CMI Entrance Exam 2017 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE Postgraduate Programme in Mathematics MSc/PhD Entrance Examination 18 May 2017 Solutions Part A (1) B. (2) B. (3) A, B. (4) A, C, D. (5) C (6) A. (7) B, D. (8) A, B, C, D. (9) B, D. (10) 3. Part B (11) (A) X is Hausdorff since the diagonal is closed. (B) For all x, y, d(x, y) ≤ d(x, x) + d(y, x) = d(y, x), so d(x, y) = d(y, x); hence d is a metric. In particular, τ 0 is the metric topology given by d. (C) Identify Bx0 , with d−1 ([0, )) ∩ ({x0 } × X). Since d−1 ([0, )) is open in X × X, we see that d−1 ([0, )) ∩ ({x0 } × X) is open in the subspace topology of ({x0 } × X) induced from the product topology of X × X, which is the same as the topology τ on X. Hence Bx0 , is open in the topology τ , so τ 0 is coarser than τ . (12) (A) f has a power-series expansion around 0 that converges everywhere on C. Since |f (0)| ≤ 0, f (0) = 0, so f (z)/z is entire, and |f (z)/z| ≤ 1. Hence, by Liouville’s theorem, f (z)/z = C for some C ∈ C, i.e., f (z) = Cz. (B) Z z dz = 0 Γ and use the fact that x = z+z 2 and y = 2ı . z−z (13) By way of contradiction, assume that there exist  > 0, an increasing sequence n1 < n2 < · · · of integers and x1 , x2 , . . . ∈ [0, 1] such that |fnk (xk ) − f (xk )| >  for every k ≥ 1. Let xki , i ≥ 1 be a convergent subsequence, converging to y ∈ [0, 1]. Construct a new sequence yj as follows: ( xki , ifj = nki yj = y, otherwise. Then the sequence fj (yj ) does not converge to f (y), a contradiction. (14) (A) If r is a positive rational number, then log(x) < xr for all real numbers x 0. (B) The two sequences below are bounded and the series is convergent: log n log log n X 1 0.1 ; 0.1 ; . n n n3/2 (15) Write φa : Fp −→ Fp for the map b 7→ ab. We check the following: (A) For each a ∈ Fp r {0}, φa is a group automorphism. (B) The map a 7→ φa is a group homomorphism: 1 7→ idFp ; φa0 a (b) = a0 ab = (φa0 ◦ φa )(b). (C) The map a 7→ φa is injective: Indeed if φa = idFp then a = 1. 1

(D) The map a 7→ φa is surjective: Let φ be any group automorphism of (Fp , +), which is a cyclic group, generated by 1. Then φ is determined by φ(1). Since φ is an automorphism, φ(1) 6= 0. Hence φ(b) = φ(1)b for every b ∈ Fp . Therefore φ = φφ(1) . (E) A bijective group homomorphism is a group isomorphism. (16) In each of the three cases, if m is a maximal ideal of R, then it is generated by (the residue class of) an irreducible polynomial dividing X 7 − 1. Further, if f (X) is an irreducible polynomial of degree d, then dimk k[X]/f (X) = d. k = Q: The irreducible factors of X 7 − 1 are (X − 1) and (X 6 + X 5 + · · · + 1). (To see that (X 6 + X 5 + · · · + 1) is irreducible over Q, write it as ((Y + 1)6 + (Y + 1)5 + · · · + 1) (where Y = X − 1) and apply the Eisenstein criterion.) Hence the dimensions are 1 and 6. k = C: Every irreducible polynomial is linear, so the dimension is 1. chark = 7: X 7 − 1 = (X − 1)7 , so the dimension is 1. (17∗ ) A has a pole if and only if 1 is an eigenvalue. If A ∈ SO3 , then its real eigenvalues are ±1, and its determinant is 1. Hence if all the eigenvalues are real, then at least one eigenvalue is 1. If it has exactly one real eigenvalue λ1 , then 1 = λ1 (a + ıb)(a − ıb), so λ1 > 0, i.e., λ1 = 1. For the second part, we need to show that for p ∈ S2 if Ap = p for some A ∈ G then for every B ∈ G, Bp is a pole for some C ∈ G. Take C = BAB −1 . (18∗ ) First suppose that Y and the all fibres f −1 (y), y ∈ Y are compact. We prove a ‘Tube lemma’: Let y ∈ Y ; if U is an open neighbourbood of f −1 (y), then there exists an open neighbourbood V of y such that f −1 (V ) ⊆ U . Proof of lemma: X \ U is closed, so f (X \ U ) is closed. Since f −1 (y) ⊆ U , y 6∈ f (X \ U ). Let Vy = Y \ f (X \ U ). One can check immediately that f −1 (Vy ) ⊆ U , finishing the proof of the lemma. Let Uλ , λ ∈ Λ be an open cover of X. For each y ∈ Y , we see, using the above lemma and the fact that f −1 (y) is compact, that there is a finite subset Λy ⊂ Λ and an open neighbourbood Vy of y such that f −1 (y) ⊆ f −1 (Vy ) ⊆ S Uλ . Since Y is λ∈Λy compact, there exist y1 , . . . , yn ∈ Y such that Y = Vy1 ∪ · · · ∪ Vyn . Thus X = f −1 (Y ) = n f −1 (Vy1 )∪· · ·∪f −1 (Vyn ) ⊆ S S Uλ , so the open cover Uλ , λ ∈ Λ has a finite subcover. i=1 λ∈Λyi Hence X is compact. Conversely, assume that X is compact and Y is Hausdorff. Then Y is compact. Let y ∈ Y . Then {y} is closed in Y , so f −1 (y) is closed in X, and hence compact, as X is compact. ∗ (19 ) The images in F := R/m of the monomials in x1 , . . . , xn form a countable spanning set 1 of F over k. If t ∈ F is transcendental over k, then { t−α | α ∈ k} is linearly independent over k, which is not possible, so F/k is algebraic. Hence F ≃ k. (20∗ ) Let R > 0 be given and B(0, R) be the open unit ball of radius R. By the mean value inequality, | sin( nz ) − sin(0)| ≤ supw∈B(0,R) | cos(w)|| nz − 0| for every z ∈ B(0, R). Hence P∞ sin nz P∞ 2 n=1 n is dominated by the series n=1 |z|/n as cos(w) is bounded on P B(0, R). Therefore the series converges uniformly on compact sets. Hence the series ∞ z n=1 sin( n )/n defines a holomorphic function. It is a standard fact that if a sequence of holomorphic functions converges uniformly on compact sets then the limit is a holomorphic function. 2