MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION SPRING 010 SOLUTIONS Algebra A1. Let F be a finite field. Prove that F [x] contains infinitely many prime ideals. Solution: The ring F [x] of polynomials with coefficients in a field F is a P.I.D. Each prime ideal is generated by a monic, irreducible polynomial. Assume there are only a finite number of prime ideals generated by the polynomials f 1,..., f n and let f(x) = 1+f 1 (x) f n (x). No f i divides f, hence f is also irreducible. This contradicts the assumption that all the prime ideals were generated by the finite set of irreducible polynomials. A. Prove that if H is a normal subgroup of G of prime index p then for all K G either (a) K H or (b) G = HK and K : K H = p. Solution: The set HK = {hk : h H, K K} is a subgroup since (hk) 1 h k = k 1 h 1 h k = (k 1 h 1 h k)k k HK because H is normal. We have H HK G. Since G : H = p, where p is prime, either (a) H = HK and so K H or (b) HK = G in which case G/H K/(K H) and hence K : K H = p. A3. Let G be a finite group and H be a Sylow p-subgroup of G. Let K be the normalizer of H in G and let L be the normalizer of K in G. (a) Show that H is a characteristic subgroup of K. (This means that if φ : K K is an automorphism of K, then φ(h) = H.) (b) Show that L = K. 1
Solution: Let H be a Sylow p-subgroup of the finite group G. The normalizer of H in G is K = {k G : k 1 Hk = H}. For part (a), let φ : K K be an automorphism of K. Then H and φ(h) are Sylow p-subgroups of K, hence conjugate, so there is k K with k 1 Hk = φ(h). Then by definition of K, φ(h) = H and therefore H is a characteristic subgroup of K. For part (b), if g L, the normlizer of K in G, then g 1 Kg = K. Define φ : K K by φ(k) = g 1 kg. Now by (a), φ(h) = H or g 1 Hg = H and hence g K. Thus L = K. Complex Analysis C1. For a > 0, evaluate sin(ax) x 4 + 1 dx. Solution: The integral is the imaginary part of e iaz z 4 + 1 dz. along the real axis. Since the denominator has degree four, this integral is πi times the sum of the residues at the two fourth roots of 1 that lie above the x-axis, namely e πi 4 = 1+i and e 3πi 4 = 1+i. Computing these residues, the integral is which simplifies to ( ia e 1+i πi 4e 3πi 4 1+i + eia ) 4e 9πi 4 πi ( e a/ e i(a/ 3π/4) e ) i(a/ 3π/4) = πe a/ sin(a/ 3π/4). The given integral is the imaginary part of this or 0. [For another valid solution observe the improper real integral converges absolutely by comparision with the integral of (1 + x 4 ) 1. Since the integrand is an odd function, the result is 0.] C. Let a 1, a, a 3,... be complex numbers with 0 < a 1 < a < and lim a n =, n 1 such that a n=1 n is finite. Let D denote C {a 1, a, a 3,...}. Show that the sum 1 h(z) = converges for all z D and that h(z) is analytic on D. z a n n=1
Solution: Let z D. Then for any N the sum of the first N functions is analytic at z since the sum of two analytic functions is analytic, so it suffices to consider the 1 sum of z a n over all n for which a n > z. For such n, on a neighborhood of z one 1 has z a n < 1 for all n. Since 1 a n n converges, by the Weierstrass M test 1 a n n z a n converges uniformly on a neighborhood of z. Hence the limit function is analytic. Adding to the first finite sum, we see that the overall sum is analytic at z. C3. (a) Show that if Im(a) > 0 and λ = 1, the function f(z) = λ z a is a conformal z ā map from the upper half plane to the open disc centered at the origin of radius 1. (b) Find an explicit conformal mapping from the quarter-plane onto the open unit disc. {z : Re(z) > 0, Im(z) > 0} Solution: For (a), since λ is a rotation it suffices to assume λ = 1. For real z, z a = z ā, so z a has magnitude 1. Hence the real axis gets sent to the unit z ā circle. Plugging in z = i for example reveals the upper half-plane gets sent to the interior. For (b), z z takes the quarter-plane to the upper half-plane, and then composing with one of the above functions gives the desired map. Logic L1. Let L be a first order language with at least one constant symbol and let φ(x) be a quantifier free formula with a single free variable x. Suppose that for any variable free terms t 1,..., t k, the set of sentences is consistent. {φ(t 1 ),..., φ(t k )} (a) Show that there is an L-structure M such that for any variable free term t we have M = φ(t). (b) Show that there is a substructure N M satisfying N = x φ(x). Solution: (a) To find a structure M such that for each variable free term t, we have M = φ(t), we must show that the set of sentences T = {φ(t) t is a variable free term } 3
