07-01. Basic definitions and examples
Definition (Ring)
- $R$,
- set
- $+$,
- binary operator
- addition
- $\times$,
- binary operator
- multiplication
$R$ is said to be ring if
- (i) $(R, +)$ is an abelian group
- (ii) $\times$ is associative: for all $a, b \in R$,
- $(a \times b) \times c = a \times (b \times c)$,
- (iii) the distributive laws hold in $R$: for all $a, b, c \in R$,
- $(a + b) \times c = (a \times c) + (b \times c)$,
- $a \times (b + c) = (a \times b) + (a \times c)$.
The ring $R$ is said to be commmutative if
- $R$ is commutative: for all $a, b \in R$,
- $a \times b = b \times a$.
The ring $R$ is said to have an identity if
- there is an element $1 \in R$ such that for all $a \in R$,
- $1 \times a = a \times 1 = a$,
Definition (zero divisor, unit)
- $R$,
- ring
$a \in R$ is said to be a zero divisor if
\[\exists b \neq 0 \text{ s.t. } ab = 0 \text{ or } ba = 0 .\]- $R$
- ring with indentity
$u \in R$ is said to be a unit in $R$ if
\[\exists v \in R \text{ s.t. } uv = vu = 1 .\]The set of units in $R$ is denoted $R^{\times}$.
Definition. integral domain
- $R$,
- commutative ring with identity $1 \neq 0$,
$R$ is said to be an integral domain if it has no zero divisors.
Integral domain allows you to solve a simple equation $ab = ac$.
Prposition 2.
- $a, b, c \in R$,
- $a$ is not zero divisor
In particular, if $R$ is integral domain, then the statement holds as well.
proof.
$ab = ac$,
\[a(b - c) = 0\]Then since $a$ is not zero divisor, $a = 0$ or $b - c = 0$.
Corollary 3.
- $R$,
- finite integral domain
Then $R$ is a field.
proof.
Let $a \neq 0 \in R$. The map $x \mapto ax$ is an injecive function by the canceling law. Since $R$ is finite, the map is also surjective. In particular, there exists $b \in R$ such that $ab = 1$. Hence $a$ is unit in $R$. Since $a$ was an arbitrary element in $R$, $R$ is a field.
Definition. subring
- $R$,
- ring
- $S \subseteq R$,
$S$ is said to be subring if
- $S$ is a subgroup of $R$,
- $S$ is closed under multiplication