2021

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UPSC Algebra

1(a) Let $m_1, m_2, \cdots, m_k$ be positive integers and $d>0$ the greatest common divisor of $m_1, m_2, \cdots, m_k$. Show that there exist integers $x_1, x_2, \cdots, x_k$ such that \[ d=x_1 m_1+x_2 m_2+\cdots+x_k m_k \]

Expert Answer

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

Step 1: Defining a Set S

Let

s = {a u + b v | u, v

are integers and

a u + b v > 0}

Step 2: Investigating Cases for

a

and

b

a > 0

, then

a = a 1 + b (0) > 0

, which implies that

a > 0

a < 0

, then

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

, which implies that

- a \in S

Similarly, if

b > 0

, then

b \in S

b < 0

, then

- b \in S

Therefore,

S \neq ϕ

and

S

contains positive integers.

By the well-ordering principle,

S

has a least element, say

d

Step 3: Proving that

d

is the GCD of

a

and

b

Now we have

d \in S

such that

d = a x + b y - - - - (1)

for some integers

x, y

Also,

d > 0

To prove that

d

is the greatest common divisor (GCD) of

a

and

b

, consider the following:

Let

a = d q + r - - - - (2)

(where

0 ⩽ r < d

) be a division of

a

d

r \neq 0

, then

r = a - d q

\begin{aligned} = a - (a x + b y) q (as d = a x + b y) \\ = a (1 - λ q) + b (- y z) > 0 (since 1 - λ q, - y q are integers) \\ r > 0, r \in S if r < d and r \in S \end{aligned}

This leads to a contradiction since

d

is the least element of

S

Therefore,

r = 0

, and we have

a = d q

This implies that

\frac{a}{d} = q

and

\frac{d}{a}

are integers.

Similarly,

\frac{d}{b}

is also an integer.

Step 4: Proving Uniqueness of GCD

Suppose

\frac{C}{c}, \frac{c}{b} \Rightarrow \frac{c}{a x + b y} \Rightarrow \frac{c}{d}

Therefore

\frac{d}{a}, \frac{d}{b}

also if

\frac{c}{a}, \frac{c}{b} \Rightarrow \frac{c}{d}

d

is gcd of

a

and

b

If possible

d^{'}

is also gcd of

a

and

b

Then

\frac{d^{'}}{a}, \frac{d^{'}}{b} \Rightarrow \frac{d}{d^{'}} \to

(3)
Similarly,

\frac{d}{c}, \frac{d}{b} \Rightarrow \frac{d^{'}}{d} \to

(4)

This implies

\frac{d}{d^{'}}

(from the divisibility relation) which, in turn, implies

d = d^{'}

(from the uniqueness of GCD).

Step 5: Extending the Argument

The above argument can be extended to more than two integers.

d = \gcd (m_{1}, m_{2}, \dots, m_{k})

, there exist integers

λ_{1}, λ_{2}, \dots, λ_{k}

such that

\begin{aligned} d = λ_{1} m_{1} + λ_{2} m_{2} + \dots + m_{k} λ_{k} \\ \Rightarrow d = λ_{1} m_{1} + λ_{2} m_{2} + \dots + λ_{k} m_{k} \end{aligned}

This concludes the proof.

Verified Answer

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2(b) Let $F$ be a field and $f(x) \in F[x]$, a polynomial of degree $>0$ over $F$. Show that there is a field $F^{\prime}$ and an imbedding $q: F \rightarrow F^{\prime}$ such that the polynomial $f^q \in F^{\prime}[x]$ has a root in $F^{\prime}$, where $f^q$ is obtained by replacing each coefficient $a$ of $f$ by $q(a)$.

Expert Answer

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

Step 1: Problem Statement

We are given a field

F

and a polynomial

f (x) \in F [x]

of degree greater than 0. The goal is to show the existence of a field

F^{'}

and an embedding

q : F \to F^{'}

such that the polynomial

f^{q} \in F^{'} [x]

has a root in

F^{'}

. Here,

f^{q}

is obtained by replacing each coefficient

a

f

q (a)

Step 2: Maximal Ideal and Field Extension

Suppose

f (x) \in F [x]

is a polynomial of degree greater than 0 over a field

F

. We consider the ideal

M = ⟨ f (x) ⟩

F [x]

. Since

f (x)

is irreducible,

M

is a maximal ideal in

F [x]

Step 3: Field Extension

We create a field extension by taking

F

modulo this maximal ideal, denoted as

\frac{F [x]}{M}

, which is itself a field.

