CMI Entrance Exam 2024 Question Paper Solutions M.Sc PhD Mathematics

CHENNAI MATHEMATICAL INSTITUTE
Postgraduate Programme in Mathematics
MSc/PhD Entrance Examination
19th May 2024

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

Part B
(11) Write D2n = ⟨σ, τ | σ n = e, τ 2 = e, τ στ −1 = σ −1 ⟩.
Let Φ ∈ Aut(D2n ). Note that Φ is determined by Φ(σ) and Φ(τ ). Since Φ(σ)n = Φ(σ n ) = e, we find
that Φ(σ) has order dividing n. Since Φ is injective, Φ(σ d ) ̸= e for d < n. Hence, the order of Φ(σ) is exactly
n. Every element of D2n is either of the form τ σ j or σ i . Note that τ σ j has order 2. Since n > 2, we find that
Φ(σ) ̸= τ σ j . Therefore, there is an index i such that Φ(σ) = σ i . The order of σ i is n/ gcd(n, i), and therefore,
i is coprime to n. Thus, there are φ(n) choices for Φ(σ). The elements in D2n that have order 2 are all of the form
τ σ j , where j ∈ {0, 1, . . . , n − 1}, or if n is even, σ n/2 . Note that since Φ(σ) ∈ ⟨σ⟩, it follows that Φ would
fail to be surjective if Φ(τ ) = σ n/2 . Thus, Φ(τ ) is an element of the form τ σ j , with j ∈ {0, 1, . . . , n − 1}.
Hence | Aut(D2n )| ≤ nφ(n).
(12) (A)
ai + b (ai + b)(−ci + d) ac + bd + (ad − bc)i ac + bd + i
= = = ,
ci + d c2 + d2 c2 + d2 c2 + d2
2
so it is in H. Given x + yi ∈ H, choose a, b, d such that x = b/d, y = 1/d and ad = 1. Now
 
a b
f = x + yi.
0 d
(B) Let K be a compact subset of H. For each M ∈ SL(2, R), choose a compact neighbourhood V (M ) of
M . Since f is open and the sets f (V (M )), M ∈ SL(2, Sn R) cover K, a finitely many subcollection, say
f (V (M1 )), . . . , f (V (Mn )) will cover K. Let L = i=1 V (Mi ); it is a compact subset of SL(2, R).
Now f −1 (K) ⊂ L · SO(2, R). Now consider the multiplication map SL(2, R) × SL(2, R) −→
SL(2, R) which is continuous. Note that L · SO(2, R) is the image of the compact subset L × SO(2, R)
of SL(2, R) × SL(2, R). Hence L · SO(2, R) is compact, so the closed subset f −1 (K) is compact.
P −1
P3
(13) (A) g∈G sg = |{(g, H) ∈ G × S | gHg = H}| = i=1 |Ni |.
(B) Consider the action of G on S by conjugation. The orbit of H1 is equal to {H1 , H2 , H3 } by hypothesis.
So the action is transitive. Hence the order of each Ni is |G|/3. Note that se = 3 where e is the identity
element of G. So by the solution to (A), sg = 0 for some g.
(14) (A) For each x ∈ X, the set Yx := {f (x) : f ∈ S} is an ideal of Z. We are done if Yx ̸= Z for some x ∈ X.
If not, then Yx = Z for all x ∈ X and hence for every x ∈ X, there exists fx ∈ S such that fx (x) = 1.
Then Y
(fx − 1R ) = 0 ∈ R.
x∈X
Expand this to write 1R as a polynomial expression in {fx | x ∈ X}. Since every polynomial expression
in {fx | x ∈ X} belongs to S and S is an additive subgroup of R, it follows that 1R ∈ S. This is a
contradiction. Hence we have Yx ⊊ Z for some x ∈ X.
(B) Assume that P is reducible in K[x]. Let g be an irreducible factor of P in K[x]. It suffices to show that
deg g = (deg P )/2. Let α a root of g. Now
[K(α) : K][K : Q] = [K(α) : Q] = [K(α) : Q(α)][Q(α) : Q],
1

