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1+1 = 0

  1. Jun 23, 2010 #1

    Fredrik

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    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

    [tex]\sum_{k=1}^n 1=0[/tex]

    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?
     
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  3. Jun 23, 2010 #2
    Re: 1+1=0

    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 [itex]1\not=0[/itex] 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:
    [tex]0 = \sum_{k=1}^n 1 > 0[/tex]
    which is a contradiction so all non-trivial ordered fields must have characteristic 0.
     
  4. Jun 23, 2010 #3

    Fredrik

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    Re: 1+1=0

    Thank you. That was crystal clear.
     
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