Michael Redman

Updated 2018 November 17

A set S is a field if it has operations + and * and distinct elements 0 and 1 such that:

The set is a commutative group under + with identity 0.

The set is closed, associative, and commutative under *, and S-0 is a group under * with identity 1

The distributive law holds, meaning for any x, a, b, in S, x(a+b)=xa+xb

By def of field F-0 is a group under multiplication. Since groups are closed under their group opeartions, x and y in F-0 implies xy in F-0 which implies xy is not 0 since 0 is not in F-0.

By def of additive identity 0+x=x. By distributive law 0(0+x)=0x=0*0+0x, which implies (since 0x has an additive inverse) that 0*0=0. This proves the proposition for x=0

For the case where x not 0, assume 0x is not zero, let 0x=y. Since x is not 0 it has a multiplicative inverse so 0x\(x^{-1}\)=y\(x^{-1}\). By def of multiplicative identity 0*1=0 and by clsoure of the group F-0 under multiplication y\(x^{-1}\) is in F-0 and therefore is not 0. So the equation is a contradiction and 0x must equal 0 to be consistent.

By existence of additive inverses there is a (-1) with (-1)+1=0, which (multiplying both sides by x, possible because of closure under multiplication) implies x((-1)+1)=0x= (by previous proposition) 0, which (by distributive law) implies (-1)x+1x=0=(-1)x+x. Adding (-x) to both sides, (-1)x+x+(-x)=-x, implies by def. of additive inverse (-1)x=-x.

By field axioms -1 has a multiplicative inverse. Assume it is x and not =-1. By def. of multiplicative inverse (-1)x=1. By earlier proposition (-1)x=-x so -x=1. Multiplying both sides by -1, and using again the proposition that (-1)a=-a, and also that -(-a)=a, x=-1, which contradicts our assumption that x not =-1. Therefore -1 is its own multiplicative inverse and (-1)(-1)=1.