# Field/Focus on R/Introduction/Section
## Definition

A set $K$ is called a field if there are two binary operations (called addition and multiplication)
    $+:K\times K\longrightarrow K{\text{ and }}\cdot :K\times K\longrightarrow K$
and two different elements $0,1\in K$, which fulfill the following properties.
* Axioms for the addition:
** Law of associativity: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a+b=b+a$ holds for all $a,b\in K$.
** $0$ is the neutral element of the addition, i.e. $a+0=a$ holds for all $a\in K$.
** Existence of the negative: For every $a\in K$, there exists an element $b\in K$ with $a+b=0$.
* Axioms of the multiplication:
** Law of associativity: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a\cdot b=b\cdot a$ holds for all $a,b\in K$.
** $1$ is the neutral element for the multiplication, i.e. $a\cdot 1=a$ holds for all $a\in K$.
** Existence of the inverse: For every $a\in K$ with $a\neq 0$, there exists an element $c\in K$ such that $a\cdot c=1$.
* Law of distributivity: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in K$.
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 $a\cdot b+c\cdot d$ instead of $(a\cdot b)+(c\cdot d)$. To further simplify the notation, the product sign is usually omitted. The special elements $0$ and $1$ 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 $K$ is a field. Then for every element $x\in K$ the element $y$ fulfilling $x+y=0$ is uniquely determined. For $x\neq 0$ the element $z$ fulfilling $xz=1$

### Proof

Let $x$ be given and suppose that $y$ and $y'$ are elements fulfilling $x+y=0=x+y'$. Then
    $y=y+0=y+(x+y')=(y+x)+y'=(x+y)+y'=0+y'=y'\,,$
which means altogether $y=y'$. For the second part see exercise.
$\Box$

## Example

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

## Lemma

Let $K$ be a field,
$a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$
$K$
* $a0=0$ (annulation rule).
* (rules for sign).
    $(-a)b=-ab=a(-b)\,$
    $(-a)(-b)=ab\,.$
    $a(b-c)=ab-ac\,.$
* From $a\cdot b=0$ one can deduce $a=0$ or $b=0$.
* ${\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).

### Proof

* We have $a0=a(0+0)=a0+a0$. Subtracting $a0$ (meaning addition with the negative of $a0$) on both sides gives the claim.
* See exercise.
* See exercise.
* See exercise.
* We prove this by contradiction, so we assume that $a$ and $b$ are both not $0$. Then there exist inverse elements $a^{-1}$ and $b^{-1}$ and hence $(ab){\left(b^{-1}a^{-1}\right)}=1$. On the other hand, we have $ab=0$ by the premise and so the annulation rule gives hence $0=1$, which contradicts the field properties.
    $(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,$
* This follows with a double induction, see exercise.
$\Box$