CMI Entrance Exam 2021 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE
Postgraduate Programme in Mathematics
MSc/PhD Entrance Examination
1st August 2021

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

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Part B
(11) Write 𝑍 for the centre of 𝐺. If 𝑍 ⊄ 𝑀, then 𝑀𝑍 > 𝑀 and by maximality 𝑀𝑍 = 𝐺. But then 𝑀
is a normal subgroup of 𝐺, a contradiction. Conversely h𝑀, 𝑁 i = 𝐺 by maximality. Also since 𝑀
and 𝑁 are abelian, we have
𝐺 = h𝑀, 𝑁 i ⊂ 𝐶𝐺 (𝑀 ∩ 𝑁 )
and hence (𝑀 ∩ 𝑁 ) ⊂ 𝑍 .

(12) 𝑓 (𝐷) ∩ 𝐷 is open in 𝐷 by Inverse mapping theorem. 𝑓 (𝐷) ∩ 𝐷 is also closed in 𝐷. Let 𝑧𝑛 = 𝑓 (𝑥𝑛 )
be a sequence in 𝑓 (𝐷) ∩ 𝐷 converging to 𝛼 ∈ 𝐷 with 𝑥𝑛 ∈ 𝐷. Then 𝛼 = 𝑓 (𝛽) for some 𝛽 ∈ 𝐷. But
the hypothesis ensures that 𝛽 ∈ 𝐷. 𝑓 (𝐷) ∩𝐷 is non-empty as 0 belongs to this. So 𝑓 (𝐷) ∩𝐷 = 𝐷.

(13) Union of connected sets having a point in common is connected. So 𝐴 is connected. Closure of
a connected set is connected. So 𝐴 is connected and, hence, 𝐴 ∈ S so 𝐴 ⊆ 𝐴.

(14) Without loss of generality, we may assume that Image(𝑓 ) ⊆ (0, ∞).
Let 𝜖 = 1. There exists a positive integer 𝑁 such that for all 𝑥, 𝑥 0 ∈ [1, ∞) with |𝑥 − 𝑥 0 | < 𝑁1 ,
|𝑓 (𝑥) − 𝑓 (𝑥 0)| < 1. Hence |𝑓 (𝑛 + 1) − 𝑓 (𝑛)| < 𝑁 for each positive integer 𝑛. Therefore 𝑓 (𝑛 + 1) <
𝑓 (1) + 𝑛𝑁 for each positive integer 𝑛. Hence
∑︁ 1 ∑︁ 1
≥
𝑛 ≥1
𝑓 (𝑛) 𝑛 ≥0 𝑓 (1) + 𝑛𝑁
which is a divergent series.

(15) On [0, 1] it is a continuous function. Need only show integral from 1 to infinity is finite. Change
∫∞
variable 𝑥 = 𝑦 100 , 𝑑𝑥 = 100𝑦 99𝑑𝑦 the integral becomes 100 𝑦𝑐 𝑒 −𝑦 𝑑𝑦 for some 𝑐 and if you take
1
∫∞
an integer 𝑁 > 𝑐 the integral is bounded by 𝑦 𝑁 𝑒 −𝑦 𝑑𝑦 which is finite.
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(16) False in both cases. Take 𝐹 = F3 the finite field of order 3 and 𝐼 1 = (𝑥 2 + 1) and 𝐼 2 = (𝑥 2 + 2𝑥 +
2). Both these are maximal ideals (check that the generating polynomials are irreducible since
they have no roots) and hence the quotients 𝑅/𝐼 1, 𝑅/𝐼 2 are both finite fields of order 9 and they
isomorphic. But 𝐼 1 ≠ 𝐼 2 . If 𝐹 = R, then for every irreducible quadratic polynomial 𝑓 (𝑋 ) ∈ 𝑅,
𝑅/(𝑓 (𝑋 )) ≃ C.

(17∗ ) (A) Let 𝑣 1, 𝑣 2 be the columns of 𝐴; they form a basis of R2 . Let 𝑢 1 = 𝑣 1 /|𝑣 1 |, 𝑎 = (𝑢 1 · 𝑣 2 ), 𝑣 20 =
𝑣 2 − 𝑎𝑢 1 , 𝑢 2 = 𝑣 20 /|𝑣 20 |. Since 𝑢 1 and 𝑢 2 form an orthonormal basis of R2 , define 𝐴𝑜 to be the
matrix with columns 𝑢 1, 𝑢 2 . Define 𝐴𝑏 to be the inverse of
 1   
|𝑣1 | 0 1 −𝑎 1 0
1 .
0 1 0 1 0 |𝑣20 |
Then 𝐴 = 𝐴𝑜 𝐴𝑏 .
(B) Suppose 𝐴 0𝐴 00 = 𝐴10𝐴100, with 𝐴 0, 𝐴10 ∈ O(2, R) and 𝐴 00, 𝐴100 ∈ B+ (2, R). Then 𝐴 00 (𝐴100) −1 ∈
O(2, R). Note that
𝑎 𝑏 𝑎0 𝑏 0
  0
𝑎𝑎 𝑎𝑏 0 + 𝑏𝑐 0
  
