# Field/Focus on R/Introduction/Section

## Definition

A set ${\displaystyle {}K}$ is called a field if there are two binary operations (called addition and multiplication)

${\displaystyle +:K\times K\longrightarrow K{\text{ and }}\cdot :K\times K\longrightarrow K}$

and two different elements ${\displaystyle {}0,1\in K}$, which fulfill the following properties.

1. Law of associativity: ${\displaystyle {}(a+b)+c=a+(b+c)}$ holds for all ${\displaystyle {}a,b,c\in K}$.
2. Law of commutativity: ${\displaystyle {}a+b=b+a}$ holds for all ${\displaystyle {}a,b\in K}$.
3. ${\displaystyle {}0}$ is the neutral element of the addition, i.e. ${\displaystyle {}a+0=a}$ holds for all ${\displaystyle {}a\in K}$.
4. Existence of the negative: For every ${\displaystyle {}a\in K}$, there exists an element ${\displaystyle {}b\in K}$ with ${\displaystyle {}a+b=0}$.
2. Axioms of the multiplication:
1. Law of associativity: ${\displaystyle {}(a\cdot b)\cdot c=a\cdot (b\cdot c)}$ holds for all ${\displaystyle {}a,b,c\in K}$.
2. Law of commutativity: ${\displaystyle {}a\cdot b=b\cdot a}$ holds for all ${\displaystyle {}a,b\in K}$.
3. ${\displaystyle {}1}$ is the neutral element for the multiplication, i.e. ${\displaystyle {}a\cdot 1=a}$ holds for all ${\displaystyle {}a\in K}$.
4. Existence of the inverse: For every ${\displaystyle {}a\in K}$ with ${\displaystyle {}a\neq 0}$, there exists an element ${\displaystyle {}c\in K}$ such that ${\displaystyle {}a\cdot c=1}$.
3. Law of distributivity: ${\displaystyle {}a\cdot (b+c)=(a\cdot b)+(a\cdot c)}$ holds for all ${\displaystyle {}a,b,c\in K}$.

It is known from school that all these axioms hold for the real numbers (and for the rational numbers) together with the natural operations.

In a field, we use the convention that multiplication connects stronger than addition. Hence, we write ${\displaystyle {}a\cdot b+c\cdot d}$ instead of ${\displaystyle {}(a\cdot b)+(c\cdot d)}$. To further simplify the notation, the product sign is usually omitted. The special elements ${\displaystyle {}0}$ and ${\displaystyle {}1}$ in a field are called (the) zero and (the) one. By definition, they have to be different.

For us, the most important examples for a field are the field of rational numbers, the field of real numbers and the field of complex numbers (to be introduced later).

## Lemma

Suppose ${\displaystyle {}K}$ is a field. Then for every element ${\displaystyle {}x\in K}$ the element ${\displaystyle {}y}$ fulfilling ${\displaystyle {}x+y=0}$ is uniquely determined. For ${\displaystyle {}x\neq 0}$ the element ${\displaystyle {}z}$ fulfilling ${\displaystyle {}xz=1}$ is also uniquely determined.

### Proof

Let ${\displaystyle {}x}$ be given and suppose that ${\displaystyle {}y}$ and ${\displaystyle {}y'}$ are elements fulfilling ${\displaystyle {}x+y=0=x+y'}$. Then

${\displaystyle {}y=y+0=y+(x+y')=(y+x)+y'=(x+y)+y'=0+y'=y'\,,}$

which means altogether ${\displaystyle {}y=y'}$. For the second part see exercise.

${\displaystyle \Box }$

## Example

We are trying to find a structure of a field on the set ${\displaystyle {}\{0,1\}}$. If ${\displaystyle {}0}$ is supposed to be the neutral element of the addition and ${\displaystyle {}1}$ the neutral element of the multiplication, then everything is already determined: The equation ${\displaystyle {}1+1=0}$ must hold, since ${\displaystyle {}1}$ has an inverse element with respect to the addition, and since ${\displaystyle {}0\cdot 0=0}$ holds, due to fact. Hence the operation tables look like

${\displaystyle {}+}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$
${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$
${\displaystyle {}1}$ ${\displaystyle {}1}$ ${\displaystyle {}0}$

and

${\displaystyle {}\cdot }$ ${\displaystyle {}0}$ ${\displaystyle {}1}$
${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$
${\displaystyle {}1}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$

With some tedious computations, one can check that this is indeed a field.

## Lemma

Let ${\displaystyle {}K}$ be a field,

and let ${\displaystyle {}a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}}$ denote elements from ${\displaystyle {}K}$. Then the following statements hold.
1. ${\displaystyle {}a0=0}$ (annulation rule).
2. ${\displaystyle {}(-a)b=-ab=a(-b)\,}$

(rules for sign).

3. ${\displaystyle {}(-a)(-b)=ab\,.}$
4. ${\displaystyle {}a(b-c)=ab-ac\,.}$
5. From ${\displaystyle {}a\cdot b=0}$ one can deduce ${\displaystyle {}a=0}$ or ${\displaystyle {}b=0}$.
6. ${\displaystyle {}{\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}}$ (general law of distributivity).

### Proof

1. We have ${\displaystyle {}a0=a(0+0)=a0+a0}$. Subtracting ${\displaystyle {}a0}$ (meaning addition with the negative of ${\displaystyle {}a0}$) on both sides gives the claim.
2. See exercise.
3. See exercise.
4. See exercise.
5. We prove this by contradiction, so we assume that ${\displaystyle {}a}$ and ${\displaystyle {}b}$ are both not ${\displaystyle {}0}$. Then there exist inverse elements ${\displaystyle {}a^{-1}}$ and ${\displaystyle {}b^{-1}}$ and hence ${\displaystyle {}(ab){\left(b^{-1}a^{-1}\right)}=1}$. On the other hand, we have ${\displaystyle {}ab=0}$ by the premise and so the annulation rule gives
${\displaystyle {}(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,}$

hence ${\displaystyle {}0=1}$, which contradicts the field properties.

6. This follows with a double induction, see exercise.
${\displaystyle \Box }$