CMI Entrance Exam 2026 Question Paper Solution M.Sc or Ph.D Mathematics

FOR CMI EXAM PREPARATION

CMI 2026
Question Paper
Solution · M.Sc or Ph.D
Mathematics
EXAM YEAR TYPE SUBJECT

CMI 2026 Question Paper Solution M.Sc or Ph.D Mathematics

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CHENNAI MATHEMATICALaINSTITUTE
Postgraduate Programme in Mathematics
MSc/PhD Entrance Examination
2nd May 2026
Important information and instructions:
(1) Questions in Part A (Questions 1 – 10) will be used for screening. There will be a
cut-off for Part A, which will not be more than 20 marks (out of 40).

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(2) Each question in Part A has one or more correct answers. Enter your answers to

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these questions into the computer as instructed. Every question is worth 4 marks. A solu-
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tion will receive credit if and only if all the correct answers are chosen, and no incorrect answer
is chosen.
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(3) Your solutions to the questions in Part B (Questions 11– 20∗ ) will be marked only
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if your score in Part A places you over the cut-off. (In particular, if your score in Part A
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is at least 20 then your solutions to the questions in Part B will be marked.)
(4) Answer 6 questions from Part B, on the pages assigned to them, with sufficient
justification. Each question is worth 10 marks. Clearly indicate which six questions you
would like us to mark in the six boxes on the front sheet. If the boxes are unfilled, we
will mark the first six solutions that appear in your answer-sheet. If you do not want a
solution to be considered, clearly strike it out.
(5) The scores in both the sections will be taken into account while making the final
decision. You are advised to spend at least 90 minutes on Part B. In order to qualify
for the PhD Mathematics interview, you must obtain at least 15 marks from among the
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starred questions 17∗ –20∗ .

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(6) Time: 3 hours.

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Notation: N, Z, Q, R, and C stand, lrespectively,
of integers, of rational numbers, of ag for the sets of non-negative integers,
real numbers, and of complex numbers. For a field F
and a positive integer n, M (F ) stands for the set of n × n matrices over F and GL (F )
n n
stands for the set of invertible n×n matrices over F . The n×n identity matrix is denoted
by In ; the field will be clear from context. When considered as topological spaces, Rn or
Cn are taken with the euclidean topology, unless otherwise stated.

Part A
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s e (1) Let X be a metric space. A subset A ⊂ X is said to be totally
 > 0, there exist a , . . . , a ∈ A such that A is contained g
l a bounded if for every

g la 1 n
a in the union of the balls
a of radius  centered at a , . . . , a . Pick the correct statement(s) from below.
1 n
(A) Every bounded set in Rn is totally bounded.
(B) Suppose that every sequence in X has a Cauchy subsequence. Then X is
totally bounded.
(C) Let f : X −→ Y be a continuous map of metric spaces. If A ⊂ X is bounded,
then f (A) is bounded in Y .
(D) Let f : X −→ Y be a uniformly continuous map of metric spaces. If A ⊂ X
is totally bounded, then f (A) is totally bounded in Y .
Solution:A, B, D. End Solution
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(A) True. Let A be bounded. Then the closure Ā of A is bounded and closed.
So Ā is compact, so totally bounded. Then so is A since it is a subset of a
totally bounded set.
(B) True. If A is not totally bounded, we can construct a sequence without a
Cauchy subsequence as follows. Choose  > 0 such that finitely many -
balls do not cover X. Choose x1 such that the -ball B1 around x1 does
not cover X. Then choose x2 6∈ B1 . If B2 is the -ball around x2 , then
X 6= B1 ∪ B2 . Choose x3 6∈ B1 ∪ B2 . Note that d(x1 , x2 ) ≥ , d(x2 , x3 ≥ 
and d(x2 , x3 ) ≥ . Continuing this way, we construct a sequence x1 , x2 , x3 , . . .
such that d(xn , xm ) ≥  for every m 6= n. So this sequence does not have a
Cauchy subsequence.
(C) False: let f : (0, 1) −→ R, f (x) = 1/x.
(D) True. Given  > 0, choose δ > 0 such that d(a, b) < δ ⇒ d(f (a), f (b)) < .
Choose a finite set of points a1 , . . . , an ∈ A such that δ-balls around ai cover
A. Then -balls around f (ai ) cover f (A).