is satisfiable, which, by the compactness theorem, follows from the assumption that T is finitely satisfiable. (b) Given our structure M from (a), let N be the substructure of M whose domain is the set of interpretations of all variable free terms, i.e., domain(n ) = {t M t is a variable free term }. Since L contains a constant term, this set is empty and is clearly closed under the interpretations of function symbols in M, so this does indeed define a substructure. Moreover, since φ is quantifier free, for any a domain(n ), and thus, by the assumption on M, L. Let L be a propositional language. N = φ(a) M = φ(a), N = x φ(x). (a) Show that any L-formula F is equivalent to an L-formula G only containing the logical connectives and. (b) Suppose F is an L-formula only containing the logical connective. Show that F is true whenever all the propositional variables in F are true. Solution: (a) We recall that by the theorem on disjunctive normal form, any L- formula F is equivalent to an L-formula H in disjunctive normal form and thus to one only containing the connectives,, and. Now, by induction on the subformulas of H, we make the following substitutions (A B) ( A B), (A B) (A B) to obtain a new formula G, in which the only occurring logical connectives are and. Since (A B) ( A B) and (A B) (A B), the theorem on substitutions of equivalent subformulas shows that H G and hence F G. (b) We prove the result by induction on the construction of F. Basis of the induction: Suppose F is a single propositional variable P. Then clearly F is true whenever all propositional variables in F, that is P, are true. Induction step: Suppose that F is the formula (A B), where A and B satisfy the induction hypothesis, that is, A is true whenever all propositional variables in A are true, and similarly for B. Assume now that all propositional variables occurring in F are true. Then, since F = (A B), all propositional variables in A and all propositional variables in B are true, whence both A and B are true. But then, by the truth table for, also F is true, which shows the induction hypothesis for F. 4
L3. Test the following argument for validity using the tableau method. Solution: Q (P R) R ( P Q) (S R) (P Q) Q (S P ) Q (P R) R ( P Q) (S R) (P Q) (Q (S P )) Q (S P ) S P Q P R P R R R ( P Q) ( P Q) P Q All branches of the tableau close and hence the argument is valid. Number Theory N1. Find the general integer solution of the Diophantine equation 3x 101y = 1. Use the Euclidean algorithm and show your work. Solution: The Euclidean algorithm gives a particular solution: 3 = 101 + 1 1 = 3 101 101 = 4 1 + 17 17 = 101 4 1 = 4 3 + 9 101 1 = 17 + 4 4 = 1 17 = +5 3 11 101 17 = 4 4 + 1 1 = 17 4 4 = 4 3 + 53 101 5
The first column has the divisions of the Euclidean algorithm. The next columns express the remainders as linear combinations of the original two numbers. A particular solution is given by x = 4, y = 53. If 3x 101y = 1 is another solution, then Since 101 and 3 are relatively prime, is an integer and is a solution for any integer k. (x + 4)3 (y + 53)101 = 0. x + 4 101 x = 4 + 101k, = y + 53 3 = k y = 53 + 3k N. (a) Let a and m be integers > 1. Prove that if a m 1 is prime then a = and m is prime. (b) Suppose that q is a prime and that q 1 is composite. divisible by some prime p < q/ such that Show that q 1 is p 1 (mod q). Solution: (a) We may factor a m 1 = (a 1)(a m 1 + + 1). If a, m > 1 then the second factor is greater than 1, so if a m 1 is prime we must have a =. If m = kl, the k 1 divides m 1, therefore if a m 1 is prime, m must be prime. (b) If q is prime and q 1 is composite let p be any prime dividing q 1 Then q 1 (mod p). Hence the multiplicative order of (mod p) divides q. Since q is prime and 1 1 (mod p), has oder q (mod p). Therefore q divides p 1, the order of the multiplicative group of integers (mod p). Now p 1 (mod q) and, since p is odd, p 1 (mod ). Therefore p 1 (mod q). Finally, if p is a factor of q 1, either p q 1 < q/ or ( q 1)/p < q 1. In either case some prime factor of q 1 is less that q/. N3. For any positive integer n, let s(n) be the sum of all positive divisors of n. Prove that if m and n are relatively prime, we have s(mn) = s(m) s(n). 6
