Section 18 Rings and Fields

18.1 Definition Rings

p.167

A ring $⟨\mathbb{R},+,\cdot ⟩$ is a set R together with two binary operations + and ·, which we call addition and multiplication, defined on R such that the following axioms are satisfied:

• R1. $⟨\mathbb{R},+,\cdot ⟩$ is an abelian group.
• R2. Multiplication is associative.
• R3. For all a, b, c ∈ R, the left distributive law, a · (b + c) = (a · b) + (a · c) and the right distributive law (a + b) · c = (a · c) + (b · c) hold.

For example, $M_n(R)$ ($n\times n$ matrix with real number entries) is a ring

Algebra Structure

+−⟲

Link to original

18.8 Properties of a ring

p.170

If R is a ring with additive identity 0, then for any a, b ∈ R we have

1. 0a = a0 = 0,
2. a(−b) = (−a)b = −(ab),
3. (−a)(−b) = ab.

18.14 Definition Commutative Ring and Unity

p.172

• A ring in which the multiplication is commutative is a commutative ring.
• A ring with a multiplicative identity element is a ring with unity;
• the multiplicative identity element 1 is called “unity.”

18.16 Definition Division Ring and Field

Overview of Algebra Structures

p.173

Let R be a ring with unity $1\ne 0$. An element u in R is a unit of R if it has a multiplicative inverse in R. If every nonzero element of R is a unit, then R is a division ring (or skew field).

NOTE

A unity is always a unit

NOTE

A field is a commutative division ring. A noncommutative division ring is called a “strictly skew field.”

Question

Why can an element on a ring not be divided by $0$

In a ring you can only “divide” by an element if it has a multiplicative inverse. But for every ring $0\cdot x=0$ for all x. If zero had an inverse y, you’d need $0\cdot y=1$ which is impossible since $0\cdot y$ is always 0, never 1. Hence 0 cannot have an inverse (except in the trivial ring where $0=1$), and so “division by zero” is undefined.

18.17 Example

In the Ring ${\mathbb{Z}}_{14}$ : the units are 1,13(-1),3,5,11(-3),9(-5) (-3)(-5)=3*5=1

2,4,6,7,8,10 are not units because gcd(m,14)>1