(2) Let V be a complex vector space. A linear operator P : V −→ V is called a
projection if P 2 = P . Pick the correct statement(s) from below.
(A) Let P be a projection operator on V . Then the rank of P is equal to the
trace of P .
(B) Let P be a projection operator on V . Then 0 is an eigenvalue of P .
(C) Let T be a linear operator on V which commutes with every projection
operator. Then every non-zero vector v ∈ V is an eigenvector of T .
(D) Let T be a linear operator on V which commutes with every projection
operator. Then T is a scalar multiple of the identity operator.
Solution:A, C, D. End Solution
(A) is true. Since P 2 = P , P is diagonalizable and the only eigenvalues are 0 and
1. So the trace of P is the multiplicity of the eigenvalue 1 and it is also the
rank of P .
(B) is false. Take P as the identity operator.
(C) is true: Let v ∈ V be non-zero. Consider the projection P on to the span of
v. Then P v = v. So T v = T P (v) = P T (v) = λv v. So v is an eigenvector
with eigenvalue λv .
(D) is true: we claim that every non-zero vector has the same eigenvalue. Let
v, w ∈ V be linearly independent vectors. Then T (v +w) = T v +T w = λv v +
λw w. Since v + w is an eigenvector of T with eigenvalue λv+w , λv+w (v + w) =
λv v + λw w. Hence λv+w = λv = λw .

(3) Consider the function f : R2 \ {(0, 0)} −→ R defined by
xy
f (x, y) = .
x2 + xy + y 2

Pick the correct statement(s) from below.
(A) f is everywhere differentiable and extends differentiably to the whole of R2 .
(B) f is everywhere differentiable and extends continuously but not differentiably
to the whole of R2 .
(C) f is everywhere differentiable but does not extend continuously to the whole
of R2 .
2

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(D) f is everywhere differentiable and extends toaa function g on the whole of R 2

such that both partial derivatives of g exist at (0, 0).
Solution:C, D. End Solution
f is a ratio of two polynomials and hence differentiable on R2 \ {0, 0}. Note
that x2 + xy + y 2 = (x + y/2)2 + 3y 2 /4 6= 0 for (x, y) 6= (0, 0). f does not have a
limit as (x, y) −→ (0, 0 (check the limits along the lines y = x and y = 0. So (A),
(B) are false and (C) is true. (D) is true if we define f (0, 0) = 0.

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(4) Let F be a field of order 212 .
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Pick the correct statement(s) from below.

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(A) F has exactly 6 subfields, including F .
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(B) Every element of F is a square (that is, for every x ∈ F , there exists an

g l aelement y ∈ F such that y 2 = x).
a(C) If K and L are subfields of F of order 8 and 64 respectively, then K ∪ L is a
subfield of F .
(D) F contains a subfield of order 128.
Solution:A, B, C. End Solution
(A) is true. Subfields of F have order 2r where r is a divisor of 12. So r =
1, 2, 3, 4, 6, 12.
(B) is true. Consider the homomorphism F \ {0} −→ F \ {0} given by x 7→ x2 .
The kernel of this map is trivial (since the characteristic of F is 2, 1 = −1).

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So this map is injective, hence bijective.

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(C) is true. Since 8 = 23 , 64 = 26 and 3 divides 6. So K ⊂ L and K ∪ L = L is a

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field.

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(D) is false. 128 = 27 and 7 does not divide 12.

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(5) Let G be a finite group ofaorder p q, where p < q are primes. Which of the
2

following statements are always true?
(A) If p is odd, then G has a normal Sylow q-subgroup.
(B) If p is odd, then G has a normal and abelian subgroup N such that G/N is
abelian.
(C) If p divides q − 1, then G has at least two Sylow p-subgroups.
(D) G is abelian.
Solution:A, B. End Solution
Let |G| = p2 q, with p < q.
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(A): we are given that p is odd.

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Let nq be the number of Sylow q-subgroups. By Sylow’s theorems:
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nq | p2 .
a
nq ≡ 1 (mod q),

ag Since nq | p2 , the possibilities are 1, p, p2 .
But p < q, so p is not congruent to 1 mod q. So nq 6= p.
If nq = p2 , then q divides p2 − 1 = (p + 1)(p − 1). Since p < q we must have
q = p + 1. But this forces p = 2 contradicting the hypothesis. So nq = 1.
Thus the Sylow q-subgroup is normal.
(B) Let N / G be the Sylow q-subgroup which is normal by the above argument.
Then
|G/N | = p2 ,
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so G/N is abelian.
(C) Take p = 2, q = 3 and G = A4 , the alternating group on 4 letters.. The
Sylow 2-subgroups of A4 have order 4. A4 has three elements of order 2 (products
of two disjoint 2-cycles) and eight elements of order 3 (3-cycles). So there is only
one subgroup of order 4. (C) is false.
(D) Take p = 2, q = 3 and G = A4 . G is not abelian.
Thus the correct answers are: (A), (B).

