UPSC Optional (Maths) Paper-02 Algebra Solution

2010

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UPSC Algebra

1(a) Let $G=\mathbb{I}-\{-1\}$ be the set of all real numbers omitting $-1$. Define the binary relation  on $G$ by $a b=a+b+a b$. Show $(G, *)$ is a group and it is abelian.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To show that

(G, *)

is a group, we need to verify the following group axioms:

Closure under $*$

For any

a, b \in G

a * b = a + b + a b

is also a real number. However, we need to ensure that

a * b \neq - 1

a * b = - 1

, then

a + b + a b = - 1

, which simplifies to

a b + a + b + 1 = 0

(a + 1) (b + 1) = 0

Since

a, b \in G

a \neq - 1

and

b \neq - 1

, so

a + 1 \neq 0

and

b + 1 \neq 0

. Therefore,

a * b \neq - 1

, and

G

is closed under

*

Associativity

For any

a, b, c \in G

(a * b) * c = (a + b + a b) * c = a + b + a b + c + (a + b + a b) c = a + b + c + a b + a c + b c + a b c

a * (b * c) = a * (b + c + b c) = a + b + c + b c + a (b + c + b c) = a + b + c + a b + a c + b c + a b c

Since

(a * b) * c = a * (b * c)

, the operation

*

is associative.

Existence of Identity Element

We need to find an element

e \in G

such that for any

a \in G

a * e = a

Let

a * e = a

, then

a + e + a e = a

, which simplifies to

e + a e = 0

e (1 + a) = 0

Since

a \neq - 1

1 + a \neq 0

, and therefore

e = 0

Existence of Inverse Element

For each

a \in G

, we need to find an element

a^{- 1} \in G

such that

a * a^{- 1} = e

Let

a * a^{- 1} = e

, then

a + a^{- 1} + a a^{- 1} = 0

, which simplifies to

a^{- 1} (1 + a) = - a

Since

a \neq - 1

1 + a \neq 0

, and therefore

a^{- 1} = - \frac{a}{a + 1}

Commutativity

For any

a, b \in G

a * b = a + b + a b = b + a + b a = b * a

Since

a * b = b * a

(G, *)

is abelian.

Conclusion

Since all these properties are satisfied,

(G, *)

is an abelian group.

Verified Answer

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1(b) Show that a cyclic group of order 6 is isomorphic to the product of a cyclic group of order 2 and a cyclic group of order 3. Can you generalize this? Justify.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Step 1: Define the Groups

Let

G

be a cyclic group of order 6, generated by an element

a

, i.e.,

G = ⟨ a ⟩ = {e, a, a^{2}, a^{3}, a^{4}, a^{5}}

Let

H

be a cyclic group of order 2, generated by an element

b

, i.e.,

H = ⟨ b ⟩ = {e, b}

Let

K

be a cyclic group of order 3, generated by an element

c

, i.e.,

K = ⟨ c ⟩ = {e, c, c^{2}}

Step 2: Define the Product Group

Consider the product group

H \times K

, consisting of elements

(b^{i}, c^{j})

where

0 \leq i < 2

and

0 \leq j < 3

Step 3: Define the Isomorphism

Define a function

ϕ : G \to H \times K

ϕ (a^{n}) = (b^{n \mod 2}, c^{n \mod 3})

Step 4: Show

ϕ

is an Isomorphism

Well-defined: $ϕ$ is well-defined because $n \mod 2$ and $n \mod 3$ are always in the correct ranges for $H$ and $K$ , respectively.
Bijective: $ϕ$ is bijective because each element $a^{n}$ in $G$ maps to a unique element $(b^{n \mod 2}, c^{n \mod 3})$ in $H \times K$ , and vice versa.
Homomorphism: $ϕ$ is a homomorphism because for any $a^{m}, a^{n} \in G$ ,

