# Ohio State University Department of Mathematics Algebra Qualifier Exam Solutions. Timothy All Michael Belfanti

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1 Ohio State University Department of Mathematics Algebra Qualifier Exam Solutions Timothy All Michael Belfanti July 22, 2013

2 Contents Spring Let G be a finite group and H a non-normal subgroup of G of index n. If H is divisible by a prime p n, then H is not a simple group Let A be an abelian subgroup of GL(n, C) of finite exponent m. Then A is a finite group and we have the sharp inequality A m n If A M n n (C) and rank(a) = 1, then det(a + I) = Tr(A) Let f(x) Q[x] be irreducible of odd degree. If α is a root of f, then Q(α) = Q(α 2k ), for all k Given a ring R, let N(R) be the set of nilpotent elements x of R (a) If R is commutative, then N(R) is an ideal of R (b) For general R, N(R) is not necessarily an ideal of R Let f(x) Q[x] be an irreducible polynomial of degree n. If K is any finite Galois extension of Q, then f(x) factors in K[x] as a product of m irreducibles all of the same degree d Autumn Let K/Q be a Galois extension whose Galois group is the quaternion group Q 8. Then K is not the splitting field of a quartic polynomial in Q[x] Let G = SL(2, F), the group of all 2 2 matrices of determinant 1 over a field F. Let G act on the set S of all one dimensional subspaces of F 2. Then G acts doubly transitively on S Let a i R, i = 1,..., n. Let f(x) = a 1 + a 2 x + + a n x n 1, and let A = (a j i+1 mod n ). Then det(a) = f(ζ 1 ) f(ζ n ) where the ζ j are the n-th roots of unity If G is a group of order 2010, then G is supersolvable If G is a finite subgroup of GL(n, Q) with order divisible by an odd prime p, then n p Let R be a commutative ring with unit (a) The nilpotent elements of R form an ideal, called the nilradical, which is contained in all prime ideals (b) The fact that the the nilradical of R is the intersection of all prime ideals of R implies the following: if A is an n n matrix with coefficients in R such that A k = 0 for some positive integer k, then the determinant and trace of A are nilpotent Spring Let G be a finite group with G = p 2 q, where p and q are primes and p < q. Then either G has a normal Sylow-q subgroup or G A 4, the alternating group on 4 letters Let p be a prime and R a ring with identity 1 0. If R = p 2, then R is commutative Let R be a principal ideal domain and let I, J R be nonzero ideals. Then IJ = I J if and only if I + J = R Let a, b C and let A : C C be defined by A(z) = az + b z, for z C. Then A is invertible if and only if a b Let A be an n n matrix over C. Then A and A t are conjugate Let K/Q be the splitting field for the polynomial f(x) = x 4 3x Then the Galois group of f is the isomorphic to the Klein-4 group. The lattice of subgroups and corresponding lattice of subfields of K can be explicitly given.. 9 i

3 CONTENTS CONTENTS Autumn Suppose that G is a group, and H is a nontrivial subgroup such that H J for every nontrivial subgroup J of G. Then H is contained in the center of G Let G be a transitive subgroup of the symmetric group on the set X. Let N be a normal non-trivial subgroup of G. Then N has no fixed points in X Let A, B be two n n complex matrices, and assume AB = BA. Then A and B have a common eigenvector Let F K L be a tower of fields, and suppose that K/F and L/K are algebraic extensions. Then L/F must be algebraic over F Let Z[i] be the ring of Gaussian integers (a) 3 is prime in Z[i] but 5 is not (b) If a prime p in Z is not prime in Z[i], then either p = 2 or p 1 mod Let n, m 1 be integers with greatest common divisor d. The ideal of Q[x] generated by x n 1 and x m 1 is principal and generated by x d Spring Let p be a prime, and let P be a finite p-group. Suppose that Q P is a proper subgroup. Then N(Q) Q Let G be a finite group with commutator subgroup G. Let N be the subgroup of G generated by the set {x 2 : x G}. Then N is a normal subgroup of G and N contains G Let A be an m n real-valued matrix and let A T denote its transpose. Then rank(a T A) = rank(a) Suppose that I is an ideal in Z[x] generated by a prime number p and a polynomial f(x). Then I is maximal if and only if f(x) is irreducible mod p There exists Galois extensions of degree four of Q with Galois groups isomorphic to Z/2Z Z/2Z and Z/4Z Let F = {a + b 7 : a, b Q} (a) Then F is a field (b) F Q[x]/(x 2 + 7) Autumn Let G be a finite group of order pq 2 where p q are primes and p doesn t divide # Aut(G). Then G is abelian Let G be an abelian subgroup of the symmetric group on a set X. Suppose that G acts transitively on X. Then the stabilizer G x is trivial for every x X Let V be a finite-dimensional vector space over Q and T : V V a linear transformation satisfying T 2 = I. Suppose V has a non-trivial proper subspace W that is invariant under T. Then 4 is the smallest that dim Q V can be Let R = Z[ m] where m is a square-free odd integer with m (a) The units of R are ± (b) 2 and 1 + m are irreducible in R (c) R is not a unique factorization domain Let K = Q(ζ) with ζ = e 2πi/8. Then G Z/8Z. The lattice of subgroups and the corresponding lattice of subfields is given below with primitive generators given for each subfield Every algebraically closed field must be infinite Spring Show that any group of order m is solvable Suppose a finite group G has normal subgroups A, B G such that A B = {e}. Assume that AB is a subgroup, ab = ba for all a A, b B, every g AB can be expressed uniquely as a product g = ab for some a A and b B, the four subgroups {e}, A, B, and AB are normal in AB. Suppose also that A and B are simple groups, and there exists a normal subgroup N AB that is not equal to one of the four subgroups listed above. Then A B Suppose I is an ideal of R = Z[x] and p I for some prime number p. Then I can be generated by 2 elements Suppose f(x) K[x] is an irreducible polynomial of degree n over the field K. Suppose L/K is a field extension with finite degree [L : K] = m. If m, n are relatively prime, then f(x) is irreducible in L[x] Let K C be the splitting field of X p 2 over Q, where p is a prime. Let α C be a root of X p 2 and let ζ be a primitive p-th root of unity ii

