UPSC Optional (Maths) Paper-02 Algebra Solution

2018

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UPSC Algebra

1(a) Let $R$ be an integral domain with unit element. Show that any unit in $R[x]$ is a unit in $R$.

Expert Answer

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

Introduction

In this problem, we are dealing with the concept of units in the context of integral domains and polynomial rings. Specifically, we are given an integral domain

R

with a unit element and are asked to prove that any unit in

R [x]

(the polynomial ring over

R

) must also be a unit in

R

Definitions

Integral Domain: An integral domain is a commutative ring $R$ with a multiplicative identity (unit element) such that the product of any two non-zero elements is non-zero.
Unit in a Ring: An element $a$ in a ring $R$ is called a unit if there exists an element $b$ in $R$ such that $a \cdot b = b \cdot a = 1$ , where $1$ is the multiplicative identity in $R$ .
Polynomial Ring $R [x]$ : The set of all polynomials with coefficients in $R$ .

Work/Calculations

Step 1: Assume

f (x)

is a Unit in

R [x]

Let’s assume that

f (x)

is a unit in

R [x]

. This means there exists a polynomial

g (x)

R [x]

such that:

f (x) \cdot g (x) = 1

Step 2: Examine the Degrees of

f (x)

and

g (x)

The degree of the polynomial

f (x) \cdot g (x)

is the sum of the degrees of

f (x)

and

g (x)

. Since

f (x) \cdot g (x) = 1

, a constant polynomial, the degree of

f (x) \cdot g (x)

is 0.

Let

deg (f (x)) = m

and

deg (g (x)) = n

deg (f (x) \cdot g (x)) = m + n = 0

Step 3: Conclude

m = n = 0

Since

m + n = 0

and

m, n \geq 0

, it must be the case that

m = n = 0

Step 4: Show

f (x)

is a Unit in

R

Since

m = 0

f (x)

is a constant polynomial, say

f (x) = a

, where

a

is in

R

. Similarly,

g (x) = b

, where

b

is in

R

From

f (x) \cdot g (x) = 1

, we have

a \cdot b = 1

This shows that

a

is a unit in

R

, as

a \cdot b = 1

Conclusion

We have shown that if

f (x)

is a unit in

R [x]

, then it must be a constant polynomial and also a unit in

R

. Therefore, any unit in

R [x]

is a unit in

R

Verified Answer

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2(a) Show that the quotient group of $(\mathbb{R},+)$ modulo $\mathbb{Z}$ is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here $R$ is the set of real numbers and $\mathbb{Z}$ is the set of integers.

Expert Answer

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

To show that the quotient group of

(R, +)

modulo

Z

is isomorphic to the multiplicative group of complex numbers on the unit circle, we will define an isomorphism between these two groups and then prove that this mapping is indeed an isomorphism.

Quotient Group Definition
The quotient group of

(R, +)

modulo

Z

is denoted as

R / Z

and consists of all cosets of the form

r + Z

where

r \in R

. Here,

r + Z

represents the set of all real numbers differing by an integer from

r

Multiplicative Group of Complex Numbers on the Unit Circle
The multiplicative group of complex numbers on the unit circle is the set of all complex numbers of the form

e^{2 π i θ}

where

θ \in R

and

i

is the imaginary unit. This group is usually denoted as

U (1)

S^{1}

Define an Isomorphism
Let’s define a function

f : R / Z \to U (1)

as follows:

f (r + Z) = e^{2 π i r}

Prove Isomorphism
To prove that

f

is an isomorphism, we need to show that

f

is a bijective homomorphism.