Solution: If m and n are relatively prime, the set of all positive divisors of mn is {ij : i m, j n, i > 0, j > 0}. Then s(mn) = i j n j = s(m)s(n). i m,j n ij = i m Real Analysis R1. Suppose that the limit L = lim g(x) exists. Prove that x 0 g(x) sin 1 f(x) = x, x 0; 0, x = 0. is continuous at 0 if and only if L = 0. Solution: Assume lim x 0 g(x) = 0. Since sin(1/x) 1 for x 0, f(x) = g(x) sin(1/x) g(x) for x 0 and hence lim f(x) = 0 and f is continuous at 0. x 0 Now assume lim g(x) exists and equals L. There is a sequence {x n } with x n 0, x 0 lim x n = 0, and sin(1/x n ) = 1. Now if f is continuous at 0, lim f(x n ) = 0 and hence n n lim g(x n) = 0, so L = 0. n Notice, however, there are functions g such that f is continuous at 0, but for which g(1/(πn)) does not exist. lim n R. Is the following function Riemann integrable on [0, 1]? (prove your claim) { x, x Q; f(x) = 0, x / Q. Solution: No. On any interval [a, b] [0, 1], inf{f(x) : a x b} = 0 and sup{f(x) : a x b} = b > 0. Therefore all lower sums L(f, P) = 0 and all upper sums are the same as for the integrable function g(x) = x, hence Therefore f is not integrable. inf{u(f, P) : P} = 1/. 7
R3. Prove that the series continuous on R. n sin( n x) converges to a function f(x) which is uniformly n=1 Solution: The sum converges pointwise absolutely to a function f(x) by comparison with the series n=1 n. Given ε > 0 fix n with n < ε. Let n f n (x) = j sin( j x). Then f(x) f n (x) j=1 j=n+1 j sin( j x) n < ε. The function sin( n x) is uniformly continuous on R since it is continuous and periodic. Therefore there exists δ > 0 with Finally x y < δ sin( n x) sin( n y) < ε. f(x) f(y) f(x) f n (x) + f n (x) f n (y) + f n (y) f(y) < 3ε. Hence f is uniformly continuous. Topology T1. Let x 0 be a point in a metric space X with metric d. Define the function f : X R by f(x) = d(x 0, x). (a) Prove that f(y) f(z) d(x, y) for all y, z X. Conclude that f is continuous. (b) Suppose that Z X is a closed subset disjoint from x 0 ; that is, x 0 Z. Prove that the infimum of f(x) on Z is positive. Solution: (a) Let y, z X then f(y) f(z) = d(x 0, y) d(x 0, z) d(y, z) by the triangle inequality. To show that f is continuous, given ɛ > 0 let δ = ɛ. Then d(x, y) < δ implies f(x) f(y) d(x, y) < δ = ɛ. 8
( b) The infimum of f(x) on Z is inf {f(y) y Z} = inf {d(x 0, y) y Z} = d(x 0, Z) Suppose that this infimum is 0. Then for each integer n > 0, choose y n Z with d(x 0, y n ) < 1/n. It follows that the sequence {y n } converges to x 0. As Z is closed, this implies x 0 Z, which contradicts our assumptions. T. Let X be a Hausdorff topological space. Define the collection of subsets T = {U = X K K X, K is compact } { } (a) Show that T satisfies the axioms of a topology. (b) For X = R with the usual metric topology, define a topology T on R as in part (a). What is the closure of the set A = {n n = 1,,...} in the T -topology? Solution: (a) The empty set is always compact, so X = X T. Also, we are given that T. Given U, V T, U, V, there are compact sets K U, K V X with U = (X K U ) and V = (X K V ). Then U V = (X K U ) (X K V ) = X (K U K V ). The union of two compact sets is compact, so U V T. Given nonempty open sets {U α α A} T, there are compacts sets K α X for which U α = (X K α ). Then U α = (X K α ) = X K α. α A α A α A The arbitrary intersection of compact sets is compact, so α A U α T. b) The open sets of T are the complements of compact sets in X = R for the given topology, so the closed sets for T consist of the compact sets in R, along with the entire space R. As the set A is not compact, the only closed set for the topology T containing A is all of X. Thus, the T -closure of A is X. T3. Show that an open continuous map from a non-empty compact space X to a connected Hausdorff space Y must be surjective. Solution: Let f : X Y be an open continuous map from X to Y. Since Y is connected, it suffices to show that f(x) is open and closed. Since X is an open set and f is an open map, f(x) is open in Y. The continuous image of a compact space is compact. Therefore f(x) is compact. But a compact subspace of a Hausdorff space is closed, since points can be separated from compact sets. Thus, f(x) is closed. 9