4 CONTENTS CONTENTS (a) If σ Gal(K/Q), then σ(α) = ζ c(σ) α and σ(ζ) = ζ d(σ) for some c(σ) Z/(p), d(σ) Z/(p) (b) There exists a group homomorphism ϕ : Gal(K/Q) GL 2 (Z/(p)) with its image contained within the set of lower triangular matrices Suppose T : R 4 R 4 is a linear transformation whose fourth power is minus the identity map (a) The possible eigenvalues of T in the field C are the primitive eighth roots of unity. The possible Jordan canonical forms for T can be given explicitly (b) There must always be proper non-trivial T invariant subspaces of R Autumn Any finite group G of order 224 has a subgroup of order None of the following groups are isomorphic: the multiplicative group of the integers modulo 13, the multiplicative group of the integers modulo 28, the alternating group on 4 letters, the group of symmetries of the regular hexagon Let A be an n n complex matrix. Then A m = 0 for some positive integer m if and only if all eigenvalues of A are Let R be a commutative ring with identity, and let A and B be ideals of R. Suppose that I is an ideal of R contained in A B. Then I A or I B Let α = 2 ( ) 2π cos 7. The minimal polynomial for α, the splitting field for that polynomial, and the Galois group for that polynomial can be explicitly given For a prime p, let F p denote the field with p elements. Let q = p n for some integer n (a) There exists (up to isomorphism) a unique field F q (b) The Galois group Gal(F q /F p ) is cyclic, generated by σ : x x p, of order n Spring The symmetric group S 5 has six Sylow-5 subgroups. This implies that S 6 contains two copies of S 5 that are not conjugate to each other For n 3, the center of the symmetric group S n is trivial Let λ = The Jordan normal form for the matrix A can be explicitly given Let K be an extension of Q contained in C such that K/Q is Galois and Gal(K/Q) is cyclic of order 4. Then i K Let k be a field and let I be a nonzero ideal in k[x], the ring of polynomials over k. Suppose that the quotient ring R = k[x]/i contains no nilpotent elements and contains no idempotents except 1. Then R is a field Let R = Z[ 3] and let S = Z[i]. There is no unital ring homomorphism φ : R S Autumn Let G be a group, and suppose that G has a normal subgroup N of order p (where p is prime). Let g be an element of G. Then N is contained in the centralizer of g p Let K be a subfield of the complex numbers C, and let p be a prime. Suppose that every proper finite extension of K in C has degree divisible by p. Then every finite extension of K in C has degree a power of p (a) (5, x 2 + 1) is not a prime ideal in Z[x] (b) (5, x 3 + x + 1) is a prime ideal in Z[x] Suppose that f(x, y) K[x, y] is a polynomial in two variables with coefficients in an infinite field K. Suppose that f(c, c) = 0 for every c K. Then x y is a factor of f(x, y) Let A be an n n matrix, with all diagonal entries equal to c and all off-diagonal entries equal to 1. Then det(a) = (c 1) n + n(c 1) n Suppose A is an n n matrix over a field K with minimal polynomial m A (x). Let f(x) K[x] be a polynomial. Then f(a) is nonsingular iff f(x) and m A (x) are relatively prime in K[x] Spring Let G be a nonabelian finite group and let n > 1. If every maximal proper subgroup of G has index n in G, then n = p is a prime and G is a p-group Let G be a group with commutator subgroup [G : G]. Let N be the subgroup of G generated by {x 2 : x G}. N is a normal subgroup of G containing [G, G] iii