$ϕ (a^{m} * a^{n}) = ϕ (a^{m + n}) = (b^{(m + n) \mod 2}, c^{(m + n) \mod 3}) = (b^{m \mod 2} * b^{n \mod 2}, c^{m \mod 3} * c^{n \mod 3}) = ϕ (a^{m}) * ϕ (a^{n})$

Generalization

This result can be generalized to say that a cyclic group of order

m n

where

m

and

n

are coprime is isomorphic to the product of a cyclic group of order

m

and a cyclic group of order

n

Justification for Generalization

Chinese Remainder Theorem: If $m$ and $n$ are coprime, then the map $ϕ : Z_{m n} \to Z_{m} \times Z_{n}$ defined by $ϕ (a) = (a \mod m, a \mod n)$ is an isomorphism.
Cyclic Groups: A cyclic group of order $m n$ is isomorphic to $Z_{m n}$ , and cyclic groups of orders $m$ and $n$ are isomorphic to $Z_{m}$ and $Z_{n}$ , respectively.
Isomorphism: Combining these, we get that a cyclic group of order $m n$ is isomorphic to the product of a cyclic group of order $m$ and a cyclic group of order $n$ .

Therefore, we can generalize that a cyclic group of order

m n

is isomorphic to the product of a cyclic group of order

m

and a cyclic group of order

n

when

m

and

n

are coprime.

Verified Answer

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2(a) Let $\left(\mathbb{R}^, \cdot\right)$ be the multiplicative group of nonzero reals and $(G L(n, I R), X)$ be the multiplicative group of $n \times n$ non-singular matrices. Show that the quotient group $G L(n, \mathbb{R}) / S L(n, \mathbb{R})$ and $\left(\mathbb{R}^ \cdot \cdot\right)$ are isomorphic where $S L(n, \mathrm{IR})=\{A \in G L(n, \mathrm{IR}) / \operatorname{det} A=1\}$. What is the centre of $G L(n$, IR ) ?

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Step 1: Define the Groups and Quotient Group

$R^{*}$ is the set of all nonzero real numbers with multiplication as the operation.
$G L (n, R)$ is the set of all $n \times n$ invertible matrices with matrix multiplication as the operation.
$S L (n, R)$ is the set of all $n \times n$ matrices with determinant 1, i.e., $S L (n, R) = {A \in G L (n, R) ∣ det (A) = 1}$ .

The quotient group

G L (n, R) / S L (n, R)

consists of cosets

A \cdot S L (n, R)

where

A \in G L (n, R)

Step 2: Define the Isomorphism

Define a function

ϕ : G L (n, R) \to R^{*}

ϕ (A) = det (A)

Step 3: Show

ϕ

Induces an Isomorphism between

G L (n, R) / S L (n, R)

and

R^{*}

Well-defined: $ϕ$ is well-defined because the determinant of any $n \times n$ invertible matrix is a nonzero real number.
Homomorphism: $ϕ$ is a homomorphism because $det (A B) = det (A) det (B)$ .
Kernel: The kernel of $ϕ$ is $S L (n, R)$ because $ϕ (A) = 1$ if and only if $det (A) = 1$ , i.e., $A \in S L (n, R)$ .
Induced Isomorphism: By the First Isomorphism Theorem, $G L (n, R) / Ker (ϕ)$ is isomorphic to the image of $ϕ$ , which is $R^{*}$ . Therefore, $G L (n, R) / S L (n, R)$ is isomorphic to $R^{*}$ .

Step 4: Find the Center of

G L (n, R)

The center

Z (G L (n, R))

G L (n, R)

consists of all matrices

A

such that

A B = B A

for all

B \in G L (n, R)

For

n > 1

, the center consists only of scalar multiples of the identity matrix, i.e.,

Z (G L (n, R)) = {c I ∣ c \in R^{*}}

For

n = 1

G L (1, R)

is essentially

R^{*}

, and every element commutes with every other element, so the center is

G L (1, R)

itself.

