Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...


 Linette Dora Foster
 4 years ago
 Views:
Transcription
1 Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups Problem Problem Problem Rings Problem Problem Problem Fields Problem Problem Problem
2 1 Groups 1.1 Problem 1 Suppose G is a finite group whose automorphism group Aut(G) has prime order. Then Aut(G) is cyclic, and so every subgroup of Aut(G) is cyclic. In particular, the inner automorphism group Inn(G) := {Φ g Aut(G) Φ g (x) = gxg 1, g G} Aut(G) is cyclic. Since G/Z(G) = Inn(G), and G/Z(G) is cyclic if and only if G is abelian, it follows that G must be abelian. By the fundamental theorem of finitely generated abelian groups, G = Z/p k 1 1 Z Z/p kn n Z for some n N, primes p 1,..., p n and positive integers k 1,..., k n. There is an injective morphism Aut(Z/p k 1 1 Z) Aut(Z/p kn n Z) Aut(Z/p k 1 1 Z Z/p kn n Z) = Aut(G) sending (f 1,..., f n ) to the product morphism f 1 f n, so H := Aut(Z/p k 1 1 Z) Aut(Z/p kn n Z) is a subgroup of Aut(G); in particular, H divides Aut(G). Letting ϕ denote the Euler totient function, Aut(Z/p k i i Z) = ϕ(pk i i ) = pk i 1 i (p i 1), so H = n i=1 p k i 1 i (p i 1) If any p i > 5, then p i 1 is necessarily composite, so we must have p i {2, 3}. If p i = 3 and k i > 1, then 6 p k i 1 i (p i 1), so we must have k i = 1 if p i = 3. Similarly, if p i = 2, and k i > 2, then 4 H, so we must have k i {1, 2} if p i = 2. If G contains any of Z/3Z Z/3Z, Z/3Z Z/4Z, Z/4Z Z/4Z, then 4 H, so G must be of one of the following forms: (Z/2Z) n, (Z/2Z) n Z/3Z, (Z/2Z) n Z/4Z where n 0. If n > 1, then Aut(G) contains Aut(Z/2Z Z/2Z) = GL 2 (F 2 ) = S 3, whence Aut(G) is not prime. Thus, the only possibilities are Z/2Z, Z/3Z, Z/2Z Z/3Z, Z/4Z, Z/2Z Z/4Z It is straightforward to see that if G is any of Z/3Z, Z/4Z and Z/6Z, then Aut(G) = 2. If G = Z/2Z, then Aut(G) is trivial, and hence is not of prime order. The only case that remains is G = Z/2Z Z/4Z. Since Aut(Z/2Z) Aut(Z/4Z) is a subgroup of Aut(G), it suffices to show that Aut(G) > 2, in which case Aut(G) is not prime, since 2 Aut(G). Indeed, it is easy to check that the assignments (1, 0) (1, 2), (0, 1) (0, 1) on generators define an automorphism of Z/2Z Z/4Z which is not in the image of Aut(Z/2Z) Aut(Z/4Z). Hence, if Aut(G) is prime, then G is isomorphic to one of 1.2 Problem 2 Z/3Z, Z/4Z, Z/6Z Let G be a finite group, and let H be a nonnormal subgroup of G of index n > 1. 2
3 (a) Suppose that H is divisible by a prime p n. Note G acts by left multiplication on the set G/H of cosets of H in G; since [G : H] = n, note G/H = n. One can view this action as a homomorphism ϕ: G S(G/H) = S n, where ϕ(g )(gh) = g gh for any g G, gh G/H. Note that ϕ(g)(h) = gh = H if and only if g H, so K := ker(ϕ) is a subgroup of H. Since K is normal in G, it must be normal in H as well. As H is not a normal subgroup of G, K must be a proper subgroup of H. If K is nontrivial, then H is not simple, and we re done. If K is trivial, then ϕ is injective, so G is isomorphic to a subgroup of S n and thus G divides S(G/H) = S n = n! by Lagrange. But G = [G : H] H = n H which is divisble by np, so n! is divisible by np, a contradiction. (b) Apparently, this is false. 