Questions: Field Definition and Examples

Question 1

Is ℤ/6ℤ a field? Why or why not?

• A. Yes — it is a commutative ring with unity, which is all a field requires
• B. No — because 6 is not prime, elements like [2] have no multiplicative inverse in ℤ/6ℤ
• C. Yes — every modular ring is a field because arithmetic always works modulo n
• D. No — because ℤ/6ℤ contains zero divisors, which disqualifies it as an integral domain, but it could still be a field

Answer: (B)

A field requires every nonzero element to have a multiplicative inverse. In ℤ/6ℤ, [2] has no inverse because gcd(2, 6) = 2 ≠ 1 — no element [k] satisfies [2]·[k] = [1]. This also means [2]·[3] = [6] = [0], revealing a zero divisor and disqualifying ℤ/6ℤ as even an integral domain. Only ℤ/pℤ for prime p is a field, because primality guarantees every nonzero element is coprime to p, so Bézout's theorem yields an inverse.

Question 2

Which of the following correctly describes the relationship between fields and integral domains?

• A. Every integral domain is a field, but not every field is an integral domain
• B. Every field is an integral domain, but not every integral domain is a field
• C. Fields and integral domains are identical — they have the same axioms
• D. Neither implies the other — they are independent algebraic structures

Answer: (B)

Every field is an integral domain: if ab = 0 in a field and a ≠ 0, multiply both sides by a⁻¹ to get b = 0, so there are no zero divisors. But the converse fails — ℤ is an integral domain (no zero divisors) yet 2 has no multiplicative inverse in ℤ, so ℤ is not a field. The integers are the canonical counterexample showing the inclusion is strict.

Question 3

Nearly every integral domain is a field.

• T. True
• F. False

Answer: False

The integers ℤ are an integral domain — they have no zero divisors — but ℤ is not a field because most elements lack multiplicative inverses. For example, 2 has no inverse in ℤ because 1/2 ∉ ℤ. A field is a strictly stronger structure: it requires multiplicative inverses for all nonzero elements, not just the absence of zero divisors.

Question 4

ℤ/pℤ is a field for every prime p.

• T. True
• F. False

Answer: True

When p is prime, every nonzero element [a] in ℤ/pℤ satisfies gcd(a, p) = 1. By Bézout's theorem, there exist integers s, t such that as + pt = 1, which means [a]·[s] = [1] in ℤ/pℤ — so [s] is the multiplicative inverse of [a]. Since every nonzero element has an inverse and ℤ/pℤ is already a commutative ring with unity, it is a field.

Question 5

Why does ℤ/pℤ form a field when p is prime but not when p is composite? Explain the key algebraic reason.

Model answer: When p is prime, every integer 1 ≤ a < p is coprime to p, so Bézout's theorem guarantees a multiplicative inverse for each nonzero element [a] in ℤ/pℤ. When p is composite, say p = mn with 1 < m, n < p, then [m]·[n] = [mn] = [0], producing a zero divisor — which immediately means the ring cannot be a field, since in a field ab = 0 forces a = 0 or b = 0.

Primality is the exact condition needed to ensure every nonzero residue class is invertible. A composite modulus always produces zero divisors from its non-trivial factorizations, which violates even the integral domain property, let alone the field property. This is why the prime condition is not just sufficient but necessary for ℤ/nℤ to be a field.