A commutative unit ring $R$ (with the multiplicative neutral element $1$) is called a **field**, if $R$ is not the zero ring and every element \(x\in R\) with $x\neq 0$ ($0$ being the additive neutral element) has a multiplicative inverse.

“Unfolding” all definitions, a *field* fulfills the following axioms:

- $(R,+)$ is a commutative group:
- Associativity of $”+”$: $x+(y+z)=(x+y)+z$ for all $x,y,z\in F.$
- Commutativity of $”+”$: $x+y=y+x$ for all $x,y\in F.$
- Neutral Element of $”+”$: There is an element $0\in F$ with $0+x=x+0=x$ for all $x\in F.$
- Inverse elements of $”+”$: For all $x\in F$ there exists an $-x\in F$ with $x+(-x)=(-x)+x=0.$

- $(R,\cdot)$ is a commutative group:
- Associativity of $”\cdot”$: $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ for all $x,y,z\in F.$
- Commutativity of $”\cdot”$: $x\cdot y=y\cdot x$ for all $x,y\in F.$
- Neutral Element of $”\cdot”$: There is an element $1\in F$ with $1\cdot x=x\cdot 1=x$ for all $x\in F.$
- Inverse elements of $”\cdot”$: For all $x\in F$ with $x\neq 0$ there exists an $x^{-1}\in F$ with $x\cdot x^{-1}=x^{-1}\cdot x=1.$

- Distributivity laws: $(x+y)\cdot z=x\cdot z + y\cdot z$ and $x\cdot (y+z)=x\cdot y + x\cdot z$ for all $x,y,z\in F.$

- In some books, you will also encounter the axiom $1\neq 0.$ This axiom will be proven later from the remaining axioms given above and the requirement that $R$ must not be the zero ring.

| | | | | created: 2014-03-28 21:11:15 | modified: 2020-07-03 18:14:40 | by: *bookofproofs* | references: [577], [6907]

[6907] **Brenner, Prof. Dr. rer. nat., Holger**: “Various courses at the University of Osnabrück”, https://de.wikiversity.org/wiki/Wikiversity:Hochschulprogramm, 2014

[577] **Knauer Ulrich**: “Diskrete Strukturen – kurz gefasst”, Spektrum Akademischer Verlag, 2001