5 CONTENTS CONTENTS 3. The rational canonical form and Jordan canonical form for the matrix A can be given explicitly Let k be a field, M the ring of n n matrices over k, and G = GL(n, k) the group of invertible elements of M. For X a subset of M, let C M (X) = {T M : ST = T S, S X}. Let C G (X) be defined analogously (a) C M (X) is a k-vector subspace of M for all subsets X of M (b) Suppose X r X 2 X 1 is a chain of subsets of G such that C 1 C 2 C r where C i = C G (X i ), 1 i r. Then r n Let R be an integral domain with quotient field F. Let p(x) R[x] be a monic polynomial. Suppose that p(x) = a(x)b(x) with a(x), b(x) monic polynomials in F[x] (a) If a(x) is not in R[x], then R is not a UFD (b) Z[2 2] is not a UFD Let f(x) Q[x] be irreducible with splitting field K over Q. Suppose that the Galois group of K/Q is abelian. Then K = Q(α) for any root α of f(x) in K Autumn (a) Let G be a group having a subgroup H of index m. Then G has a normal subgroup N whose index [G : N] is a multiple of m and a divisor of m! (b) Let f(x) Q[x] be an irreducible polynomial of degree n, and suppose that the Galois group of f is A n. Let α be a root of f(x) in C. Prove : there is no field F strictly between Q and Q(α) Let p and q be primes with p < q. If G is a group with G = p 2 q, then either G has a normal subgroup of order q or G A Let R be a commutative ring with (a) Every maximal ideal of R is prime b. If R is a PID, then every prime ideal is maximal c. There exists an example of a commutative ring R with 1, and an ideal I R such that I is prime but not maximal Let F be a field of characteristic 0 containing the n-th roots of unity, and let b F. Let K be an extension of F generated by an n-th root β of b (a) K is the splitting field over F of the polynomial x n b (b) G = Gal(K/F) is cyclic of order dividing n Let V be a finite-dimensional vector space over a field F. Suppose T : V V is a cyclic linear transformation, i.e., there exists v V such that {v, T v,..., T n 1 v} is a basis for V over F (n is the dimension of V). Let S : V V be a linear transformation that commutes with T. If S is a polynomial in T, then S = p(t) for some polynomial p(x) in F[x] Let R be the ring of n n matrices over a field F. Every nonzero element of R is either a unit or a zero-divisor Spring Let F be a field and let p(x) be a non-constant, not necessarily irreducible polynomial with coefficients in F. There exists an extension field E/F where p(x) has a root Let V be a non-trivial vector space, let T : V V be a linear transformation, and suppose that there are no T -stable subspaces of V other than 0 and V itself. The ring of all linear endomorphisms of V which commute with T forms a division ring In the ring Q[X, Y] there exists a finite set of generators for the ideal I := {f(x, Y) Q[X, Y] : f(i, i) = 0} Let K/k be a finite Galois extension of fields. Let N : K k be defined by N(α) := σ σ(α) (a) The norm is a homomorphism of groups (b) Let K/k be an extension of degree n of finite fields. Let q be the cardinality of k. Then N(α) = α (qn 1)/(q 1) for α K (c) Let K/k be as in part(b). Then N : K k is surjective (a) There are two abelian groups of order (b) A group of order 28 cannot be simple (c) There exists two non-isomorphic non-abelian groups of order iv

6 CONTENTS CONTENTS 6. Let C be the complex numbers and set V := C 4. Let S and T be two endomorphisms of V, whose matrices with respect to the canonical orthonormal basis are as given. There exists a basis of V which simultaneously diagonalizes T and S Autumn Let V be a vector space over a field F. Let P : V V be a projection (a) V = im P ker P (b) If V is finite dimensional and P and Q are two commuting projections, then V has a basis with respect to which both P and Q are represented by diagonal matrices Let V be a finite dimensional vector space over a field F, T : V V a linear transformation. Then V can be written as a direct sum V = U W where U, W are subspaces, stable under T, such that T U is invertible and T W is nilpotent Suppose that G is a finite group and N is a normal subgroup of order p, where p is the smallest prime dividing the order of G. Then N is contained in the center of G (a) S 5 is generated by (12) and (12345) (b) Let f(x) be an irreducible quintic in Q[x] with exactly two non-real roots. Then the Galois group of f is S Let I be the ideal generated by 3 and x 3 + x in Z[x] (a) I is not a principal ideal (b) I is not a prime ideal Let R be a commutative ring with identity, and let J be the intersection of all maximal ideals of R. Then 1 + J = {1 + x : x J} is a subgroup of the group of units of R Spring Let G be a finite simple group having a subgroup H of prime index p. Then p is the largest prime divisor of G Let C denote the complex numbers and let G = GL(2, C) be the group of all invertible 2 2 matrices with complex entries (a) Any finite abelian subgroup A of GL(2, C) is diagonalizable (b) There exists an infinite abelian subgroup of GL(2, C) that is not diagonalizable (c) There exists a finite subgroup H of GL(2, C) that is not diagonalizable Let p be a prime number and let n be a natural number. Let Q denote the field of rational numbers and let V = Q n denote the n-dimensional vector space of column vectors with entries from Q. Let A be the given matrix. Let T : V V be the linear transformation T (v) = Av for all v V (a) The minimal polynomial of T is x n p (b) V has no proper, non-trivial T-invariant subspaces (a) Let R be a PID and let P be a non-zero prime ideal of R. Then R/P is a field (b) There exists a UFD S and a non-zero prime ideal Q of S such that S/Q is not a field Let R be a commutative ring with identity. For an ideal I of R, define V I = {P : P is a prime ideal of R and I P}. Let I and J be ideals of R. Let IJ denote the ideal of R generated by {ab : a I, b J} (a) V I V J = V I+J (b) V I V J = V IJ = V I J Let p be a prime and let F = F p be the finite field of cardinality p. Let a F with a 0. Let E be the splitting field of g(x) = x p x + a over F (a) Let α E be a root of g. Then the other roots of g are α + 1, α + 2, (b) The polynomial g(x) is irreducible over F (c) The automorphism group of E/F is cyclic of order p Autumn Let α = and let K = Q(α) (a) The minimal polynomial of α over Q is m(x) = x 4 10x , and K is the splitting field of m(x) v

