# Field/Linear algebra/Introduction/Section
## Definition

A commutative ring $K$ is called a field if $R\neq 0$
$0$
In all details, this means the following:

## 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 $0,1\in K$ that fulfill the following properties.
* Axioms for the addition:
** Associative law: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in K$.
** Commutative law: $a+b=b+a$ 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:
** Associative law: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in K$.
** Commutative law: $a\cdot b=b\cdot a$ 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 $a\cdot c=1$.
* Distributive law: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in K$.
The properties described in fact for rings (and the conventions) hold in particular for fields. Using the concept of a group, we may say that a field is a set with two binary operations $+$ and $\cdot$ and two fixed elements $0\neq 1$, such that $(K,+,0)$ and $(K\setminus \{0\},\cdot ,1)$ are commutative groups, and that the distributivity law holds.
For an element $x\in K$ and a natural number $n\in \mathbb {N}$, we define $nx$ to be the $n$-fold sum of $x$ with itself. Here we put $0x=0$. For
    $n1_{K}=\underbrace {1_{K}+1_{K}+\cdots +1_{K}} _{n{\text{ summands}}}\,,$
we also write simply $n_{K}$ or just $n$. This means that we can find every natural number in every field (also in every ring). However, this assignment is not necessarily injective, and it is possible that $2=0$ or $7=0$ holds in a field (see the examples below). For a negative integer $n$, we set
    $nx=(-n)(-x)\,,$
where $-x$ denotes the negative of $x$ in the field. Due to exercise, everything fits well together. For example, one may consider $(-n)(-x)$ as the $-n$-fold sum of $-x$ with itself, or as the product of $-x$ and $-n$, where this means the $-n$-fold sum of $1_{K}$ with itself.
Due to fact, we know that for every $a\in K$, $a\neq 0$, the element $z$ fulfilling $az=1$ is unique. It is called the inverse of $a$ and denoted by $a^{-1}$.
For $a,b\in K$, $b\neq 0$, we write
    $a/b:={\frac {a}{b}}:=ab^{-1}\,.$
The terms on the left are abbreviations for the term on the right.
For a field element $a\in K$ and $n\in \mathbb {N}$, we denote its $n$-th power by $a^{n}$; this is defined as the $n$-fold product of $a$ with itself ($n$ is the number of factors). Moreover, we set $a^{0}=1$, and, for $a\neq 0$ and a negative integer $n$, we interpret $a^{n}$ as ${\left(a^{-1}\right)}^{-n}$.
A "strange“ field is given in the following example. This field with two elements is important in computer science and in coding theory; it will not play a big role here. It shows that it is not helpful for every field to imagine its elements on the number line.

## 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.

## Example

On the set $\{0,1,2,3,4,5,6\}$ (with seven elements) one can define a field structure using
Without any further theory, it is very tedious to so show that this is indeed a field.

## Lemma

Let $K$ denote a field. Then $a\cdot b=0$ implies that $a=0$ or
$b=0$

### Proof

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 annihilation rule gives
    $(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,$
hence $0=1$, which contradicts the field properties.
$\Box$
* ↑ This implies in particular that the multiplication can be restricted to give a binary operation on $K\setminus \{0\}$. This follows from the field axioms, as we will see below.