Solutions to the August 2008 Qualifying Examination Any student with questions regarding the solutions is encouraged to contact the Chair of the Qualifying Examination Committee. Arrangements will then be made for the student to meet with a faculty member who can answer their questions. 1
Abstract Algebra 1. (a) A subgroup H of G is normal if for all g G, g 1 Hg H. (b) Note that (ghg 1 h 1 ) 1 = hgh 1 g 1, which is of the same form, so an element in N can be written as n = (g 1 h 1 g1 1 h 1 1 ) (g m h m gm 1 h 1 m ) for some g i, h i G. Let a G. Since a 1 na = a 1 (g 1 h 1 g1 1 h 1 1 ) (g m h m gm 1 h 1 m )a = (a 1 (g 1 h 1 g1 1 h 1 1 )a)(a 1 (g 2 h 2 g2 1 h 1 2 )a) (a 1 (g m h m gm 1 h 1 m )a), it suffices to look at one term. We drop the suffix. a 1 ghg 1 h 1 a = a 1 gaa 1 haa 1 g 1 aa 1 h 1 a = (a 1 ga)(a 1 ha)(a 1 ga) 1 (a 1 ha) 1 N. (c) Since φ(g) H is a subgroup of H, it is also abelian. Using the composition, it suffices to assume that φ is onto. Since φ(aba 1 b 1 ) = φ(a)φ(b)φ(a) 1 φ(b) 1 = e H (H is abelian), we have N ker(φ). We then have the natural maps provided by the first homomorphism theorem G G/N G/ ker(φ) = H. 2. (a) A unit in R is an element with a two sided multiplicative inverse, i.e., a R is a unit if there exists b R such that ab = 1(= ba). (b) Consider the ideal ur. If this is not R, then ur is contained in a maximal ideal M. But then u ur M S, a contradiction. Thus R = ur, so 1 ur, i.e., there is an element b R such that 1 = ub(= bu). Hence u is a unit. (c) One can do this problem using what is called the determinant trick, but we ll give an inductive proof, using induction on the number of generators of M. Note that any quotient of R by ar, where a M, preserves the hypotheses. If M = ar then we are given that ar = a 2 R, so we have an equation a = ra 2 for some r R. Thus a(1 ra) = 0. Using the notation of part (b), 1 ra R S (since if 1 ra M then 1 M), so is a unit. Hence a = 0, so M = 0. Now assume that M = (a 1,..., a n ) with n > 1, and the result is true for 1 k < n. In R/a 1 R we still have M/a 1 R = (M/a 1 R) 2, and the maximal ideal is at most n 1 generated, so M + a 1 R = 0 in R/a 1 R. Thus M = a 1 R, and we are done by the case k = 1. 3. (a) Since Z 2 only has 2 elements we can assume that the coefficient of x 3 is 1. Also, the constant term is 1, or else x divides the polynomial. This leaves 2 possible coefficients for x 2 and x, so there are only 4 possibilities. Since we have a degree 3 polynomial over a field, if it factors it must have a degree one factor. Since we have ruled out a factor of x, the only other possible linear factor is x 1. Thus we have an irreducible polynomial if and only if 1 is not a root (given our other preconditions). The only polynomials of degree 3 that are irreducible are then x 3 + x 2 + 1 and x 3 + x + 1. (b) Let n be the largest exponent of y occuring in p(x, y). Then we can write 0 = p(x, e x ) = e nx p n (x) + e (n 1)x p n 1 (x) + + p 0 (x), where each p i (x) Q[x]. If n = 0, then y does not occur in p(x, y), so the hypothesis already gives p(x, y) = p(x) = 0. Assume n 1. Then p n (x) = (e x p n 1 (x) + e nx p 0 (x)). Since e x always dominates a polynomial in x, if we take the limit on the right as x,
we get 0. Thus p n (x) = 0. But this says that y n does not appear in p(x, y), a contradiction (this proof works in R[x, y] too.) 4. (a) It is clear that Q[ 2 + 3] Q[ 2, 3]. If we show that 2, 3 L = Q[ 2+ 3] then we will be done. We have ( 2+ 3) 2 = 13+2 6 L, so 6 L. Then 6( 2+ 3) = 3 2+2 3 L. Then 3( 2+ 3)+(3 2+2 3) = 3 L (which then gives 2 L also). (b) By part (a) we must find the dimension of Q[ 2, 3] as a Q-vector space. By Eisenstein, x 2 2 is the minimal polynomial of 2 over Q, so Q[ 2] is 2 dimensional over Q. Since the minimal polynomial of 3 over Q is x 2 3 (by Eisenstein), the minimal polynomial of 3 over Q[ 2] must divide this. The only non-trivial divisors are linear, so we are really asking if 3 Q[ 2]. Since 1, 2 is a basis for Q[ 2] over Q, we would have 3 = a + b 2 with a, b Q, or 3 = a 2 + 2b 2 + 2ab 2. Hence ab = 0 and 3 = a 2 + 2b 2. Thus a = 0 and 3 = 2b 2, or b = 0 and 3 = a 2. Neither 3, nor 3/2 are rational, so this is not the case. Thus Q[ 2, 3] is a 2-dimensional Q[ 2] vector space, and hence Q[ 2, 3] is a 2 2 = 4 dimensional Q vector space. You could also show directly that 1, 2, 3, 6 is a basis over Q.
