MASTERS EXAMINATION IN MATHEMATICS

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1 MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION FALL 2010 Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth 20 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 2010 Algebra A1. Let G = Z/2 Z/6 Z/12. Show that G can not be generated by two elements. A2. Let G be a finite group of order n. Let H be a proper subgroup of G if index m. If n does not divide m!, show that G is not a simple group. A3. Let p be a prime number. Show the polynomial is irreducible in Q[x]. Φ p (x) = x p 1 + x p x + 1 C1. Show that z =3 e z2 z z dz 2πe9. Complex Analysis C2. Find all possible values of ( 3 + i) i and log(5 + 5i). In both cases indicate which of the values is the principal value. C3. Let D be the wedge in the upper right quadrant of the complex plane between the x axis and the line y = tan(θ)x, where 0 < θ < π. Explicitly find a conformal map that 2 takes D bijectively to the interior of an open disk containing the origin. 2

3 PURE MATHEMATICS OPTION FALL 2010 Logic L1. Let L be a first order language, M an L-structure and N M a substructure. (a) What does it mean for N to be an elementary substructure of M? Suppose L = {R}, where R is a binary relation symbol and let M = Z, R M and N = 2Z, R N be the L-structures defined by where 2Z is the set of even integers. (b) Is N a substructure of M? R M = {(n, m) Z Z n < m}, R N = {(n, m) 2Z 2Z n < m}, (c) Is N an elementary substructure of M? L2. Let L be a first order language. (a) Suppose φ(x, y) is an L-formula whose free variables are among x, y. Write an L-sentence σ that is true in an L-structure M if and only if there is a unique pair (a, b) of elements of M such that M = φ(a, b). For every formula ψ(x, ȳ) in L, we denote by!x ψ(x, ȳ) the formula where z is some variable not among x, ȳ. (b) Which of the following formulas are equivalent? x ( ψ(x, ȳ) z (ψ(z, ȳ) x = z) ), σ,!x!y φ(x, y),!y!x φ(x, y) L3. Test the validity of the following argument using the tableau-method proof system. Arguments not formalized in the tableau-method, e.g., using truth tables, will receive a maximum of 10 points. A (B C) (C A) D (B D) A D 3

4 PURE MATHEMATICS OPTION FALL 2010 Number Theory N1. Find all x that satisfy the following 3x 2 mod 13, 4x 1 mod 7, 2x 3 mod 11 Express your answer in the form x a mod b. N2. (a) State the law of quadratic reciprocity for a pair of odd primes. ( ) 85 (b) Use the law of quadratic reciprocity to compute. 101 N3. Show that X 2 1 mod p n has exactly two solutions mod p n for each odd prime p. [Suggestion: First consider the case n = 1. Then assume X satisfies X 2 1 mod p n 1, and show that there is a unique Y mod p n such that Y X mod p n 1 and Y 2 1 mod p n. ] Real Analysis R1. (a) Define lim sup a n for a sequence {a n } n=1 of real numbers. (b) Let {a n } n=1, {b n } n=1 be two sequences bounded from above. Prove that lim sup(a n + b n ) lim sup(a n ) + lim sup(b n ). (c) Give an example with strict inequality in (b). R2. For which values of x does the series n=1 (3x + 1) n n log n converge? absolutely converge? R3. Let f be a non-negative continuous function on [0, 1] with with that f(x) = 0 for all x [a, b]. 1 0 f(x) dx = 0. Prove 4

5 PURE MATHEMATICS OPTION FALL 2010 Topology T1. Let X be a topological space, and let A X be a connected subset of X. Let B = cl(a) denote the closure of A in X. Show that B is connected. T2. Show that a topological space X is Hausdorff if and only if the diagonal is closed for the product topology. (X) = {(x, x) x X} X X T3. Suppose that f : S 1 X is an injective continuous map from the circle to a Hausdorff space X. Show that the subspace f(s 1 ) is homeomorphic to S 1. 5

