Field (mathematical concept): Difference between revisions

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== Formal definition ==
== Formal definition ==


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


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


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


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


'''Example''':
<blockquote>
Just like in [[ring|rings]], $+_F,\cdot_F$ do not have to take the usual definitions of $+,\cdot $.
</blockquote>
 
<blockquote>
The smallest field is $(\{0_F,1_F\},+_F,\cdot_F)$
</blockquote>
 
<blockquote>
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.
</blockquote>
 
<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"
{| class="wikitable"
|-
|-
! $F$
! $F$
! $(F,+,\cdot )$ is field?
! Note
|-
| $\mathbb{N}$
| No, $a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
|-
|-
| $\mathbb{Z}$
|| $\mathbb{Q}$
| No, $a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$
| Rational number field, $1_F=1,0_F=0$
|-
|-
| $\mathbb{Q}$
|| $\mathbb{R}$
| Yes
| Real number field, $1_F=1,0_F=0$
|-
|-
| $\mathbb{R}$
| | $\{(a,b)|a,b \in \mathbb{F}\}$
| Yes
| real 2D vector field, $1_F=(1,1),0_F=(0,0)$
|}
|}


<blockquote>
'''Counterexamples''':
Alternatively, a field can be defined as a [[ring]] with additional properties:


* $a \cdot b = b \cdot a$ ('''multiplicative commutativity''')
{| class="wikitable"
* $\exists 1_F \in F: a \cdot 1_F=a$ ('''multiplicative identity''')
|-
* $\forall a \neq 0_F: \exists (a^{-1}) \in F, a\cdot (a^{-1})=1_F$ ('''multiplicative inverse''')
! $F$
 
! Why $(F,+_F,\cdot_F )$ is not a field?
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.
|-
 
| $\mathbb{N}$
<ref>{{#cite:Q2202}}</ref><ref>{{#cite:Q2204}}</ref>
| $a=1 \in \mathbb{N}: \nexists (-1) \in \mathbb{N}$
</blockquote>
|-
| $\mathbb{Z}$
| $a=2 \in \mathbb{N}: \nexists (2^{-1}) \in \mathbb{Z}$
|}

Latest revision as of 15:12, 16 October 2026

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