CMI Entrance Exam 2025 Question Paper Solution M.Sc PhD Mathematics

CHENNAI MATHEMATICAL INSTITUTE
Postgraduate Programme in Mathematics
MSc/PhD Entrance Examination
24th May 2025

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

PART B
(11) (A) ⇒ (B) : Let 𝐾 be a nontrivial minimal subgroup of 𝐺. Then 𝐾 ∩ 𝐻 is
nontrivial and by minimality 𝐾 ∩ 𝐻 = 𝐾. This gives the first statement of (B).
Let 0 ≠ 𝑥 ∈ 𝐺. Then ⟨𝑥⟩ ∩ 𝐻 ≠ 0, so that 𝑛𝑥 ∈ 𝐻 for a positive integer 𝑛. Here
⟨𝑥⟩ denotes the subgroup generated by 𝑥 in 𝐺. This gives the second statement of
(B). (B) ⇒ (A) : Let 𝐾 be a nontrivial subgroup of 𝐺. To prove that 𝐾 ∩ 𝐻 ≠ 0,
it is enough to show that ⟨𝑎⟩ ∩ 𝐻 ≠ 0 for some nonzero element 𝑎 ∈ 𝐾. So let
us assume that 𝐾 = ⟨𝑎⟩ is cyclic. If 𝐾 is finite then it will contain a minimal
nontrivial subgroup which is contained in 𝐻 by the first statement of (B). So
𝐾 ∩ 𝐻 ≠ 0. Suppose that 𝐾 is infinite. By hypothesis, 𝑛𝑎 ∈ 𝐻 for a positive
integer 𝑛. Since 𝑎 has infinite order, we conclude that 0 ≠ 𝑛𝑎 ∈ 𝐻 ∩ 𝐾.
(12) (A) Since 𝐼𝐽 ⊆ 𝐼 ∩ 𝐽, 𝑍(𝐼 ∩ 𝐽 ) ⊆ 𝑍(𝐼𝐽 ). To prove the containment in the other
direction, let 𝑎 ∈ 𝑍(𝐼𝐽 ) and 𝑓 ∈ 𝐼 ∩ 𝐽. We want to show that 𝑓(𝑎) = 0.
Note that 𝑓 2 ∈ 𝐼𝐽. Hence 𝑓(𝑎)2 = 0. Therefore 𝑓(𝑎) = 0.
(B) 𝐼(𝑎) is the kernel of the surjective ring homomorphism 𝒞(ℝ) ⟶ ℝ, 𝑓 ↦ 𝑓(𝑎).
(C) If 𝑓 and 𝑔 have compact support, so does 𝑓 + 𝑔. If 𝑓 has compact support
and 𝑔 is any continuous function, 𝑔𝑓 has compact support. Hence the given
set is an ideal. If it were not a proper ideal, it would contain the unit, i.e.,
the constant function 1, which is not compactly supported. For every 𝑎 ∈ ℝ,
there exists a compactly supported continuous function 𝑓 such that 𝑓(𝑎) ≠ 0.
(D) False. Choose a maximal ideal 𝔭 containing the ideal 𝐼 given in (C). Then
𝑍(𝔭) ⊆ 𝑍(𝐼) = ∅.
(13) The relation ℎ(𝑓(𝑥)+𝑔(𝑦)) = 𝑥𝑦, for all points 𝑥, 𝑦 ∈ ℝ imply that ℎ is surjective.
If 𝑓(𝑥) = 𝑓(𝑦) then 𝑥 = ℎ(𝑓(𝑥) + 𝑔(1)) = ℎ(𝑓(𝑦) + 𝑔(1)) = 𝑦, hence 𝑓 is injective.
Also since 𝑓 is continuous, 𝑓 is strictly monotone. If 𝑓 is bounded above then
lim𝑥↦∞ 𝑓(𝑥) = 𝛼 exists, but