Homomorphism
A function $f$ is a homomorphism if for all $a, b \in R / Z$ :

f (a + b) = f (a) \cdot f (b)

Let’s verify this property for our function

f

f ((r_{1} + Z) + (r_{2} + Z)) = f (r_{1} + r_{2} + Z) = e^{2 π i (r_{1} + r_{2})} = e^{2 π i r_{1}} \cdot e^{2 π i r_{2}} = f (r_{1} + Z) \cdot f (r_{2} + Z)

Injective (One-to-One)
A function $f$ is injective if different elements in the domain map to different elements in the codomain.

f (r_{1} + Z) = f (r_{2} + Z)

, then:

e^{2 π i r_{1}} = e^{2 π i r_{2}} ⟹ r_{1} - r_{2} \in Z ⟹ r_{1} + Z = r_{2} + Z

Surjective (Onto)
A function $f$ is surjective if every element in the codomain has a preimage in the domain.

For any

e^{2 π i θ} \in U (1)

θ + Z \in R / Z

is a preimage, since:

f (θ + Z) = e^{2 π i θ}

Conclusion
Since

f

is a bijective homomorphism, it is an isomorphism. Thus, the quotient group

R / Z

is isomorphic to the multiplicative group of complex numbers on the unit circle,

U (1)

S^{1}

Verified Answer

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3(a) Find all the proper subgroups of the multiplicative group of the field $\left(\mathcal{Z}_{13},+_{13}, \times_{13}\right)$, where $+{ }_{13}$ and $x_{13}$ represent addition modulo 13 and multiplication modulo 13 respectively.

Expert Answer

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

To find all the proper subgroups of the multiplicative group of the field

(Z_{13}, +_{13}, \times_{13})

, we need to consider the elements and their orders in this group.

The multiplicative group of a finite field is formed by the nonzero elements of that field under multiplication. In this case, we are working with

Z_{13}

, so the nonzero elements are

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Finding the Generator
The generator of the multiplicative group

Z_{13}^{*}

2

. This means that every element in the multiplicative group

Z_{13}^{*}

can be represented as a power of

2

modulo

13

, confirming that

Z_{13}^{*}

is indeed a cyclic group.

The cyclic group generated by

2

is as follows:

{2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1}

Finding the Proper Subgroups
Now, let’s find all the proper subgroups of this cyclic group. A proper subgroup is a subgroup that is not equal to the entire group. The proper subgroups of the cyclic group generated by

2

are as follows:

Subgroup generated by $2$ : ${1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7}$ (This is the entire group, so it is not a proper subgroup)
Subgroup generated by powers of $2$ with step $2$ : ${1, 4, 3, 12, 9, 10}$ (Order $6$ )
Subgroup generated by powers of $2$ with step $3$ : ${1, 8, 12, 5}$ (Order $4$ )
Subgroup generated by powers of $2$ with step $4$ : ${1, 3, 9}$ (Order $3$ )
Subgroup generated by powers of $2$ with step $6$ : ${1, 12}$ (Order $2$ )
Subgroup generated by powers of $2$ with step $12$ : ${1}$ (Trivial subgroup)

Conclusion
The proper subgroups of the multiplicative group of the field

(Z_{13}, +_{13}, \times_{13})

are:

${1, 4, 3, 12, 9, 10}$ (Order $6$ )
${1, 8, 12, 5}$ (Order $4$ )
${1, 3, 9}$ (Order $3$ )
${1, 12}$ (Order $2$ )
${1}$ (Trivial subgroup)

Each of these subgroups is closed under multiplication modulo

13

and contains the identity element

1

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2018

1(a) Let \(R\) be an integral domain with unit element. Show that any unit in \(R[x]\) is a unit in \(R\).

Expert Answer

2(a) Show that the quotient group of \((\mathbb{R},+)\) modulo \(\mathbb{Z}\) is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here \(R\) is the set of real numbers and \(\mathbb{Z}\) is the set of integers.

Expert Answer

3(a) Find all the proper subgroups of the multiplicative group of the field \(\left(\mathcal{Z}_{13},+_{13}, \times_{13}\right)\), where \(+{ }_{13}\) and \(x_{13}\) represent addition modulo 13 and multiplication modulo 13 respectively.

Expert Answer

Get Complete Solution Mathematics Optional Paper | UPSC

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

UPSC Previous Years Maths Optional Papers with Solution | Paper-01

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