1.3 Problem 3 (a) Let G = x be cyclic group and H a subgroup of G. Since every element of H is a positive power of x, there exists minimal positive k N such that x k H. We claim H = x k. Indeed, for any n k N such that x n H, apply Euclidean division to write n = kq + r for some k N and 0 r < k. Then x n (x k ) q = x n kq = x r H, and since r < k, the minimality of k implies r = 0, i.e. k n. (b) The same counting methods in S06.G2(a) show that GL 2 (Z/pZ) = (p 2 1)(p 2 p) elements and hence SL 2 (Z/pZ) = GL 2(Z/pZ) (Z/pZ) = (p 1)2 p(p + 1) p 1 = (p 1)(p)(p + 1) (c) Let l be an odd prime. By the second Sylow theorem, any two Sylow lsubgroups of SL 2 (Z/pZ) are conjugate, hence isomorphic, so it suffices to exhibit a Sylow lsubgroup which is cyclic for each odd prime l. Note that p, p 1, p + 1 have no odd prime factors in common, so if l q divides SL 2 (Z/pZ) = (p 1)p(p + 1), then l q divides one of p 1, p or p+1. Since the subgroups of a cyclic group are bijective with the divisors of the order of that cyclic group, it is enough to find cyclic subgroups of SL 2 (Z/pZ) of order p 1, p and p + 1, as this will exhibit the required cyclic Sylow lsubgroup for all possible l. Note that (Z/pZ) is a cyclic group of order p 1 which embeds diagonally into SL 2 (Z/pZ), so this takes care of the case p 1. Any Sylow psubgroup has order p and thus must be cyclic. The case of p + 1 is the trickiest. Let l q be the largest power of l dividing p + 1, and note that GL 2 (Z/pZ) = (p 1) 2 p(p + 1), so l q must also be the largest power of l dividing GL 2 (Z/pZ), i.e. any subgroup of order l q of GL 2 (Z/pZ) is a Sylow l subgroup of GL 2 (Z/pZ). Let F = F p 2 be the finite field with p elements; recall F is a 2dimensional vector space over Z/pZ. Then F is cyclic (see F06.F1 below) of order p 2 1 = (p 1)(p + 1). For any fixed α F, the left multiplication map L α : F F sending x F to αx is a vector space isomorphism, so after fixing a basis B for F over Z/pZ, F can be embedded in GL 2 (Z/pZ) by sending α F to the invertible matrix representing L α with respect to B. Since SL 2 (Z/pZ) is the kernel of 3
4 det: GL 2 (Z/pZ) (Z/pZ), it is a normal subgroup of GL 2 (Z/pZ). The subgroup of GL 2 (Z/pZ) isomorphic to F has a cyclic subgroup P of order l q, which as remarked above must be a Sylow lsubgroup of GL 2 (Z/pZ). By F05.G2, P SL 2 (Z/pZ) is a Sylow lsubgroup of SL 2 (Z/pZ) which must also be cyclic, so we re done. 2 Rings 2.1 Problem 1 Let m be a prime ideal of Z[X], and let ϕ: Z Z[X] be the inclusion map. The preimage of any prime ideal under a ring homomorphism is a prime ideal, so ϕ 1 (m) = Z m is a prime ideal of Z, whence Z m = {0} or Z m = pz for some prime p Z. Suppose Z m = pz. Then p m, so pz[x] m, whence p, m = m. In particular, Z[X]/m = Z[X]/ p, m = (Z[X]/pZ[X])/m = (Z/pZ)[X]/m where m is the image of the ideal m under the projection π : Z[X] Z[X]/pZ[X] (this is indeed an ideal since π is surjective). Since m is a prime ideal, Z[X]/m is a domain, so m must also be a prime ideal of (Z/pZ)[X]. As Z/pZ is a field, Z/pZ[X] is a PID, and any nonzero prime ideal of a PID is generated by an irreducible element. Hence, either m = {0}, in which case m = pz[x], or m is principal, generated by a (monic) irreducible polynomial g(x) (Z/pZ)[X]. In the latter case, let f(x) m such that π(f(x)) = g(x). Note f must exist, since m is the image of m under π. We may also assume f is monic, as pz[x] m and the leading coefficient of f is congruent to 1 mod p since g(x) is monic. Further, f(x) is irreducible over Z[X], or else g(x) would admit a nontrivial factorization over (Z/pZ)[X] by reducing the given factorization of f(x) mod p. We claim m = p, f(x). Indeed, the inclusion p, f(x) m is trivial by the choice of f. Let q(x) m; then