7 CONTENTS CONTENTS (b) There is an automorphism σ Gal(K/Q) such that σ(α) = α where α = 5 5. Also, σ( 5) = 5 and σ 2 (α) = α (c) Gal(K/Q) is cyclic Let A be an n n matrix with complex entries and let λ 1, λ 2,..., λ n be its eigenvalues, counted with multiplicity. Let p(x) be any polynomial in C[x] (a) The determinant of p(a) is n k=1 p(λ k ) (b) The trace of p(a) is n k=1 p(λ k ) Let F be a field and let f, g be distinct irreducible polynomials of F[x] (a) There are only finitely many ideals of the ring F[x]/(f 2 g) and they are described below (b) There does not exist a surjective ring homomorphism from F[x]/(f 2 ) to F[x]/(f 2 g) Let G be a finite group and let p be the smallest prime divisor of G. Then every subgroup H of index p in G is normal in G Let G be a group such that G/Z is cyclic where Z is the center of G. Then G is abelian Suppose that {E 1,..., E t } is a set of non-zero 4 4 matrices with complex entries satisfying: E 2 i = E i and E i E j = E j E i = 0 for all 1 i < j t. Then t Spring Let G be a finite group with G = p a n, where p is a prime and a and n are natural numbers with p < n < 2p.. 47 (a) If a > 1, then G has a normal p-subgroup P with P > (b) There exists a group G with G = pn with p < n < 2p such that G does not have a normal Sylow-p subgroup (a) Let A and B be the two commuting n n real matrices with A 2 = B 2 = I, where I is the identity n n matrix. Then A and B are simultaneously diagonalizable (b) If H is a Klein 4-subgroup of GL(2, R), then H is conjugate to a unique diagonal subgroup of GL(2, R) (c) GL(2, R) contains no subgroup isomorphic to A 4, the alternating group on {1, 2, 3, 4} Let V be a finite-dimensional vector space over F 2, the field of two elements. Let T : V V be a non-singular linear transformation such that T 5 = I but T I (a) The minimal polynomial for T has degree either 4 or (b) The smallest possible dimension of V for which there are two non-similar non-identity linear transformations T : V V and S : V V with T 5 = I = S 5 is Let c be a complex number and let R be the ring of all complex numbers which can be written as a polynomial in c with rational coefficients. Then c is algebraic if and only if R is a field Let p be a prime number and let R be the set of all rational number with denominator prime to p. R is a subring of Q. 49 (a) The units of R are those (reduced) fractions a/b such that a is coprime to p (b) R is a principal ideal domain (c) R has a unique maximal ideal M. A generator for M is p, and R/M is isomorphic to Z/pZ Let F be the splitting field of x over Q (a) The Galois group of F over Q is isomorphic to the Klein-4 group (b) The quadratic subfields of F are Q(i), Q( 2) and Q( 2) Autumn There are five non-isomorphic groups of order Let G be a finite group, let x, y be distinct elements of G of order 2, and let H = x, y (a) H is a dihedral group (b) Suppose x and y are not conjugate in G. Then the order of H is divisible by 4 and the center of H is non-trivial Suppose that P is a prime ideal of a commutative ring R (with identity) and let I and J be ideals of R. Let IJ be the ideal of R generated by all products ij where i I, j J (a) Suppose that IJ P; then either I P or J P (b) Suppose that I J P; then either I P or J P (a) Every integral domain which is also a finite set is actually a field vi

8 CONTENTS CONTENTS (b) Every non-trivial prime ideal of F p [x] is actually maximal (a) Let L C be a finite Galois extension of Q with Galois group G. Suppose that G has odd order. Then L R. 51 (b) The converse is false (c) There exists an extension L C of degree 3 that is not contained in R Let A, B be complex n n matrices such that AB = BA (a) Suppose that v C n is an eigenvector of A with eigenvalue λ. Then Bv is either 0 or an eigenvector of A with eigenvalue λ (b) Suppose that A has n distinct eigenvalues. Then B is diagonalizable Spring Let p be a prime dividing the order of a finite group G. Then G contains an element of order p Let G be a group. For a, b G put a b = b 1 ab, [a, b] = a 1 b 1 ab (a) [a, bc] = [a, c][a, b] c for a, b, c G (b) Suppose that G = AB, where A, B are subgroups of G. Let [A, B] denote the subgroup of G generated by all [a, b], where a A, b B. Then [A, B] is a normal subgroup of G (c) Suppose, in addition, that A and B are abelian. Then G/[A, B] is abelian and [G, G] = [A, B] Let R = k[x, Y], where X, Y are independent indeterminants over the field k. Let p = p(x) be an irreducible polynomial in k[x]. Put F = k[x]/(p) (a) pr is a prime ideal of R but not a maximal ideal of R (b) The maximal ideals of F[Y] are those ideals principally generated by polynomials f that are irreducible in F[Y]. 54 (c) All the maximal ideals of R containing pr are those ideals generated by (f, p) where f mod p is irreducible in F[Y] There are 6, up to similarity by real matrices, 7 7 real matrices having characteristic polynomial x 3 (x 2 + 1) Let K be a splitting field of x 3 3 over Q (a) The Galois group of K over Q is isomorphic to S (b) There exists a subfield E of C such that [E( 3 3) : E] = There exists α C such that (a) Q(α) is a normal extension of degree 4 over Q whose Galois group over Q is cyclic (b) Q(α) is a normal extension of degree 5 over Q Autumn Let G be ag roup defined by G = a, b a 4 = b 3 = 1, ba = ab (a) The order of G is (b) For each divisor d of G, the number of elements of order d in G is Let a, b be a 5-cycles defined by a = (12345), b = (13524) (a) The centralizer of a in S 5 is the cyclic group generated by a (b) a and b are not conjugate in A Let A be the given matrix (a) The characteristic polynomial of A is h(x) = x 2 (x 1) 2 (x + 1)(x 2), and the minimal polynomial of A is m(x) = x 2 (x 1)(x + 1)(x 2) (b) The eigenvalues of A are 0, 1, 1, (c) Let V λ be the λ-eigenspace. Bases for these spaces can be given explicitly (d) A is not diagonalizable None of the rings listed are isomorphic: M 2 (F) where F is the field with 4 elements, M 2 (Z 4 ), Z 16 Z 16, Z 4 Z Let K be the splitting field of x 4 3 over Q (a) The Galois group of K/Q is isomorphic to the dihedral group D (b) The quadratic subfields of K over Q are Q( 3), Q(i), and Q( 3) with corresponding subgroups σ 2, στ, σ, and σ 2, τ, respectively, under the Galois correspondence (with σ and τ as in part (a)) (c) There is a unique subfield F of K such that [F : Q] = 4 and F is Galois over Q Let F be a finite field with q elements and a F. Assume that q 1 mod 3. Then x 3 a is irreducible over F if and only if a (q 1)/ vii

