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 ${}0,1\in K$ 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 ${}a,b,c\in K$.

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 ${}xz=1$

is also uniquely determined.

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 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 ${}a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$ 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. ${}{\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (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.

$\Box$