In ℤ/6ℤ, we have 2 × 1 ≡ 2 × 4 (mod 6), yet 1 ≠ 4. This failure of the cancellation law is best explained by:
Aℤ/6ℤ lacks a multiplicative identity, so cancellation is undefined
Bℤ/6ℤ is not commutative, and the cancellation law requires commutativity
Cℤ/6ℤ has zero divisors (2 × 3 ≡ 0 mod 6), and the cancellation law fails precisely when zero divisors are present
DCancellation is only valid for elements with multiplicative inverses, and 2 has no inverse in ℤ/6ℤ
The cancellation law and the absence of zero divisors are logically equivalent in a commutative ring with unity. In ℤ/6ℤ, 2 × 3 ≡ 0 with both 2 and 3 nonzero — so 2 is a zero divisor. This is precisely why 2 × 1 ≡ 2 × 4 doesn't imply 1 = 4: subtracting gives 2 × (1 − 4) ≡ 2 × (−3) ≡ 0, but −3 ≢ 0. The zero-divisor relationship is the root cause, and this equivalence is the central theorem of integral domain theory.
Question 2 Multiple Choice
Which of the following is an integral domain?
Aℤ/6ℤ (integers mod 6)
Bℤ/4ℤ (integers mod 4)
Cℤ/5ℤ (integers mod 5)
Dℤ/9ℤ (integers mod 9)
ℤ/nℤ is an integral domain if and only if n is prime. Since 5 is prime, ℤ/5ℤ has no zero divisors and is actually a field — and every field is an integral domain. In contrast: ℤ/6ℤ has 2 × 3 ≡ 0, ℤ/4ℤ has 2 × 2 ≡ 0, and ℤ/9ℤ has 3 × 3 ≡ 0. Each of these has zero divisors, so none is an integral domain.
Question 3 True / False
Most integral domain is a field.
TTrue
FFalse
Answer: False
The integers ℤ are the canonical counterexample. ℤ is a commutative ring with unity and has no zero divisors — so it is an integral domain. But ℤ is not a field because most integers lack multiplicative inverses within ℤ (there is no integer n such that 2n = 1). Every field is an integral domain, but not every integral domain is a field. The hierarchy runs: fields ⊂ integral domains ⊂ commutative rings with unity.
Question 4 True / False
Every field is an integral domain.
TTrue
FFalse
Answer: True
In a field, every nonzero element has a multiplicative inverse. Suppose ab = 0 with a ≠ 0 in a field. Then a has an inverse a⁻¹, and multiplying both sides gives b = 0. This proves fields have no zero divisors, satisfying the definition of an integral domain. Since fields are also commutative rings with unity, they meet all requirements. Fields ⊂ integral domains, not the other way around.
Question 5 Short Answer
Why does the absence of zero divisors in an integral domain guarantee that the cancellation law holds?
Think about your answer, then reveal below.
Model answer: Suppose ac = bc with c ≠ 0. Subtracting gives (a − b)c = 0. Since c ≠ 0 and there are no zero divisors, we must have a − b = 0, so a = b. The cancellation law follows directly from the no-zero-divisors condition. Conversely, if the ring had zero divisors — say dc = 0 with d ≠ 0 and c ≠ 0 — then 0·c = d·c but 0 ≠ d, so cancellation fails. The two conditions are logically equivalent.
The deeper point is that cancellation is what makes divisibility arguments work the way we expect from the integers: if 6 = 2×3 and 6 = 2×k, we can conclude k = 3. In a ring with zero divisors, this reasoning breaks down entirely. The entire theory of GCDs, prime factorization, and irreducibility depends on being able to cancel — which is why these concepts are defined in integral domains, not arbitrary commutative rings.