📌 Question:
“Give me an example that a ring doesn’t satisfy the commutativity property.”
✅ Answer:
The classic example is the ring of 2×2 real matrices: \(M_2(\mathbb{R})\)
This ring satisfies:
- Associative addition and multiplication
- Distributivity of multiplication over addition
But it does not have commutative multiplication.
➕ What Do “Addition” and “Multiplication” Really Mean in Abstract Algebra?
In abstract algebra, “addition” and “multiplication” are not fixed operations like + and × on real numbers. Instead, they are abstract binary operations — their behavior is defined by the algebraic structure they’re part of, such as groups, rings, or fields.
The terms “addition” and “multiplication” are conventional names that signal which algebraic laws the operations follow — not necessarily how they are computed.
🧩 Addition: Group Operation
In rings and fields, the addition operation (often written as +) must satisfy the abelian group axioms:
| Property | Requirement |
|---|---|
| Closure | \(a + b \in S\) |
| Associativity | (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)(a+b)+c=a+(b+c) |
| Identity element | There exists \(0 \in S\)such that a + 0 = a |
| Inverse element | For every \(a \in S\), there exists \(-a \in S\) such that a + (-a) = 0 |
| Commutativity | a + b = b + a |
This is always true in:
- Groups under addition (if specified)
- Rings
- Fields
✖️ Multiplication: Varies by Structure
The multiplication operation (written as ⋅ or just juxtaposed) has weaker or stronger requirements depending on the structure:
| Structure | Multiplication Must Be… |
|---|---|
| Semigroup | Associative |
| Monoid | Associative + Identity |
| Ring | Associative, distributes over addition |
| Commutative Ring | Same as ring + multiplication is commutative |
| Field | Commutative, associative, identity, and every nonzero element has an inverse |
So in a ring, multiplication must be:
- Associative
- Distributive over addition:
- \(a \cdot (b + c) = a \cdot b + a \cdot c \quad \text{(left distributivity)} \)
- \((a + b) \cdot c = a \cdot c + b \cdot c \quad \text{(right distributivity)}\)
But a ring does not require:
- Commutativity of multiplication
- Multiplicative inverses
- Multiplicative identity (though many rings do have one)
⚠️ Important Distinction
The key difference is:
- Addition must always satisfy the group structure (abelian group)
- Multiplication has more flexible rules, depending on the structure:
- In rings, multiplication isn’t even required to be invertible or commutative
- In fields, it must be commutative, and every nonzero element must have a multiplicative inverse
a field is a group and a field is also a ring, is this correct?
ChatGPT said:
Great question! You’re very close, but let’s clarify the relationships properly:
✅ Corrected Understanding:
A field is:
- An abelian group under addition
- A commutative group (except 0) under multiplication
- A commutative ring with unity in which every nonzero element has a multiplicative inverse
So the correct phrasing would be:
✅ A field is:
- A ring (commutative, with 1)
- An abelian group under addition:
(F,+)(F, +)(F,+) is a group with identity 0 - A commutative group under multiplication for all nonzero elements:
(F∖{0},⋅)(F \setminus \{0\}, \cdot)(F∖{0},⋅) is a group with identity 1 - Satisfies distributivity:
a(b+c)=ab+aca(b + c) = ab + aca(b+c)=ab+ac
❌ So to clarify:
| Statement | Is it true? |
|---|---|
| A field is a group | ❌ Not exactly — it’s two groups: one under +, one under * (without 0) |
| A field is a ring | ✅ Yes — a field is a special kind of ring (commutative with 1, and every nonzero element invertible) |