MASTERS EXAMINATION IN MATHEMATICS

MASTERS EXAMINATION IN MATHEMATICS PURE MATH OPTION, Spring 018 Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth 0 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 it. Any student whose answers need clarification may be required to submit to an oral examination. Algebra A1. Let G be a finite Abelian group of order n m, where m is odd. If the subgroup of order n is not cyclic, show that the product of all elements in G is the identity. Hint: evaluate separately the product of elements of order two. Solution. Let g G be an element of order g >. Then we have g g 1, so both g and its inverse g 1 appear in the product and they cancel each other. By the fundamental theorem of finite abelian groups, it is enough to consider groups G of the form Z/Z Z/Z in which the number of factors is at least two (by hypothesis). For such groups one easily concludes (e.g. by induction). A. Express both 13 and 5 + i as a product of irreducible elements of Z[i]. Solution. We have 13 = (3 + i)(3 i) and 5 + i = (3 i)(1 + i). Each of the factors is irreducible in Z[i] since the corresponding norm is a prime integer. A3. Prove the following assertions. (i) We have the equality Q[ 3, 5] = Q[ 3 + 5]. (ii) Let a and b be two rational numbers, such that a 0. The fields Q[ a] and Q[ b] are equal if and only if there exists c Q such that a = bc. Solution (i) The inclusion Q[ 3, 5] Q[ 3 + 5] is absolutely immediate. reverse inclusion, we notice that we have ( 3 + 5) 3 = 18 3 + 14 5 1 For the

by a quick computation. Since 14 3 14 5 equally belongs to Q[ 3 + 5]. Their sum will equally be an element of Q[ 3 + 5]. However, the sum in question equals 4 3 and we are done. Solution (ii) By hypothesis we have a = r1 + r b for some rational numbers r 1 and r. This gives a = r1 + br + r 1 r b. If r1 0 and r 0, we conclude that b Q and it follows that a Q in which case we are done. If r 1 = 0, then we define c := r and we have equally finished. If r = 0, then a Q and so Q[ a] = Q. Since Q[ a] = Q[ b], we infer that b is equally a rational number, and again the conclusion follows. Complex Analysis C1. Using residues, evaluate the following integral. Solution. x 1 + x 4 dx The degree of the denominator exceeds the degree of the numerator by, so the residue theorem implies that this integral is equal to ( ) z πi Res 1 + z ; z 4 k z k : 1+z 4 k =0, Im(z k)>0} The denominator z 4 + 1 is zero when z 4 = 1 = e πi, so z = e π 4, e 3π 4, e 5π 4, e 7π 4. The first two are in the upper half-plane so the integral will equal ( ( ) ( )) z πi Res 1 + z ; e π z 4 4 + Res 1 + z ; e 3π 4 4 For simple poles, the residue of p(z)/q(z) at z k is given by p(z k )/q (z k ), so in the case at hand 1 it will equal 4z k. Hence the integral is equal to πi (e iπ 4 + e 3iπ 4 ) = πi( 1 i 1 1 i 1 ) = πi ( i 1 ) = π = π. C. Find the number of zeroes the function f(z) = e z + 5z + z has in z C : z 1}, counted according to multiplicity. Solution. On z = 1, one has e z + z e z + z = e Rez + 1 e + 1 5 = 5z. Thus by Rouche s theorem, e z + z + 5z and 5z have the same number of zeroes inside z < 1, counted according to multiplicity, namely.

