MASTERS EXAMINATION IN MATHEMATICS

Transcription

1 MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION FALL 2007 Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth the same number of points. All answers will be graded, but the score for the examination will be the sum of the scores of your best 8 solutions. Use separate answer sheets for each question. DO NOT PUT YOUR NAME ON YOUR ANSWER SHEETS. When you have finished, insert all your answer sheets into the envelope provided, then seal and print your name on it. Any student whose answers need clarification may be required to submit to an oral examination. 1

2 PURE MATHEMATICS OPTION FALL 2007 Algebra A1. Find the number of elements of order 7 in a simple group of order 168. Solution: Let G be a simple group of order 168 = The number of Sylow 7-subgroups n 7 divides = 24 and has the form n 7 = 7l+1 by the Sylow Theorems. Therefore n 7 = 1 or n 7 = 8. Since G is simple, n 7 1; else G would have a normal subgroup of order 7. Therefore n 7 = 8. Every element of order 7 is in a Sylow 7-subgroup, each such subgroup has 6 elements of order 7, and two different Sylow 7-subgroups have no elements in common except for (e). Therefore there are 8 6 = 48 elements of order 7 in G. A2. Let D be an integral domain. (a) Suppose that f(x), g(x) D[x] are not zero. Show that Deg(f(x)g(x)) = Degf(x) + Degg(x). (b) Show that the ideal I = (x + 1, 7) of Z[x] is not principal. (c) Show that the quotient Z[x]/I is a field. How many elements does it have? Solution: (a) Since f(x), g(x) 0, we may write f(x) = a 0 + a m x m and g(x) = b 0 + +a n x n, where a m, b n 0. Thus f(x)g(x) = a 0 b 0 +(a 0 b 1 +a 1 b 0 )x+ +a m b n x m+n. Since D is an integral domain a m b n 0. Therefore Deg f(x)g(x) = m+n = Deg f(x)+ Deg g(x). (b) Suppose that I = (x + 1, 7) is principal and let p(x) Z[x] be a generator. Then p(x) divides 7 which means that p(x) = ±1, ±7 by part (a), and thus since p(x) divides x + 1 it follows that p(x) = ±1 by the same. Therefore I = Z[x]. But I is proper; the image of I under the reduction of scalars map π : Z[x] Z 7 [x] has image (x + 1), which is a proper ideal of Z 7 [x], contradiction. Therefore I is not principal. (c) Note there is a ring homomorphism π : Z[x] Z 7 determined by π(n) = n + 7Z and π(x) = 1. Note that π is onto and x + 1, 7 Ker π. Since I has at most 7 cosets, π induces an isomorphism Z[x]/I Z 7. Therefore the former has 7 elements. A3. Show that a finite integral domain D is a field. [Hint: For a non-zero a D consider the function f : D D defined by f(x) = ax for all x D.] 2

3 Solution: Suppose a D is not zero. We must show that a has a multiplicative inverse in D. Consider the function f : D D defined by f(x) = ax for all x D. Then f is one-one; for if x, x D and f(x) = f(x ) then ax = ax or equivalently a(x x ) = 0 from which x x = 0 since D is an integral domain and a 0. Since f is one-one and D is finite, f must be onto. Therefore f(x) = 1 for some x D. We have shown ax = 1. Since xa = ax we conclude that a has a two-sided inverse in D. Complex Analysis C1. Determine the number of zeroes (with multipicities) of the polynomial f(z) = z 4 + 5z inside the circle z = 2. Solution: We may write f(z) = g(z) + h(z), where g(z) = 5z 3 and h(z) = z On the circle z = 2 we have g(z) = 40 and h(z) = 27. In particular h(z) < g(z) on the circle, and so by Rouché s theorem, f(z) has the same number of zeroes inside the circle, counted with multiplicity, as g. It is clear that g(z) has a zero of order 3 at 0 and no other zeroes. Hence f(z) has three zeroes (with multiplicity) inside the circle. C2. Let R be any real number > 1, and let S denote the semi-circle z = R, I(z) 0, described counterclockwise. Set e iz I = z dz. (a) Show that S I 2πR R 2 1. (b) Use residue calculus to find the exact value of Solution: I + R R e it t dt. (a) For any point z = x+iy in S we have e iz = e y 1 and z 2 +1 z 2 1 = R 2 1. Hence eiz 1 for any z in S. It follows that z 2 +1 R 2 1 I 1 R 2 1 lengths = 2πR R 2 1. (b) We have I + R R e it t dt = e iz D z dz, 3

