UPSC Optional (Maths) Paper-02 Algebra Solution

2016

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UPSC Algebra

1(a) Let $\mathbb{K}$ be a field and $\mathbb{K}[X]$ be the ring of polynomials over $\mathbb{R}$ in a single variable $X$. For a polynomial $f \in \mathbb{K}[X]$, let $ (f) $ denote the ideal in $\mathbb{K}[X]$ generated by $ f $. Show that $ (f) $ is a maximal ideal in $\mathbb{K}[X]$ if and only if $ f $ is an irreducible polynomial over $\mathbb{K}$.

Expert Answer

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

To show that

(f)

is a maximal ideal in

K [X]

if and only if

f

is an irreducible polynomial over

K

, we need to prove two directions:

If $(f)$ is a maximal ideal, then $f$ is irreducible.
If $f$ is irreducible, then $(f)$ is a maximal ideal.

Direction 1:

(f)

is a maximal ideal

\Rightarrow

f

is irreducible

Step 1: Assume

(f)

is a maximal ideal

Let’s assume that

(f)

is a maximal ideal in

K [X]

Step 2: Show that

f

is irreducible

Suppose, for the sake of contradiction, that

f

is not irreducible. Then

f = g h

for some

g, h \in K [X]

with

\deg (g) < \deg (f)

and

\deg (h) < \deg (f)

Let

I = (g)

. Then

f \in I

because

f = g h

, and

I

contains

(f)

. However,

I \neq K [X]

because

\deg (g) < \deg (f)

This contradicts the assumption that

(f)

is a maximal ideal, because we’ve found an ideal

I

strictly between

(f)

and

K [X]

Therefore,

f

must be irreducible.

Direction 2:

f

is irreducible

\Rightarrow

(f)

is a maximal ideal

Step 1: Assume

f

is irreducible

Let’s assume that

f

is an irreducible polynomial in

K [X]

Step 2: Show that

(f)

is a maximal ideal

Let

I

be an ideal in

K [X]

such that

(f) \subseteq I \subseteq K [X]

Since

f

is in

I

, there exists a polynomial

g \in I

such that

f = g

Since

f

is irreducible,

g

must either be a unit or associate of

f

. If

g

is a unit, then

I = K [X]

. If

g

is an associate of

f

, then

I = (f)

Therefore,

(f)

is a maximal ideal.

Summary

We have shown that if

(f)

is a maximal ideal, then

f

is irreducible, and conversely, if

f

is irreducible, then

(f)

is a maximal ideal. Hence,

(f)

is a maximal ideal in

K [X]

if and only if

f

is an irreducible polynomial over

K

Verified Answer

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2(b) Let $ p $ be a prime number and $ \mathbb{Z}_{p} $ denote the additive group of integers modulo $ p $. Show that every non-zero element of $ \mathbb{Z}_{p} $ generates $ \mathbb{Z}_{p} $.

Expert Answer

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

To show that every non-zero element of

Z_{p}

generates

Z_{p}

, we need to prove that for any non-zero element

a

Z_{p}

, the cyclic subgroup generated by

a

, denoted

⟨ a ⟩

, is equal to

Z_{p}

Step 1: Define the Cyclic Subgroup

⟨ a ⟩

The cyclic subgroup

⟨ a ⟩

generated by

a

is defined as:

⟨ a ⟩ = {a^{n} \mod p : n \in Z}

Step 2: Show that

⟨ a ⟩

is a Subset of

Z_{p}

First, let’s establish that

⟨ a ⟩

is a subset of

Z_{p}

. This is trivially true since

⟨ a ⟩

consists of elements that are congruent modulo

p

, and therefore, each element in

⟨ a ⟩

is an element in

Z_{p}

Step 3: Show that

⟨ a ⟩

Contains

p

Distinct Elements

Let’s assume that

a

is a non-zero element in

Z_{p}

. We want to show that

a, 2 a, 3 a, \dots, (p - 1) a

are all distinct modulo

p

Suppose, for the sake of contradiction, that

i a \equiv j a \mod p

for some

i \neq j

where

1 \leq i, j \leq p - 1

This implies

(i - j) a \equiv 0 \mod p

, or

p

divides

(i - j) a

Since

p

is a prime and

p ∤ (i - j)

(because

i \neq j

), it must be the case that

p

divides

a

, which contradicts the assumption that

a

is a non-zero element of

Z_{p}

Therefore,

a, 2 a, 3 a, \dots, (p - 1) a

are all distinct modulo

p

, and

⟨ a ⟩

contains

p

distinct elements.

Step 4: Conclude that

⟨ a ⟩ = Z_{p}

Since

⟨ a ⟩

is a subset of

Z_{p}

and contains

p

distinct elements, it must be the case that

⟨ a ⟩ = Z_{p}

Summary

We have shown that for any non-zero element

a

Z_{p}

, the cyclic subgroup

⟨ a ⟩

generated by

a

is equal to

Z_{p}

. Therefore, every non-zero element of

Z_{p}

generates

Z_{p}

Verified Answer

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3(a) Let $ K $ be an extension of a field $ F $; Prove that the elements of $ K $, which are algebraic over $ F $, form a subfield of $ K $. Further, if $ F \subset K \subset L $ are fields, $ L $ is algebraic over $ K $ and $ K $ is algebraic over $ F $, then prove that $ L $ is algebraic over $ F $.