ℎ(𝛼 + 𝑔(1)) = lim ℎ(𝑓(𝑥) + 𝑔(1)) = lim 𝑥 = ∞.
𝑥↦∞ 𝑥↦∞
1

So 𝑓 is unbounded above, and similarly unbounded below. Thanks to continuity
𝑓 maps onto ℝ. So 𝑓 is a bijection. So 𝑓(𝑥) + 𝑔(0) takes all values in ℝ. But
ℎ(𝑓(𝑥) + 𝑔(0)) = 0 for all 𝑥, which is a contradiction.
(14) ∫ 𝑓(𝑥)𝑔(𝑛𝑥)𝑑𝑥 = 𝑛1 ∫ 𝑓 ( 𝑛𝑦 ) 𝑔(𝑦)𝑑𝑦 = 𝑛1 ∑𝑛𝑘=1 ∫
1 𝑛 𝑘+1 1
𝑓 ( 𝑛𝑦 ) 𝑔(𝑦)𝑑𝑦 = 𝑛1 ∑𝑛𝑘=1 ∫ 𝑓 ( 𝑘+𝑧
𝑛 ) 𝑔(𝑧)𝑑𝑧,
0 0 𝑘 0
by periodicity of 𝑔. By replacing 𝑔 by 𝑔 + 𝑐 for some constant 𝑐, we can assume
𝑔 is non-negative. Using continuity and mean value theorem for integrals, we
can write the above sum as 𝑛1 ∑𝑛𝑘=1 𝑓 ( 𝑘+𝑧 𝑛 ) ∫ 𝑔(𝑧)𝑑𝑧 for 𝑧𝑘 ∈ (0, 1). Now
𝑘 1
0
converges to the Riemann integral
1 𝑛 𝑘+𝑧𝑘 1
𝑛 ∑ 𝑘=1
𝑓 ( 𝑛 ) ∫ 𝑓(𝑥)𝑑𝑥.
0
(15) (A) True. Let 𝑎 ∈ ℝ be such that 𝑓(𝑎) > 0 and 𝑏 ∈ ℝ be such that 𝑓(𝑏) < 0.
2 2

Let ℓ be the line perpendicular to the segment joining 𝑎 and 𝑏 passing through
the midpoint of the segment. For each 𝑝 ∈ ℓ, there is a straight-line path
joining 𝑎 with 𝑝 and then 𝑝 with 𝑏. On each such path there is a point 𝑞 such
that 𝑓(𝑞) = 0.
(B) True. Let 𝑈 = (𝑎, 𝑏). Suppose that 𝑓(𝑈 ) is open. Note that 𝑓(𝑈 ) is bounded.
Then 𝜕𝑓(𝑈 ) is uncountable. On the other hand, since 𝜕𝑈 has exactly two
elements, 𝜕𝑓(𝑈 ) is finite.
(16) (A) False. The point 𝑧 = 0 is an essential singularity. To see this, suppose
that for some 𝑛 ≥ 1, 𝑔(𝑧) = 𝑧 −𝑛 𝑓(𝑧) is such that lim𝑧→0 𝑔(𝑧) = ∞, then
𝑧 = 0 is a pole of 𝑔(𝑧). Let 𝑘 be the order of the pole of 𝑔(𝑧) at 𝑧 = 0
and write 𝑔(𝑧) = 𝑧 −𝑘 ℎ(𝑧) where ℎ(𝑧) is analytic in |𝑧| < 𝑟 for some 𝑟 ≤ 1
with ℎ(𝑧) ≠ 0. Then 𝑓(𝑧) = 𝑧 𝑛−𝑘 ℎ(𝑧). If 𝑛 ≥ 𝑘, then 𝑓(𝑧) is analytic at
𝑧 = 0 and hence continuous at 𝑧 = 0, contrary to our assumption on 𝑓(𝑧).
If 𝑛 < 𝑘, then 𝑛 − 𝑘 < 0. As ℎ(0) ≠ 0, we have lim𝑧→0 𝑓(𝑧) = ∞, again a
contradiction.
(B) False. Write 𝑎𝑛 for the fractional part of 𝑒𝑛!. Note that 𝑎𝑛 ⟶ 0. Hence
𝑓(𝑒2𝜋𝑖𝑎𝑛 ) = 0.
(17 ) (A) Write 𝛾(𝑡) = 𝐼𝑛 + 𝑡𝐴 + 𝑂(𝑡2 ) where 𝐴 = 𝛾 ′ (0) ∈ 𝑀𝑛 . Now consider 𝛾𝐴 .
∗