Conclusion

The quotient group

G L (n, R) / S L (n, R)

is isomorphic to

R^{*}

, and the center of

G L (n, R)

{c I ∣ c \in R^{*}}

for

n > 1

and

G L (1, R)

for

n = 1

Verified Answer

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3(a) Consider the polynomial ring $Q[x]$. Show $p(x)=x^3-2$ is irreducible over $Q$. Let $I$ be the ideal in $Q[x]$ generated by $p(x)$. Then show that $Q[x] / I$ is a field and that each element of it is of the form $a_0+a_1 t+a_2 t^2$ with $a_0, a_1, a_2$ in $Q$ and $t=x+I$.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Step 1: Show

p (x) = x^{3} - 2

is Irreducible over

Q

To show that

p (x) = x^{3} - 2

is irreducible over

Q

, we can use Eisenstein’s criterion with the prime

p = 2

. The coefficients of

p (x)

are

1, 0, 0, - 2

, and we observe:

All coefficients except the leading one are divisible by 2.
The constant term, -2, is not divisible by $2^{2}$ .

Thus, by Eisenstein’s criterion,

p (x)

is irreducible over

Q

Step 2: Show

Q [x] / I

is a Field

Let

I

be the ideal generated by

p (x)

Q [x]

. To show that

Q [x] / I

is a field, it suffices to show that

I

is a maximal ideal.

Since

p (x)

is irreducible over

Q

, it is also a prime element in

Q [x]

. Therefore,

I

is a prime ideal. In a principal ideal domain like

Q [x]

, a non-zero prime ideal is also a maximal ideal. Hence,

I

is a maximal ideal, and

Q [x] / I

is a field.

Step 3: Describe the Elements of

Q [x] / I

Let

t = x + I

Q [x] / I

. We want to show that every element in

Q [x] / I

can be written in the form

a_{0} + a_{1} t + a_{2} t^{2}

where

a_{0}, a_{1}, a_{2} \in Q

Consider an arbitrary polynomial

f (x)

Q [x]

. We can perform polynomial division to write

f (x)

as:

f (x) = q (x) (x^{3} - 2) + r (x)

where

q (x)

is the quotient and

r (x)

is the remainder such that

\deg (r (x)) < \deg (x^{3} - 2)

Q [x] / I

f (x) + I = r (x) + I

because

q (x) (x^{3} - 2)

is in

I

The remainder

r (x)

can have a degree at most 2, so

r (x) = a_{0} + a_{1} x + a_{2} x^{2}

. Therefore, in

Q [x] / I

f (x) + I = a_{0} + a_{1} t + a_{2} t^{2}

Conclusion

The polynomial $p (x) = x^{3} - 2$ is irreducible over $Q$ .
$Q [x] / I$ is a field.
Each element of $Q [x] / I$ can be written in the form $a_{0} + a_{1} t + a_{2} t^{2}$ where $a_{0}, a_{1}, a_{2} \in Q$ and $t = x + I$ .

Verified Answer

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3(b) Show that the quotient ring $\mathbb{Z}[i] /(1+3 i)$ is isomorphic to the ring $\mathbb{Z} / 10 \mathbb{Z}$ where $\mathbb{Z}[i]$ denotes the ring of Gaussian integers.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Step 1: Define the Ring and Ideal

Let

Z [i]

be the ring of Gaussian integers, which consists of all numbers of the form

a + b i

where

a, b \in Z

and

i

is the imaginary unit.

Let

I

be the ideal generated by

1 + 3 i

Z [i]

Step 2: Define the Quotient Ring

The quotient ring

Z [i] / I

consists of cosets

(a + b i) + I

where

a + b i \in Z [i]

Step 3: Define the Isomorphism

We aim to show that

Z [i] / I

is isomorphic to

Z / 10 Z

Define a function

ϕ : Z [i] \to Z / 10 Z

ϕ (a + b i) = a + 3 b \mod 10

Step 4: Show

ϕ

is an Isomorphism

Well-defined: $ϕ$ is well-defined because $a + 3 b$ is an integer for all $a, b \in Z$ .
Homomorphism: $ϕ$ is a homomorphism because