so 2 deg(g) = deg(P )[K(α) : Q(α)]. Since deg g < deg P , it follows that [K(α) : Q(α)] < 2 so
[K(α) : Q(α)] = 1. Hence deg(g) = deg(P )/2.
A(z)
(15) Write f (z) for the meromorphic function B(z) .
We first note that
A(αi )
= Resαi f (z).
B ′ (αi )
Hence we need to show that
X k
Resαi f (z) = 0
i=1
A(X)
Solution 1: Expand B(X) as partial fractions
k
A(X) X ai
=
B(X) i=1
X − αi
Qk
We can assume that B is monic, i.e., B = i=1 (X − αi ). Hence
k
X Y
A(X) = ai (X − αj ).
i=1 j̸=i
Pk
from which it follows that the coefficient of X k−1 in A(X) = i=1 ai . Since deg A(X) < k − 1, it follows
Pk
that i=1 ai = 0. On the other hand,
 
k
X aj 
ai = Resαi  .
j=1
z − αj

Hence
k
X k
X
Resαi f (z) = ai = 0.
i=1 i=1
Solution 2:
For each real number R > maxi |αi |, define
Z
IR := f (z)dz,
CR

where CR is the circle of radius R with centre at 0, oriented counter-clockwise. Hence
k
X
IR = 2πi Resαi f (z).
i=1

On the other hand, since deg A(X) < deg B − 1, it follows that
Z
2π
|IR | ≤ |f (z)|dz ≤
CR R
Hence
k
X
Resαi f (z) = 0.
i=1
(16) (A) Let r < r′ be rational numbers and write r − r′ = m n for some positive integers m, n. Then f (r) =
f (r + n1 ) = f (r + n2 ) = · · · = f (r′ ). Therefore there exists c ∈ R such that f (r) = c for all r ∈ Q.
Now let r ∈ R. Then there exists a sequence rk ∈ Q, k ≥ 1 converging to r. Since f is continuous,
f (r) = limk f (rk ) = c. Hence f is a constant function.
(B) Let α = inf n ann . Then for any ϵ > 0, there exists N such that aN < N (α+ϵ). Let β = max{a1 , · · · aN }.
Let n > N . Write n = N q + r with 0 ≤ r < N . By the sub-additivity of an ,
an ≤ qaN + ar ≤ qaN + β
and hence
an qaN β qN (α + ϵ) β
α≤ ≤ + < + −→ α + ϵ
n n n n n
2

since qN
n −→ 1 as n −→ ∞, Hence
an
lim = α.
n
(17∗ ) Consider the set X = {(v1 , v2 ) | v1 , v2 are linearly independent vectors in V }. Then |X| = (pn −1)(pn −p).
For a two-dimensional subspace W of V , define XW := {(v1 , v2 ) ∈ X | v1 , v2 is a basis of W }. For each
two-dimensional subspace W of V , GL2 (Fp ) acts transitively and without fixed points on XW . Moreover X =
∪W XW where W runs over all the two-dimensional subspaces of V . Hence the orbits of the action on X are
in bijective correspondence with the set of two-dimensional Fp -subspaces of V . Hence the number of two-
dimensional Fp -subspaces of V is
(pn − 1)(pn − p)
.
(p2 − 1)(p2 − p)
(18∗ ) Let s, t : N2 −→ R be the restrictions of the first and second projections R2 −→ R. We now show that the
R-subalgebra of R generated by s, t is isomorphic to a polynomial ring in two variables. To do this, it suffices to
show that the natural map R[X, Y ] −→ R, X 7→ s, Y 7→ t is injective. Let f (X, Y ) be in the kernel of this
map, i.e, f (s, t) ≡ 0. We want to show that f (X, Y ) is the zero polynomial. By way of contradiction, assume
Pd
that it is non-zero. Write f (X, Y ) = k=0 fk (X)Y k for some suitable d. For each 0 ≤ k ≤ d, fk (X) has
only finitely many zeros. Hence there exists n ∈ N such that f (n, Y ) is a non-zero polynomial. Therefore there
exists m ∈ N such that f (n, m) ̸= 0. Thus f (s, t) is non-zero at (n, m), a contradiction. Therefore f (X, Y )
is the zero polynomial.
(19∗ ) We show that the set S is closed and bounded. First, we show that S is bounded. Let λ ∈ S, there exists a
non-zero vector v ∈ Rn and a matrix A = (ai,j ) ∈ X such that Av = λv. Then, we find that
X
|λ||vi | = | ai,j vj | ≤ max{|ai,j | : 1 ≤ i, j ≤ n} × max{|vj | | 1 ≤ j ≤ n}.
j

