field (algebra)

TLDR: Field is an algebra in which you can $+,-,\times ,÷$ (except by $0$).

Field

A field $(F,+,\cdot )$ is a commutative ring ring with 1, where the multiplicative identity $1\in F$ (neutral element) is not equal to the additive identity $0$ (neutral element), i.e. $1\ne 0$.
Furthermore, every non-zero element of $F$ has an inverse, called the unit or invertible element.

An alternative equivalent definition is given by the following two properties:
(1) $(F,+),(F\setminus \{0\},\cdot )$ are abelian groups
(2) the distributive property hold (as in the ring).

→

Field axioms

The field axioms state that for all $a,b,c\in F$:
a) Addition is commutative: $a+b=b+a$
b) Addition is associative: $(a+b)+c=a+(b+c)$
c) Distributive property: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$
d) Distinct additive and multiplicative identities: $\exists 0,1\in F:\forall a\in F:a+0=a$ and $a\cdot 1=a$, with $1\ne 0$
e) Every element has an additive inverse: $\forall a\in F,\exists -a\in F:a+(-a)=0$, and every non-zero element has a multiplicative inverse: $\forall a\in F\setminus \{0\},\exists a^{-1}\in F:a\cdot a^{-1}=1$

EXAMPLE

$(\mathbb{Q},+,\cdot ),(\mathbb{R},+,\cdot ),(\mathbb{C},+,\cdot ),({\mathbb{Z}}_p,+,\cdot ),p\in \mathbb{P}$ are fields.
$p$ has to be prime because if for ${\mathbb{Z}}_m$, $m=a\cdot b$, then $a\cdot b\equiv 0(\mathrm{mod}m)$, meaning there are zero divisors. Elements with multiplicative inverses cannot be zero divisors, so not every non-zero element can have a multiplicative inverse when $m$ is composite (→ (1) from above is violated).
$\mathbb{N}$ lacks additive and multiplicative inverses.

${\mathbb{Z}}_2=(\{0,1\},+,\cdot )$

The congruence class modulo 2 with addition and multiplication defined as:

$+$	0	1
0	0	1
1	1	0

$\cdot$	0	1
0	0	0
1	0	1

TODO show it’s a field by checking the axioms

Finite integrity rings are fields

(1) Every field is an integrity ring.
(2) If $R$ is an integrity ring and finite ($|R|\le \infty$), then $R$ is a field.

A field with a partial order is complete if every $A\subset \Omega$ that is bounded from above has $\sup A\in \Omega$

The field of real numbers $(\mathbb{R},+,\cdot )$ is a complete field with linear order.
In fact, it is the unique complete field with linear order (up to isomorphism).
$\mathbb{Q}$ is not a complete field, as e.g. $\{x^2\le 2\}$ has no supremum in $\mathbb{Q}$.

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