12.Ring and field

Overview

Group theory: have only one binary operation, can't describe relationships between multiple operations.

Ring and field: have two binary operation, they're based on group theory.

Definition

Let $<A,+,\cdot >$ be an algebraic system, and let “$+$” and “$\cdot$” be binary operations. If the following conditions are satisfied:

(1) $<A,+>$ is an abelian group;

(2) $<A,\cdot >$ is a semigroup;

(3) The left and right distributive laws are established: multiplication is distributive over addition, i.e., for $\mathrm{\forall }a,b,c\in A$,

$$a\cdot (b+c)=a\cdot b+a\cdot c,$$$$(b+c)\cdot a=b\cdot a+c\cdot a.$$

then $<A,+,\cdot >$ is called a ring. If the multiplication in $<A,+,\cdot >$ is also commutative, then $A$ is called a commutative ring.

The definition of a ring shows: A ring is an algebraic system with two operations, where addition is an abelian group, multiplication is a semigroup, and the distributive law is satisfied.

Without causing confusion, the distributive law can be written directly in the usual notation for addition and multiplication. For example, the distributive law can be directly written as

$$a(b+c)=ab+ac,(b+c)a=ba+ca$$

Note that multiplication and addition here are just two symbols and do not necessarily mean traditional addition and multiplication.

Typical rings

Integers, rational numbers, real numbers, and complex numbers form rings under the usual addition and multiplication operations, i.e., $<\mathbb{Z},+,\cdot >$, $<\mathbb{Q},+,\cdot >$, $<\mathbb{R},+,\cdot >$, $<\mathbb{C},+,\cdot >$ are all rings.

The algebraic system $<M_n(R),+,\times >$ is a ring, where $M_n(R)$ represents the set of $n$-order real matrices, and the operations “$+$” and “$\times$​” denote matrix addition and multiplication, respectively.

Important appointment

In a ring, there are two operations, and each operation may have special elements like identity and inverse.

For convenience, we make the following conventions:

Let $<A,+,\cdot >$ be a ring. The identity element of the additive group $<A,+>$ is usually denoted as 0, called the zero element (this is for multiplication). The inverse of an element $a$ with respect to addition is denoted as $-a$, called the additive inverse. If the multiplicative semigroup $<A,\cdot >$ has an identity, denote it as 1. If some element $a$ has an inverse in the multiplicative semigroup $<A,\cdot >$, it is denoted as $$a^{-1}$$.

It is clear that the identity and inverse in the ring refer to multiplication, while those in addition are called zero element and additive inverse. - The definition of multiples and powers of an element is given by: $$na = \underbrace{a + a + \cdots + a}{n\ times}\ , a^n = \underbrace{aa\cdots a}{n \text{ times}}$$ and satisfies

$$(na)b=a(nb)=nab,a^n\cdot a^m=a^{n+m},(a^n{)}^m=a^{nm}$$

For $\mathrm{\forall }a,b\in A$: - Since there is an additive inverse, subtraction can be defined as: $a-b=a+(-b)$

Property with respect to zero: $0a=a0=0,a\in A$

Property with respect to the additive inverse: $(-a)b=a(-b)=-ab,(-a)(-b)=ab$

The distributive law of subtraction also holds: $a(b-c)=ab-ac,(a-b)c=ac-bc$​

It is apparent that 0 acts as the identity element for addition, but as the zero element for multiplication.

Zero divisor

Let $A$ be a ring, for $a,b\in A$ where $a\ne 0,b\ne 0$, if $ab=0$, then $a$ is called a left zero divisor, $b$ is called a right zero divisor. If an element is both a left and right zero divisor, it is called a zero divisor.

For example, in $<{\mathbb{Z}}_6,+,\cdot >$, $\bar{2}\cdot \bar{3}=0$ (here $0$ refers to the multiplicative zero element, which is indeed $0$), so both $\bar{2}$ and $\bar{3}$​ are zero divisors.

$\bar{2}$ and $\bar{3}$ represent elements divided by 6 with a remainder of 2 and 3, respectively, and when they are multiplied by 6 there is no remainder, so the remainder 0,0 happens to be the zero element in the ring.

A sufficient and necessary condition for a ring to have no right and left zero factors is that the elimination law of multiplication holds.

Special rings and field

Let $<A,+,\cdot >$ be a ring.

(1) If $A\ne \{0\}$, commutative, with an identity element 1 and no zero divisors, then $A$ is called an integral domain.

(2) If $A$ has at least two elements $0$ and $1$, and $A^{\ast }=A-\{0\}$ forms a multiplicative group, then $A$ is called a division ring.

(3) If $A$ is a commutative division ring, then $A$ is called a field.

$\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$ are all integral domains, and $\mathbb{Q},\mathbb{R},\mathbb{C}$ are fields. Polynomial rings $\mathbb{Z}[x],\mathbb{Q}[x],\mathbb{R}[x],\mathbb{C}[x]$ are also integral domains. ${\mathbb{Z}}_n$ is not an integral domain when $n$ is not a prime number, as there are zero divisors. But when $n$ is a prime number, ${\mathbb{Z}}_n$ is a field.

Finite field

The necessary and sufficient condition for $<{\mathbb{Z}}_n,+,\cdot >$ to be a field is that $n$ is a prime number.

Proof:

Necessity: Prove by contradiction.

Suppose $n$ is not a prime number, then $n=n_1{n}_2$, $n_1\ne 1$, $n_2\ne 1$, then $\overline{n_1}\cdot \overline{n_2}=\bar{0}$ with $\overline{n_1}\ne \bar{0}$, $\overline{n_2}\ne \bar{0}$. Therefore, $\overline{n_1}$ and $\overline{n_2}$ are zero divisors, which contradicts the definition of a field.

Sufficiency: Suppose $n=p$ is a prime number, then ${\mathbb{Z}}_n\ne \{0\}$. For any $\bar{k}\in {\mathbb{Z}}_p^{\ast }$, since $(k,p)=1$, there exist $a,b\in \mathbb{Z}$ such that $ak+bp=1$, yielding $\bar{a}\bar{k}=\bar{1}$, so ${\bar{k}}^{-1}=\bar{a}$. Therefore, for any $\bar{k}\in {\mathbb{Z}}_p^{\ast }$, $k$ has an inverse, so ${\mathbb{Z}}_n^{\ast }$ is a group, making ${\mathbb{Z}}_n$ a field.

A field with a finite number of elements is called a finite field.

The field ${\mathbb{Z}}_p$ of prime order is the simplest finite field.