(B) Every tangent vector of 𝐺 is also a tangent vector of GL2 , so there exists
𝐴 ∈ 𝑀2 such that 𝛾𝐴 ′
(0) = 𝐴 is the tangent vector. 𝐼2 = (𝐼2 + 𝑡𝐴 +
𝑂(𝑡 )) (𝐼2 + 𝑡𝐴 + 𝑂(𝑡 )) = 𝐼2 + 𝑡(𝐴∗ + 𝐴) + 𝑂(𝑡2 ), so 𝐴∗ = −𝐴. Since
2 ∗ 2

det 𝛾𝐴 (𝑡) = 1 for all 𝑡, Tr 𝐴 = 0.
𝑖𝑎 𝑧
(C) Φ(𝐼2 ) = (1, 0, 0, 0) and the tangent space there is ℝ3 . Let 𝐴 = [ ]∈
−𝑧 ̄ −𝑖𝑎
𝑉. Then
1 + 𝑖𝑎𝑡 + 𝑂(𝑡2 ) 𝑧𝑡 + 𝑂(𝑡2 )
Φ(𝛾𝐴 (𝑡)) = Φ ([ ])
−𝑧𝑡̄ + 𝑂(𝑡2 ) 1 + 𝑖𝑎𝑡 + 𝑂(𝑡2 )
= (1 + 𝑂(𝑡2 ), 𝑎𝑡 + 𝑂(𝑡2 ), ℜ(𝑧)𝑡 + 𝑂(𝑡2 ), ℑ(𝑧)𝑡 + 𝑂(𝑡2 )).
Therefore (𝐷Φ)𝐼2 (𝐴) = (𝑎, ℜ(𝑧), ℑ(𝑧)). Note that the map (𝐴, 𝐵) ⟶ [𝐴, 𝐵]
is ℝ-bilinear. An ℝ-basis for 𝑉 is
𝑖 0 0 1 0 𝑖
{[ ],[ ],[ ]} .
0 −𝑖 −1 0 𝑖 0
Call these 𝐴1 , 𝐴2 , 𝐴3 respectively. (𝐷Φ)𝐼2 (𝐴𝑖 ) is the 𝑖th standard basis
vector 𝑒𝑖 of ℝ3 . Moreover, [𝐴1 , 𝐴2 ] = 𝐴3 , [𝐴2 , 𝐴3 ] = 𝐴1 , [𝐴3 , 𝐴1 ] = 𝐴2 .
Hence the induced multiplication on ℝ2 is the vector cross product.
(18 ) Since 𝑃 is irreducible, 𝔽𝑞 (𝛼) = 𝔽𝑞2𝑑 for each root 𝛼 of 𝑃. Thus the splitting field
∗

of 𝑃 over 𝔽𝑞 is 𝔽𝑞2𝑑 . One has that Gal(𝔽𝑞2𝑑 /𝔽𝑞 ) ≃ ℤ/(2𝑑), generated by the
2

Frobenius element 𝜎 ∶ 𝑥 ↦ 𝑥𝑞 . Since the Galois group is cyclic, all roots of 𝑃 are
of the form 𝜎𝑖 (𝛼), where 𝛼 is a chosen root. Then we set
𝑄1 ∶= ∏ (𝑥 − 𝜎𝑖 (𝛼))
𝑖even

and
𝑄2 ∶= ∏ (𝑥 − 𝜎𝑖 (𝛼)),
𝑖odd
we have that 𝑃 = 𝑄1 𝑄2 . Each of the polynomials 𝑄𝑖 have coefficients which are
symmetric polynomials in their roots, and it is easy to see that these coefficients
are fixed by 𝜎2 . The fixed field of ⟨𝜎2 ⟩ is 𝔽𝑞2 and thus 𝑄𝑖 has coefficients in 𝔽𝑞2 .
Since all roots of 𝑄𝑖 lie in a single Galois orbit for ⟨𝜎2 ⟩, 𝑄𝑖 is irreducible in 𝔽𝑞2 [𝑥].
(19∗ ) If 0 < 𝛼 < 1 is rational 𝜔 ∶= 𝑒2𝜋𝑖𝛼 is a root of unity. So (𝑟𝜔)𝑚! = 𝑟𝑚! for
all 𝑚 ≥ 𝑛 for 𝑛 ∶= 𝑜(𝜔). Summing from the the first 𝑁 th terms we obtain
𝑠 = ∑𝑚≥𝑛 𝑟𝑚! and it suffices to show this sum diverges to ∞. Consider the sum
up to 𝑚 + 1 terms of the above series: 𝑠𝑚 = 𝑟𝑛! + ⋯ + 𝑟𝑚! .
Let 𝑁 ≥ 1. Choose 𝑟 so that 𝑟𝑁! ≥ 1/2. This is possible since 𝑡 ↦ 𝑡𝑁! is
continuous at 1 and so lim𝑟→1− 𝑟𝑁! = 1. For this choice of 𝑟, we have 𝑠𝑁 ≥
(𝑁 − 𝑛 + 1)/2. As 𝑛 = 𝑜(𝜔) is fixed, we see that 𝑠𝑁 → ∞ as 𝑁 → ∞.
(20 ) (A) Cover ℝ2 by unit squares 𝑅𝑚,𝑛 with sides parallel to the coordinate axes
∗

