CMI Entrance Exam 2019 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE Postgraduate Programme in Mathematics MSc/PhD Entrance Examination 15 May 2019 (1) A, B. (2) D. (3) A, C. (4) C, D. (5) A, B, C. (6) B, C, D. (7) C. (8) C. (9) C, D. (10) 0. Part B (11) T is not Hausdorff; it is the co-finite topology on C. For any polynomial f , f −1 (A) is a closed set for every finite set A, so f is continuous. n P (12) n≥0 n z is convergent on {z ∈ C : |z| < 1}. a For each real number  > 0, there exists N such that for all n ≥ N , |an | < . Then we can write X |z|N |F (z)| ≤ C + |z|n = C + (1 − |z|) n≥N for some C ∈ R that does not depend on z. Hence for z ∈ C with |z| < 1, (1 − |z|)|F (z)| can be made to take values arbitrarily close to , for any  > 0, by taking |z| −→ 1. By way of contradiction assume that G = F . Let ζ ∈ C with |ζ| = 1 is a pole of G. Let M be the order of the pole at ζ. Write c−M c−1 G(z) = M + ··· + + G1 (z) (z − ζ) (z − ζ) where G1 (z) is an analytic function. As z −→ ζ, (1 − |z|)|G(z)| = |(z − ζ)G(z)| exhibits one of the following behaviours: if M > 1, then it approaches infinity; if M = 1 (which implies that c−1 6= 0), it approaches c−1 6= 0. This is a contradiction. (13) Assume that {|n| : n ∈ Z} is bounded. Let N be such that |n| ≤ N for every n ∈ Z. Let x, y ∈ R. Without loss of generality, |x| ≥ |y|, and we want to show that |x + y| ≤ |x|. n   n X n |x + y| ≤ | ||x|r |y|n−r r r=0 ≤ (n + 1)N |x|n for every n 1 1 Hence |x + y| ≤ N n (n + 1) n |x| for every n, so |x + y| ≤ |x|. hR i1 1 n (14) Write an = 0 |f (x)|n dx . Let M = sup{|f (x)| : 0 ≤ x ≤ 1}. Then an ≤ M for every n, so lim sup an ≤ M . Since [0, 1] is compact, for every  > 0, there exists an interval I ⊆ [0, 1] of positive length such that M −  ≤ |f (x)| ≤ M for every x ∈ I . Then Z 1 n 1 1 an ≥ |f (x)| dx ≥ [(M − )n · length(Ie )] n = (M − )(length(Ie )) n . n I Hence lim inf an ≥ M −  for every  > 0; therefore lim inf an ≥ M , so lim an = M . (15) Without loss of generality, we may assume that V 6= 0. Let M, N ∈ V be non-zero elements. Let λ be an eigenvalue of N M −1 . Then det(λM − N ) = det M det(λIn − N M −1 ) = 0. However, (λM − N ) ∈ V , so λM − N = 0. 1

(16) If p2 divides n, then there is a non-cyclic group Z/pZ × Z/pZ × Z/(n/p2 )Z. There are exactly pq such matrices, and theyQform a nonabelian group. If p1 , . . . , pr are the distinct prime divisors of n, then φ(n) = ti=1 (pi − 1), which is coprime to n, by above. (17∗ ) Let g ∈ G be of order 2. Then the minimal polynomial of g divides X 2 −1. If char F 6= 2, then the minimal polynomial of g is X + 1 or X 2 − 1. In either case, g is diagonalizable and the conjugacy class of g is determined by the number of −1s on the diagonal; there must be at least one −1. Hence |X| = n. If char F = 2, then X 2 − 1 = (X − 1)2 , which must be the minimal polynomial of g. (It cannot be X−1.) Hence g is not diagonalizable, and the conjugacy class of g is determined by the number of 2 × 2 Jordan blocks (with eigenvalue 1). Hence |X| = b n2 c. (18∗ ) Consider the sequences (4n+1)π 2 and nπ 1 . Both converge to 0, but f ( (4n+1)π 2 ) = 1 and f ( nπ ) = 0 for every n ≥ 1, so f does not extend to Y . Let h : X −→ X1 , t 7→ (t, sin 1t ). 1 It is a homeomorphism. Since X1 ⊆ [0, 1] × [−1, 1], Y1 is compact. Identifying X with X1 using h, we get that Y1 is a compactification of X. Further, f = π2 ◦ h, where π2 is the projection Y1 −→ R on to the second component. Note that f extends to the map π2 (after identifying X with X1 using h). (19∗ ) Let α, 3α, β3 , . . . , βn be the roots of f , so f (0) = 3α2 β3 · · · βn . If f (0) = 1, then 13 = α2 β3 · · · βn satisfies a monic irreducible polynomial g ∈ Z[X]. Since g is irreducible in Q[X], g = (X − 31 ), which is a contradiction. Hence f (0) 6= 1. Note that K = Q(α, 3α, β3 , . . . , βn ) and that every field automorphism of K permutes these generators of K. Hence there are at most n! distinct field automorphisms of K, so σ is of finite order, which we denote by m. Then α = σ m (α) = σ(σ m−1 (α)) = 3(3m−1 α) = 3m α), so α = 0. ∗ (20 ) Since [0, x] is compact and f is continuous, one can use max, so g is well-defined. Note that f is uniformly continuous, so for every  > 0, there exists δ > 0 such that for every x, y with |x−y| < δ |f (x)−f (y)| < . Let u < v < u+δ. Then f (u)− ≤ f (x) ≤ f (u)+ for every x ∈ [u, v], so g(v) ≤ g(u)+. On the other hand g(v) ≥ g(u), so |g(v)−g(v)| < . Hence g is continuous. 2