9 CONTENTS CONTENTS Spring (a) The alternating group A 4 of degree 4 has order (b) A 4 does not contain a subgroup of order Let K = Q(ζ) where ζ = e 2πi/7. The minimal polynomial of ζ is m(x) = x 6 + x 5 + x 4 + x 3 + x 2 + x (a) K is a Galois extension of Q (b) For each integer t, 1 t 6, there is a unique automorphism σ t such that σ t (ζ) = ζ t (c) The Galois group of K/Q is isomorphic to Z/7Z (a) There exists a commutative ring R and an ideal I of R such that R/I is an integral domain and R is not an integral domain (b) There exists an integral domain R and an ideal I of R such that R/I is not an integral domain (c) There exists integral domains R and S such that R S is not an integral domain Let M be an n n matrix with complex entries, and let V = C n. Suppose that the minimal polynomial of M is m(x) = (x 1)(x 2)(x 3). Let p 1 (x) = (x 2)(x 3), p 2 (x) = (x 1)(x 3), p 3 (x) = (x 1)(x 2). Note that the greatest common divisor of p 1, p 2, p 3 is 1, so that one can find polynomials q 1, q 2, q 3 in C[x] such that q 1 (x)p 1 (x) + q 2 (x)p 2 (x) + q 3 (x)p 3 (x) = 1. Let A k = q k (M)p k (M) for k = 1, 2, (a) The set of A i are orthogonal, idempotent, and are a partition of I (b) Let V k be the range of A k for k = 1, 2, 3. Then V = V 1 V 2 V (c) V k = ker(m ki) for k = 1, 2, Let G be a nontrivial p-group (a) The center of G is nontrivial (b) Let M be a maximal subgroup of G. Then [G : M] = p (c) A maximal subgroup of G is normal in G Let τ be an automorphism of the real numbers R (a) τ fixes Q (b) τ is order preserving (c) τ is the identity automorphism Autumn Let N be a non-identity normal subgroup of a finite p-group P. Then N intersects the center of P non-trivially Let Q be the rational numbers (a) Q[x]/(x a)q[x] is isomorphic to Q for any a Q (b) Suppose that I is an ideal of Q[x] such that Q[x]/I Q. Then I = (x a)q[x] for some a Q Let A be the given 3 3 matrix (a) The characteristic polynomial of A is h(x) = (x 3) 3 whereas the minimal polynomial of A is m(x) = (x 3) (b) Let V = Q 3, and let T be the linear transformation from V V associated with A. Then V = V 1 V 2 where V 2 = e 2, Ae 2 and V 1 = e 1, and e 1, e 2, e 3 are the standard orthonormal basis vectors (c) The rational canonical form of A can be explicitly given (d) The matrix P such that P 1 AP is in rational canonical form can be explicitly given Let T : V V be a linear transformation on a finite dimensional vector space V. Suppose that W 1 and W 2 are eigenspaces of T with distinct eigenvalues λ 1 and λ 2. Let W = W 1 + W 2. Then W = W 1 W Let K/F be a Galois extension of fields, of degree n; let p be a prime number dividing n, and write n = p k m where p does not divide m (a) There is an intermediate field E, F E K, such that E has degree m over F (b) Suppose that E is Galois over F. Then E is the unique extension of F in K of degree m Let f(x) = x p x 1 F p [x], and let k be the splitting field of f; here p is a prime and F p is the finite field with p elements. Let σ(x) = x p be the Frobenius automorphism of k (a) Fix a root α of f in k. Then {α, α + 1, α + 2,... α + p 1} are all the roots of f (b) k = F p (α) (c) The order of σ is p viii