(6) Let (an )n≥1 be a sequence of positive real numbers. For each k ∈ N, define
Sk : R −→ R by
k
X
Sk (x) = an sin(nx).
n=1
Let D = {x ∈ R : limk−→∞ Sk (x) exists in R}, and f : D −→ R be defined by
f (x) = limk−→∞ Sk (x).
Pick the correct statement(s) from below.
(A) If an = √1n , then D = R.
(B) If an = n12 and g : (0, 1] −→ R is defined by g(x) = SN (x) (x), where N (x) is
the greatest integer less than or equal to 1/x, then lim supx−→0+ g(x)
x
< ∞.
(C) If an = n13 , then f is Lipschitz continuous on D, i.e.,
( )
|f (x) − f (y)|
sup : x, y ∈ D, x 6= y < ∞.
|x − y|
(D) If an = n1 , then f is continuous on (0, 2π) ∩ D.
Solution:A, C, D. End Solution
(A) True. Note that Sk (x) is the k-th partial sum of the series ∞n=1 an sin(nx).
P

If x ∈ 2πZ, then all terms vanish. Assume x ∈ / 2πZ. Using the geometric
series
k
1 − eikx
einx = eix
X
,
n=1 1 − eix
we obtain
k
2 1
einx ≤
X
|Sk (x)| ≤ ix
= .
n=1 |1 − e | | sin(x/2)|
Since an ↓ 0, by Dirichlet’s test, the series converges for all x ∈ R.
(B) False. Let us fix x ∈ (0, 1] and denote N (x) simply by N .
N
g(x) X sin(nx) 1
= .
x n=1 nx n
For 1 ≤ n ≤ N , we have 0 < nx ≤ N x ≤ 1. Since t 7→ sint t is continuous on
[0, 1], we have sin(nx)
nx
≥ mint∈[0,1] sint t := c > 0. Thus
N (x)
g(x) X 1
≥c .
x n=1 n

Since the harmonic series diverges, we have lim supx−→0+ g(x)
x
= ∞.
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(C) True. Since ∞
P
n=1 n
∞
a < ∞, the series
P
aa sin(nx) converges
n=1 n uniformly
∞
for all x ∈ R by the Weierstrass M-test. We note that n=1 nan < ∞. By
P

the mean value theorem, for all x, y ∈ R,
|an (sin(nx) − sin(ny))| ≤ an · n|x − y|.
Summing,
∞
X
|f (x) − f (y)| ≤ |x − y| nan ,
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n=1

m
so f is Lipschitz continuous.

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(D) True. The argument of part (A) implies D = R. Let x0 ∈ (0, 2π). There
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exists δ > 0 such that x0 ∈ Iδ = [δ, 2π − δ]. It is enough to show that f
l a
g l a
is continuous on Iδ . We apply Dirichlet’s test for uniform convergence on
ag
a Iδ . The sequence an = n1 decreases monotonically to 0. By part (A), for all
x ∈ Iδ , the partial sums of sin(nx) satisfy
N
X 1
sin(nx) ≤ .
n=1 sin(δ/2)
Thus, the partial sums are uniformly bounded on Iδ . By Dirichlet’s test,
sin(nx)
the series ∞ converges uniformly on Iδ . Because each term sin(nx)
P
n=1 n n
is continuous on R, the uniform convergence ensures that the limit function
f (x) is continuous on Iδ .
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(7) Let k be a field and let R = k[X]/(X n ) with n ≥ 2. Which of the following
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statements about R-modules are always true?
(A) Every finitely generated R-module is a direct sum of cyclic modules.
(B) Every R-module is free.
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(C) Every cyclic R-module is isomorphic to R or R/(X ) for some 1 ≤ m ≤ n.
(D) Every R-module is cyclic.
Solution:A, C. End Solution
n
Write X for the image of X in R. Then X = 0, so multiplication by X defines
a nilpotent operator on any finitely generated R-module M .
Every finitely generated R-module M is a finite-dimensional vector space over
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k, equipped with a nilpotent linear operator given by multiplication by X. By
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linear algebra, such a module decomposes into a direct sum of Jordan blocks.

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Each Jordan block corresponds to a module of the form
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s e R/(X ).
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a From the above, every finitely generated R-module is a direct sum of cyclic
m
modules of the form R/(X ). (A) is true.
(B) is false. Counterexample: Consider the ideal (X) of R which has a natural
structure of an R-module. It is not free: it were free, it would be isomorphic to
R, but it contains torsion elements since
n−1
X ·X = 0.
Thus it is not free.
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Let M = Rf be a cyclic module. Then M ∼
= R/ Ann(f ). Since every ideal of
m
R is of the form (X ), we get

m
M∼
= R or R/(X )

for some 1 ≤ m ≤ n. Thus, (C) is true.
(D) is false: for example, the R-module R/(X) ⊕ R/(X) is not cyclic.

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(8) Let f (z) = ∞ n=0 an z be a power series with a radius of convergence exactly equal
P

to 1. Suppose f (z) can be analytically continued to a meromorphic function on
the open disk D = {z ∈ C : |z| < 2}, and its only singularity in this region is a
c−2 c−1
pole of order 2 at z = 1 with principal part (z−1) 2 + z−1 , where c−2 , c−1 6= 0.

