Definition: Field

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.

Algorithms: 1
Branches: 2
Chapters: 3 4 5 6 7
Corollaries: 8 9
Definitions: 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
Examples: 47 48
Explanations: 49
Lemmas: 50
Proofs: 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Propositions: 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Theorems: 91 92 93 94