C3. Evaluate the following integral, where C is the circle z C : z = 10} oriented positively. C sin z dz (z + π) 10 Solution. By the extended Cauchy s integral formula, since the pole at z = π is inside the circle, the integral will equal d 9 πi 1 9! dz sin(z) 9 z= π Since d9 sin(z) = cos(z), the answer is therefore πi 1 π cos( π) = i dz 9 9! 9! Number Theory N1. Suppose F n = n + 1. Show that for all m > n, F n F m. N. Determine all n with φ(n) n. N3. State the definition of the µ-function. Prove that 1 n = 1; µ(d) = 0 n > 1. Find an asymptotic formula for d n as X. You may use the identity µ(d) d d X a,b X gcd(a,b)=1 ab = 6 ( ) 1 π + O. X Real Analysis R1. For the following sequence of functions f n (x) = arctan (nx) find its pointwise limit on the real line R. Is this limit uniform on R? Is this limit uniform on [1, )? Explain your answer. 3

Solution. First note that as y, arctan(y) π, and as y, arctan(y) π, and arctan(0) = 0. So, π, x > 0 lim arctan(nx) = h(x) = 0, x = 0 n π, x < 0 Since h is discontinuous at x = 0, the limit cannot be uniform on any interval containing 0, hence on R. However on [1, ) we can use the fact that arctan(y) is a monotonically increasing function. So, if x 1, then arctan(n) f n (x) < π and the sequence arctan(n) converges to π as n. So, for any ε > 0 there exists an N > 0 such that for all n > N and for all x 1 we have f n (x) π arctan(n) π < ε. This proves that the sequence converges uniformly to the constant function π on [1, ). R. State the definition of convergent series and of absolutely convergent series. Prove that that if n=1 a n is an absolutely convergent series then the series n=1 a n converges. Give an example of a convergent series n=1 b n such that n=1 b n diverges. Solution. Since a n absolutely converges, we have a n < +. We also have a n 0. Thus there exists N so that for all n > N we have a n < 1 and therefore a n a n. Hence we can compare positive series a n = n=1 N a n + n=1 a n N+1 N a n + n=1 n=n+1 a n < +. For the counter example, consider b n = ( 1) n n 1/. This is an alternating series with terms monotonically decreasing to zero in absolute value. Such series converge by Leibnitz criterion. But b n = 1 n=1 1 n = +. R3. Let f : (0, ) (0, ) be a twice differentiable function and assume that f (x) < 0 for all x (0, ). Prove that f(x) is nondecreasing on (0, ). Prove that f(x) is uniformly continuous on (0, ). 4

Solution. Since f (x) < 0 it follows that f (x) is decreasing on (0, ). Fix arbitrary 0 < x 0 < x. By MVT there exists z (x 0, x) so that f(x) = f(x 0 ) + f (z)(x x 0 ) < f(x 0 ) + f (x 0 )(x x 0 ) In particular f (x 0 ) 1/(x x 0 ) and sending x we deduce f (x 0 ) 0. This applies to any x 0 (0, ). So f(x) is non-decreasing. But it cannot be constant on any interval (a, b), because f would vanish on (a, b) contradicting f < 0. Therefore f(x) is increasing. Let ɛ > 0 be given. Let m = inff(x) x > 0}. There exists a (0, ) so that m f(a) < m + ɛ/. Let k = f (a) and let δ = ɛ/k. For any x 1 < x in (0, a] we have m f(x 1 ) < f(x ) f(a) < m + ɛ/. So f(x ) f(x 1 ) < ɛ/. For any x 1 < x with x x 1 < δ and a x 1 by MVT we have for some z (x 1, x ) f(x ) f(x 1 ) = f (z)(x x 1 ) k(x x 1 ) < kδ < ɛ/. Finally for any x 1 < a < x with x x 1 < δ we combine the two estimates f(x ) f(x 1 ) = f(x ) f(a) + f(a) f(x 1 ) < ɛ/ + ɛ/ = ɛ. Topology T1. Suppose X is a compact metric space, and f : X X an isometric embedding. Show that f is surjective. Solution. Suppose for contradiction that f is not surjective. Let x X f(x). Since x is not in the compact set f(x), the distance δ = d(x, f(x)) is positive. Define a sequence (x n ) recursively by x 0 = x and x n = f(x n 1 ). Since f(x k ) f(x) for all k > 0, we have d(x 0, x k ) δ for all k > 0. Since f is an isometric embedding, it follows by applying this map i times to x 0 and x k that d(x i, x i+k ) = d(x 0, x k ) δ. Therefore, no subsequence of (x n ) is Cauchy. However, a compact metric space is sequentially compact, and so (x n ) has a convergent subsequence, which is therefore Cauchy. This is the desired contradiction. T. Let X and Y be topological spaces, and let A i } be a finite collection of closed subsets that cover X. Show that f : X Y is continuous if f Ai is continuous for each i. Give an example to show the finiteness assumption is necessary. Solution. Recall that continuity is equivalent to the condition that the preimage of each closed set is closed. Let f Ai be continuous for all i. Suppose the collection A i } contains n sets. Then for a closed set C Y we have n n f 1 (C) = f 1 (C) A i = (f Ai ) 1 (C) i=1 In the first equality above, we have used that the sets A i cover X. By continuity of f Ai, each set (f Ai ) 1 (C) is closed in the subspace topology of A i. However, since A i is closed in 5 i=1

