Ring

A ring \(R\) is a set together with two binary operations \(+\) and \(\times\) (called addition and multiplication) satisfying the following axioms:

1. \((R, +)\) is an abelian group
2. \(\times\) is associative : \((a \times b )\times c = a \times (b \times c) \) for all \(a,b,c \in R\)
3. the distributive laws hold in \(R\) : for all \(a,b,c \in R\) $$ (a + b)\times c = (a \times c) + (b \times c) $$ and $$ a \times (b + c) = (a \times b) + (a \times c) $$

Commutative ring

The ring \(R\) is commutative if multiplication is commutative.

Ring with identity

The ring \(R\) is said to have an identity (or contain a \(1\)) if there is an element \(1 \in R\) with $$ 1 \times a = a \times 1 = a \space\space \text{for all } a \in R. $$

Division ring, field

A ring \(R\) with identity \(1\), where \(1 \ne 0\), is called a division ring or skew field if every nonzero element \(a \in R\) has a multiplicative inverse, i.e., there exists \(b \in R\) such that \(ab = ba = 1\). A commutative division ring is called a field.

Proposition 1

Let \(R\) be a ring. Then,

1. \(0a = a0 = 0\) for all \(a \in R\)
2. \((-a)b = a(-b) = -(ab)\) for all \(a,b \in R\)
3. \((-a)(-b) = ab\) for all \(a,b \in R\)
4. if \(R\) has an identity \(1\), then the identity is unique and \(-a = (-1)a\).

Proof

Let \(a\) and \(b\) arbitrary elements of the ring \(R\).

$$ 0a = (0 + 0)a = 0a + 0a $$ and thus \(0a = 0\). Similarly, $$ a0 = a(0 + 0) = a0 + a0 $$ and thus \(a0 = 0\). Therefore, $$ 0a = a0 = 0 $$

$$\tag*{$\blacksquare$}$$

Using the distributive laws and what we’ve just shown, we have $$ (-a)b + ab = (-a + a)b = 0a = 0 $$ and $$ a(-b) + ab = a(-b + b) = a0 = 0 $$ That is, both \((-a)b\) and \(a(-b)\) are the inverse of \(ab\). Equivalently, $$ (-a)b = a(-b) = -(ab) $$

$$\tag*{$\blacksquare$}$$

We’ve shown \((-a)b = -(ab)\). It follows that $$ (-a)(-b) -(ab) = (-a)(-b) - a(b) = -a(-b + b) = -a0 = 0 $$ Therefore, \(-(ab)\) is the inverse of \((-a)(-b)\). That is, \((-a)(-b) = ab\).

$$\tag*{$\blacksquare$}$$

Suppose that \(R\) has identities \(x\) and \(y\). Then, by definition of identity,

$$ x = xy = y $$

Therefore, the identity is unique.

Since the identity is unique, \(1\) is the only element with which we can write \(a = 1a\). It follows that $$ a + (-1)a = 1a + (-1)a = (1 - 1)a = 0a = 0 $$

That is, \((-1)a\) is the inverse of \(a\). Or, equivalently, \(-a = (-1)a\).

$$\tag*{$\blacksquare$}$$

Zero divisor

Let \(R\) be a ring. A nonzero element \(a\) of \(R\) is called a zero divisor if there is a nonzero element \(b\) in \(R\) such that either \(ab=0\) or \(ba=0\).

Unit

Let \(R\) be a ring. Assume \(R\) has an identity \(1 \ne 0\). An element \(u\) of \(R\) is called a unit in \(R\) if there is some \(v\) in \(R\) such that \(uv = vu = 1\). The set of units in \(R\) is denoted \(R^\times\).

Subring

Let \(R\) be a ring. A subring of \(R\) is a subgroup of \(R\) that is closed under multiplication.