CMI Entrance Exam 2016 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE
Postgraduate Programme in Mathematics
MSc/PhD Entrance Examination
18 May 2016
Solutions

Part A
(1) A,D.
(2) A.
(3) A,C.
(4) B, C.
(5) A.
(6) B.
(7) A, B, C, D.
(8) A, C, D.
(9) 1.
(10) 1.

Part B
(11) In polar coordinates, the region U is given by U 0 := (1, 2) × [0, 2π). Write p0 and q 0 for
the points in U 0 corresponding to p and q respectively. Let γ 0 : [0, 1] −→ U 0 be given by
t 7→ t · p0 + (1 − t) · q 0 . This is continuous, and is such that γ 0 (0) = p0 and γ 0 (1) = q 0 and
such that γ 0 is differentiable on (0, 1). Let f be the map (converting polar coordinates
to Cartesian coordinates),
R × [0, 2π) −→ R2 , (r, θ) 7→ (r cos θ, r sin θ).
Let γ = f ◦ γ 0 . Since f is differentiable, γ has the desired properties.
(12) By way of contradiction, suppose that IJ is a prime ideal. In an integral domain, a
product of two non-zero ideals is non-zero, so IJ is a non-zero prime ideal. In a PID,
every non-zero prime ideal is maximal, so IJ is maximal. Since IJ ⊆ I ∩ J, we conclude
that I = IJ = J. Let a be a generator for I and J; hence IJ is generated by a2 .
Therefore we see that a ∈ (a2 ). Write a = ba2 , so a(1 − ba) = 0. Since a 6= 0, ba = 1,
i.e., a is a unit, so I is not a proper ideal, a contradiction.
(13) Write z = x + ıy with x, y ∈ R and f (z) = u(x, y) + ıv(x, y), where u and v are maps
from R2 to R. Then v(0, t) = 0 = v(t, 0) for every t ∈ R. Therefore
∂v ∂v
(0, 0) = 0 = (0, 0).
∂x ∂y
Since f is entire,
∂u ∂u
(0, 0) = 0 = (0, 0).
∂x ∂y
Therefore f 0 (0) = 0.
(14) Let  > δ, and consider the closed ball
B := {y ∈ Rn |d(x, y) ≤ }.
There exists y ∈ A such that d(x, y) ≤  so
δ = inf{d(x, y) | y ∈ A ∩ B}.
Since A∩B is closed and bounded, it is compact. Consider the function f : A∩B −→ R,
y 7→ d(x, y). It is continuous, so it attains its infimum and supremum, i.e., there exists
y ∈ A ∩ B ⊆ A such that δ = d(x, y).

(15) (A): Fix a basis v1 , . . . , vn of V . Let n1 , . . . , nd be such that T ni vi = vi . Let n = lcm{ni |
1 ≤ i ≤ d}. Then T n (v) = v for every v ∈ V  , so, over C, the minimal polynomial of T
1 1
has distinct roots. For (B), take T = over F2 .
0 1
(16) [Q(ω) : Q] = 2, since the minimal polynomial √ of ω is X 2 + X + 1. A basis √ √of Q(ω)
over Q is {1, 3 3 3 3
√ ω}. The minimal polynomial √ √of 2 over √ Q√is X − 2, so {1, 2, 4} is a
basis of Q( 3 2) over Q. Therefore {1, 3 2, 3 4, ω, ω 3 2, ω 3 4} span F as Q-vector-space.
Since X√ 3 − 2 is irreducible over Q(ω), [F : Q(ω)] = 3, and, therefore [F : Q] = 6. Hence
√ √ √
{1, 2, 3 4, ω, ω 3 2, ω 3 4} is a Q-basis for F . Since ω 2 = −(1 + ω) and ω 3 = 1, we see
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that the matrix of µ with respect to the above basis (in the given order) is
 