π(q(x)) = j(x)g(x) for some j(x) (Z/pZ)[X]. Taking l(x) Z[X] such that π(l(x)) = j(x), it follows then that π(q(x) l(x)f(x)) = π(q(x)) j(x)g(x) = 0 (Z/pZ)[X], i.e. there exists s(x) Z[X] such that q(x) l(x)f(x) = p s(x). Hence, q(x) = f(x) l(x) + p s(x) p, f(x), so m = p, f(x). Now suppose Z m = {0}. Since Z[X] is a domain, {0} is a prime ideal of Z[X], so suppose m {0} and let g(x) m nonzero. Since Z[X] is a UFD, g(x) can be uniquely factored into a product of irreducibles. Using the primality of m inductively on this factorization implies that m contains some irreducible factor f(x) of g(x). As m Z = {0}, we may further assume that f is nonconstant, and since f is irreducible, the content of f must be 1. We claim that m = f(x). Let 0 h(x) m. Note that f(x) divides h(x) over Z[X] if and only if f(x) divides h(x) over Q[X]. Indeed, content is multiplicative by Gauss lemma, and so f(x)q(x) = h(x) for some q(x) Q[X] implies cont(f)cont(g) = cont(h). Since cont(h) Z and cont(f) = 1, this implies cont(g) = cont(h) Z. Since cont(k(x)) Z implies k(x) Z[X] for any k Q[X], this proves the claim. Suppose, then, that f(x) does not divide h(x) over Q[X]. Since Q[X] is a PID and f(x) is irreducible and h(x) is nonzero, this means that f, h generate the unit ideal in Q[X], i.e. there exist p(x), r(x) Q[X] such that 4
5 f(x)p(x) + h(x)r(x) = 1. But clearing the denominators of p(x), r(x) by multiplying by a sufficiently large integer n > 0 implies nf(x)p(x) + nh(x)r(x) = n m, a contradiction since m Z = {0} by hypothesis. Thus, f(x) divides h(x) over Q[X], so f(x) divides h(x) over Z[X] as argued above, whence m = f(x). Hence, the prime ideals of Z[X] are of the following form: (1) {0} (2) pz[x] for some prime p Z (3) f(x) for some irreducible polynomial f(x) Z[X] (4) p, f(x) for some prime p Z and irreducible polynomial f(x) Z[X] such that π(f) (Z/pZ)[X] is also irreducible 2.2 Problem 2 Let R be a Noetherian domain. (a) Let x R be a nonzero nonunit. Recall that y R divides x if and only if y x. Suppose x is not irreducible; then we may write x = yz for y, z R nonzero nonunits, whence x y. Since R is a domain, this containment is proper. Indeed, if y x, then y = vx for some v R, so x = zvx. Hence, (1 zv)x = 0, and as x 0, this implies 1 zv = 0, i.e z is a unit, a contradiction. If y and z are both irreducible elements of R, then we are done; if not, suppose that (WLOG) y is reducible. Then we can similarly write y as a product of nonzero nonunits wv, extending the chain of ideals x y w. Continuing inductively, this division process must terminate in finitely many steps for any divisor of x, as the chain of ideals thus furnished must stabilize since R is Noetherian. This proves the claim. (b) Let Ω be the set of nonzero ideals I of R such that I does not contain a product of nonzero prime ideals. If Ω is nonempty, then it contains a maximal element since R is Noetherian; denote this element A. A cannot be prime by hypothesis, so there exist x, y R such that xy A, but x, y / A. Then (A + x )(A + y ) = A + xa + ya + xy A. Since x, y / A, A is properly contained in both A x := A + x and A y := A + y. By the maximality of A, then A x, A y / Ω, so both A x and A y contain a finite product of nonzero prime ideals. The total (finite) product of these nonzero prime ideals is then contained in A x A y A, a contradiction, so Ω must be empty. (c) Let x R be an irreducible element, and let I = x. By part (b), I contains a finite product of nonzero prime