Step 4: Define Embedding

q

Define the embedding

q : F \to \frac{F [x]}{M}

q (a) = a + M

. This embedding

q

is a homomorphism.

Step 5: Ker

q

and Injectiveness

We consider the kernel of

q

, i.e., elements

a \in F

such that

q (a) = 0 + M

. This implies

a + M = M

, which further implies

a \in M = ⟨ f (x) ⟩

. Therefore,

a = f (x) q (x)

for some

q (x) \in F [x]

. Since

f (x)

is irreducible and

a

is a polynomial of degree greater than 0, this relation holds only when

a = 0

. Consequently,

ker q = {0}

q

is injective.

Step 6: Isomorphism and Subfield

Hence,

F

is isomorphic to

q (F)

. We can view

F^{'}

as containing

F

by identifying

a \in F

with

q (a)

and vice versa.

Step 7: Construct

Ψ

Define

Ψ : F [x] \to \frac{F [x]}{M}

such that

Ψ (f (x)) = f (x) + M

Ψ

is a natural homomorphism.

Step 8: Show

α

as a Root

Let

α = Ψ (x) = x + M

. We claim that

α

is a root of

f^{q}

F

Step 9: Evaluate

f^{q}

Evaluate

f^{q}

as follows:

\begin{aligned} Ψ (f^{q}) & = Ψ (α_{0} + α_{1} x + α_{2} x^{2} + \dots + α_{n} x^{n}) \\ = Ψ (α_{0}) + Ψ (α_{1}) Ψ (x) + \dots + Ψ (α_{n}) Ψ (x^{n}) \end{aligned}

Step 10: Use

p^{(x)}

Note that

Ψ (p^{(x)}) = p^{(x)} + M = p^{(x)} + ⟨ p^{(x)} ⟩ = M =

zero of

F

Step 11: Zero of

F^{'}

Thus, the zero of

F^{'}

is:

\begin{aligned} Zero of F^{'} = Ψ (α_{0}) + Ψ (α_{1}) Ψ (x) + \dots + Ψ (α_{n}) Ψ (x^{n}) \\ = q (a_{0}) + q (a_{1}) Ψ (x) + \dots + q (a_{n}) Ψ (x^{n}) \end{aligned}

Step 12: Zero is a Root

Since

F ≅ q (F)

, the zero is also

a_{0} + a_{1} α + a_{2} α^{2} + \dots + a_{n} α^{n}

, which is equal to

f (a) = f^{q}

Step 13: Conclusion

Hence,

a

is a root of

f (a)

F

. By replacing each coefficient

a

f

q (a)

, we obtain

f^{q}

Therefore, there exists a field

F^{'}

and an embedding

q : F \to F^{'}

such that the polynomial

f^{q} \in F^{'} [x]

has a root in

F^{'}

, where

f^{q}

is obtained by replacing each coefficient

a

f

q (a)

Verified Answer

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4(a) Show that there are infinitely many subgroups of the additive group $\mathbb{Q}$ of rational numbers.

Expert Answer

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

The additive group of rational numbers, denoted by

Q

, consists of all rational numbers under the operation of addition. To show that there are infinitely many subgroups of

Q

, we can construct an infinite family of subgroups.

Construction of Subgroups:

Consider the set

n Q

, where

n

is any positive integer. This set consists of all rational numbers that can be written in the form

n q

, where

q

is any rational number. Formally,

n Q = {n q ∣ q \in Q}

Properties:

Identity Element: $0$ is in $n Q$ because $0 = n \times 0$ .
Closure under Addition: If $a, b \in n Q$ , then $a + b = n q_{1} + n q_{2} = n (q_{1} + q_{2})$ is also in $n Q$ .
Closure under Inverse: If $a \in n Q$ , then $- a = - n q = n (- q)$ is also in $n Q$ .

Therefore,

n Q

is a subgroup of

Q

for each positive integer

n

Infinitude of Subgroups:

Since

n

can be any positive integer, and there are infinitely many positive integers, we have constructed an infinite family of subgroups

n Q

Q

Conclusion:

Hence, we have shown that there are infinitely many subgroups of the additive group

Q

of rational numbers.

Verified Answer

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2021

1(a) Let \(m_1, m_2, \cdots, m_k\) be positive integers and \(d>0\) the greatest common divisor of \(m_1, m_2, \cdots, m_k\). Show that there exist integers \(x_1, x_2, \cdots, x_k\) such that \[ d=x_1 m_1+x_2 m_2+\cdots+x_k m_k \]

Expert Answer

Expert Answer

4(a) Show that there are infinitely many subgroups of the additive group \(\mathbb{Q}\) of rational numbers.

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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