Pick the correct statement(s) from below.
(A) limn−→∞ (an+1 − an ) = 0.
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(B) The series ∞ n=0 |an |(3/2) converges.
P

(C) Let Γ be the contour |w| = 3/2 traversed counterclockwise. Then

1 I
f (w) dw 6= 0.
2πi Γ

(D) Let Rn be the radius of convergence of the Taylor expansion of f (z) centered
at zn = eiπ/n . Then limn−→∞ nRn = π.
Solution:C, D. End Solution
c−2 c−1 P∞ n
(A) False. We can write f (z) = h(z) + (z−1) 2 + z−1 , where h(z) = n=0 bn z is
holomorphic on |z| < 2. Using the power series for (z − 1)−1 and (z − 1)−2 ,
we obtain an = c−2 (n + 1) − c−1 + bn . Since the radius of convergence of h(z)
is at least 2, bn −→ 0. Thus, limn−→∞ (an+1 − an ) = c−2 6= 0.
(B) False. The radius of convergence of the series ∞ |a |z n is equal to the
P
P∞n
n=0
n
radius of convergence of n=0 an z , which is 1. So n=0 |an |(3/2)n cannot
P∞

converge.
(C) True. The integral is equal to c−1 6= 0 by Cauchy’s Residue Theorem.
(D) True. The distance from zn = eiπ/n to the nearest singularity z = 1 is
π
Rn = |eiπ/n − 1| = 2 sin( 2n ). Therefore, limn−→∞ nRn = π.

(9) Let {an }n≥0 be a sequence of real numbers such that limn−→∞ an = 0. Pick the
correct statement(s) from below.P
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(A) The limit limx−→1− ((1 − x) ∞ n=0 an x ) as x approaches 1 from below exists
and is equal to 0.
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(B) The limit limx−→1+ ((1 − x) ∞ n=0 an x ) as x approaches 1 from above exists
P

and is equal to 0.
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(C) The function f (x) = (1 − x) ∞ n=0 an x is defined and continuous in the open
P

interval (0, 1).
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(D) If the function f (x) = (1 − x) ∞ n=0 an x is defined and continuous in the
P

open interval (0, 1), then |an | ≤ 1/n for all n ≥ 1.
Solution:A, C. End Solution
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a |a | < /2. For any M > N
Let  > 0 be given. There exists N ∈ N such that n
and 0 < x < 1, we have the estimate
M N M
an xn | ≤ |(1 − x) an xn | + |(1 − x) an xn |
X X X
|(1 − x)
n=0 n=0 n=N +1
N M
 n
an xn | + |(1 − x)
X X
≤ |(1 − x) x |
n=0 n=N +1 2

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N

m an xn | + /2.
X
≤ |(1 − x)

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| ≤ /2 for (1 − x) < δ. So se
n=0

m (A) holds. Statement (C) holds because a x is a power seriesla
ethat
Clearly, there exists δ > 0 such that |(1 − x) a xP N n
P
n

s
n=0

a g n
∞
one sees
l a
n=0 n

ag(B) and (D) are false for a = 1/ n.
(centered at 0) that converges in (0,√1) by a similar estimate to the one above.
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(10) A sequence {Ck } of real 2 × 2 matrices is said to converge to a 2 × 2 matrix C,
written Ck −→ C, if the entries of Ck converge to the corresponding entries of
C. Now suppose that A is a real 2 × 2 matrix such that the sequence {Ak }k≥1
converges to the zero matrix: Ak −→ 0.
Pick the correct statement(s) from below.
(A) The eigenvalues of A are always real.
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(B) The eigenvalues λj of A must satisfy |λj | < 1, j = 1, 2.
(C) The eigenvalues of A P must be distinct.
k
e m
s
(D) The geometric series ∞ k=0 A converges to (I − A) .
−1

Solution:B, D. End Solution
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!
0 − 12
(A) is false: take A = 1 .
2
0
(B) is true: let λ ∈ C be an eigenvalue of A and let v be the corresponding
eigenvector. So Ak v = λk v converges to the zero vector as k −→ ∞. So
λk −→ 0. !
1
0
(C) is false: take A = 2 1 .
0 2
(D) is true: let Sk denote the sequence of partial sums: Sk = I + A + · · · + Ak .
Then (I − A)Sk = I − Ak+1 . So (I − A)Sk converge to I andPhence the
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series k (I − A)Ak converges to I. This shows that the series k Ak also
P

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converges, say to B. Then (I − A)B = I and hence B = (I − A)−1 .

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a Part B

(11) Let k = Z/pZ where p is a prime. Determine all the ring automorphisms of the
polynomial ring k[X].
Solution:
Step 1: Action on constants.
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Since ϕ is a ring automorphism, it restricts to an automorphism of the subring
k. But k = Z/pZ has no nontrivial automorphisms, so
ϕ(a) = a for all a ∈ k.