and centres at (𝑚 + 1/2, 𝑛 + 1/2) with (𝑚, 𝑛) ∈ ℤ2 . It is easily seen that
the rectangles {𝑅𝑚,𝑛 ∣ (𝑚, 𝑛) ∈ ℤ2 } can be arranged in a sequence 𝑆0 =
𝑅0,0 , 𝑆1 , 𝑆2 , … , 𝑆𝑛 , … so that (i) 𝑆𝑘 ≠ 𝑆𝑙 if 𝑘 ≠ 𝑙, (ii) 𝑆𝑖 ∩ 𝑆𝑖+1 is a common
edge of both 𝑆𝑖 , 𝑆𝑖+1 .
Suppose that 𝑓0 ∶ 𝐼 → 𝐼 2 is a surjective continuous map. We may arrange so
that 𝑓0 (0) = (0, 0) = 𝑓(1). (Otherwise let 𝛼 ∶ 𝐼 → 𝐼 2 describe the straight-
line segment from 𝑓(0) to 0 and 𝛽 ∶ 𝐼 → 𝐼 2 , the straight-line segment joining
𝑓(1) to (0, 0) and consider the concatenation 𝑓 ∶= 𝛼 ⋅ 𝑓0 ⋅ 𝛽 ∶ 𝐼 → 𝐼 2 , which
describes 𝛼, 𝑓0 , 𝛽 with thrice the speed in succession. Then 𝑓 maps 0, 1 ∈ 𝐼
to (0, 0).)
Consider 𝜙 ∶ ℝ → ℝ2 is defined as follows: 𝜙(𝑡) = (0, 0) if 𝑡 ≤ 0, 𝜙(𝑡) =
𝑓0 (𝑡), 0 ≤ 𝑡 ≤ 1, and , inductively, having defined 𝜙 on [𝑗 − 1, 𝑗] such that
𝜙([𝑗 − 1, 𝑗]) = 𝑆𝑗−1 = 𝑅𝑚,𝑛 , with 𝜙(𝑗 − 1) = (𝑚, 𝑛), we proceed to define
𝜙 on [𝑗, 𝑗 + 1] as follows. By our hypothesis, 𝑆𝑗 = 𝑅𝑘,𝑙 where (𝑘, 𝑙) ∈
{(𝑚, 𝑛 + 1), (𝑚, 𝑛 − 1), (𝑚 + 1, 𝑛), (𝑚 − 1, 𝑛)}. Let 𝜎 ∶ 𝐼 → 𝑅𝑚,𝑛 ∪ 𝑅𝑘,𝑙 be
the straight line segment that joins (𝑚, 𝑛) to (𝑘, 𝑙). Then define
𝜎(2𝑡), if 0 ≤ 𝑡 ≤ 1/2
𝜙(𝑗 + 𝑡) = {
(𝑘, 𝑙) + 𝑓0 (2𝑡 − 1), if 1/2 ≤ 𝑡 ≤ 1
Then 𝜙 ∶ ℝ → ℝ2 is a continuous surjection. (Continuity follows from pasting
lemma and surjectivity follows from the fact that ⋃ 𝑅𝑚,𝑛 = ℝ2 .)
(B) The complement of the graph of 𝜙 is path connected. Suppose that both
𝑝0 = (𝑥0 , 𝑦0 , 𝑧0 ) , 𝑝1 = (𝑥1 , 𝑦1 , 𝑧1 ) are not in Γ. If 𝑥0 = 𝑥1 , then 𝑝0 , 𝑝1
belong the plane 𝑃 with equation 𝑥 = 𝑥0 and Γ ∩ 𝑃 is a singleton. So 𝑃 \Γ
is a punctured plane, which is path connected.
Suppose that 𝑥0 < 𝑥1 . Then Γ ∩ {(𝑡, 𝑦, 𝑥) ∣ 𝑥0 ≤ 𝑡 ≤ 𝑥1 } is compact. So
we can choose a 𝑐 > 0 large such that Γ is contained in the half space
{(𝑥, 𝑦, 𝑧) ∣ 𝑧 < 𝑐}. Now the straight-line segment joining 𝑞0 = (𝑥0 , 0, 𝑐) and
3

𝑞1 = (𝑥1 , 0, 𝑐) does not meet Γ. Applying first case (twice) we get a path
from 𝑝0 to 𝑞0 to 𝑞1 to 𝑝1 not meeting Γ.

4