Fields

The sets $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ behave in a similar manner. They all have addition, subtraction, multiplication, and division that satisfy certain properties. This common behavior is summarised in the notion of fields. Thus, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are examples of fields.

Definition (Field)

A set $F$ is called a field if it is endowed with binary operations $+$ and $\ast$ and satisfies the following axioms.

1. $F$ is closed under $+$ (closure): $\mathrm{\forall }a,b\in F$, $a+b\in F$.
2. $+$ is commutative: $\mathrm{\forall }a,b\in F$, $a+b=b+a$.
3. $+$ is associative: $\mathrm{\forall }a,b,c\in F$, $a+(b+c)=(a+b)+c$.
4. There is an additive identity element $0$ with respect to $+$: $\mathrm{\exists }0\in F$ such that $\mathrm{\forall }a\in F$, $a+0=a$.
5. There is an inverse of each element with respect to $+$: $\mathrm{\forall }a\in F,$$\mathrm{\exists }-a\in F$ such that $a+(-a)=0$.
6. $F$ is closed under $\ast$ (closure): $\mathrm{\forall }a,b\in F$, $a\ast b\in F$.
7. $\ast$ is commutative: $\mathrm{\forall }a,b\in F$, $a\ast b=b\ast a$.
8. $\ast$ is associative: $\mathrm{\forall }a,b,c\in F$, $a\ast (b\ast c)=(a\ast b)\ast c$.
9. There is a multiplicative identity $1$ w.r.t. $\ast$: $\mathrm{\exists }1\in F$ such that $\mathrm{\forall }a\in F$, $a\ast 1=a$.
10. There is an inverse of each element except $0$ w.r.t. $\ast$: $\mathrm{\forall }a\in F\setminus \{0\},$$\mathrm{\exists }{a}^{-1}$ such that $a\ast a^{-1}=1$.
11. Distributive law: $\mathrm{\forall }a,b,c\in F$, $a\ast (b+c)=a\ast b+a\ast c$.

We usually call $+$ and $\ast$ ``addition'' and ``multiplication'', respectively.

Remark. Multiplication $\ast$ may be denoted by $\cdot$ or $\times$; or these symbols may be omitted (as in $ab$ for $a\ast b$).  □

Example. 

1. $\mathbb{Q}$ is a field.
2. $\mathbb{R}$ is a field.
3. $\mathbb{C}$ is a field.

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Example. 

1. $\mathbb{N}$ is not a field. Because $0\notin \mathbb{N}$. Even if we adopt the convention that $0\in \mathbb{N}$, the additive inverse of $1$ is $-1\notin \mathbb{N}$.
2. $\mathbb{Z}$ is not a field. Because the multiplicative inverse of $2$ is $1/2\notin \mathbb{Z}$.

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Example. Let $A=\{1,2,3\}$. $A$ is not closed under the usual addition and multiplication of numbers. For example, $2+3=5\notin A$ and $2\times 3=6\notin A$. □

What's nice about defining the field is that we can prove some properties common to all fields so that we don't have to repeat the same proof for each of, say, $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$.

Proposition 

Let $F$ be a field. We have the following:

1. For all $f\in F$, $f\cdot 0=0$.
2. For all $h,k\in F$, if $hk=0$, then $h=0$ or $k=0$.

Proof. 

1. By the definition of $0$ (additive identity), for any $f$ we have $$f\cdot 0=f\cdot (0+0).$$ By the distributive law, $$f\cdot 0=f\cdot 0+f\cdot 0.$$ Add the inverse of $f\cdot 0$ on both sides, $$-(f\cdot 0)+f\cdot 0=-(f\cdot 0)+(f\cdot 0+f\cdot 0).$$ By the associativity of addition, $$-(f\cdot 0)+f\cdot 0=(-(f\cdot 0)+f\cdot 0)+f\cdot 0=0+f\cdot 0.$$ By the definitions of an additive inverse and $0$, $$0=f\cdot 0.$$ And we are done.
2. Suppose $hk=0$. If $h=0$, we are done. Suppose $h\ne 0$. Then its multiplicative inverse $h^{-1}$ exists. Then we have, by using Part 1, $$h^{-1}(hk)=h^{-1}0=0.$$ But we have $$\begin{array}{rlr}{h}^{-1}(hk)& =(h^{-1}h)k& \text{(associativity of multiplication)}\\ & =1\cdot k& \text{(definition of multiplicative inverse)}\\ & =k& \text{(definition of multiplicative identity)}.\end{array}$$ Thus, $k=0$.

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