Expert Answer

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

Part 1: Prove that the elements of

K

that are algebraic over

F

form a subfield of

K

To show that the set of elements in

K

that are algebraic over

F

form a subfield, we need to prove that this set is closed under addition, multiplication, and contains multiplicative inverses for non-zero elements. Let’s denote this set as

A

Step 1: Closure under Addition

Let

a, b \in A

. Then

a

and

b

are roots of some polynomials

f (x), g (x) \in F [x]

respectively. We need to show that

a + b

is also algebraic over

F

Consider the polynomials

f (x - b)

and

g (x - a)

. Their product

h (x) = f (x - b) g (x - a)

is a polynomial in

F [x]

. The root of

h (x)

x = a + b

, which means

a + b

is algebraic over

F

Step 2: Closure under Multiplication

Let

a, b \in A

. Then

a

and

b

are roots of some polynomials

f (x), g (x) \in F [x]

respectively. We need to show that

a b

is also algebraic over

F

Consider the polynomial

h (x) = f (x / b) g (x / a)

. This is a polynomial in

F [x]

. The root of

h (x)

x = a b

, which means

a b

is algebraic over

F

Step 3: Existence of Multiplicative Inverses for Non-zero Elements

Let

a \in A

and

a \neq 0

. Then

a

is a root of some polynomial

f (x) \in F [x]

. We need to show that

a^{- 1}

is also algebraic over

F

Consider the polynomial

g (x) = x^{n} f (1 / x)

, where

n

is the degree of

f (x)

. This is a polynomial in

F [x]

. The root of

g (x)

x = a^{- 1}

, which means

a^{- 1}

is algebraic over

F

Step 4:

0

and

1

are in

A

The zero polynomial and the constant polynomial

x - 1

show that

0

and

1

are algebraic over

F

Summary for Part 1

We have shown that

A

is closed under addition, multiplication, and contains multiplicative inverses for non-zero elements. Also,

0

and

1

are in

A

. Therefore,

A

is a subfield of

K

Part 2: If

F \subset K \subset L

and

L

is algebraic over

K

and

K

is algebraic over

F

, then

L

is algebraic over

F

Step 1: Take an element

a \in L

Let

a

be an element in

L

Step 2: Show that

a

is algebraic over

K

Since

L

is algebraic over

K

a

is a root of some polynomial

f (x) \in K [x]

Step 3: Show that

a

is algebraic over

F

The coefficients of

f (x)

are in

K

and are algebraic over

F

. Let these coefficients be

k_{1}, k_{2}, \dots, k_{n}

. Each

k_{i}

is a root of some polynomial

g_{i} (x) \in F [x]

Consider the polynomial

h (x) = f (x, k_{1}, k_{2}, \dots, k_{n})

where we treat

k_{1}, k_{2}, \dots, k_{n}

as variables. The polynomial

h (x)

has

a

as a root and its coefficients are in

F (k_{1}, k_{2}, \dots, k_{n})

Since each

k_{i}

is algebraic over

F

F (k_{1}, k_{2}, \dots, k_{n})

is a finite extension of

F

and hence is algebraic over

F

. Therefore, the coefficients of

h (x)

are algebraic over

F

, and

a

is algebraic over

F

Summary for Part 2

We have shown that any element

a \in L

is algebraic over

F

. Therefore,

L

is algebraic over

F

Final Summary

We have proven that the elements of

K

that are algebraic over

F

form a subfield of

K

and that if

F \subset K \subset L

and

L

is algebraic over

K

and

K

is algebraic over

F

, then

L

is algebraic over

F

Verified Answer

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4(a) Show that every algebraically closed field is infinite.

Expert Answer

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

To show that every algebraically closed field

K

is infinite, we will proceed by contradiction. We will assume that

K

is a finite field and algebraically closed, and then show that this leads to a contradiction.

Step 1: Assume

K

is a Finite Algebraically Closed Field

Let’s assume that

K

is a finite field with

q

elements and is algebraically closed. Being algebraically closed means that every non-constant polynomial

f (x) \in K [x]

has a root in

K

Step 2: Consider a Polynomial of Degree

q

Consider the polynomial

f (x) = (x - a_{1}) (x - a_{2}) \dots (x - a_{q})

, where

a_{1}, a_{2}, \dots, a_{q}

are all the elements of

K

Step 3: Show that

f (x)

has No Root in

K

Notice that

f (a)

for any

a \in K

is given by:

f (a) = (a - a_{1}) (a - a_{2}) \dots (a - a_{q})

Since

a

is one of the

a_{i}

‘s, one of the factors in this product is zero, making

f (a) = 0

Step 4: Contradiction

We have shown that

f (x)

has

q

distinct roots in

K

, which means that

f (x)

should be a polynomial of degree at least

q

. However,

f (x)

is a polynomial of degree

q

, and we assumed that

K

is algebraically closed, meaning that every non-constant polynomial has a root in

K

This is a contradiction because we have found a polynomial

f (x)

that has no root in

K

even though

K

is assumed to be algebraically closed.

Summary

We have reached a contradiction by assuming that

K

is a finite algebraically closed field. Therefore, every algebraically closed field must be infinite.

Verified Answer

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2016

Expert Answer

2(b) Let \( p \) be a prime number and \( \mathbb{Z}_{p} \) denote the additive group of integers modulo \( p \). Show that every non-zero element of \( \mathbb{Z}_{p} \) generates \( \mathbb{Z}_{p} \).

Expert Answer

Expert Answer

4(a) Show that every algebraically closed field is infinite.

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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