# Fields

Michael Redman

Updated 2019 April 19

# Definition of Field

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.

S is closed, associative and commutative under *, 1*x=x for all x in S, and every x not 0 has an inverse $$x^{-1}$$ such that x$$x^{-1}$$=1

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

# If x,y in a field F, then xy=0 implies at least one of x or y is 0; or equivalently (contrapositive) x and y not zero implies xy not zero.

Let xy=0 and neither x nor y = 0. Then xy=xy+xy=x(y+y) implies (since x not 0 and has a multiplicative inverse) y=y+y implies (by existence of additive inverse, adding -y to each side) 0=y which is a contradiction.

# 0x=0

Let 0x=y with y not 0 for some x. Then 0x=y implies (0+0)x=y implies 0x+0x=y implies y+y=y implies y=0 which is a contradiction.

# -x=(-1)x

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.

# -1 $$\ne$$ 0

By def of additive inverse (-1)+1=0. If (-1)=0 then 0+1=0 which contradicts the field axiom that 0 and 1 are distinct.

Note "-1"=1 in the 2-element field consisting of only 0 and 1.

# $$(-1)^2$$=1

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, which implies (-1)(-x)=-1, which implies (by above proposition that (-1)a=-a, and proposition from "Groups" page that -(-a)=a) x=-1 which is a contradiction.