Field & Field Extensions MCQ Quiz in मल्याळम - Objective Question with Answer for Field & Field Extensions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept: 

For a finite field of order pn  (where p is a prime and n is a positive integer), the proper subfields correspond to the divisors of n .

Given:

Let F be a field of order 16384.

To find : The number of proper subfields of F 

Explanation: 

The order of F , which is 16384 , is a power of 2 (since 16384 = 2¹⁴ ).

For a finite field of order pⁿ (where p is a prime and n is a positive integer), the proper subfields correspond to the divisors of n .

Here, n = 14 ,

so we find the divisors of 14 = 1, 2, 7 .

Each divisor represents a subfield with order 2^d where d is a divisor of 14 ,

except for d = 14 itself (which would give the field F itself, not a proper subfield).

Thus, the number of proper subfields is the number of divisors of 14 excluding 14 itself, which gives 3 proper subfields.

Answer: The number of proper subfields of F is 3.

Hence Option(2) is the correct answer.

Explanation:

A finite field
F has order \(q=p^n\), where

$p$ is a prime number (the characteristic of the field), and $n$ is a positive integer.

The group of nonzero elements \(F^×\) forms a cyclic group of order $q-1$.

If \(F^×\) has a subgroup of order 17, then 17 must divide $q-1$ (as per Lagrange's theorem).

Therefore,
q-1 = k \cdot 17 for some integer $k$, which implies $q=17k+1$.

 \(q=p^n\)(where
p is prime and $n \geq 1$).

To minimize $q$, choose the smallest $k$ such that
q = 17k + 1 is a power of a prime.

For $k=1$: $q = 17 \cdot 1 + 1 = 18$ (not a power of a prime).

For k=2:$q = 17 \cdot 2 + 1 = 35$ (not a power of a prime).

For
k = 3:
q = 17 \cdot 3 + 1 = 52 (not a power of a prime).

For $k=4$:
q = 17 \cdot 4 + 1 = 69 (not a power of a prime).

For
k = 5:
q = 17 \cdot 5 + 1 = 85 (not a power of a prime).

For $k=6$: $q = 17 \cdot 6 + 1 = 103$.

Thus, the smallest value of
q satisfying the criteria is
q = 103.

Solution:

f(x) = x2 - 2

The roots of the polynomial f(x) are \( \pm \sqrt{2}\). Since all the roots of f(x) are distinct, f(x)=x²-2 is separable.
The Galois group of the separable polynomial f(x)=x^2-2 is the Galois group of the splitting field of f(x) over \(\mathbb{Q}\). Since the roots of f(x) are \(\pm \sqrt{2}\), the splitting field of f(x) is \( \mathbb{Q}(\sqrt{2})\). Thus, we want to determine the Galois group

\(\operatorname{Gal}(\mathbb{Q}(\sqrt{2}) / \mathbb{Q}) .\)

Let \(\sigma \in \operatorname{Gal}(\mathbb{Q}(\sqrt{2}) / \mathbb{Q})\). Then the automorphism \sigma permutes the roots of the irreducible polynomial f(x) = x²-2. Thus \(\sigma(\sqrt{2})\) is either \(\sqrt{2}\) or \(-\sqrt{2}\). Since \(\sigma\) fixes the elements of \(\mathbb{Q}\), this determines \sigma completely as we have

\(\sigma(a+b \sqrt{2})=a+b \sigma(\sqrt{2})=a \pm \sqrt{2}\) .

The map \(\sqrt{2} \mapsto \sqrt{2}\) is the identity automorphism 1 of \(\mathbb{Q} \sqrt{2}.\)
The other map \( \sqrt{2} \mapsto-\sqrt{2}\) gives non-identity automorphism \( \tau\). Therefore, the Galois group \( \operatorname{Gal}(\mathbb{Q}(\sqrt{2}) / \mathbb{Q})=\{1, \tau\}\) is a cyclic group of order 2 .