6 MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION FALL 2010 SOLUTIONS Algebra A1. Let G = Z/2 Z/6 Z/12. Show that G can not be generated by two elements. Solution: G is a direct sum of cyclic groups each of even order. There is a homomorphism from each of these summands onto Z/2. Therefore G maps onto H = Z/2 Z/2 Z/2. If G where generated by two elements, then H would be also. If a, b H, the subgroup generated by a, b consist of elements of the form xa + yb where x, y {0, 1}. There are at most four such elements while H has order 8. Therefore G cannot be generated by two elements. A2. Let G be a finite group of order n. Let H be a proper subgroup of G if index m. If n does not divide m!, show that G is not a simple group. Solution: G acts by left multiplication on the set S of cosets gh of H in G, that is, there is a homomorphism φ : G Perm(S) sending g 1 to the permutation sending gh to g 1 gh. ker φ is normal in G. Since φ acts nontrivially, ker φ G, and, since n does not divide m!, ker φ 0. Therefore G is not simple. A3. Let p be a prime number. Show the polynomial is irreducible in Q[x]. Solution: Note that Φ p (x) = xp 1 x 1. Let Φ p (x) = x p 1 + x p x + 1 f(u) = Φ p (u + 1) = (u + 1)p 1. u Then f(u) satisfies the Eisenstein s Criterion, hence f(u) is irreducible in Z[u] and therefore Φ p (x) is irreducible in Z[x]. We conclude that Φ p (x) is irreducible in Q[x] by Gauss s Lemma. 1

7 C1. Show that z =3 e z2 z z dz 2πe9. Complex Analysis Solution: The integral is at most the length of the contour times the maximum magnitude of the integrand on the contour. The length of the contour is 6π, and the integrand is at most e Re( z2 ) z 2 2 z e9 3. Hence the integral is at most 2πe9. C2. Find all possible values of ( 3 + i) i and log(5 + 5i). In both cases indicate which of the values is the principal value. Solution: 3 + i = 2e iπ 6 = e ln 2+ iπ 6 = e ln 2+i π 6 +2πki for any integral k. Hence ( 3+i) i = e i ln 2 π 6 2πk for any integral k, or e π 6 2πk [cos(ln 2) + i sin(ln 2)]. The case k = 0 gives the principal value. For the second part, (5+5i) is (5 2)e iπ 4, so the possible logarithms are ln(5 2) + i( π + 2kπ) for integers k, with k = 0 giving the principal value of the 4 logarithm. C3. Let D be the wedge in the upper right quadrant of the complex plane between the x axis and the line y = tan(θ)x, where 0 < θ < π. Explicitly find a conformal map that 2 takes D bijectively to the interior of an open disk containing the origin. Solution: z z π/θ takes D to the upper half plane. The conformal map (z i)/(z +i) takes the upper half plane to the unit disc. Hence (z π/θ i)/(z π/θ + i) works. Logic L1. Let L be a first order language, M an L-structure and N M a substructure. (a) What does it mean for N to be an elementary substructure of M? Suppose L = {R}, where R is a binary relation symbol and let M = Z, R M and N = 2Z, R N be the L-structures defined by where 2Z is the set of even integers. (b) Is N a substructure of M? R M = {(n, m) Z Z n < m}, R N = {(n, m) 2Z 2Z n < m}, (c) Is N an elementary substructure of M? 2

8 Solution: (a) N is an elementary substructure of M if and only if for all formulas φ(x) and parameters a in N, we have N = φ(a) M = φ(a). (b) N is a substructure of M since the domain N of N is a subset of the domain M of M and moreover R N = N 2 R M. (c) N is not an elementary substructure of M, since, for example, while L2. Let L be a first order language. M = x (R0x & Rx2), N = x (R0x & Rx2). (a) Suppose φ(x, y) is an L-formula whose free variables are among x, y. Write an L-sentence σ that is true in an L-structure M if and only if there is a unique pair (a, b) of elements of M such that M = φ(a, b). For every formula ψ(x, ȳ) in L, we denote by!x ψ(x, ȳ) the formula where z is some variable not among x, ȳ. (b) Which of the following formulas are equivalent? x ( ψ(x, ȳ) z (ψ(z, ȳ) x = z) ), σ,!x!y φ(x, y),!y!x φ(x, y) Solution: (a) Let σ be the following formula: x y ( φ(x, y) & z v (φ(z, v) (x = z & y = v) ). (b) None of the three formulas are equivalent. To see this, we can let φ(x, y) be the formula Rxy, where R is a binary relation symbol. We define two {R}-structures M and N with common domain {1, 2, 3} as follows R M = { (1, 1), (1, 2), (1, 3), (2, 1) } Then while R N = { (1, 1), (2, 1), (3, 1), (1, 2) } M = σ, M =!x!y φ(x, y), M =!y!x φ(x, y), N = σ, N =!x!y φ(x, y), N =!y!x φ(x, y), showing that the formulas are inequivalent. 3

