Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much
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 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 information18. Cyclotomic polynomials II
18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients
More informationFinite Fields. [Parts from Chapter 16. Also applications of FTGT]
Finite Fields [Parts from Chapter 16. Also applications of FTGT] Lemma [Ch 16, 4.6] Assume F is a finite field. Then the multiplicative group F := F \ {0} is cyclic. Proof Recall from basic group theory
More information(Rgs) Rings Math 683L (Summer 2003)
(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More informationA connection between number theory and linear algebra
A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.
More information2 ALGEBRA II. Contents
ALGEBRA II 1 2 ALGEBRA II Contents 1. Results from elementary number theory 3 2. Groups 4 2.1. Denition, Subgroup, Order of an element 4 2.2. Equivalence relation, Lagrange's theorem, Cyclic group 9 2.3.
More informationANALYSIS OF SMALL GROUPS
ANALYSIS OF SMALL GROUPS 1. Big Enough Subgroups are Normal Proposition 1.1. Let G be a finite group, and let q be the smallest prime divisor of G. Let N G be a subgroup of index q. Then N is a normal
More informationAlgebraic structures I
MTH5100 Assignment 110 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationInformation Theory. Lecture 7
Information Theory Lecture 7 Finite fields continued: R3 and R7 the field GF(p m ),... Cyclic Codes Intro. to cyclic codes: R8.1 3 Mikael Skoglund, Information Theory 1/17 The Field GF(p m ) π(x) irreducible
More informationMATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM
MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is
More informationPage Points Possible Points. Total 200
Instructions: 1. The point value of each exercise occurs adjacent to the problem. 2. No books or notes or calculators are allowed. Page Points Possible Points 2 20 3 20 4 18 5 18 6 24 7 18 8 24 9 20 10
More informationName: Solutions Final Exam
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] For
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 informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
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 informationMATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions
MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,
More informationIntroduction to finite fields
Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in
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 informationFurther linear algebra. Chapter II. Polynomials.
Further linear algebra. Chapter II. Polynomials. Andrei Yafaev 1 Definitions. In this chapter we consider a field k. Recall that examples of felds include Q, R, C, F p where p is prime. A polynomial is
More informationGALOIS THEORY AT WORK: CONCRETE EXAMPLES
GALOIS THEORY AT WORK: CONCRETE EXAMPLES KEITH CONRAD 1. Examples Example 1.1. The field extension Q(, 3)/Q is Galois of degree 4, so its Galois group has order 4. The elements of the Galois group are
More informationModern Computer Algebra
Modern Computer Algebra Exercises to Chapter 25: Fundamental concepts 11 May 1999 JOACHIM VON ZUR GATHEN and JÜRGEN GERHARD Universität Paderborn 25.1 Show that any subgroup of a group G contains the neutral
More informationϕ : Z F : ϕ(t) = t 1 =
1. Finite Fields The first examples of finite fields are quotient fields of the ring of integers Z: let t > 1 and define Z /t = Z/(tZ) to be the ring of congruence classes of integers modulo t: in practical
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 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 informationSolutions of exercise sheet 8
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 8 1. In this exercise, we will give a characterization for solvable groups using commutator subgroups. See last semester s (Algebra
More informationChapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples
Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter
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 informationHomework problems from Chapters IVVI: answers and solutions
Homework problems from Chapters IVVI: answers and solutions IV.21.1. In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence
More informationLECTURE NOTES IN CRYPTOGRAPHY
1 LECTURE NOTES IN CRYPTOGRAPHY Thomas Johansson 2005/2006 c Thomas Johansson 2006 2 Chapter 1 Abstract algebra and Number theory Before we start the treatment of cryptography we need to review some basic
More informationGALOIS THEORY. Contents
GALOIS THEORY MARIUS VAN DER PUT & JAAP TOP Contents 1. Basic definitions 1 1.1. Exercises 2 2. Solving polynomial equations 2 2.1. Exercises 4 3. Galois extensions and examples 4 3.1. Exercises. 6 4.
More informationGroups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002
Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary
More informationALGEBRA PH.D. QUALIFYING EXAM September 27, 2008
ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely
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 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 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 information50 Algebraic Extensions
50 Algebraic Extensions Let E/K be a field extension and let a E be algebraic over K. Then there is a nonzero polynomial f in K[x] such that f(a) = 0. Hence the subset A = {f K[x]: f(a) = 0} of K[x] does
More information12 16 = (12)(16) = 0.
Homework Assignment 5 Homework 5. Due day: 11/6/06 (5A) Do each of the following. (i) Compute the multiplication: (12)(16) in Z 24. (ii) Determine the set of units in Z 5. Can we extend our conclusion
More informationGALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2)
GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2) KEITH CONRAD We will describe a procedure for figuring out the Galois groups of separable irreducible polynomials in degrees 3 and 4 over
More informationA Generalization of Wilson s Theorem
A Generalization of Wilson s Theorem R. Andrew Ohana June 3, 2009 Contents 1 Introduction 2 2 Background Algebra 2 2.1 Groups................................. 2 2.2 Rings.................................
More informationCOURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA
COURSE SUMMARY FOR MATH 504, FALL QUARTER 20178: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties
More informationMath 312/ AMS 351 (Fall 17) Sample Questions for Final
Math 312/ AMS 351 (Fall 17) Sample Questions for Final 1. Solve the system of equations 2x 1 mod 3 x 2 mod 7 x 7 mod 8 First note that the inverse of 2 is 2 mod 3. Thus, the first equation becomes (multiply
More informationThe Galois group of a polynomial f(x) K[x] is the Galois group of E over K where E is a splitting field for f(x) over K.
The third exam will be on Monday, April 9, 013. The syllabus for Exam III is sections 1 3 of Chapter 10. Some of the main examples and facts from this material are listed below. If F is an extension field
More informationGroup Theory. 1. Show that Φ maps a conjugacy class of G into a conjugacy class of G.
Group Theory Jan 2012 #6 Prove that if G is a nonabelian group, then G/Z(G) is not cyclic. Aug 2011 #9 (Jan 2010 #5) Prove that any group of order p 2 is an abelian group. Jan 2012 #7 G is nonabelian nite
More informationSome practice problems for midterm 2
Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is
More information