Question:
Let \(R=\{0,1,2,3\}\), under addition and multiplication modulo \(4\) is
Let \(R=\{0,1,2,3\}\), under addition and multiplication modulo \(4\) is
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\(\mathbb{Z}_n\) is a field only when \(n\) is prime. Since \(4\) is not prime, \(\mathbb{Z}_4\) is only a ring.
Updated On: Jul 3, 2026
- A field
- an integral domain
- a ring
- a skew field
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The Correct Option is C
Solution and Explanation
Concept:
The set \[ R=\{0,1,2,3\} \] with addition and multiplication modulo \(4\) is the ring of integers modulo \(4\), denoted by \[ \mathbb{Z}_4 \]
Step 1: Check ring structure. Under addition modulo \(4\), \(\mathbb{Z}_4\) is an abelian group. Under multiplication modulo \(4\), multiplication is associative and distributive over addition. Thus, \[ \mathbb{Z}_4 \] is a ring.
Step 2: Check whether it is an integral domain. An integral domain has no zero divisors. But in \(\mathbb{Z}_4\), \[ 2\neq 0 \] and \[ 2\cdot 2=4\equiv 0\pmod 4 \] So \(\mathbb{Z}_4\) has zero divisors. Therefore, it is not an integral domain.
Step 3: Check whether it is a field. Every field is an integral domain. Since \(\mathbb{Z}_4\) is not an integral domain, it is not a field.
Step 4: Final answer. \[ \boxed{\text{a ring}} \]
The set \[ R=\{0,1,2,3\} \] with addition and multiplication modulo \(4\) is the ring of integers modulo \(4\), denoted by \[ \mathbb{Z}_4 \]
Step 1: Check ring structure. Under addition modulo \(4\), \(\mathbb{Z}_4\) is an abelian group. Under multiplication modulo \(4\), multiplication is associative and distributive over addition. Thus, \[ \mathbb{Z}_4 \] is a ring.
Step 2: Check whether it is an integral domain. An integral domain has no zero divisors. But in \(\mathbb{Z}_4\), \[ 2\neq 0 \] and \[ 2\cdot 2=4\equiv 0\pmod 4 \] So \(\mathbb{Z}_4\) has zero divisors. Therefore, it is not an integral domain.
Step 3: Check whether it is a field. Every field is an integral domain. Since \(\mathbb{Z}_4\) is not an integral domain, it is not a field.
Step 4: Final answer. \[ \boxed{\text{a ring}} \]
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