Field & Field Extensions MCQ Quiz in தமிழ் - Objective Question with Answer for Field & Field Extensions - இலவச PDF ஐப் பதிவிறக்கவும்

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 , where

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

The group of nonzero elements  forms a cyclic group of order $q-1$.

If  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 $q=17k+1$.

 (where $p$ is prime and $n\ge 1$).

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 . 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 . Since the roots of f(x) are , the splitting field of f(x) is . Thus, we want to determine the Galois group

Let . Then the automorphism \sigma permutes the roots of the irreducible polynomial f(x) = x²-2. Thus  is either  or . Since  fixes the elements of , this determines \sigma completely as we have

 .

The map  is the identity automorphism 1 of 
The other map  gives non-identity automorphism . Therefore, the Galois group  is a cyclic group of order 2 .