Field (mathematical concept)

A field is a set closed under two binary operations, with some special properties by definition.

It is a subclass of rings.

Formal definition

Let $F:|F| \ge 2$ be a set, and $+_F,\cdot_F$ two binary operations defined over $F$.

If $\forall a,b,c \in F$:

• $a+_Fb=b+_Fa;a \cdot_F b = b \cdot_F a$ (commutativity)
• $(a+_Fb)+_Fc=a+_F(b+_Fc);(a \cdot_F b)\cdot_F c = a \cdot_F (b \cdot_F c)$ (associativity)
• $\exists 0_F,1_F \in F: a+_F0_F=a, a \cdot_F 1_F=a$ (identity)
• $\exists (-a) \in F: a +_F (-a)=0_F; \forall a \neq 0_F: \exists (a^{-1}) \in F, a\cdot_F (a^{-1})=1_F$ (inverse)
• $a\cdot_F (b+_Fc)=a\cdot_F b +_F a\cdot_F c$ (distributivity)

Then $(F,+_F,\cdot_F)$ is a field.

Just like in rings, $+_F,\cdot_F$ do not have to take the usual definitions of $+,\cdot $.

The smallest field is $(\{0_F,1_F\},+_F,\cdot_F)$.

Alternatively, a field can be defined as a ring with additional properties:

• $a \cdot_F b = b \cdot_F a$ (multiplicative commutativity)
• $\exists 1_F \in F: a \cdot_F 1_F=a$ (multiplicative identity)
• $\forall a \neq 0_F: \exists (a^{-1}) \in F, a\cdot_F (a^{-1})=1_F$ (multiplicative inverse)

Also, note unlike for rings, there is no need to specify distributivity for each slot. The slots for multiplication are symmetric in a field, because of multiplicative commutativity.

Examples

For each of the $F$ below, $(F,+_F,\cdot_F )$, where $+_F,\cdot_F=+,\cdot$ unless specified, is a field:

$F$	Note
$\mathbb{Q}$	Rational number field, $1_F=1,0_F=0$
$\mathbb{R}$	Real number field, $1_F=1,0_F=0$
$\{(a,b)|a,b \in \mathbb{F}\}$	real 2D vector field, $1_F=(1,1),0_F=(0,0)$

Counterexamples:

$F$	Why $(F,+_F,\cdot_F )$ is not a field?
$\mathbb{N}$	$a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
$\mathbb{Z}$	$a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$