Ring, Field and Integral Domain

Rings

Ring

A ring $R$ is a set equipped with two laws of composition $+$ and $\cdot$, called addition and multiplication, satisfying the following axioms:

• $(R,+)$ is an abelian group with identity $0$, called zero.
• The multiplication is associative: $a(bc)=(ab)c$ for all $a,b,c\in R$.
• Distributivity: $(a+b)c=ac+bc$ for all $a,b,c\in R$.

If $R$ has identity $1$ with respect to multiplication we say that $R$ is unital.

Commutative Ring

A ring $R$ is commutative if the multiplication is commutative.

e.g. The integers $\mathbb{Z}$ is a commutative unital ring.

Attention

We will restrict to commutative unital rings and just call them rings.

Subring

A subset $S$ in a ring $R$ is called a subring if $S$ is closed under addition, subtraction, multiplication and contains $1$.

Unit

An element of a ring $R$ is called a unit if it is invertible with respect to multiplication. The set of invertible elements is a group called the the group of units in $R$ and denotes $R^{\times }$.

Field and Domain

Zero Divisor & Integral Domain

A zero divisor in a ring $R$ is a non-zero element $a$ such that $ab=0$ for some non-zero $b\in R$. A ring without zero divisors is called an integral domain. In other words, an integral domain is a ring in which the product of any two non-zero elements is non-zero.

Field

A field is a ring $F$ in which $1\ne 0$ and every non-zero element is invertible, that is $F^{\times }=F\backslash \{0\}$.

e.g. The rational numbers $\mathbb{Q}$ is a field.

Proposition

Any field is an integral domain.

Proof Suppose $ab=0$, $a$ can be either $0$ or a unit, if it is not $0$, then $a^{-1}ab=b=0$. $\square$

Proposition

${\mathbb{Z}}_n$ is a field if and only if $n$ is a prime.

Proof Suppose $n$ is prime, then every non-zero element $[a]\in {\mathbb{Z}}_n$ satisfies $a^{n-1}\equiv 1\mathrm{mod}n$ by Fermat’s Little Theorem. Therefore $[a{]}^{-1}=[a{]}^{n-2}$. Conversely, if $n$ is not prime, then there exists $[a],[b]\in {\mathbb{Z}}_n$ such that $[a][b]=[n]=0$ but $a,b\ne 0$, so ${\mathbb{Z}}_n$ is not even an integral domain. $\square$

Cancellation Law

An integral domain $R$ satisfies the cancellation law: if $ab=ac$ and $a\ne 0$ then $b=c$.

Proof Suppose $ab=ac$ and $a\ne 0$. Then $ab-ac=a(b-c)=0$. Since $a\ne 0$ and $R$ is an integral domain, $b-c=0$. $\square$

Divisibility in Rings

Let $R$ be a ring. An element $a\in R$ divides $b\in R$ if there exists $c\in R$ with $b=ac$.

Ordered Field

Ordered Field

An ordered field is a field $F$ along with a subset $P$ of $F$, called the positive subset, with the following properties:

• if $a\in F$, then $a\in P$ or $a=0$ or $-a\in P$.
• if $a\in P$, then $-a\notin P$.
• if $a,b\in P$, then $a+b\in P$ and $ab\in P$.

Equivalently, a field $F$ together with a total order $\le$ on $F$ is an ordered field if the order satisfies the following properties for all $a,b,c\in F$:

• $a\le b⟹a+c\le b+c$.
• $0\le a,0\le b⟹0\le ab$.

Proposition

The positive subset $P$ is closed under multiplicative inverse. i.e. Suppose $F$ is an ordered field with positive subset $P$. Then $1\in P$ and $a^{-1}\in P$ for all $a\in F$.