= ∈ O(2, R)
0 𝑐 0 𝑐0 0 𝑐𝑐 0
if and only if
 0 0     −1
𝑎 𝑏 𝑎 𝑏
= .
0 𝑐0 0 𝑐
Therefore 𝐴 00 = 𝐴100, and, hence, 𝐴 0 = 𝐴100.
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(C) By above, there is a well-defined function
𝜓 : GL(2, R) −→ O(2, R) × B+ (2, R) 𝐴 ↦→ (𝐴𝑜 , 𝐴𝑏 )
that is the inverse of 𝜙.
We first show that the map 𝐴 ↦→ (𝐴𝑏 ) −1 is continuous. From the description of (𝐴𝑏 ) −1
given above, we see that its entries are rational functions of the entries of 𝐴, with non-zero
denominators (since 𝐴 ∈ GL(2, R). Hence the map 𝐴 ↦→ (𝐴𝑏 ) −1 is continuous. Hence the
maps 𝐴 −→ 𝐴𝑜 = 𝐴(𝐴𝑏 ) −1 and 𝐴 −→ 𝐴𝑏 are continuous. Therefore 𝜓 is continuous.
𝜙 is continuous: The entries of 𝐴 0𝐴 00 are polynomial functions of the entries of 𝐴 0 and of 𝐴 00.

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(18∗ ) 𝜙 satisfies 𝑇 𝑝 − 1 = 0. Hence the only eigenvalue of 𝜙 is 1. Let 𝑣 1 be an eigenvector for the
eigenvalue 1. Then 𝜙 induces an invertible linear transformation of 𝑉 /h𝑣 1 i; proceed by induction
on dimension.

(19∗ ) (A) First,
√5 [Q(𝜁 5 ) : Q] = 4 since the polynomial 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1 is irreducible over Q. Next,
5
[Q( 2) : Q] = 5 since 𝑥 − 2 is irreducible over Q. For both use the Eisenstein criterion.
So [𝐾 : Q] is divisible by both 4 and 5. On the other hand, we have [𝐾 : Q(𝜁 5 )] ≤ 5. So we
conclude [𝐾 : Q] = 20.
(B) If 𝚤 ∈ Q(𝜁 5 ), then 𝛼 := 𝚤𝜁 5 ∈ Q(𝜁 5 ). Note that 𝛼 20 = 1 and 𝛼 𝑛 ≠ 1 for 1 ≤ 𝑛 ≤ 19. So 𝛼 is a
primitive 20th root of unity and it is contained in Q(𝜁 5 ). But we know that [Q(𝜁 5 ) : Q] = 4
and by the fact given in the hint, we have [Q(𝛼) : Q] > 4. So 𝚤 ∉ Q(𝜁 5 ).
(C) Suppose 𝚤 ∈ 𝐾. Consider the tower Q ⊂ Q(𝜁 5 ) ⊂ Q(𝜁 5, 𝚤) ⊂ 𝐾. The degree of the first
extension is 4; the degree of the second extension is 2 (since 𝚤 ∉ Q(𝜁 5 )). So [𝐾 : Q] is
divisible by 8. This is not possible since [𝐾 : Q] = 20.

(20∗ ) (A) Write 𝑓 (𝑧) =
Í 𝑛 . Then
𝑛 ≥0 𝑎𝑛 𝑧
𝑓 (𝑧) − 2𝑧 𝑓 (𝑧) − 𝑧 2 𝑓 (𝑧) = 𝑎 0 + 𝑎 1𝑧 + 𝑎 2𝑧 2 + 𝑎 3𝑧 3 + · · ·
− 2𝑎 0𝑧 − 2𝑎 1𝑧 2 − 2𝑎 2𝑧 3 + · · ·
− 𝑎 0𝑧 2 − 𝑎 1𝑧 3 − 𝑎 2𝑧 4 + · · ·
= 𝑎 0 + (𝑎 1 − 2𝑎 0 )𝑧
Let
𝑎 0 + (𝑎 1 − 2𝑎 0 )𝑧
𝑔(𝑧) =
1 − 2𝑧 − 𝑧 2 Í
This is analytic in a neighbourhood of the origin. Let 𝑛 ≥0 𝑏𝑛 𝑧𝑛 be the Taylor series expan-
sion of 𝑔(𝑧) around 0. Then (1 − 2𝑧 − 𝑧 2 )𝑓 (𝑧) = (1 − 2𝑧 − 𝑧 2 )𝑔(𝑧), from which we see that
𝑎𝑛 = 𝑏𝑛 for every 𝑛.
(B) Note that
1
𝑔(𝑧) = .
1 − 2𝑧 − 𝑧 2
Let 𝛾 1, 𝛾 2 be the roots of 1 − 2𝑧 − 𝑧 2 . Note that 𝛾 1 ≠ 𝛾 2 . Then there exist 𝛽 1, 𝛽 2 such that
𝛽1 𝛽2
𝑔(𝑧) = + .
𝑧 − 𝛾1 𝑧 − 𝛾2
Therefore
(−1)𝑛 𝑛!𝛽 1 (−1)𝑛 𝑛!𝛽 2
𝑔 (𝑛) (𝑧) = + .
(𝑧 − 𝛾 1 )𝑛+1 (𝑧 − 𝛾 2 )𝑛+1
Hence
𝛽1 𝛽2
𝑎𝑛 = − 𝑛+1 − 𝑛+1 .
𝛾1 𝛾2
1
Now replace 𝛽𝑖 by −𝛽𝑖 and 𝛾𝑖 by 𝛾𝑖 .
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