Taking the maximum value of |vi |, we thus find from the above that
|λ| ≤ max{|ai,j | : 1 ≤ i, j ≤ n}.
2
Since X is a compact subset of Mn (R) ≃ Rn , it is bounded. Hence, there exists D > 0 such that max{|ai,j | :
1 ≤ i, j ≤ n} ≤ D for all A ∈ X. Thus, we have shown that S is bounded.
In order to show that S is closed, take a sequence λ1 , λ2 , . . . , λi , . . . in S which converges to λ ∈ C. We
show that λ ∈ S. For each λi , there is a non-zero vector vi and Ai ∈ X such that Ai vi = λi vi . Assume
without loss of generality that |vi | = 1 for all i. Since X is compact, there is a subsequence Ani such that Ani
converges to A ∈ X. Since vni all have norm 1, it follows that after passing to a subsequence if necessary, we can
assume without loss of generality that vi converge to a vector v of norm 1. Thus, we find that Av = λv. Since
A ∈ X, it follows that λ ∈ S. This shows that S is closed. Being a closed and bounded subset of C, we find that
S is compact.
2
(20∗ ) (A) |y + (Aψ)(y)| = |y − (y + ψ(y))2 | ≤ |y| + |(y + ψ(y))2 | ≤ ϵ + λ4 ≤ λ2 .
(B) If we compose the two functions, we get the identity map. More precisely, y = x + x2 = (y + ψ(y)) +
(y + ψ(y))2 = (y + ψ(y)) − ψ(y) = y.
(C)
d(Aψ1 , Aψ2 ) = sup{|(y + ψ1 (y))2 − (y + ψ2 (y))2 | : y ∈ [−ϵ, ϵ]}
= sup{|(2y + ψ1 (y) + ψ2 (y))(ψ1 (y) − ψ2 (y))| : y ∈ [−ϵ, ϵ]}
≤ λd(ψ1 , ψ2 )
(D) Let n, k be positive integers. Then

d(An ϕ, An+k ϕ) ≤ λd(An−1 ϕ, An−1+k ϕ) ≤ . . . ≤ λn d(ϕ, Ak ϕ) ≤
λn d(ϕ, Aϕ) + d(Aϕ, A2 ϕ) + d(A2 ϕ, A3 ϕ) + · · · + d(Ak−1 ϕ, Ak ϕ) ≤



λn
λn 1 + λ + · · · + λk−1 d(ϕ, Aϕ) ≤ λn (1 + λ + · · · )d(ϕ, Aϕ) =


d(ϕ, Aϕ).
1−λ
Hence this is a Cauchy sequence. Since X is complete, the sequence has a limit. (Proof that X is complete:
It suffices to show that X is closed in C 1 ([−ϵ, ϵ]), since closed subsets of complete spaces are complete.
Consider the continuous function
F : C 1 ([−ϵ, ϵ]) −→ C 1 ([−ϵ, ϵ]) ϕ 7→ id + ϕ.
3

Composing this with the sup-norm function gives a continuous map G : C 1 ([−ϵ, ϵ]) −→ R. Then
X = G−1 ([0, λ2 ]).)
(E) Let ϕ ∈ X. Since A is continuous, we see that
 
A lim An ϕ = lim An+1 ϕ = lim An ϕ.
n n n
n
Hence take ψ = limn A ϕ.

4