9 L3. Test the validity of the following argument using the tableau-method proof system. Arguments not formalized in the tableau-method, e.g., using truth tables, will receive a maximum of 10 points. Solution: A (B C) (C A) D (B D) A D A (B C) (C A) D (B D) A D C A (C A) D D C A D D (B D) B D A B C B C Since the tableau does not close after all formulas have been checked, the argument is invalid. In this case, we can see that A, B, C, D will render all premises true, but the conclusion false. A Number Theory N1. Find all x that satisfy the following 3x 2 mod 13, 4x 1 mod 7, 2x 3 mod 11 Express your answer in the form x a mod b. 4

10 Solution: We start by simplifying the equations. One easily checks that mod 13, mod 7, mod 11. We rewrite the equations as We first consider the two top equations x 5 mod 13, (1) x 2 mod 7, (2) x 7 mod 11. (3) x 5 mod 13, x 2 mod 7. An x that satisfies the first equation is of the form x = k for an integer k. We insert this in the second equation to obtain k 2 mod 7. This equation simplifies to 6k 4 mod 7, or equivalently 3k 2 mod 7. Since mod 7, we get k 3 mod 7. This means that k = 3 + 7l for an integer l. Consequently x = k = (3 + 7l) = l. We insert this value x = l in equation (3) to obtain l 7 mod 11. This immediately give 13 7l 7 mod 11, and then 2 7l 7 mod 11 and then 2l 1 mod 11 and finally l 6 mod 11. Write l = m for an integer m. We have x = (6 + 11m) = m. The desired solution is x 590 mod N2. (a) State the law of quadratic reciprocity for a pair of odd primes. ( ) 85 (b) Use the law of quadratic reciprocity to compute. 101 Solution: (a) If p and q are distinct odd primes, the ( ) ( ) p q = ( 1) 1 2 (p 1) 1 2 (q 1). q p (b) ( ) ( 85 5 = ( 101 = = 1 5 ) ( ) ) ( ) 101 = 17 ( 1 5 ) ( ) as ( 16 17) = 1. 5

11 N3. Show that X 2 1 mod p n has exactly two solutions mod p n for each odd prime p. [Suggestion: First consider the case n = 1. Then assume X satisfies X 2 1 mod p n 1, and show that there is a unique Y mod p n such that Y X mod p n 1 and Y 2 1 mod p n. ] Solution: Let f(x) = X 2 1. If a Z/p and f(a) = 0 mod p, then (a 1)(a+1) = 0 mod p. Since Z/p has no zero divisors, a = 1 or a = 1. Assume inductively that if a Z/p n and f(a) = 0, then a = ±1 mod p n. Let b Z/p n+1 and f(b) = 0 mod p n+1. Then f(b) 0 mod p n and hence b ±1 mod p n. Therefore b = a + tp n where a = ±1. Now f(b) = (a+tp n ) 2 1 2atp n mod p n+1. Sincef(b) 0 mod p n+1, t 0 mod p. Thus b ±1 mod p n+1. Real Analysis R1. (a) Define lim sup a n for a sequence {a n } n=1 of real numbers. (b) Let {a n } n=1, {b n } n=1 be two sequences bounded from above. Prove that lim sup(a n + b n ) lim sup(a n ) + lim sup(b n ). (c) Give an example with strict inequality in (b). Solution: (a) If {a n } is bounded then the sequence A n = sup{a n, a n+1,... } is nonincreasing and bounded; therefore converges. Its limit is one of the equivalent definitions of lim sup a n. (b) For each n, let A n = sup{a i i n}, B n = sup{b i i n}, C n = sup{a i + b i i n}. For i n one has a i A n, b i B n giving a i + b i A n + B n. Hence C n A n + B n. Passing to the limit in n one obtains the claim. (c) For example: 0, 1, 0, 1, 0, 1,... and 1, 0, 1, 0, 1,.... R2. For which values of x does the series n=1 (3x + 1) n n log n converge? absolutely converge? 6

