Fields: Characteristic 0 and Beyond

[Fredrik]

Most definitions of "field" I've seen include the requirement 1≠0. One book didn't, and instead defined the "characteristic" of a field as the smallest non-negative integer n such that

\sum_{k=1}^n 1=0

The books that include the requirement 1≠0 then go on to define an "ordered field", and a "complete ordered field". ("Complete" in the sense that every set that has an upper bound has a least upper bound). Then they claim that all complete ordered fields are isomorphic. This seems to overlook the possibility that 1+1=0, or that 1+1+...+1=0 for some number of 1s on the left. Does the definition of a complete ordered field (including the axiom 1≠0) imply that the field has characteristic 0?

 

[rasmhop]

If 1=0 in a field, then for every element x of the field we have:
x = 1x = 0x = 0
so the only field in which we can have 1=0 is the trivial field {0}. Thus we can assume 1\not=0 and our theory will still cover all non-trivial fields. This is why most books don't care about the case 1=0. If you are interested in it you can often check it manually.

Suppose F is a non-trivial ordered field with characteristic n > 0. Then 1>0 since if 1<0 we have -1 > 0, but then 1=(-1)^2 >0 which is a contradiction. Since 1 > 0, we have 1+1 > 0, 1+1+1 > 0, ... Continuing on we get:
0 = \sum_{k=1}^n 1 &gt; 0
which is a contradiction so all non-trivial ordered fields must have characteristic 0.

 

[Fredrik]

Thank you. That was crystal clear.