Step 2: Determination by the image of X.
The automorphism ϕ is completely determined by ϕ(X), since every polynomial
is a polynomial in X with coefficients in k.
Let
ϕ(X) = f (X) ∈ k[X].
Step 3: Degree considerations.
Since ϕ is an automorphism, it must be surjective. This implies that
deg(ϕ(X)) = 1.
Thus f (X) must be a linear polynomial:
f (X) = aX + b, a, b ∈ k.

Step 4: Invertibility condition.
For ϕ to be invertible, we must have a 6= 0. Indeed, if a = 0, then ϕ(X) = b is
constant, and ϕ is not injective.
Conversely, if a 6= 0, define
ψ(X) = a−1 (X − b).
Then one checks that ψ is the inverse of ϕ, so ϕ is an automorphism.
Step 5: Conclusion.
All ring automorphisms of k[X] are of the form
ϕ(X) = aX + b, a ∈ k × , b ∈ k.
Thus,
Aut(k[X]) ∼= {X 7→ aX + b : a ∈ k × , b ∈ k},
which is the affine group over k.
End Solution

(12) Let f (z) be a non-constant meromorphic function on C with only finitely many
poles. Assume that lim f (1/z) = 2026. Let m denote the number of poles of f
z−→0
and let t be the order of the zero of f (z) − 2026 at ∞. Determine the number of
solutions (counted with multiplicity) to f (z) = 2026 in terms of m and t.
Hint: First show that f is a rational function.
Solution:
Step 1: f is a rational function. Proof: Let a1 , . . . , am be the distinct poles of
f . For each 1 ≤ i ≤ m, write
−1
X bi,k
f (z) = + fi (z)
k=−Ni z − ai

where fi (z) is holomorphic in a neighbourhood of ai . Write
m −1
X X bi,k
P (z) = .
i=1 k=−Ni z − ai
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a
Then f (z) − P (z) does not have poles on C. Further,
lim [f (1/z) − P (1/z)] = c,
z−→0

so f (z) has a removable singularity at ∞. Hence f is a rational function.
Step 2: Write
P (z)
f (z) =
Q(z)
where P and Q are polynomials and Q is monic. Therefore f (z) = 2026 if and
m
m
only if P (z)−2026Q(z) = 0. Let n = deg P, m = deg Q. Since lim f (1/z) = 2026
.co
.co m
z−→0
is finite, n = m.
m s e
e
Step 3: Let t be the order of the zero of f (z) − 2026 at ∞. Hence deg(P (z) −
s l a
a ag
2026Q(z)) = m − t. Therefore there are exactly m − t solutions (counted with
g l
multiplicity) for the equation f (z) = 2026. 
a End Solution

(13) Let A be a set (possibly uncountable). For each finite subset F of A, let SF
be a non-empty set of functions from F to {0, 1}. Assume that the collection
{SF | F ⊆ A, F finite} is consistent, i.e., {s|F | s ∈ SG } = SF for every pair
F ⊆ G of finite subsets of A. Show that there exists a function f : A −→ {0, 1}
such that f |F ∈ SF for every finite subset F of A.
Solution:Step 1: For each α ∈ A, let
m
Xα = {0, 1}
.co
with the discrete topology. Let
s eYm
g Xla= X .
a
α
α∈A

with the product topology. It is compact by Tychonoff’s theorem.
Step 2: For each finite F ⊂ A, define
CF = {f ∈ X | f |F ∈ SF }.
Each CF is closed and non-empty.
m
[ m
\
Step 3: Let F1 , . . . , Fm be finite subsets of A and let F = Fi . Then CFi =
i=1 i=1
CF 6= ∅. Since X is compact, there exists
m
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\

m
f∈ CF .

m
F ⊆A
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F finite

las Then f |F ∈ SF for every finite subset F of A.
ag 

ag End Solution

(14) (A) (5 marks) Let M2 (R) ∼ = R4 be the space of real 2 × 2 matrices with the Eu-
clidean topology. Show that the set of matrices with two distinct eigenvalues
(in C) is open in M2 (R) .
(B) (5 marks) Let n be a positive integer. Let V be an n-dimensional vector space
over a finite field with q elements. Calculate the number of k-dimensional
subspaces of V for 1 ≤ k ≤ n.
Solution:
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(A) Let
!
a b
A= .
c d
The characteristic polynomial of A is
pA (t) = t2 − (a + d)t + (ad − bc).
The eigenvalues are distinct if and only if the discriminant of this quadratic
is nonzero:
∆(A) = (a + d)2 − 4(ad − bc).
Simplifying,
∆(A) = (a − d)2 + 4bc.
Thus A has two distinct eigenvalues if and only if
∆(A) 6= 0.
Now observe that ∆(A) is a polynomial in the entries a, b, c, d, hence defines
a continuous function
∆ : M2 (R) −→ R.
Therefore, the set of matrices with distinct eigenvalues is
{A ∈ M2 (R) : ∆(A) 6= 0} = ∆−1 (R \ {0}).
Since R \ {0} is open and ∆ is continuous, this set is open in M2 (R).
(B) To count the number of k-dimensional subspaces, first count the number of
ordered bases {v1 , . . . , vk } of cardinality k. Since V has q n elements, there
are q n − 1 choices for v1 , q n − q choices for v2 and so on. So the number of
ordered bases of cardinality k is (q n − 1)(q n − q) · (q n − q k−1 ).
Now a given k-dimensional subspace of V has many ordered bases. In fact,
since its dimension is k, the number of ordered bases of W is precisely (q k −
1)(q k − q) · (q k − q k−1 ).
(q n −1)(q n −q)·(q n −q k−1 )
So the number of k-dimensional subspaces of V is (q k −1)(q k −q)·(q k −q k−1 ). .

