Field (mathematical concept): Difference between revisions

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Then $(F,+_F,\cdot_F)$ is a '''field'''.
Then $(F,+_F,\cdot_F)$ is a '''field'''.
'''Example''':
{| class="wikitable"
|-
! $F$
! $(F,+_F,\cdot_F )$ is field?
|-
| $\mathbb{N}$
| No, $a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
|-
| $\mathbb{Z}$
| No, $a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$
|-
| $\mathbb{Q}$
| Yes
|-
| $\mathbb{R}$
| Yes
|}


<blockquote>
<blockquote>
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<ref>{{#cite:Q2202}}</ref><ref>{{#cite:Q2204}}</ref>
<ref>{{#cite:Q2202}}</ref><ref>{{#cite:Q2204}}</ref>
== Examples ==
For each of the $F$ below, $(F,+_F,\cdot_F )$, where $+_F,\cdot_F=+,\cdot$ unless specified, is a field:
{| class="wikitable"
|-
! $F$
! 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''':
{| class="wikitable"
|-
! $F$
! Why $(F,+_F,\cdot_F )$ is not a field?
|-
| $\mathbb{N}$
| $a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
|-
| $\mathbb{Z}$
| $a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$
|}

Latest revision as of 15:12, 16 October 2026

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

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, since the slots for multiplication are symmetric in a field because of multiplicative commutativity.

[1][2]

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?
$\mathbb{N}$ $a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
$\mathbb{Z}$ $a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$