X, we have that (f Ai ) 1 (C) is closed in X. Thus the formula above describes f 1 (C) as a finite union of closed sets in X, which is therefore closed. This shows f is continuous. The restriction to a finite collection A i } is essential: Consider X = [0, 1], Y = R, and the function 0 if x = 0 f(x) = 1 otherwise. This function is not continuous because the preimage of the open set ( 1, 1) is 0}, which is not open. Consider the collection of closed sets A 0 = 0} and A n = [ 1, 1] for n Z with n > 0. Then n the collection A n } n Z 0 covers [0, 1], and for each n the restriction of f to A n is constant and therefore continuous. T3. If X and Y are topological spaces, say X Y if there exists a continuous surjection from X to Y. Let Y be the set a, b}. Let X be a nonempty topological space. Prove: (1) If Y has the trivial topology then X Y iff X has at least two points. () If Y has the discrete topology then X Y iff X is disconnected. (3) If Y has the topology, a}, Y } then X Y iff X has a non-trivial topology. Solution (1). Suppose X Y. Let f : X Y be a continuous surjection. Then the subsets of X given by f 1 (a) and f 1 (b) are disjoint and nonempty. Thus X has at least two points. Suppose X has at least two points. Choose a point x a X. Define f : X Y by a if x = x a f(x) = b otherwise Since X has at least two points, there exists x b x a with f(x b ) = b, and f is surjective. Since Y has the trivial topology, every function X Y is continuous. Hence X Y. Solution (): Suppose X Y. Let f : X Y be a continuous surjection. Define U = f 1 (a) and V = f 1 (b). Since a}, b}} is a cover of Y by disjoint open sets, and f is continuous, the sets U, V are disjoint, open, and cover X. By surjectivity of f, both U and V are nonempty. Thus U, V give a separation of X, and X is not connected. Suppose X is not connected. Let U, V be a separation of X. Define a function f : X Y by a if x U f(x) = b otherwise The respective preimages under f of the open sets, a}, b}, Y are, U, V, X. All of the latter sets are open, so f is continuous. Since U and V are nonempty, f is surjective. Thus X Y. Solution (3): Suppose X Y. Let f : X Y be a continuous surjection. Define U = f 1 (a). Since a} is open in Y, we have that U is open in X. Since f is surjective, the 6