End Solution

(15) Let V be a finite-dimensional vector space over a field and let GL(V ) denote the
group of invertible linear transformations V −→ V . Let V = ⊕ni=1 Vi , where Vi are
subspaces of V . Let G ⊂ GL(V ) be a subgroup such that for all g ∈ G, g(Vi ) ⊂ Vi
for all 1 ≤ i ≤ n. Suppose further that the centraliser of G in GL(V ) is contained
in the center of GL(V ). Show that n = 1 and V1 = V .
Hint: Consider the projections πi : V −→ Vi and show that every element of G
commutes with each πi .
Solution:For v ∈ V , let v = v1 + · · · + vn be the unique decomposition of v
determined by V = ⊕ni=1 Vi . Then πi (v) = vi .
Let g ∈ G. Then gπi (v) = g(vi ) ∈ Vi by hypothesis.
On the other hand, since g(vi ) ∈ Vi for each i, g(v) = g(v1 ) + · · · + g(vn ) is
the unique decomposition of g(v) associated to V = ⊕ni=1 Vi and g(vi ) is the i-th
projection of g(v). So πi g(v) = πi (g(v1 ) + · · · + g(vn )) = g(vi ).
Hence each πi is in the centralizer of G and so in the center of GL(V ). So each
πi is a scalar multiple of the identity map.
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Let π = c I for some c in the field. Since π =aπ , we get c = c . Then c = 1
i i i
2
i i
2
i i i
or ci = 0. In the latter case, πi = 0, so Vi = 0. If ci = 1, then πi = I. So there is
only one non-zero Vi and it is equal to V .
End Solution

(16) Let F be a finite field with q elements.
(A) (7 marks) Show that every element of F can be written as a sum of two
squares. That is, for every a ∈ F , show that there exist b, c ∈ F such that
m
.co
a = b2 + c 2 .
m
.co m
(B) (3 marks) For a ∈ F , let N (a) denote the cardinality of the set {(b, c) ∈ F 2 |
2 2
N (a) = q 2 .
s e
em
b + c = a}. Show that
P

s
a∈F
l
gof a
(A)laFirst assume that the characteristic of F is 2. Then in fact every element
Solution:
a
ag there 2
F is a square: the map F −→ F, a 7→ a is a bijection of F . So for all a ∈ F ,
exists b ∈ F such that a = b . 2

Now let the characteristic of F be odd. Let F × denote the multiplicative
group of non-zero elements of F . The homomorphism F × −→ F × , a 7→ a2
has kernel of order 2. So exactly half the elements of F × are squares and F
has exactly q−1
2
+ 1 = q+1
2
squares. Let a ∈ F . The set {a − b2 | b ∈ F } also
q+1
has cardinality equal to 2 . The two sets {b2 | b ∈ F } and {a − b2 | b ∈ F }
cannot be disjoint, since their cardinalities add up to q + 1 > |F |. So there
exists c ∈ F such that a = b2 + c2 . m
(B) N (a) = 1= 1 = |F 2 | = q 2 .
.co
P P P P

em
a∈F a∈F (b,c)∈F 2 (b,c)∈F 2

End Solution
la s
b2 +c2 =a

∗ 2
ag
(17 ) Let K : [0, 1] −→ R be a continuous function. Let C([0, 1]) denote the set of all
real-valued continuous functions on [0, 1].
Let f ∈ C([0, 1]). For x ∈ [0, 1], define
Z 1
T (f )(x) := K(x, y)f (y) dy.
0

Also, set ||f k∞ := sup |f (x)|.
m
.co
x∈[0,1]

m (A) Show that T (f ) ∈ C([0, 1]).