$ϕ ((a + b i) + (c + d i)) = ϕ (a + c + (b + d) i) = a + c + 3 (b + d) \mod 10 = (a + 3 b) + (c + 3 d) \mod 10 = ϕ (a + b i) + ϕ (c + d i)$
Kernel: The kernel of $ϕ$ is $I$ because $ϕ (a + b i) = 0$ if and only if $a + 3 b \equiv 0 \mod 10$ , i.e., $a + b i$ is a multiple of $1 + 3 i$ .
Surjective: $ϕ$ is surjective because for any $n \in Z / 10 Z$ , we can find $a + b i$ such that $ϕ (a + b i) = n$ . For example, $ϕ (n) = n$ .
Induced Isomorphism: By the First Isomorphism Theorem, $Z [i] / Ker (ϕ)$ is isomorphic to the image of $ϕ$ , which is $Z / 10 Z$ . Therefore, $Z [i] / I$ is isomorphic to $Z / 10 Z$ .

Conclusion

The quotient ring

Z [i] / (1 + 3 i)

is isomorphic to the ring

Z / 10 Z

Verified Answer

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2010

1(a) Let \(G=\mathbb{I}-\{-1\}\) be the set of all real numbers omitting \(-1\). Define the binary relation \(\) on \(G\) by \(a b=a+b+a b\). Show \((G, *)\) is a group and it is abelian.

Expert Answer

1(b) Show that a cyclic group of order 6 is isomorphic to the product of a cyclic group of order 2 and a cyclic group of order 3. Can you generalize this? Justify.

Expert Answer

Expert Answer

3(a) Consider the polynomial ring \(Q[x]\). Show \(p(x)=x^3-2\) is irreducible over \(Q\). Let \(I\) be the ideal in \(Q[x]\) generated by \(p(x)\). Then show that \(Q[x] / I\) is a field and that each element of it is of the form \(a_0+a_1 t+a_2 t^2\) with \(a_0, a_1, a_2\) in \(Q\) and \(t=x+I\).

Expert Answer

3(b) Show that the quotient ring \(\mathbb{Z}[i] /(1+3 i)\) is isomorphic to the ring \(\mathbb{Z} / 10 \mathbb{Z}\) where \(\mathbb{Z}[i]\) denotes the ring of Gaussian integers.

Expert Answer

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2010

1(a) Let \(G=\mathbb{I}-\{-1\}\) be the set of all real numbers omitting \(-1\). Define the binary relation \(*\) on \(G\) by \(a * b=a+b+a b\). Show \((G, *)\) is a group and it is abelian.

Expert Answer

1(b) Show that a cyclic group of order 6 is isomorphic to the product of a cyclic group of order 2 and a cyclic group of order 3. Can you generalize this? Justify.

Expert Answer

Expert Answer

3(a) Consider the polynomial ring \(Q[x]\). Show \(p(x)=x^3-2\) is irreducible over \(Q\). Let \(I\) be the ideal in \(Q[x]\) generated by \(p(x)\). Then show that \(Q[x] / I\) is a field and that each element of it is of the form \(a_0+a_1 t+a_2 t^2\) with \(a_0, a_1, a_2\) in \(Q\) and \(t=x+I\).

Expert Answer

3(b) Show that the quotient ring \(\mathbb{Z}[i] /(1+3 i)\) is isomorphic to the ring \(\mathbb{Z} / 10 \mathbb{Z}\) where \(\mathbb{Z}[i]\) denotes the ring of Gaussian integers.

Expert Answer

Get Complete Solution Mathematics Optional Paper | UPSC

UPSC Previous Years Maths Optional Papers with Solution | Paper-02

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CONTENTS

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1(a) Let \(G=\mathbb{I}-\{-1\}\) be the set of all real numbers omitting \(-1\). Define the binary relation \(\) on \(G\) by \(a b=a+b+a b\). Show \((G, *)\) is a group and it is abelian.