The integers mod 6 — the set {0, 1, 2, 3, 4, 5} with addition and multiplication mod 6 — is a ring. Is it a field?
AYes, because every element has an additive inverse
BNo, because the element 2 has no multiplicative inverse mod 6
CYes, because 6 is a positive integer and Z/6Z is finite
DNo, because Z/6Z is not commutative
A field requires every *nonzero* element to have a multiplicative inverse. In Z/6Z, the element 2 has no inverse: there is no integer n with 2n ≡ 1 (mod 6), because 2 and 6 share the common factor 2. The deeper reason is that 6 is not prime, so the ideal 6Z is not maximal, and the quotient Z/6Z is not a field. Option A confuses additive inverses (which exist in any ring) with multiplicative inverses (the field condition). Option D is wrong because Z/6Z is commutative.
Question 2 Multiple Choice
The characteristic of a field is the smallest positive integer n such that n·1 = 0. A student claims a field could have characteristic 6 (since 6 = 2·3 and 6·1 = 0 in Z/6Z). What is wrong with this reasoning?
ANothing — fields can have any positive integer characteristic
BZ/6Z is not a field, so it cannot serve as evidence for field characteristics
CThe characteristic of a field must be prime or zero, and the field axioms themselves force this
DCharacteristic is only defined for infinite fields like Q or R
The student's example fails because Z/6Z is not a field (2 has no inverse). More importantly, the field axioms themselves rule out composite characteristic. If the characteristic were n = ab with 1 < a, b < n, then (a·1)(b·1) = ab·1 = 0 in the field — but a field has no zero divisors (if xy = 0 then x = 0 or y = 0, since x has an inverse). This contradicts a·1 ≠ 0 and b·1 ≠ 0, which follow from n being the *minimum* positive integer giving 0. So characteristic must be 0 or prime.
Question 3 True / False
Nearly every field has characteristic 0.
TTrue
FFalse
Answer: False
False. Finite fields like Z/pZ (integers mod a prime p) have characteristic p, a prime number. In Z/5Z, adding 1 to itself 5 times gives 0: 1+1+1+1+1 = 5 ≡ 0 (mod 5). Fields of characteristic 0 — like Q, R, and C — are infinite, because in characteristic 0 the elements 1, 1+1, 1+1+1, ... are all distinct. But characteristic-0 fields do not exhaust all fields.
Question 4 True / False
In a field, the element zero (the additive identity) should have a multiplicative inverse.
TTrue
FFalse
Answer: False
False — and this is why the field axiom says *every nonzero element* has a multiplicative inverse, not every element. If 0 had a multiplicative inverse x, then 1 = 0·x = 0, which would make the multiplicative identity equal to the additive identity and collapse the field to the trivial ring {0}. Fields are defined to have at least two elements (0 ≠ 1), so 0 is explicitly excluded from the multiplicative inverse requirement.
Question 5 Short Answer
Why is Z/pZ a field when p is prime, but Z/nZ is not a field when n is composite? Connect the answer to the relationship between maximal ideals and fields.
Think about your answer, then reveal below.
Model answer: Z/nZ is a field if and only if nZ is a maximal ideal of Z, which happens if and only if n is prime. When p is prime, any ideal of Z/pZ containing a nonzero element must contain an invertible element, which forces it to equal all of Z/pZ — so Z/pZ has no proper nonzero ideals and every nonzero element has a multiplicative inverse. When n is composite, say n = ab, the element a is nonzero in Z/nZ but has no inverse (since gcd(a,n) > 1), so the ideal generated by a is proper and nonzero, showing nZ is not maximal and Z/nZ is not a field.
The key theorem is: a commutative ring R is a field ⟺ R has no proper nonzero ideals ⟺ R is a quotient by a maximal ideal. For Z, the ideals are exactly nZ for non-negative integers n. The ideal nZ is maximal if and only if n is prime. This connects the abstract field definition directly to the primality condition: the multiplicative inverses exist mod n precisely because prime moduli make every nonzero residue coprime to the modulus.