m .co em
(B) Show that there exists a constant M > 0, depending only on K, such that
s
s e kT (f )k ≤ M kf k .
g la
la a
∞ ∞

ag (C) Let F = {f ∈ C([0, 1]) : kf k ≤ 1}. Show that the family {T (f ) : f ∈ F}
∞
is uniformly bounded and equicontinuous.
(D) Deduce that any sequence (fn ) in F has a subsequence (fnk ) such that T (fnk )
converges uniformly on [0, 1].
Solution:
(A) Let xn −→ x. Then
Z 1
|T (f )(xn ) − T (f )(x)| = (K(xn , y) − K(x, y))f (y) dy .
m 0
.
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Since |f (y)| ≤ kf k∞ ,
Z 1
|T (f )(xn ) − T (f )(x)| ≤ kf k∞ |K(xn , y) − K(x, y)| dy.
0
Because K is continuous on the compact set [0, 1]2 , it is uniformly continuous,
so
sup |K(xn , y) − K(x, y)| −→ 0.
y
Hence the integral tends to 0, and T (f ) is continuous.
(B) We estimate:
Z 1 Z 1
|T (f )(x)| ≤ |K(x, y)| |f (y)| dy ≤ kf k∞ |K(x, y)| dy.
0 0
Taking supremum over x,
Z 1
kT (f )k∞ ≤ kf k∞ · sup |K(x, y)| dy.
x∈[0,1] 0

Since K is continuous on a compact set, the supremum is finite. Call it M .
Then
kT (f )k∞ ≤ M kf k∞ .
(C) Uniform boundedness: If kf k∞ ≤ 1, then by part (b),
kT (f )k∞ ≤ M,
so the family is uniformly bounded.
Equicontinuity: For x1 , x2 ∈ [0, 1],
Z 1 Z 1
|T (f )(x1 ) − T (f )(x2 )| ≤ |K(x1 , y) − K(x2 , y)| |f (y)| dy ≤ |K(x1 , y) − K(x2 , y)| dy.
0 0
The right-hand side does not depend on f , and since K is uniformly contin-
uous,
sup |K(x1 , y) − K(x2 , y)| −→ 0 as x1 −→ x2 .
y

Hence the family is equicontinuous.
(D) Let (fn ) be a sequence in F. Then (T (fn )) is uniformly bounded and equicon-
tinuous by part (c).
By the Arzelà–Ascoli theorem, there exists a subsequence (T (fnk )) which
converges uniformly on [0, 1].
End Solution

(18∗ ) Let K be a field and consider the ring of formal power series
(∞ )
n
X
K[[X]] = an X : an ∈ K .
n=0
n
(A) Show that an element f (X) = ∞ n=0 an X ∈ K[[X]] is invertible if and only
P

if a0 6= 0.
(B) Show that every nonzero element f ∈ K[[X]] can be written uniquely in the
form
f (X) = X m u(X),
where m ≥ 0 and u(X) is a unit.
(C) Deduce that every ideal of K[[X]] is of the form (X n ) for some n ≥ 0. In
particular, K[[X]] is a principal ideal domain.
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(D) Determine all prime ideals of K[[X]]. a
Solution:
n
(A) (⇒) Suppose f (X) is invertible, so there exists g(X) = ∞
n=0 bn X such that
P

f (X)g(X) = 1. Looking at the constant term of the product,
a0 b0 = 1,
so a0 6= 0.
(⇐) Suppose a0 6= 0. We construct g(X) = bn X n such that f (X)g(X) = 1.
P

m
.co
Equating coefficients gives:

o m a b =1 ⇒ m
and m
c
for.n ≥ 1,
0 0 b0 = a−1
0 ,

s e
s e l a
a ag
n

l
X
a b = 0.
g
k n−k

a This determines b recursively:
n
k=0

n
X
bn = −a−1
0 ak bn−k .
k=1

Thus g(X) exists, so f (X) is invertible.
n
(B) Let f (X) = ∞ n=0 an X 6= 0, and let m be the smallest index such that
P

am 6= 0. Then

m
 

.co
f (X) = X m am + am+1 X + am+2 X 2 + · · · .

em
Set
s
Then u(0) = a 6= 0, so byla
u(X) = am + am+1 X + · · · .

ag index m.
m part (a), u(X) is a unit. Uniqueness follows from
the uniqueness of the smallest
(C) Let I ⊂ K[[X]] be a nonzero ideal. Choose a nonzero element f ∈ I with
minimal m as in part (b), so f = X m u with u a unit. Then X m = f u−1 ∈ I,
so (X m ) ⊂ I.
Conversely, if g ∈ I, write g = X n v with v a unit. By minimality of m, we
must have n ≥ m, so g ∈ (X m ). Hence I = (X m ).
Thus every ideal is principal.
(D) From part (c), every ideal is of the form (X n ).
m
.co
An ideal (X n ) is prime if and only if whenever f g ∈ (X n ), then f ∈ (X n ) or
m g ∈ (X n ).

m .co If n ≥ 2, then X · X n−1 = X n ∈ (X n ) only when n ≤ 2, but neither factor
s e m
e la
lies in (X n ) for n ≥ 2. Hence (X n ) is not prime for n ≥ 2. Note that (X) is

las a prime ideal since K[[X]]/(X) ∼ = K is an integral domain.
ag
ag Thus the only nonzero prime ideal is (X). The zero ideal (0) is also prime
since K[[X]] is an integral domain.
Therefore, the prime ideals are: (0) and (X).
End Solution

