CMI Entrance Exam 2014 Question Paper Solution M.Sc Maths

MSC/PHD MATHEMATICS 2014 SOLUTIONS

Part A

1. B, D.

2. A, B.

3. B.

4. B, C.

5. C, D.

6. B, C.

7. A.

8. A, B, D.

9. 2.

10. 2.
Part B
3
(11) The set of rank 2 matrices in M2×3 (R) is open: Consider the map f : M2×3 (R) → R
given by sending a matrix to the triple of its 2 × 2 minors. This is a continuous map.
3
The set of rank 2 matrices is the inverse image of the set {(x1 , x2 , x3 ) ∈ R | xi 6=
0 for some 1 ≤ i ≤ 3}. This set is open in R3 , hence the set of rank 2 matrices is open
in M2×3 (R).

(12) (A) The kernel of φ is a proper ideal in F . Hence it is zero, as there are no nonzero
proper ideals in a field. Since φ(1) = 1, using properties of a field homomorphism
we conclude that φ(r) = r for every r ∈ Q. Now if r ∈ Q and x ∈ F , then
φ(rx) = φ(r)φ(x) = rφ(x). Thus φ is a homomorphism of vector spaces over Q.
Since φ is injective and dimension of F over Q is finite, it follows that φ is also
surjective. Thus φ is a field isomorphism.

1

(B) Consider the map φ : F × → F × defined by φ(x) = x2 . Then φ is a group
homomorphism and the set of squares in F × is the image of φ. The kernel of φ
is the set of all x ∈ F × such that x2 = 1. Since F is a field this equation has at
most 2 solutions. Further since the characteristic of F is different from 2, it has
exactly two solutions. Hence the kernel of φ contains two elements and we have
an isomorphism F × /ker(φ) ∼ = im(φ). Hence the cardinality of im(φ) is half of the
cardinality of F × .
(13) (a) For A ∈ Mn (C), det(A) is a polynomial in the entires of A, so all the multiple
partial derivatives exist and are continuous.
(b) Let Aij ∈ Mn−1 (C) be the matrix with ith row and jth column removed from A.
Then the total derivative d(det) is the matrix in Mn (C) whose (i, j)-th entry is
(−1)i+j det(Aij ).
(c) d(det)=0 if and only if det(Aij ) = 0 for all i, j. This is equivalent to rank(Aij ) ≤
n − 2 for all i, j. We prove that this is equivalent to rank(A) ≤ n − 2.
Suppose rank(A) ≤ n − 2. This means the rows (or columns) of A have at most
n − 2 linearly independent vectors. After removing a row and column there will
be at most n − 2 linearly independent vectors. Hence rank(Aij ) ≤ n − 2 for all
i, j. Conversely if rank(A) > n − 2, the rows of A have at least n − 1 linearly
independent vectors. Choose n − 1 linearly independent row vectors and form
a matrix of size (n − 1) × n. Then n − 1 columns of this matrix are linearly
independent. So we can remove one column and the remaining (n − 1) × (n − 1)
matrix will still have (n − 1) linearly independent vectors. So there exits a pair
(i, j) such that rank(Aij ) = n − 1.
Hence d(det)(A)=0 if and only if rank(A) ≤ n − 2.

(14) Let F0 = {i | ai 6= 0} then F0 = ∪∞n=1 Fn . Each Fn is a finite set, otherwise
X X N
sup{ ai : F ⊆ R finite subset} ≥ sup{ ai : F ⊆ Fn finite subset} ≥ ∀N ∈ N.
i∈F i∈F
n

We can not replace countability by finiteness, since any convergent series with infinitely
many non-zero entries will satisfy the condition. Take for example ai = i12 for i ∈ N
and ai = 0 for i ∈ R \ N.
(15) Since the order of G is divisible by 2, G has an element x of order 2, by Cauchy’s
theorem. Since x 6= 1 and φ(x) = 1, φ is not injective. As φ is a function from a finite
set to itself, it can not be surjective as well.
(16) Let zn = 2πin. Then |zn | → ∞ and ezn → 1. Suppose g(z) = f (ez ) is a non-constant
polynomial. We have g(2πni) = f (1) for all n ∈ N. So g − f (1) has infinitely many
zeros and hence g = f (1). Thus f is equal to the constant f (1) on C \ {0}, which is
the range of the function ez . It follows by continuity that f is constant on all of C.

2

(17*) (A) On (R) is compact : Consider the map f : Mn×n (R) → Mn×n (R) given by f (A) =
t −1
AA − In . This map is continuous and On (R) is closed since it is equal to f (0).
t
On (R) ⊂ Mn×n (R) is bounded: the condition AA = In expressed in terms  of the 
a b
entries of A gives (for example when n = 2) the following: let A = .
c d
2 2 2 2
The we get a + b = 1 = c + d , and ac + bd = 0. This implies |a|, |b|, |c|, |d| ≤ 1.
(B) Consider det : On (R) → {1, −1}. This is continuous and surjective. This shows
that On (R) is disconnected.
(C) On (C) is not compact, since it is not bounded as a subset of Mn×n (C). The
conditions expressed above for a, b, c, d when the entries are complex numbers
show this. For example, fixing a, |a|
0, there are solutions for b such that
2 2
a + b = 1. Similarly for c, d.

(18*) f is given entire. ExpandPit as a power series in a neighbourhood of 0. f = ∞
P n
n=0 rn .z .
We are given: f (aj ) = ∞ n
n=0 rn .aj = bj for all j. Taking limits as j → ∞ on either
side and using uniform convergence, we see that f (0) = lim bj . But we are given
j→∞
bj
lim k = 0 for all k ≥ 0. Taking k = 0, we get lim bj = 0, hence r0 = f (0) = 0.
j→∞ aj j→∞
P∞ n b
Now write f (z) = n=1 rn .z . Proceeding as before and using lim ajk = 0 for k = 1,
j→∞ j
we get r1 = 0. Similarly rj = 0 for all j and hence f = 0 in a neighbourhood of 0.
Since f is entire, we get f = 0 everywhere and bn = 0 for all n.

(19*) Following Cayley’s theorem, use the action of G on the set of left cosets G/H. This
n
gives a homomorphism f : G → SN , where N = m = |G/H|.
|S |
Since |SN | = N ! < 2n, it implies that 2N < n = |G|. If f is injective, then by the
above inequality, f is actually an isomorphism. In this case, G is not simple because
SN is not, as it has the alternating group AN as a normal subgroup. On the other
hand, if f is not injective, it has a non-trivial kernel K ⊂ G. K is a proper nonzero
normal subgroup of G, hence again G is not simple.

(20*) Let φ : R → R be a continuous function such that |φ(x)| → ∞ as |x| → ∞. Then by
the formula for the limit of composition of two maps, one gets f ◦ φ ∈ C0 (R) for every
f ∈ C0 (R).
Suppose that f ◦ φ is infinitely differentiable for every f in C0∞ (R). Let a ∈ R. The
image φ[a − 1, a + 1] is a compact interval. Choose f ∈ C0∞ (R) such that f = 1 on
φ[a − 1, a + 1]. Then f ◦ φ = φ on (a − 1, a + 1). Thus φ agrees with the infinitely
differentiable function f ◦ φ on the neighbourhood (a − 1, a + 1) of a. Hence φ is
infinitely differentiable in a neighbourhood of a. As a is an arbitrary point of R, φ is
differentiable.

3