4 where D is the closed contour made up of the semi-circle S and the interval [ R, R] described from left to right. The only pole of the integrand e iz z = enclosed by D occurs at z = i and has residue Hence the integral is equal to π e. e iz (z i)(z + i) e i i (i + i) = e 1 2i. C3. Let U denote the open set in the complex plane consisting of all complex numbers which are not on the non-positive real axis. Thus U includes all strictly positive numbers and all complex numbers that are not real. Define conformal maps of U onto (a) the upper half-plane I(z) 0, and (b) the unit disk z < 1. Solution: (a) The map may be defined by z iz 1/2. Here z 1/2 denotes the principal branch of the square root, defined to take the value r 1/2 e iθ/2 when z = re iθ and π < θ < π. (b) Here we may define the map by z z1/2 1 z 1/ This is the composition of the map in (a) with the conformal map z z i z + i from the upper half-plane to the unit disk. Logic L1. Propositional formulas F and G are truth equivalent, denoted F G, if their truth value is the same for every truth evaluation of the variables. Show without using truth tables. Solution: ((Q R) R) ( R) ((P (Q S)) P ) LHS ( ( Q R) R) ((Q R) R) ( R) RHS ( R) (( P (Q S)) P ) ( R) 4

5 L2. Is the sentence a) satisfiable? ( x r(x, x)) ( x y z((r(x, y) r(y, z)) r(x, z))) ( x y r(x, y)) b) satisfiable in a finite structure? Solution: a) YES: (N, <) is a model b) NO: Assume that there is a finite model S and let a 0 be any element. For every i = 0, 1,..., let a i+1 be such that r(a i, a i+1 ) holds in S; such elements exist by the third conjunct of the sentence. By the second conjunct, r(a i, a j ) holds for every i < j. As S is finite, some elements in the sequence are repeated, i.e., a i = a j for some i < j, and thus r(a i, a i ) holds in S for some i, contradicting the first conjunct. L3. Is the following a valid argument? ( x r 1 (x)) ( x r 2 (x)) x(r 1 (x) r 2 (x)) Solution: YES: Assume that the conclusion is false in some structure S. Then there is an element a such that r 1 (a) and r 2 (a) hold in S. But then S = ( x r 1 (x)) and S = ( x r 2 (x)), hence the premise is also false in S. Number Theory N1. Let p and q be odd primes, and suppose that p = 2q + 1. Prove that if a is a positive integer with 1 < a < p 1, then p a 2 is a primitive root modulo p. Solution φ(p) = 2q so x is a primitive root mod p iff x 2 1 (mod p) and x q 1 (mod p). Now (x q ) 2 1 (mod p), so x q 1 x q 1 mod p If x = p a 2, then x q ( a 2 ) q = ( 1) q a 2q = ( 1) q a p 1 ( 1) q = 1 (mod p), since q is odd, and so x q 1 (mod p). Now x 2 = ( a 2 ) 2 = a 4 1 (mod p) if and only if a 2 1 (mod p) or a 2 1 (mod p). The only square roots a of 1 mod p between 1 and p 1 are 1 and p 1, but 1 < a < p 1, so a 2 1 (mod p) is impossible. Since p = 2q + 1 and q is odd, p 3 (mod 4), and 1 is not a square mod p, so a 2 1 (mod p) is also impossible. In fact, a cannot have order 4 because 4 does not divide p 1, the order of the mutiplicative group of integers mod p. Thus the second alternative is also impossible, and so p a 2 is a primitive root. 5

6 N2. a) Prove the Chinese Remainder Theorem: If gcd(m, n) = 1, then for all pairs of integers (a, b) there is a unique c (mod mn) such that: Solution c a (mod m) c b (mod n) Since gcd(m, n) = 1, we know that there are integers x, y such that: Note that this equation implies that Then we can take c = xma + ynb, then: xm + yn = 1 yn 1 (mod m) xm 1 (mod n) xmb + yna yna a (mod m) xmb + yna xmb b (mod n) If c is another solution, then m (c c ) and n (c c ), and so mn c c since gcd(m, n) = 1. b) Solve the following system of congruences: x 5 (mod 6) 7x 1 (mod 10) 7x 11 (mod 15) Solution Note that 7x 1 (mod 10) x 3 (mod 10), while 7x 11 (mod 15) x 8 (mod 15). (Note that gcd(7, 10) = gcd(7, 15) = 1.) The lowest common multiple of {6, 10, 15} is 30, and the integers between 1 and 30 congruent to 5 mod 6 are {5, 11, 17, 23, 29} of which only 23 is congruent to 3 mod 10 and 8 mod 15. Hence the solution is x 23 (mod 30) N3. Let φ be the Euler totient or phi function. a) Find φ(27075) 6