(19∗ ) (A) (7 marks) Let G be a finite, non-abelian group satisfying the following prop-
erty: if H1 , H2 are distinct proper maximal subgroups of G, then H1 ∩ H2 is
trivial. Show that G cannot be simple, i.e., show that G has a non-trivial,
proper normal subgroup.
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Recall that a proper subgroup H of G is called maximal if no other proper
subgroup contains H.
(B) (3 marks) Let G be a finite, non-abelian group such that every proper sub-
group of G is abelian. Show that G cannot be simple.
Solution:
(A) Assume for a contradiction that G is simple. Let H be a maximal subgroup
of G. Then H cannot be trivial (since G is not abelian). Further the normalizer
N (H) of H cannot be G since G is simple. So maximality forces N (H) = H.
Suppose that the order of H is m and the order of G is n. The number of
conjugates of H in G is n/m and each of them is also maximal, since H is maximal.
Let S denote the union of all the conjugates of H in G. By hypothesis, the
intersection of any two of these conjugates is trivial. So S has exactly (m−1)n
m
+1 =
n
n− m + 1 elements.
n
Since H is proper, |S| = n − m + 1 < n. Since m ≥ 2, we also have |S| =
n
n− m + 1 ≥ n2 + 1.
Now let x ∈ G \ S. Then x must lie in a proper maximal subgroup of G, say
H 0 . By the same argument as above the union S 0 of all the conjugates of H 0 also
has at least n2 + 1 elements. Note that a conjugate of H and a conjugate of H 0
intersect trivially. So S ∩ S 0 = {e}. But then S ∪ S 0 has at least n + 1 elements
which is a contradiction.
(B) Assume that G is simple for a contradiction. We show that the intersection
of any two maximal proper subgroups of G is trivial. Then we will be done by
part (A). Let H1 , H2 be distinct maximal subgroups of G. By hypothesis, H1 , H2
are abelian. Let x ∈ H1 ∩ H2 . Then x commutes with every element in H1 and in
H2 and hence with every element in H1 H2 = G. So x is in the center of G. If G is
simple then its center is trivial (since it is not abelian). So x = e. End Solution

(20∗ ) Let f : Rn −→ Rn be a continuously differentiable function. Suppose that there
exists a constant c > 0 such that for all x, y ∈ Rn , the following inequality holds:
kf (x) − f (y)k ≥ ckx − yk.
Prove that f is a global diffeomorphism from Rn onto Rn .
Solution:To prove that f is a global diffeomorphism, we must show that f is
bijective and that both f and its inverse f −1 are continuously differentiable. Since
f ∈ C 1 (Rn ), the Inverse Function Theorem implies that if f is bijective and its
derivative Df (x) is invertible at every x ∈ Rn , then f −1 is also C 1 , making f a
diffeomorphism. We divide the proof into four steps.
Step 1: Injectivity. Suppose f (x) = f (y) for some x, y ∈ Rn . Then 0 = kf (x) −
f (y)k ≥ ckx − yk which implies x = y. Therefore, f is globally injective.
Step 2: Local Invertibility. Fix an arbitrary point x ∈ Rn and let v ∈ Rn be a
non-zero vector.
f (x + tv) − f (x)
Df (x)v = lim .
t−→0 t
Hence
kf (x + tv) − f (x)k ck(x + tv) − xk
kDf (x)vk = lim ≥ lim = ckvk.
t−→0 |t| t−→0 |t|
Since c > 0, kDf (x)vk > 0 for all v 6= 0. This implies that the kernel of Df (x) is
trivial. Thus, Df (x) is invertible for all x ∈ Rn .
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a diffeomorphism everywhere.
By the Inverse Function Theorem, f is a local
Consequently, f is an open map; so f (Rn ) is an open set in Rn .
Step 3: Closed Image. We now show that f (Rn ) is a closed set. Let {yk } be a
sequence in the image f (Rn ) that converges to some limit y ∈ Rn . We need to
show that y ∈ f (Rn ).
Since yk ∈ f (Rn ), there exist xk ∈ Rn such that f (xk ) = yk . Since {yk } is a
Cauchy sequence and
1 1
kxj − xk k ≤ kf (xj ) − f (xk )k = kyj − yk k,
m
.co
c c
m
.co m
n
it follows that {xk } is a Cauchy sequence in R . Hence the sequence {xk } con-
verges to some x ∈ Rn . Since f is continuously differentiable, it is continuous.
m s e
s e
Therefore f (x) = limk−→∞ f (xk ) = y. Thus, y = f (x).
l a
l a
Step 4: Surjectivity and Conclusion. Since f (Rn ) is both open and closed in Rn ,
g ag
a
f (Rn ) = Rn . Since f is a bijective map of class C 1 with a non-singular Jacobian
everywhere, it is a global diffeomorphism.
End Solution

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