Linear Algebra A. (a) Straightforward. (b) For λ = 0, a basis for its eigenspace is {(1, 1, 1)}. For λ = 1, a basis for its eigenspace is {(1, 1, 0), (1, 0, 1)}. (c) A is diagonalizable since A has 3 linearly independent eigenvectors. In fact, P 1 AP = diag{0, 1, 1} where P = 1 1 1 1 1 0 1 0 1 B. (a) A map T : V W is a linear transformation if T (r 1 v 1 + r 2 v 2 ) = r 1 T (v 1 ) + r 2 T (v 2 ) for all vectors v 1, v 2 V and all scalars r 1, r 2 Q. (b) Let dim(v ) = m and dim(w ) = n. Fix two bases of V and W. Let A be the m n matrix representing the linear transformation T. Since T is surjective, m n and the rank of A is n. W.l.o.g., we may assume that the first n row vectors of A are linear independent. Denote the corresponding n n matrix by B. Then B is invertible. Let S : W V be the linear transformation corresponding to the n m matrix (B 1 0). Then T S is the identity map on W since it corresponds to the matrix (B 1 0) A which is the n n identity matrix. C. (a) Let v X. Then, (w, v) = 0 for all w X. Since Y X, (w, v) = 0 for all w Y. So v Y. Thus X Y. (b) For v Y, define a linear transformation α(v) Hom(X/Y, Q) by α(v)(w) = (w, v) for all w X. Note that α(v) is well-defined since α(v)(w) = (w, v) = 0 whenever w Y. Thus, we obtain a linear map α : Y Hom(X/Y, Q) sending v Y to α(v) Hom(X/Y, Q). Note that α(v) = 0 if and only if α(v)(w) = (w, v) = 0 for all w X, i.e., if and only if v X. So Ker(α) = X, and α induces an injection β : Y /X Hom(X/Y, Q). Finally, note that the vector spaces Y /X and Hom(X/Y, Q) have the same dimension (dim X dim Y ). Therefore, the injection β must be an isomorphism. D. Since A 2 i = A i, the minimal polynomial of A i divides λ(λ 1). So each A i is diagonalizable. Since the matrices A 1,..., A k commute, they can be simultaneously diagonalized. So there exists an invertible matrix P such that P 1 A i P is diagonal for all i = 1,..., k. Note that the entries of P 1 A i P are either 0 or 1. Since k P 1 A i P = I n, we may assume (w.l.o.g.) that the entries on the diagonal of
P 1 A i P are (r 1 +... + r i 1 ) many 0 s, followed by r i many 1 1, and followed by all the zeroes again (in particular, the rank of A i is r i ). So k a i (P 1 A i P ) = diag{a 1,..., a 1,..., a k,..., a k } (there are precisely r i many a i s on the diagonal). It follows that ( k ) ( k ) k det a i A i = det a i (P 1 A i P ) = where r i is the rank of A i. E. (a) Let {v, T (v),..., T n 1 (v)} be a basis of V. Since ST = T S, we have S ( T i (v) ) = T i( S(v) ). So the linear map S is uniquely determined by S(v). Note that S(v) = n 1 ( i=0 a it i (v) for some scalars a 0, a 1,..., a n 1. Then, S(v) = n 1 i=0 a it i) (v). Hence S = n 1 i=0 a it i. (b) We will work with matrices instead of linear operators. Let A be the matrix representing T. Let diag{n 1,..., N r } be the rational canonical form of A, and let p 1,..., p r be the corresponding invariant polynomials. Then, p 1... p r. Note that A (hence T ) has a cyclic vector if and only if r = 1. Assume r > 1. We will draw a contradiction. Let B = diag{n 1, 0,..., 0}. Then AB = BA. By the assumption, B = p(a) for some polynomial p. In particular, N 1 = p(n 1 ) and 0 = p(n r ). Since p r is the minimal polynomial of N r, we have p r p. So p 1 p, and 0 = p(n 1 ) = N 1 which is absurd. a r k k