ideals P 1,..., P k, and by hypothesis each P i contains a principal prime ideal p i for some p i R. Then in particular, p 1 p 2 p k P 1 P 2 P k I, so p 1 p 2 p k I. Then p 1 p 2 p k = rx for some r R, whence rx p 1, so x p 1 or r p 1 by the primality of p 1. In the latter case, this implies x = sp 1 for some s R. By the irreducibility of R and primality of p 1, this must mean s R, so I = p 1, which proves the claim. Suppose that r p 1, i.e. r = sp 1 for some s R. Since R is a domain, cancellation implies p 2 p 3 p k = sx. If we continue this process, then either 5
6 I = p i for some 1 < i < n, or we eventually obtain p k = lx for some l R. If x / p k, then l p k, so l = qp k for some q R. Then cancellation once more implies xq = 1, i.e. x is a unit, a contradiction, so x p k. Applying the same reasoning as above, x = p k, so x is prime. 2.3 Problem 3 Let R be a commutative ring, and let M be a finitely generated Rmodule. Suppose there exists a positive integer n and a surjective Rmodule homomorphism ϕ: M R n. Then there is an exact sequence 0 ker(ϕ) M ϕ R n 0 which is split, because R n is a free Rmodule, hence projective. Thus, M = R n ker(ϕ). Since M is finitely generated, there exists a positive integer k and a surjective Rmodule homomorphism R k M. Hence, there is a sequence of surjective Rmodule homomorphisms R k M π ker(ϕ) where π is the canonical projection M = R ker(ϕ) ker(ϕ). Hence, ker(ϕ) is finitely generated by k elements, namely the images of the k generators of R k under this sequence of morphisms. 3 Fields 3.1 Problem 1 Let F be a finite field of positive characteristic p; recall, then, that F = p k for some k N. To show that F is cyclic, two lemmas are useful here: Lemma Let ϕ denote the Euler totient function. For any n N, d n ϕ(d) = n. Proof. Since Z/nZ has exactly one (cyclic) subgroup of order d for each d n, and there are ϕ(d) generators of this cyclic subgroup, it follows that Z/nZ has ϕ(d) elements of order d for each d dividing n. Then d n ϕ(d) = Z/nZ = n. Lemma Suppose G is a finite group such that there are at most d elements in G of order dividing d for each d N +. Then G is cyclic. Proof. Let G := n, and let ρ(d) denote the number of elements of order d in G. By Lagrange s theorem, the order of any element of G must divide n, so ρ(d) = 0 if d does not divide n. We therefore have n = d n ρ(d) = d n ρ(d). If ρ(d) is nonzero, then G has at least one element of order d, which generates a cyclic subgroup C d of G of order d. Since all d elements of C d have order dividing d, these must be all elements of order dividing d in G. In particular, the ϕ(d) generators of C d constitute all elements of order d in G, so ρ(d) = ϕ(d) or ρ(d) = 0. By the lemma above, n = d n ρ(d) d n ϕ(d) = n 6
7 so ρ(d) = ϕ(d) for every d n. In particular, ρ(n) = ϕ(n) > 0, so there is at least one element of G with order n, whence G is cyclic. Corollary Any finite multiplicative subgroup of a field F is cyclic. In particular, if F is finite, then F is cyclic. Proof. The polynomial X d 1 F [X] has at most d roots, so any finite multiplicative subgroup of F meets the hypotheses of the lemma above. The argument above is due to J.P. Serre: see A Course in Arithmetic for more details. Now, by Lagrange s theorem, every element of F has multiplicative order dividing F = p k 1, so every nonzero element of F is a root of X pk 1 1 F [X], whence every element of F is a root of f(x) = X pk X F [X]. Since f has at most p k = F roots over F, every element of F is a root of f of multiplicity of one, and these are all roots of f. Hence, F is a splitting field for the separable polynomial f over Z/pZ, so F is a Galois extension of Z/pZ. 3.2 Problem 2 See S04.F1, part (c). 