CMI Entrance Exam 2015 Question Paper Solution M.Sc Maths

2015 MSc/PhD Mathematics Solutions Part A (1). B, D. (2). B. (3). A, B, C, D. (4). A, C. (5). C, D. (6). A, C. (7). B, C. (8). 2. (9). 0. (10). 0. Part B (11). After changing coordinates, we may assume that f vanishes in an open neighbourhood of the origin. Hence ∂n   f (0, 0) = 0 ∂xi ∂y n−i for every n ≥ 0 and 0 ≤ i ≤ n. Therefore all the coefficients of f are zero, so f = 0. (12). (a). Let v 6= 0 ∈ ker A, so (In + A)v = v; hence 1 is an eigenvalue of (In + A). Conversely, suppose that (In + A)v = λv for some nonzero v. If Av = 0, then v = λv, so λ = 1. Otherwise, multiplying on the left by A, we see that Av = λ(Av); hence, again, λ = 1. (b). A2 + 1 = 0, so the minimal polynomial of A divides t2 + 1 which has distinct roots. Hence the minimal polynomial of A has distinct roots, so A is diagonalizable. (13). For r ∈ R, r > 0, let Cr := {z ∈ C : |z| = r}, Ur := {z ∈ C : |z| < r} and r = min{|f (z)| : z ∈ Cr }. As f it nonconstant, the origin is an isolated zero of f . Choose r > 0 such that the origin is the only zero in Cr ∪ Ur . As the set {z : |f (z)| < r } intersects Ur but not Cr , it must be in Ur , by the second condition on f . Hence the origin is the only zero of f . Now letting r −→ ∞, we see that for all z 6∈ Ur , |f (z)| ≥ r . Hence f has a pole at infinity, i.e., there exists a non-negative integer m (z −1 ) is analytic at z = 0. Suppose that f P∞ mi such thatmz f−1 is a power-series expression i=n ai z . Since z f (z ) is analytic at z = 0, we see that ai = 0 for all i ≥ m. Hence f is a polynomial that vanishes exactly at z = 0. Thus f = cz n for some positive interger n and non-zero c ∈ C. (14). For each positive integer k, using a), choose Nk such that  |bk − am,k | < ∀m ≥ Nk . 2 Let mk = N1 + · · · + Nk . Then mk is increasing. Now, using b), choose a positive integer K, such that  |amk ,k − 1| < ∀k ≥ K. 2 By triangle inequality |bk − 1| <  ∀k ≥ K. (15). May assume that f is homogeneous of degree d. Induct on d. If xd appears in f , then so does y d (with the same coefficient. Hence we may write f = α(xd + y d ) + xyf1 (x, y), where α could be zero. Note that f1 (x, y) = f1 (y, x), so by induction, there exists g1 such that f1 (x, y) = g1 (x + y, xy). Write xd + y d = (x + y)d − xyf2 (x, y); f2 (x, y) = f2 (y, x), so by induction, there exists g2 such that f2 (x, y) = g2 (x + y, xy). (16). Note that for every a ∈ [0, 1] and for every open subset U of X containing f −1 (a), there is an open neighbourhood W (for example, take W = [0, 1] \ f (X \ U )) of a such that f −1 (W ) ⊆ U . Now let Ui , i ∈ I be an open S covering of X. For every a ∈ [0, 1], there exists a finite subset Ia of I such that f −1 (a) ⊆ i∈Ia Ui . Let Wa be an open neighbourhood of a such 1

that f −1 (Wa ) ⊆ i∈Ia Ui . There exist finitely many a1 , . . . , an such that [0, 1] = nj=1 Waj . S S Then X = nj=1 f −1 (Waj ) = nj=1 i∈Ia Ui . This is a finite subcovering of the given open S S S j covering. (17∗ ). For each such P , let RP be the smallest subring of Q that contains { p1 |p ∈ P }. If P 6= Q, then RP 6= RQ . The set of all the primes is countably infinite, and the power-set of a countably infinite set is uncountable. Hence Q has uncountably many subrings. (18∗ ). Let |x| ≤ M . Since sin x ≤ x, X sin x X 1 n f (x) = ≤M < ∞, n n2 n≥1 n≥1 and hence f is well-defined. P sin x Also the same argument shows, if we set fN = N ≥n≥1 n n , then fN converges uniformly to f as N 7→ ∞ in the interval (−M, M ), and hence f is continuous. Since M is arbitrary f is continuous everywhere. x cos n Since cos nx ≤ 1, g is well-defined, and P Again let |x| ≤ M and we set g = n≥1 n2 x 0 = P cos n fN N ≥n≥1 n2 uniformly converges to g, by the same arguments. Since fN converges uniformly to f and since fN 0 converges uniformly to g in the interval (−M, M ), f 0 = g. Since M is arbitrary, f is differentiable everywhere. (19∗ ). By the usual computation, pm(m−1)/2 is the largest power of p dividing the order of GLm (Fp ). Also, one may check directly that U is a subgroup of GLm (Fp ) and its order is pm(m−1)/2 . Therefore, U is a Sylow p-subgroup of GLm (Fp ). Now, by Sylow’s Theorem 2 and its Corollary, there exists A ∈ GLn (Fp ) such that AGA−1 ⊂ U . (20∗ ). Let A ∈ Mm×n (C) be the matrix that consists of 1s on the diagonal on rows and columns 1, . . . , k and 0s elsewhere. Then the space of all matrices of rank k inside Mm×n (C) is X := {B1 AB2 : B1 ∈ GLm (C); B2 ∈ GLn (C)} ; we need to show that this space is connected. Fix B2 ∈ GLn (C). Then the space XB2 := {B1 AB2 : B1 ∈ GLm (C)} is connected since it is a continuous image of the connected space GLm (C). Similarly, the space Y := {AB2 : B2 ∈ GLn (C)} is connected. Since [ X=Y ∪ XB2 B2 ∈GLn (C) and each XB2 intersects Y , we see that X is connected. 2