Field (mathematical concept)

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Revision as of 13:10, 16 October 2026 by Rongzhou (talk | contribs) (Created page with "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 $+,\cdot$ two binary operations defined over $F$. If $\forall a,b,c \in F$: * $a+b=b+a;a \cdot b = b \cdot a$ ('''commutativity''') * $(a+b)+c=a+(b+c);(a \cdot b)\cdot c = a \cdot (b \cdot c)$ ('''associativity''') * $\exists 0_F,1_F \in F: a+0_F=a, a \cdot 1_F=a$ ('''identity''') * $\e...")
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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 $+,\cdot$ two binary operations defined over $F$.

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

  • $a+b=b+a;a \cdot b = b \cdot a$ (commutativity)
  • $(a+b)+c=a+(b+c);(a \cdot b)\cdot c = a \cdot (b \cdot c)$ (associativity)
  • $\exists 0_F,1_F \in F: a+0_F=a, a \cdot 1_F=a$ (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)
  • $a\cdot (b+c)=a\cdot b + a\cdot c$ (distributivity)

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

Example:

$F$ $(F,+,\cdot )$ 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

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