3.3 Problem 3 Let f(x) be an irreducible polynomial over F, and K/F a normal extension. Let L/K/F be the Galois closure of K over F, and let G := Gal(L/F ). Suppose f(x) = f 1 (x) f n (x) K[x], where f i is irreducible over K for each i = 1,..., n. Let α be a root of f 1, and β be a root of f i for some fixed i 1. Since G acts transitively on the roots of f, there exists σ G such that σ(α) = β. Since K/F is normal, σ K is an automorphism of K, so σ(f 1 (x)) K[x] is an irreducible polynomial having σ(α) = β as a root. Hence, f i (x) divides σ(f 1 (x)) over K[x], but both are irreducible, and so must be equal. Then deg(f i (x)) = deg(σ(f 1 (x))) = deg(f 1 (x)), so every irreducible factor of f(x) has the same degree, namely deg(f 1 (x)) = [K(α) : K]. As an easy corollary, this shows that if f has a root in K, then f splits over K, since f has an irreducible factor of degree 1 over K. 7
φ(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 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 informationAlgebra Qualifying Exam Solutions. Thomas Goller
Algebra Qualifying Exam Solutions Thomas Goller September 4, 2 Contents Spring 2 2 2 Fall 2 8 3 Spring 2 3 4 Fall 29 7 5 Spring 29 2 6 Fall 28 25 Chapter Spring 2. The claim as stated is false. The identity
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 informationMINKOWSKI THEORY AND THE CLASS NUMBER
MINKOWSKI THEORY AND THE CLASS NUMBER BROOKE ULLERY Abstract. This paper gives a basic introduction to Minkowski Theory and the class group, leading up to a proof that the class number (the order of the
More informationALGEBRA QUALIFYING EXAM SPRING 2012
ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.
More 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 informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
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 informationSolutions of exercise sheet 6
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 6 1. (Irreducibility of the cyclotomic polynomial) Let n be a positive integer, and P Z[X] a monic irreducible factor of X n 1
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 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 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 informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More informationbut no smaller power is equal to one. polynomial is defined to be
13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said
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 informationALGEBRA QUALIFYING EXAM PROBLEMS
ALGEBRA QUALIFYING EXAM PROBLEMS Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND MODULES General
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 informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
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 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 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 informationComputations/Applications
Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x
More informationGraduate Preliminary Examination
Graduate Preliminary Examination Algebra II 18.2.2005: 3 hours Problem 1. Prove or give a counterexample to the following statement: If M/L and L/K are algebraic extensions of fields, then M/K is algebraic.