10 CONTENTS CONTENTS (d) f(x) is irreducible Spring Let G be a group, H and K normal subgroups of G (a) There is a homomorphism Φ : G G/H G/K with kernel H K (b) If G/H and G/K are abelian, then G/(H K) is abelian Let G be a finite group and P a Sylow-p subgroup of G. Let K be a subgroup of G which contains N G (P), the normalizer of P in G. Then N G (K) = K Let D = {q Q : q = a/b, gcd(p, b) = 1} (a) D is a subring of Q (b) If d D and d pd, then d is invertible in D (c) There is a homomorphism D onto Z/pZ with kernel pd Let A be the given 4 4 matrix (a) There is a vector v R 4 such that { v, A v, A 2 v, A 3 v} is a basis for R (b) With v as above, if p(x) R[x] and p(a) v = 0, then deg p(x) > (c) With v as above, there is a polynomial q(x) of degree 4 such that q(a) v = (d) For q(x) as in (c), we have that q(x) = m(x), the minimal polynomial for A (a) There are 2 similarity classes of 8 8 real matrices A with minimal polynomial m(x) = (x 1) 3 (x + 1) 2 and characteristic polynomial h(x) = (x 1) 5 (x + 1) (b) A representative from each class in part (a) can be explicitly given Let K be the splitting field of x 12 1 over Q (a) K = Q( 3, i) (b) The Galois group G of K/Q is isomorphic to the Klein-4 group (c) There are 3 subgroups of G and, consequently, 3 corresponding subfields of K under the Galois correspondence. 71 Autumn Let G be a finite group and p a prime dividing the order of G. Assume that a Sylow-p subgroup P of G is cyclic.. 73 (a) P has a unique subgroup of order p (b) Suppose P x 1 Px {1} for all x G. Then G possesses a nonidentity normal p subgroup Let Q + be the additive group of rational numbers (a) Suppose M is a maximal subgroup of Q +. Then Q + /M is a group of prime order (b) Q + does not possess maximal subgroups Let V and W be vector spaces over a field F and T : V W a linear transformation. Let {v 1, v 2,..., v m } be a basis of V (a) T is surjective if and only if T (v 1 ),..., T(v m ) span W (b) There exists a T such that T (v i ) 0 for all i = 1, 2,..., m and ker T {0} Let M n (C) denote the ring of n n matrices with complex entries, C[x] the ring of all polynomials with complex coefficients, and let A be the given 3 3 matrix. Let φ : C[x] M n (C) be the unique ring homomorphism for which φ(x) = A and let φ(α) = αi for all α C (a) The characteristic polynomial of A is h(x) = (x + 2)(x 1)(x 3) (b) ker φ = h(x) (c) The dimension of im φ is 3 as a vector space over C (a) Let K = Q( 3 2). Then Aut(K) = {id} (b) There is an extension L C of K such that Aut(L) {id} and the degree [L : K] is as small as possible. The extension L is not uniquely determined Let F p be field of p elements (a) There is an irreducible polynomial of degree 2 over F p (b) The polynomial x is irreducible over F (c) Using (b), we may construct a field of 9 elements ix

11 CONTENTS CONTENTS Spring Let G be the commutator subgroup of a group G, that is, the subgroup generated by all elements of the form aba 1 b (a) G is a normal subgroup of G (b) G/G is abelian (c) Let N be a normal subgroup of G such that G/N is abelian. Then G N True or false (a) Not every subgroup of order 60 has a homomorphic image of order different from 60 and (b) There are 5 non-isomorphic abelian groups of order p 4, p a prime number (c) There are no simple groups of order (d) If a, b are elements of a finite group, then ab and ba have the same order (e) If every subgroup of a finite group is normal, then the group is abelian For each group G with G 4, there is a Galois extension K/Q with Galois group isomorphic to G Let K be an extension field of finite degree n over a field F (a) Let f(x) F[x] be the irreducible polynomial for some α K. Then the degree f divides n (b) If M is a subring of K and F M K, then M is a field (c) Let α, β K and let [F(α) : F] = l and [F(β) : F] = m where l and m are relatively prime. Then [F(α, β) : F] = lm Let f(x), g(x) Z[x] (a) Suppose f(x) = g(x)h(x). If a prime number p divides f(x), then either p divides g(x) or p divides h(x). 78 (b) If f(x) Z[x] is reducible in Q[x] then it is reducible in Z[x] (c) The converse of part (b) is not true (a) The Cayley-Hamilton theorem states that if h(x) is the characteristic polynomial of a n n matrix A with coefficients in a commutative ring, then h(a) = (b) Let A be an n n matrix over the field C of complex numbers. If Tr(A i ) = 0 for all i {1, 2,...}, then A has 0 as an eigenvalue (c) Let A be a 3 3 matrix over the field C of complex numbers. If Tr(A) = Tr(A 2 ) = Tr(A 3 ) = 0, then A 3 = x

