Field/Focus on R/Introduction/Section

Definition

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

$+:K\times K⟶K\text{ and }\cdot :K\times K⟶K$

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 $a\ne 0$, 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 $a,b,c\in K$.

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 $x\ne 0$ the element $z$ fulfilling $xz=1$

is also uniquely determined.

Proof

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

$y=y+0=y+(x+y^{\prime })=(y+x)+y^{\prime }=(x+y)+y^{\prime }=0+y^{\prime }=y^{\prime },$

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

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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.

Lemma

Let $K$ be a field,

and let $a,b,c,a_1,\dots ,a_r,b_1,\dots ,b_s$ denote elements from $K$. Then the following statements hold.

1. $a0=0$ (annulation rule).

2. $(-a)b=-ab=a(-b)$

   (rules for sign).

3. $(-a)(-b)=ab.$

4. $a(b-c)=ab-ac.$

5. From $a\cdot b=0$ one can deduce $a=0$ or $b=0$.
6. $\left(\sum _{i=1}^r{a}_i\right)\left(\sum _{k=1}^s{b}_k\right)=\sum _{1\le i\le r,1\le k\le s}{a}_i{b}_k$ (general law of distributivity).

Proof

1. We have $a0=a(0+0)=a0+a0$. Subtracting $a0$ (meaning addition with the negative of $a0$) 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 $a$ 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

   $(ab)\left(b^{-1}{a}^{-1}\right)=0\left(b^{-1}{a}^{-1}\right)=0,$

   hence $0=1$, which contradicts the field properties.

6. This follows with a double induction, see exercise.

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