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 information9. Finite fields. 1. Uniqueness
9. Finite fields 9.1 Uniqueness 9.2 Frobenius automorphisms 9.3 Counting irreducibles 1. Uniqueness Among other things, the following result justifies speaking of the field with p n elements (for prime
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 informationAlgebra Ph.D. Entrance Exam Fall 2009 September 3, 2009
Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009 Directions: Solve 10 of the following problems. Mark which of the problems are to be graded. Without clear indication which problems are to be graded
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 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 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 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 informationAlgebra Prelim Notes
Algebra Prelim Notes Eric Staron Summer 2007 1 Groups Define C G (A) = {g G gag 1 = a for all a A} to be the centralizer of A in G. In particular, this is the subset of G which commuted with every element
More informationOhio State University Department of Mathematics Algebra Qualifier Exam Solutions. Timothy All Michael Belfanti
Ohio State University Department of Mathematics Algebra Qualifier Exam Solutions Timothy All Michael Belfanti July 22, 2013 Contents Spring 2012 1 1. Let G be a finite group and H a nonnormal subgroup
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 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 information22M: 121 Final Exam. Answer any three in this section. Each question is worth 10 points.
22M: 121 Final Exam This is 2 hour exam. Begin each question on a new sheet of paper. All notations are standard and the ones used in class. Please write clearly and provide all details of your work. Good
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 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 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 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 informationPractice Algebra Qualifying Exam Solutions
Practice Algebra Qualifying Exam Solutions 1. Let A be an n n matrix with complex coefficients. Define tr A to be the sum of the diagonal elements. Show that tr A is invariant under conjugation, i.e.,
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 informationHomework Problems, Math 200, Fall 2011 (Robert Boltje)
Homework Problems, Math 200, Fall 2011 (Robert Boltje) Due Friday, September 30: ( ) 0 a 1. Let S be the set of all matrices with entries a, b Z. Show 0 b that S is a semigroup under matrix multiplication
More informationx 3 2x = (x 2) (x 2 2x + 1) + (x 2) x 2 2x + 1 = (x 4) (x + 2) + 9 (x + 2) = ( 1 9 x ) (9) + 0
1. (a) i. State and prove Wilson's Theorem. ii. Show that, if p is a prime number congruent to 1 modulo 4, then there exists a solution to the congruence x 2 1 mod p. (b) i. Let p(x), q(x) be polynomials
More informationMAT 535 Problem Set 5 Solutions
Final Exam, Tues 5/11, :15pm4:45pm Spring 010 MAT 535 Problem Set 5 Solutions Selected Problems (1) Exercise 9, p 617 Determine the Galois group of the splitting field E over F = Q of the polynomial f(x)
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 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 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 informationA SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC
A SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC SIDDHI PATHAK Abstract. This note gives a simple proof of a famous theorem of Burnside, namely, all groups of order n are cyclic
More informationSolutions of exercise sheet 4
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every
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 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 informationContradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G.
1. Galois Theory 1.1. A homomorphism of fields F F is simply a homomorphism of rings. Such a homomorphism is always injective, because its kernel is a proper ideal (it doesnt contain 1), which must therefore
More informationHigher Algebra Lecture Notes
Higher Algebra Lecture Notes October 2010 Gerald Höhn Department of Mathematics Kansas State University 138 Cardwell Hall Manhattan, KS 665062602 USA gerald@math.ksu.edu This are the notes for my lecture
More informationFrank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups:
Frank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups: Definition: The external direct product is defined to be the following: Let H 1,..., H n be groups. H 1 H 2 H n := {(h 1,...,
More informationAlgebra Ph.D. Preliminary Exam
RETURN THIS COVER SHEET WITH YOUR EXAM AND SOLUTIONS! Algebra Ph.D. Preliminary Exam August 18, 2008 INSTRUCTIONS: 1. Answer each question on a separate page. Turn in a page for each problem even if you
More informationElements of solution for Homework 5
Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ
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 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 informationAugust 2015 Qualifying Examination Solutions
August 2015 Qualifying Examination Solutions If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems,
More informationMath 120: Homework 6 Solutions
Math 120: Homewor 6 Solutions November 18, 2018 Problem 4.4 # 2. Prove that if G is an abelian group of order pq, where p and q are distinct primes then G is cyclic. Solution. By Cauchy s theorem, G has
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 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 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going
More informationTwo subgroups and semidirect products
Two subgroups and semidirect products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset
More informationJohns Hopkins University, Department of Mathematics Abstract Algebra  Spring 2009 Midterm
Johns Hopkins University, Department of Mathematics 110.401 Abstract Algebra  Spring 2009 Midterm Instructions: This exam has 8 pages. No calculators, books or notes allowed. You must answer the first
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 informationTotal 100
Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4
More informationAlgebra Exam Syllabus
Algebra Exam Syllabus The Algebra comprehensive exam covers four broad areas of algebra: (1) Groups; (2) Rings; (3) Modules; and (4) Linear Algebra. These topics are all covered in the first semester graduate
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 informationSample algebra qualifying exam
Sample algebra qualifying exam University of Hawai i at Mānoa Spring 2016 2 Part I 1. Group theory In this section, D n and C n denote, respectively, the symmetry group of the regular ngon (of order 2n)
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 informationAlgebra Exam, Spring 2017
Algebra Exam, Spring 2017 There are 5 problems, some with several parts. Easier parts count for less than harder ones, but each part counts. Each part may be assumed in later parts and problems. Unjustified
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 informationPRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.
PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.
More informationGalois Theory TCU Graduate Student Seminar George Gilbert October 2015
Galois Theory TCU Graduate Student Seminar George Gilbert October 201 The coefficients of a polynomial are symmetric functions of the roots {α i }: fx) = x n s 1 x n 1 + s 2 x n 2 + + 1) n s n, where s
More informationQuizzes for Math 401
Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that
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 informationModern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.
1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6
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 informationHomework 4 Solutions
Homework 4 Solutions November 11, 2016 You were asked to do problems 3,4,7,9,10 in Chapter 7 of Lang. Problem 3. Let A be an integral domain, integrally closed in its field of fractions K. Let L be a finite
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 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 informationFall /29/18 Time Limit: 75 Minutes
Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHUID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages
More informationI216e Discrete Math (for Review)
I216e Discrete Math (for Review) Nov 22nd, 2017 To check your understanding. Proofs of do not appear in the exam. 1 Monoid Let (G, ) be a monoid. Proposition 1 Uniquness of Identity An idenity e is unique,
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 informationALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011
ALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved
More informationName: Solutions Final Exam
Instructions. Answer each of the questions on your own paper. 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] All of
More informationAlgebra qual study guide James C. Hateley
Algebra qual study guide James C Hateley Linear Algebra Exercise It is known that real symmetric matrices are always diagonalizable You may assume this fact (a) What special properties do the eigenspaces
More informationMath 210B: Algebra, Homework 1
Math 210B: Algebra, Homework 1 Ian Coley January 15, 201 Problem 1. Show that over any field there exist infinitely many nonassociate irreducible polynomials. Recall that by Homework 9, Exercise 8 of
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 informationQUALIFYING EXAM IN ALGEBRA August 2011
QUALIFYING EXAM IN ALGEBRA August 2011 1. There are 18 problems on the exam. Work and turn in 10 problems, in the following categories. I. Linear Algebra 1 problem II. Group Theory 3 problems III. Ring
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 informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #24 12/03/2013
18.78 Introduction to Arithmetic Geometry Fall 013 Lecture #4 1/03/013 4.1 Isogenies of elliptic curves Definition 4.1. Let E 1 /k and E /k be elliptic curves with distinguished rational points O 1 and
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 informationThe Outer Automorphism of S 6
Meena Jagadeesan 1 Karthik Karnik 2 Mentor: Akhil Mathew 1 Phillips Exeter Academy 2 Massachusetts Academy of Math and Science PRIMES Conference, May 2016 What is a Group? A group G is a set of elements
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
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 informationReal Analysis Prelim Questions Day 1 August 27, 2013
Real Analysis Prelim Questions Day 1 August 27, 2013 are 5 questions. TIME LIMIT: 3 hours Instructions: Measure and measurable refer to Lebesgue measure µ n on R n, and M(R n ) is the collection of measurable
More information