7 Solution = = = = Hence φ(27075) = 2 (4 5) (18 19) = b) Find all integers n, such that φ(n) = 12. Be sure to justify your answer. Solution Note that p n, p prime implies that (p 1) φ(n), and since the divisors of 12 are {1, 2, 3, 4, 6, 12}, the possible primes dividing n are 2, 3, 5, 7, 13. Also, if p k divides n, then p k 1 (p 1) divides n, so the only p for which p 2 n is possible are 2 and 3, and the only p for which p 3 n is possible is 2. For each possible prime p dividing n, if p k is the largest power of p dividing n we have a factorization n = p k m with gcd(p, m) = 1 and 12 = p k 1 (p 1)φ(m). Notice that if n is odd then φ(n) = φ(2n), and that using the same analysis as above (i.e. checking for which primes p, p 1 φ(m)), we find that: Putting this all together, we get that: φ(m) = 1 m = 1 or 2 φ(m) = 2 m = 3 or 4 φ(m) = 3 is impossible φ(m) = 6 m = 9 or 18 p k φ(p k ) φ(m) m n = p k m or 2 13 or or 4 21 or is impossible or 4 27 (impossible) or or 18 gcd(m, n) impossible or 18 (impossible) m odd m = 13 or or 42 Hence the possible values of n are {13, 21, 26, 28, 36, 42} Real Analysis 7

8 R1. Consider the power series 1 n=1 2 n n xn. (i) For which values of x R does the series converge (absolutely? conditionally?) (ii) Show that on [ 1, 1] the convergence is uniform. n x Solution (i) The ratio of the absolute value of two successive terms is which n+1 2 tends to x as n. Hence the series converges absolutely for x < 2. 2 If x = +2, the series is the harmonic series, 1, which diverges by comparison or the n integral test. If x = 2 the series converges, but only conditionally, by the Leibnitz test. (ii) On [ 1, 1] the absolute value of the nth term, a n 1/2 n, a converging geometric series, so the original series converges uniformly by the Weierstrass M-test. R2. Let f be continuous on R. Given δ > 0 let g(x) = 1 2δ δ δ f(x + t) dt. (i) Prove g has a continuous derivative on R. [Start with the change of variable u = x + t.] (ii) Given a closed interval [a, b] and an ε > 0, show it is possible to choose δ > 0 so that g(x) f(x) < ε for all x [a, b]. Solution (i) Substituting u = x + t in the integral we get g(x) = 1 2δ x+δ x δ f(u) du. By the fundamental theorem of calculus, g (x) = 1 {f(x + δ) f(x δ)} which is 2δ continuous. (ii) g(x) f(x) = 1 δ f(x + t) f(x) dt 1 f(x + t) f(x) dt. Now f is 2δ δ 2δ uniformly continuous on [a 1, b + 1], so given ε > 0, there is a δ with 0 < δ 1 such that t δ = f(x + t) f(x) < ɛ. Therefore g(x) f(x) < ε for x [a, b]. δ δ R3. Let f : [0, 1) R be a differentiable function with f (x) 2007 for all x [0, 1). (i) For any sequence x n 1 with 0 x n < 1 prove that {f(x n )} n=1 converges. (ii) Deduce that a = lim x 1 f(x) exists, and so f extends to a continuous function f : [0, 1] R. 8

