Question

The integral domain of which cardinality is not possible:
A. 5
B. 6
C. 7
D. 10

• A. A and B only
• B. A and C only
• C. B and D only
• D. C and D only

Hint:

A crucial fact to memorize for abstract algebra questions: The order of a finite field is always a prime power (\(p^n\)). Since every finite integral domain is a field, its order must also be a prime power. This immediately rules out any integer that is not of the form \(p^n\).

Answer 1

The Correct Option is C

Step 1: Core Concept:
This problem concerns finite rings and their structure. An integral domain is a commutative ring with a multiplicative identity and no zero-divisors (if \(ab=0\), then \(a=0\) or \(b=0\)). A fundamental theorem links finite integral domains and fields.

Step 2: Essential Knowledge:
Key theorems: 1. Every finite integral domain is a field. 2. The order (cardinality) of a finite field is a prime power, \( p^n \) for prime \( p \) and integer \( n \ge 1 \). Therefore, a finite integral domain's cardinality must be a prime power. The task is to identify non-prime powers.

Step 3: Step-by-step Analysis:
Evaluate the given cardinalities:
A. 5: 5 is prime, expressible as \( 5^1 \). This is a prime power, thus possible. (Field \( \mathbb{Z}_5 \)).
B. 6: 6 factors as \( 2 \times 3 \). Not a prime power. Therefore, impossible.
C. 7: 7 is prime, expressible as \( 7^1 \). This is a prime power, thus possible. (Field \( \mathbb{Z}_7 \)).
D. 10: 10 factors as \( 2 \times 5 \). Not a prime power. Therefore, impossible. The question targets the impossible cardinalities, which are 6 and 10.

Step 4: Solution:
The cardinalities that are not possible for an integral domain are 6 (B) and 10 (D).