sets U and f 1 (b) are nonempty and disjoint, and in particular U X. Thus the topology of X includes the nonempty open set U X, and the topology is non-trivial. Suppose X has a non-trivial topology. Let U be a open set in X that is not or X. Define a function f : X Y by a if x U f(x) = b otherwise The respective preimages under f of the open sets, a}, Y are, U, X. All of the latter sets are open, so f is continuous. Since U is nonempty and U X, the function f is surjective. Thus X Y. Logic L1. Recall that for a structure M with universe M, a set A M n is definable in M if there is a parameter p M k for some k and n + k-ary formula φ, so that Let N = (N; 0, 1, +, ). A = a A n M = φ[ p, a]}. (a) Show the following are definable in N. 7} n n is a perfect square} (m, n) m < n} (m, n) m divides n} n n is prime} (b) Show there is a subset of N which is not definable in N. Solution.Part (a): 7} is definable by x = y and the parameter 7 (it also definable without parameters by x = 1 + 1... + 1 seven times). n n is a perfect square} is definable by y(x = y y). (m, n) m < n} is definable by z(x + z = y z 0). (m, n) m divides n} is definable by φ div (x, y) := z(x z = y) n n is prime} is definable by φ(y) := y 1 x(φ div (x, y) (x = 1 x = y)). Part (b): For any definable X N, let φ X, k = k X, and p X N k witness that it is definable. Note that X (φ X, k X, p X ) is an injection into a countable set. So there are only countably many definable subsets of N. Since the powerset of N is uncountable, there are subsets which are not definable. L. Recall a graph is a structure G = (V ; E) such that E is a symmetric and irreflexive binary relation. Solution. Part (a). The following are the axioms: symmetric: x y(xry yrx); 7

irreflexivity: x (xrx); For every even k, λ k := ( x 1 )( x )...( x k )( x k+1 ) 1 i k x irx i+1 x 1 x k+1. For part (b), suppose for contradiction that the class of bipartite graphs is finitely axiomatizable. Then there is some sentence φ, such that A = φ iff A is a bipartite graph. Let Σ be the axiomatization defined in part (a). Then Σ = φ. By compactness it follows that there is some finite subset of Σ, such that = φ. Let n be large enough such that if λ k, then k < n. Let A be a graph with an odd-length cycle of size at least n. Then A =, and so A = φ. Contradiction with the assumptions that φ axiomatizes bipartite graphs. A cycle of length n is a sequence v 1,..., v n+1 V so that for all 1 i n, v i, v i+1 E; v 1 = v n+1 ; and for 1 i < j n, v i v j. A theorem of graph theory states that a graph is bipartite if and only if it contains no cycles of odd length. (a) Give a theory in the language L = R}, R a binary relation symbol, that axiomatizes the class of bipartite graphs. (b) Show the class of bipartite graphs is not finitely axiomatizable. L3. Let A = (A; A ) and B = (B; B ) be countable dense linear orders without endpoints. (a) Show A and B are isomorphic. (b) Show the theory of dense linear orders without endpoints is complete. Solution. (a) Enumerate A = a n n < ω} and B = b n n < ω}. Define an order preserving function f : A B as follows. Let f(a 0 ) be any element in B At even stages, n: if f(a n ) is already defined, do nothing. Otherwise, let D be the set of elements a, such that f(a) has been defined, and set B < := f(a) a D, a < a n } and B > := f(a) a D, a > a n }. By the inductive assumption for all b 1 B < and b B >, we have b 1 < b. Since B is a dense linear order, pick b B, such that for all b B <, b < b and for all b B >, b < b. Set f(a n ) = b. At odd stages, n + 1: if b n is in the range of the partial function defined so far, go to the next stage. Otherwise, let D be the set of elements a, such that f(a) has been defined and let R be the range, and let A < := a D f(a) < b n } and A > := a D b n < f(a)}. By the inductive assumption for all a 1 A <, a A >, a 1 < a. Since A is a dense linear order let a A be such that for all a A <, a < a and for all a A >, a < a. Set f(a ) = b n. Then by the even stages in the construction, dom(f) = A, and the odd stages ensure that ran(f) = B. Inductively we maintained that f is order preserving, and so it is an isomorphism. (b). The theory of dense linear orders without endpoints has no finite models, and by part (a) it is ℵ 0 -categorical. So, by the Los-Vaught Test test, it is complete 8