12 Spring Let G be a finite group and H a non-normal subgroup of G of index n. If H is divisible by a prime p n, then H is not a simple group. Proof. Let H denote the set of left-cosets of H in G not including H itself. Let H act by left-multiplication on the set H. This induces a homomorphism ϕ : H S n 1 where S n 1 is the Symmetric group on n 1 letters. Since n 1 < n p #H, it follows that ker ϕ is non-trivial. Suppose ker ϕ = H. Fix h H, then h(gh) = gh for all g G. So (g 1 hg)h = H for all g G, that is, g 1 hg H for all g G. It follows that H is normal in G. Assuming H is not normal in G, then it must be that ker ϕ < H is a proper subgroup. Since H contains ker ϕ, a non-trivial proper normal subgroup, we have that H is not simple. 2. Let A be an abelian subgroup of GL(n, C) of finite exponent m. Then A is a finite group and we have the sharp inequality A m n. Proof. Let T A. Since m is the exponent of A, it follows that m T (x), the minimal polynomial for T, is a divisor of x m 1. Since x m 1 has distinct roots, so does m T (x). So T is diagonalizable. We now show that if A is a collection of commuting linear transformations on a vector space V over C, then A can be simultaneously diagonalized. The proof will proceed by induction on dim C V. The base case is vacuously true, so let s assume that A can be diagonalized on any vector space V such that T : V V for all T A and dim C V n 1. Now, find T A such that T has multiple eigenvalues. If no such T exists then every element of A is already in diagonal form (since m T (x) = x ζ for all T A). Let ζ be an eigenvalue for T and let V ζ = { v C : T v = ζ v}. Note that TS v = ST v = ζs v S A. Since A preserves V ζ, the induction hypothesis applies. So there exists a basis of vectors v 1,ζ,..., v n(ζ),ζ of V ζ that diagonalizes A on V ζ. Now, the set of basis vectors m T (ζ)=0 { v 1,ζ,..., v n(ζ),ζ } diagonalizes A on C n by the fact that C n = V ζ. m T (ζ)=0 Now, assume A has been diagonlized, and let T A. Since the exponent of A is m, the eigenvalues of T must be contained within the m-th roots of unity. So there are only m possibilities for each of the diagonal entries of T. It follows that A is finite, in fact, #A m n with equality when A is (or is conjugate to) the subgroup of matrices with all possible combinations of m-th roots of unity on the diagonal. 3. If A M n n (C) and rank(a) = 1, then det(a + I) = Tr(A)

13 Spring 2012 Proof. Let V 0 be the null space of A. Since rank(a) = 1, it must be that dim C V 0 = n 1. Let v 1,..., v n 1 be a basis for V 0, and let v n V 0 so that v 1,..., v n is a basis for C n. Let P be the change of basis matrix that puts A in terms of the basis v 1,..., v n, i.e., PAP 1 = (v i,j ) where Note that A v i = v 1,i v i + + v n,i v i. det(a + I) = det(p(a + I)P 1 ) = det(pap 1 + I). Since PAP 1 + I is in upper triangular form, it s easy to see that det(pap 1 ) + I = v n,n + 1. It follows that det(a + I) = det(pap 1 + I) = v n,n + 1 = Tr(PAP 1 ) + 1 = Tr(A) + 1, since Tr is invariant under conjugation. 4. Let f(x) Q[x] be irreducible of odd degree. If α is a root of f, then Q(α) = Q(α 2k ), for all k 0. Proof. We proceed by induction. Note that Q(α 2 ) Q(α), moreover, α satisfies m 1 (x) = x 2 α 2 Q(α 2 )[x]. So [Q(α) : Q(α 2 )] = 1, 2. Since f(x) is of odd degree, so is [Q(α) : Q]. Since degrees are multiplicative, it follows that [Q(α) : Q(α 2 )] = 1, that is, Q(α) = Q(α 2 ). Suppose Q(α 2k 1 ) = Q(α) for some k N. In the same manner as before, α 2k 1 satisfies m k (x) = x 2 α 2k Q(α 2k )[x]. So [Q(α 2k 1 ) : Q(α 2k )] = 1, 2. Again, since [Q(α) : Q] is odd it follows that [Q(α 2k 1 ) : Q(α 2k )] = 1, that is, Q(α 2k ) = Q(α). 5. Given a ring R, let N(R) be the set of nilpotent elements x of R. (a) If R is commutative, then N(R) is an ideal of R. Proof. Let a, b N(R) where a k = 0 and b n = 0. Then (ab) max{n,k} = a max{n,k} b max{n,k} = 0, similarly (ra) k = r k a k = 0, since R is commutative. It remains to check that N(R) is closed under addition. By the binomial theorem, m ( ) m (a + b) m = a m j b j. j Let m min{k, n}/ Then either m j k or j n for all j = 0,..., m. So (a + b) m = 0. j=0 (b) For general R, N(R) is not necessarily an ideal of R. Proof. Let R = M 2 (C), and let a = [ ] and b = [ ] Then a 2 = b 2 = 0. On the other hand, a + b = [ ] An easy computation shows that (a + b) 2 = (a + b), hence (a + b) n = (a + b) for all n N. So a + b N(R). 2