9 Solution (i) By the mean value theorem, f(x n ) f(x m ) = f (ξ)(x n x m ) for some ξ (x m, x n ), hence f(x n ) f(x m ) 2007 x n x m. Given ε > 0 let δ = ε/2007. Then there exists k such that m, n k = x n x m < δ, since lim x n converges. Therefore f(x n ) f(x m ) ε so that f(x n ) is a Cauchy sequence. Hence lim f(x n ) converges. (ii) First, lim f(x n ) = l is independent of the choice of sequence satisfying the conditions of (i). If y n were another such sequence, then the sequence x 1, y 1, x 2, y 2,... would converge to 1, the sequence f(x 1 ), f(y 1 ), f(x 2 ), f(y 2 ),... would converge, say to l, and the two subsequences, f(x n ) and f(y n ), would converge to the same limit, l. Define f(1) = l extending f to [0, 1]. Then f(lim x n ) = f(1) = l = lim f(x n ) = lim f(x n ), hence f is continuous on [0, 1]. Topology T1. Let X be a connected complete metric space and let U be a non-empty open subset of X, regarded as a metric space with the restricted metric. Prove that if U is complete then U = X. Solution: Since X is connected, it suffices to show that U is open and closed. We are given that U is open. To show that U is closed we will show that U contains all of its limit points. Let p be a limit point of U. Since X is a metric space, there is a sequence (u n ) of points of U such that u n p as n. Any convergent sequence is Cauchy. Thus (u n ) is Cauchy in X and, since U has the restricted metric it is also Cauchy in U. Since U is complete, there is a point u U such that u n u in U, and hence also in X. Since limits of sequences are unique in a metric space, we have u = p. Thus p U, and U is closed, as required. T2. Let (X) = {(x, x) x X} X X 1. Show that R 2 (R) has two connected components. 2. Show that if f : R R is a homeomorphism, then f is monotonic (Hint: consider the restriction of the function f f to the set {(x, y) x < y}.) Solution: 1. Let R 2 + = {(x, y) R 2 x < y} and R 2 = {(x, y) R 2 x > y}. Then R 2 (R) is the disjoint union of R 2 and R 2 +. It suffices to show that each of these sets is connected. Consider the map g : R 2 R 2 given by g(x, y) = (x, x y). Then g( (R)) = R {0}, and g is a homeomorphism, since it is continuous and an inverse to itself. Moreover, g(r 2 +) = R (0, ), and g(r 2 ) = R (, 0). Thus each of R 2 ± is connected, being homeomorphic to a product of intervals. This shows that R 2 (R) has two components. 9

10 2. Suppose that f : R R is a homeomorphism. The map f f : R 2 R 2 is defined by f f((x, y)) = (f(x), f(y)). We have f f(r 2 (R)) R 2 (R), since f(x) f(y), for all x y. Thus, it takes components of R 2 (R) to components of R 2 (R). In particular, either f f(r 2 ) R 2 or f f(r 2 ) R 2 +. In the first case we have f(x) < f(y) whenever x < y. In the second case we have f(x) > f(y) whenever x < y. Thus, f is monotonic. T3. Let B(R) denote the set of all bounded sequences of real numbers with the uniform topology. Recall that the uniform topology is given by the sup metric ρ, which is defined as follows. If X = (x n ) n N and Y = (y n ) n N are points of B(R) then the distance ρ(x, Y ) is ρ(x, Y ) = sup( x n y n ). n N Prove that B(R) is a complete metric space. Solution Let (X i ) i N be a Cauchy sequence with respect to the metric ρ on B(R). For each i N let X i = (x n,i ) n N. First we show that there exists M R such that x n,i < M for all n, i N. For each i N, since X i is bounded, there exists M i such that x i,n < M i. Since the sequence (X i ) i N is Cauchy, there exists N N such that ρ(x i, X N ) < 1 for all i > N. In particular this means that M i M N + 1 for all i > N. Thus we may take M to be the maximum of M 1, M 2,..., M N 1, M N + 1. Now, since (X i ) i N is Cauchy, we have x n,i x n,j sup x n,i x n,j = ρ(x i, X j ). n N This shows that (x n,i ) i N is a Cauchy sequence for each n. Since R is complete, we may define x n, = lim x n,i for each n N. Moreover, x n, M for all n since i x n,i < M for all n, i N. Now consider the bounded sequence X = (x n, ) n N. Given ɛ > 0, there exists I N such that ρ(x i, X j ) < ɛ for all i, j I. Then, for i I, we have x n,i x n, = lim x n,i x n,j ɛ. Thus we see that ρ(x i, X ) ɛ for i I. In other words, j lim X i = X. This shows that (B(R), ρ) is a complete metric space. i 10