The set of even integers 2Z = {..., −4, −2, 0, 2, 4, ...} with ordinary addition and multiplication — which of the following is true?
A2Z is a field — every nonzero even integer has a multiplicative inverse in 2Z
B2Z is a ring with unity — the element 2 serves as the multiplicative identity
C2Z is a ring without unity — all ring axioms are satisfied, but no multiplicative identity exists in 2Z
D2Z is not a ring — it fails closure under multiplication since products of even integers are not always even
2Z satisfies all ring axioms: (2Z, +) is an abelian group (closed, associative, identity 0, inverses, commutative); multiplication is associative; distributivity holds. But there is *no multiplicative identity in 2Z* — no even integer e such that e·n = n for all even n. The integer 1 would work, but 1 ∉ 2Z. This makes 2Z a ring *without unity*. Option B is wrong: 2·4 = 8 ≠ 4, so 2 is not a multiplicative identity. Option D is wrong: 2×2 = 4 ∈ 2Z, so closure under multiplication holds — even integers are always even.
Question 2 Multiple Choice
Which of the following is NOT required by the ring axioms?
AAddition is commutative
BEvery element has an additive inverse
CMultiplication is commutative
DMultiplication distributes over addition from both sides
Commutativity of multiplication is *not* required by the ring axioms — only associativity and distributivity are required for multiplication. The set of 2×2 real matrices M₂(R) is the standard example: it is a ring (with all required axioms satisfied), but matrix multiplication is not commutative (AB ≠ BA in general). A ring where multiplication *is* commutative is specifically called a *commutative ring*. Options A, B, and D are all required: (R, +) must be an abelian group (commutativity of addition, A; additive inverses, B), and distributivity (D) is the key axiom linking the two operations.
Question 3 True / False
In any ring R, the identity 0·a = 0 for all a ∈ R is a theorem that follows from the ring axioms — it does not need to be assumed as an additional axiom.
TTrue
FFalse
Answer: True
True. Proof using only ring axioms: 0·a = (0 + 0)·a = 0·a + 0·a (by distributivity). Adding the additive inverse of 0·a to both sides (which exists since (R,+) is a group): 0 = 0·a. No additional assumptions are needed. This demonstrates the power of the axiom system — properties that seem obvious often fall out as theorems. The analogous result a·0 = 0 follows from the other distributive law. Zero always 'kills' multiplication, distinguishing the additive identity from any possible multiplicative identity.
Question 4 True / False
Nearly every ring is expected to contain a multiplicative identity element (unity), just as most ring is expected to contain an additive identity element (zero).
TTrue
FFalse
Answer: False
False. An additive identity (0) is required — (R, +) must be an abelian group, which mandates an identity element. But a multiplicative identity (1) is *not* required by the ring axioms. Rings possessing a multiplicative identity are called 'rings with unity' or 'unital rings'; those without are valid rings lacking this additional structure. The even integers 2Z are the canonical example: all ring axioms are satisfied, yet no even integer serves as a multiplicative identity. Many important algebraic structures are rings without unity.
Question 5 Short Answer
What is a zero divisor in a ring, and why does the ring of integers Z contain no zero divisors?
Think about your answer, then reveal below.
Model answer: A zero divisor is a nonzero element a in a ring R such that there exists a nonzero element b with ab = 0. In Z, if ab = 0 for integers a and b, then at least one must be zero — this is the fundamental property of integer multiplication. Z is therefore an integral domain (a commutative ring with unity and no zero divisors). By contrast, in Z/6Z, the elements 2 and 3 are both nonzero yet 2·3 = 6 ≡ 0 (mod 6), so both are zero divisors.
Zero divisors are what distinguish general rings from integral domains, and integral domains from fields. The absence of zero divisors enables cancellation: if ab = ac and a ≠ 0, then b = c (multiply both sides by a's inverse in the appropriate sense). This fails in Z/6Z: 2·3 = 2·0 (mod 6) but 3 ≠ 0. Understanding zero divisors is key to the algebraic hierarchy: fields (every nonzero element has a multiplicative inverse) ⊂ integral domains (no zero divisors, commutative, with unity) ⊂ commutative rings ⊂ rings.