−1 0 0 1 0 0
 0 −1 0 0 1 0
 
0
 0 −1 0 0 1 .
−1 0 0 0 0 0
 
 0 −1 0 0 0 0
0 0 −1 0 0 0
(17∗ ) Suppose that G is not dense. Let l := inf{x ∈ G | x > 0}. We first show that l > 0
and that l ∈ G. If l = 0, then there exists a small open neighbourhood U of l = 0 that
contains non-zero elements of G; Now, for every g ∈ G, {g + u | u ∈ U } is an open
neighbourhood of g that contains elements of G other than g, so G is dense; hence l > 0.
Now, if l 6∈ G, then there exists  > 0 such that (l, l + ) ⊆ G. We may assume, without
loss of generality, that  < l. Let x, y ∈ (l, l + ), with x < y. Then y − x ∈ G and
0 < y − x < . This contradicts the choice of l, so l ∈ G.
Now let g ∈ G, and let n be the largest integer such that nl ≤ g < (n + 1)l. Hence
0 ≤ g − nl < l, so by the minimality of l, g = nl, i.e., G = Z · l.
(18∗ ) Let φ : X −→ C be the constant function taking the value 1. Then for every g ∈ G,
(g · φ)(x) = φ(g −1 (x)) = 1 = φ(x) for every x ∈ X, so g · φ = φ for every g ∈ G. Let
X
F 0 := {f ∈ F | f (x) = 0}.
x∈X

It is a subspace of F . To show that F = F 0 ⊕ Chφi, we need to show 0
P that F ∩ Chφi = 0
0 0
and that F = F + Chφi. If αφ ∈ F for some α ∈ C, then 0 = x∈X (αφ)(x) = α|X|,
so α = 0, i.e., F 0 ∩ Chφi = 0. Let f ∈ F . Set α = |X| 1 P
x∈X f (x). Then f − αφ ∈ F ,
0

so F = F 0 + Chφi.
∗
(19 ) (A) Write A = (aij ) and B = (bij ). If aij > 1 then for every k, bjk = 0, for otherwise,
the (ik)th entry of AB would be greater than 1. However, if there exists j such
that for every k bjk = 0, then B is not invertible. Hence aij ∈ {0, 1} for every i, j.
Similarly bij ∈ {0, 1} for every i, j. Now suppose that aij = aik = 1 with j < k.
Then exactly one of bji , bki is 1 and for l 6= i, bjl = bkl = 0. Therefore either the
jth row or the kth row of B is zero, a contradiction. Hence, for every i, there is a
unique ki such that aiki = 1. Therefore A is a permutation matrix. Since B = A−1 ,
B is a permutation matrix.
(B) Let T : Cn −→ Cn be the linear transformation given by ei 7→ Aei , where the ei are
the standard basis of Cn . Let v ∈ Cn be a non-zero vector that is not an eigenvector
of A (and of T ). (Such a vector exists by the hypothesis on A.) Then v and Av are
linearly independent. Extend it to a basis v, Av, v3 , . . . , vn of Cn . In this basis, T
is given by a matrix B that satisfies B1,1 = 0. Since A and B represent the same
linear transformation (in two different bases), they are similar to each other.
(20∗ ) Define f : S 1 × {1, −1} −→ S 1 by setting f (w, 1) = Sqrt(w) and f (w, −1) = − Sqrt(w).
It is continuous. Surjectivity: Let w ∈ S 1 . If Sqrt(w2 ) = w, then f (w, 1) = w; otherwise,
Sqrt(w2 ) = −w, and hence f (w, −1) = w. Injectivity: If f (w1 , 1) = f (w2 , −1), then
w1 = (f (w1 , 1))2 = (f (w2 , −1))2 = w2 ; however f (w, 1) 6= f (w, −1) for any w. Therefore
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there do not exist w1 , w2 such that f (w1 , 1) = f (w2 , −1). If f (w1 , 1) = f (w2 , 1) then
w1 = (f (w1 , 1))2 = (f (w2 , 1))2 = w2 ; similarly if f (w1 , −1) = f (w2 , −1) then w1 =
w2 . Since S 1 × {1, −1} and S 1 are Hausdorff and compact, f is a homeomorphism, a
contradiction, since S 1 × {1, −1} is not connected, while S 1 is.

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