14 Spring Let f(x) Q[x] be an irreducible polynomial of degree n. If K is any finite Galois extension of Q, then f(x) factors in K[x] as a product of m irreducibles all of the same degree d. Proof. Let θ 1,..., θ n be the roots of f and F the splitting field for F. We have the following diagram: KF K(θ i ) F K Q(θ i ) K Q(θ i ) Q Let f(x) = q 1 (x) q m (x) where q i K[x] are irreducible (since K[x] is a UFD and f is separable, f must factor into a square free product of irreducibles). Renumber the roots, if necessary, so that θ i is a root of q i (x) and θ j is a root of q j (x) for some fixed 1 i < j n. Note that deg q j = [K(θ j ) : K]. It remains to show that [K(θ j ) : K] = [K(θ i ) : K]. Since K and F are Galois over Q, so is KF. Let σ Gal(KF/Q) such that σ : θ i θ j (let σ Gal(F/Q) such that σ (θ i ) = θ j, let σ Gal(KF/Q) such that σ F = σ ). Now, since K is Galois, σ(k) = K. Hence σ(k Q(θ i )) = K Q(θ j ). So as claimed. [K(θ i ) : K] = [Q(θ i ) : K Q(θ i )] = [Q(θ j ) : K Q(θ j )] = [K(θ j ) : K], 3

15 Autumn Let K/Q be a Galois extension whose Galois group is the quaternion group Q 8. Then K is not the splitting field of a quartic polynomial in Q[x]. Proof. Suppose, for contradiction, that K is the splitting field of a quartic polynomial, say f. Note that Gal(K/Q) is completely determined by its action on the four roots of f, hence Q 8 Gal(K/Q) S 4, the symmetric group on four letters. Since #Q 8 = 8, it follows that Q 8 is a Sylow 2-subgroup of S 4. Since the Sylow 2-subgroups of S 4 are all conjugate, hence isomorphic, it follows that Q 8 D 8 S 4, the dihedral group of order 8. But Q 8 has more elements of order 4 than D 8, so this is impossible. Hence K is not the splitting field of a quartic polynomial. 2. Let G = SL(2, F), the group of all 2 2 matrices of determinant 1 over a field F. Let G act on the set S of all one dimensional subspaces of F 2. Then G acts doubly transitively on S. Proof. Let v, w S, v, w S such that v w and v w. Write v = (v 1, v 2 ), v = (v 1, v 2 ), etc. From the above hypotheses, we have the matrices [ ] [ ] v1 w Q = 1 v and P = 1 w 1 v 2 w 2 v 2 w 2 are invertible. Let A GL(2, F) such that A = PQ 1. Then A v = v and A w = w. It follows that Q 1 AQ = Q 1 P is the transformation defined by A in terms of the basis v, w. Let R := det A 0 ]. 0 1 [ 1 Then Q 1 PR is the transformation that maps v 1 det A v and w w in terms of the basis v, w. So the matrix PRQ 1 is the matrix such that [ ] [ ] v PRQ 1 v1 1 [ ] [ ] = det A v v and PRQ 1 w1 w = w 2 w. det A 2 Note that det PRQ 1 = det PQ 1 det R = 1. Hence, SL(2, F) acts doubly transitively on S. 3. Let a i R, i = 1,..., n. Let f(x) = a 1 + a 2 x + + a n x n 1, and let A = (a j i+1 mod n ). Then det(a) = f(ζ 1 ) f(ζ n ) where the ζ j are the n-th roots of unity. 4

16 Autumn 2011 Proof. Let v i be the column vector (1, ζ i,..., ζ n 1 i ). Then it s straightforward to verify that A v i = f(ζ i ) v i. Let ζ be a primitive n-th root of unity. We can assume, without loss of generality, that ζ i = ζ i 1. Consider the matrix B with columns v i : Note that B T = B, moreover, ζ ζ 2 ζ n 1 B = 1 ζ 2 ζ 4 ζ 2(n 1) ζ n 1 ζ 2(n 1) ζ (n 1)(n 1) v i v j = n 1 k=0 { 0 if n i + j 2 ζ (i+j 2)k = n else. It follows that n n B 2 = 0 0 n 0, n 0 0 whence det B 0 (in fact det(b) = n n/2 up to a fourth root of unity). So the v i form a basis for C n. The matrix A put in terms of this basis is the diagonal matrix with the diagonal entries f(ζ i ). So det(a) = f(ζ 1 )f(ζ 2 ) f(ζ n ), as claimed. 4. If G is a group of order 2010, then G is supersolvable. Proof. Note that 2010 = Let n 67 denote the number of Sylow-67 subgroups of G. Since n 67 1 mod 67 and n 67 30, it follows that n 67 = 1. Let P 67 be the normal Sylow-67 subgroup of G. Let G = G/P 67, and let n 5 denote the number of Sylow-5 subgroups of G. By the same reasoning as above, n 5 = 1 or 6. Suppose n 5 = 1. Let P 5 be the normal Sylow-5 subgroup of G, and let P 67,5 be the pre-image of P 5 via the natural map G G. Since P 5 is normal in G, it follows that P 67,5 is normal in G. In the exact same manner, we let G = G/P 67,5, P 3 the (necessarily) normal Sylow-3 subgroup of G, and P the preimage of P 3 via the natural map G G. Then P 67,5,3 is normal in G, and the chain proves that G is supersolvable. {e} P 67 P 67,5 P 67,5,3 G Suppose n 5 = 6. Let n 3 be the number of Sylow-3 subgroups of G. Then n 3 = 1 or 10. By counting elements, it s impossible for n 5 = 6 and n 3 = 10. So it must be that n 3 = 1. Proceeding in the exact same manner as the preceding paragraph (and using the exact same notation) we get a chain of normal subgroups {e} P 67 P 67,3 P 67,3,5 G proving that G is supersolvable. 5. If G is a finite subgroup of GL(n, Q) with order divisible by an odd prime p, then n p 1. 5

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