12 Solution: There are four cases: (1) x < 2/3 or x > 0, (2) 2/3 < x < 0, (3) x = 2/3, (4) x = 0. In case (1) the ratio test for the absolute values a n = 3x + 1 n /n log n gives a n+1 lim a n So a n 0 and the series diverges. = 3x + 1 lim (n + 1) log(n + 1) n log n In case (2) the same ratio test gives absolute convergence. = 3x + 1 > 1. In case (3) the series is ( 1) n /n log n; it converges conditionally by Leibnitz rule, because 1/n log n 0 monotonically. In case (4) the series is 1/n log n = +. This can be deduced from Cauchy criterion comparing the series to 2 k a 2 k = const 1, or by the integral criterion with change k of variables u = e x. So the series converges for x [ 2/3, 0) with conditional convergence at x = 0. R3. Let f be a non-negative continuous function on [0, 1] with with that f(x) = 0 for all x [0, 1]. 1 0 f(x) dx = 0. Prove Solution: Assume f(c) > 0 for some c [0, 1]. By continuity of f there is an interval J of length l > 0 with f(x) > f(c)/2 for x J. Then any partition of [0, 1] has a refinement P with the lower sum L(f, P) lf(c)/2 Thus 1 f(x) dx lf(c)/2 > 0. 0 This gives a contradiction. Topology T1. Let X be a topological space, and let A X be a connected subset of X. Let B = cl(a) denote the closure of A in X. Show that B is connected. Solution: Given open sets U V = with B U V we must show that either B U or B V. It is given that A is connected, and A U V, so either A U or A V. Assume A U. Then U V = implies that A (X V ). As (X V ) is closed, we also have B = cl(a) (X V ). Thus, B V =. Since B U V we conclude that B U. The case where A V is similar. T2. Show that a topological space X is Hausdorff if and only if the diagonal is closed for the product topology. (X) = {(x, x) x X} X X 7

13 Solution: Assume that X is Hausdorff. Let x, y X with (x, y) (X). Then x y, so by the Hausdorff assumption, there exists open sets U, V X with x U, y V and U V =. If (U V ) (X) then there exists some point (z, z) U V, so that z U V, a contradiction. Thus, U V (X (X)) is an open neighborhood in the product topology for the point (x, y). This shows that the complement of (X) is an open set, hence (X) is closed for the product topology. Conversely, suppose that (X) is closed, and x, y X satisfy x y. Then (x, y) (X (X)) which is open for the product topology. Thus, there exists open sets U, V X with (x, y) U V (X (X)). Then x U and y V with U V =. This shows that X is Hausdorff. T3. Suppose that f : S 1 X is an injective continuous map from the circle to a Hausdorff space X. Show that the subspace f(s 1 ) is homeomorphic to S 1. Solution: The circle is compact, since it is a closed and bounded subspace of R 2. The subspace f(s 1 ) is Hausdorff, since it is a subspace of a Hausdorff space. Thus f is a continuous bijection from the compact space S 1 to the Hausdorff space f(s 1 ). Any continuous bijection g from a compact space to a Hausdorff space is a homeomorphism. To show this it suffices to show that any such bijection g is a closed map. (If it is closed then it is also open since, being a bijection, it maps complements of closed sets to complements of closed sets.) But a closed subspace of a compact space is compact and a continuous image of a compact space is compact. Thus g maps closed sets to compact sets. Since compact subspaces of a Hausdorff space are closed, it follows that g maps closed sets to closed sets. 8