Rings: Definition, Axioms, Examples

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2014)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~18 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

The ring axioms are the gateway to the entire second half of the algebra syllabus: ideals, quotient rings, PIDs, UFDs, and polynomial rings all presuppose a clean command of what a ring is and how to verify the axioms. UPSC 2014 asked a 15-mark Section B question that required working through the ring axioms for a specific set with defined operations. The most common trap is forgetting to verify every axiom — particularly additive inverses and distributivity — rather than just closure.

Minimum Theory

Definition

A ring $(R, +, \cdot)$ is a set $R$ with two binary operations $+$ (addition) and $\cdot$ (multiplication) satisfying:

R1. $(R, +)$ is an abelian group:

• (R1a) Closure: $a + b \in R$.
• (R1b) Associativity: $(a + b) + c = a + (b + c)$.
• (R1c) Identity: $\exists\, 0 \in R$ such that $a + 0 = a$.
• (R1d) Inverses: $\exists\, {-a} \in R$ such that $a + (-a) = 0$.
• (R1e) Commutativity: $a + b = b + a$.

R2. Multiplication is associative: $(ab)c = a(bc)$.

R3. Distributivity (both sides): $a(b + c) = ab + ac \quad \text{and} \quad (a + b)c = ac + bc.$

Additional Structure

Standard Examples

$(\mathbb{Z}, +, \cdot)$: Commutative ring with unity 1. No zero divisors. Not a field (most elements have no multiplicative inverse in $\mathbb{Z}$).

$(\mathbb{Z}_n, +_n, \cdot_n)$: Ring under addition and multiplication mod $n$. Has zero divisors iff $n$ is composite. A field iff $n$ is prime.

$(2\mathbb{Z}, +, \cdot)$ (even integers): Commutative ring without unity (no even integer serves as multiplicative identity).

$M_n(R)$ (matrix ring): Ring under matrix addition and multiplication. Non-commutative for $n \ge 2$. Has zero divisors (singular matrices).

$R[x]$ (polynomial ring): If $R$ is a commutative ring, so is $R[x]$. $\deg(fg) = \deg f + \deg g$ when $R$ is an integral domain.

Key Consequences of the Axioms

From the axioms alone (without assuming commutativity), one can prove:

• $0 \cdot a = a \cdot 0 = 0$ for all $a$ (the additive identity is a multiplicative absorber).
• $(-a)b = a(-b) = -(ab)$.
• $(-1)a = -a$ if $R$ has unity.

Question Archetypes

verify-ring-axioms (1 question; 2014)

Recognition Cues

• “Show that $(R, +, \cdot)$ is a ring” or “verify whether the following is a ring”
• Operations are given explicitly (possibly non-standard)
• 15 marks: every axiom must be checked

Solution Template

1. Additive abelian group: Check closure, associativity, identity ($0$), inverses ($-a$), commutativity — in that order.
2. Multiplicative associativity: Check $(ab)c = a(bc)$.
3. Distributivity: Check both $a(b+c) = ab + ac$ and $(a+b)c = ac + bc$.
4. State whether the ring has unity; if so, exhibit it.
5. State whether the ring is commutative.

Worked Example

2014 Paper 2, 2014-P2-Q5b (15 marks)

Let $R = 2\mathbb{Z} = \{ \ldots, -4, -2, 0, 2, 4, \ldots \}$ be the set of all even integers, with the usual addition and multiplication of integers. Show that $R$ is a commutative ring without unity.

Solution.

We verify all ring axioms for $(2\mathbb{Z}, +, \cdot)$.

R1. $(2\mathbb{Z}, +)$ is an abelian group.

• Closure: If $a = 2m$ and $b = 2n$ (with $m, n \in \mathbb{Z}$), then $a + b = 2(m+n) \in 2\mathbb{Z}$. $\checkmark$
• Associativity: Inherited from $(\mathbb{Z}, +)$: $(a+b)+c = a+(b+c)$. $\checkmark$
• Additive identity: $0 = 2 \cdot 0 \in 2\mathbb{Z}$ and $a + 0 = a$. $\checkmark$
• Additive inverses: If $a = 2m \in 2\mathbb{Z}$ then $-a = -2m = 2(-m) \in 2\mathbb{Z}$ and $a + (-a) = 0$. $\checkmark$
• Commutativity: $a + b = b + a$ inherited from $\mathbb{Z}$. $\checkmark$

R2. Multiplication is associative.

If $a, b, c \in 2\mathbb{Z}$ then $(ab)c = a(bc)$ since $\mathbb{Z}$ is associative under multiplication. $\checkmark$

R3. Distributivity.

For all $a, b, c \in 2\mathbb{Z} \subseteq \mathbb{Z}$: $a(b+c) = ab + ac \quad \text{and} \quad (a+b)c = ac + bc$ hold in $\mathbb{Z}$, hence in $2\mathbb{Z}$. $\checkmark$

Also, multiplication is closed: $a = 2m, b = 2n \Rightarrow ab = 4mn = 2(2mn) \in 2\mathbb{Z}$. $\checkmark$

Commutativity of multiplication: $ab = ba$ for all $a, b \in 2\mathbb{Z} \subseteq \mathbb{Z}$. So $2\mathbb{Z}$ is a commutative ring. $\checkmark$

No unity. Suppose for contradiction that $e \in 2\mathbb{Z}$ is a multiplicative identity. Then $e \cdot 2 = 2$, giving $e = 1$. But $1 \notin 2\mathbb{Z}$ (since 1 is odd). Contradiction. Hence $2\mathbb{Z}$ has no unity.

Conclusion. $(2\mathbb{Z}, +, \cdot)$ is a commutative ring without unity.

$\boxed{2\mathbb{Z} \text{ is a commutative ring without unity}}\;\blacksquare$

Common Traps

• Skipping the closure of multiplication — it must be checked separately (the fact that $(R,+)$ is closed under addition does not imply $R$ is closed under multiplication).
• Only checking one of the two distributive laws — both $a(b+c)$ and $(a+b)c$ must be verified in a non-commutative setting.
• Confusing “no unity” with “no identity element for $+$” — the additive identity $0$ is always present; it is the multiplicative identity $1$ that may be absent.
• Asserting commutativity of multiplication by “inheritance from $\mathbb{Z}$” without noting that the set $R$ is a subset of $\mathbb{Z}$ and inherits the operation.

Marks-Aware Writing

15 marks (Section B):

• Additive group (R1a–e): ~6 marks (roughly 1 mark per axiom; write each explicitly).
• Multiplicative associativity (R2): ~2 marks.
• Distributivity (R3): ~3 marks.
• Commutativity / unity question: ~2 marks.
• Clarity and conclusion: ~2 marks.

Never skip an axiom. Even if the result is “obvious by inheritance from $\mathbb{Z}$”, state which property is being inherited.

Practice Set

Only one historical question on this atom (shown above).