Field/Linear algebra/Introduction/Section

Definition

A commutative ring ${}K$ is called a field if ${}R\neq 0$

and if every element different from ${}0$ has a multiplicative inverse.

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.

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

The graph of the real function that assigns to a number ${}a\neq 0$ its inverse ${}a^{-1}$. This mapping is not defined in ${}0$, nor can it be extended in a continuous way.

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 the 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  (1). Hence, the operation tables look like

${}+$	${}0$	${}1$
${}0$	${}0$	${}1$
${}1$	${}1$	${}0$

and

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

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

${}+$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}0$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}1$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$	${}0$
${}2$	${}2$	${}3$	${}4$	${}5$	${}6$	${}0$	${}1$
${}3$	${}3$	${}4$	${}5$	${}6$	${}0$	${}1$	${}2$
${}4$	${}4$	${}5$	${}6$	${}0$	${}1$	${}2$	${}3$
${}5$	${}5$	${}6$	${}0$	${}1$	${}2$	${}3$	${}4$
${}6$	${}6$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$

${}\cdot$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}0$	${}0$	${}0$	${}0$	${}0$	${}0$	${}0$	${}0$
${}1$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}2$	${}0$	${}2$	${}4$	${}6$	${}1$	${}3$	${}5$
${}3$	${}0$	${}3$	${}6$	${}2$	${}5$	${}1$	${}4$
${}4$	${}0$	${}4$	${}1$	${}5$	${}2$	${}6$	${}3$
${}5$	${}0$	${}5$	${}3$	${}1$	${}6$	${}4$	${}2$
${}6$	${}0$	${}6$	${}5$	${}4$	${}3$	${}2$	${}1$

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.

$\Box$