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Ordered fields

  1. Nov 16, 2004 #1


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    Let F be a field ordered by <
    Let F(p) be ordered by <'
    Let F(q) be ordered by <''
    (These may be either algebraic or transcendental extensions)

    Suppose that F(p) and F(q) are isomorphic as fields (with p mapping to q)

    Suppose that the restrictions of <' and <'' to F coincide with <.

    Suppose that for any b in F:
    b <' p iff b <'' q

    Then, we may conclude that F(p) and F(q) are isomorphic as ordered fields.


    In the case that F(p) has a positive element smaller than any positive element of F, I think the proof is straightforward, so I've been working entirely on the remaining case... but I'm having difficulty: I keep tripping over myself because everything seems so obvious. :frown: Anyone have any ideas?

    In case you're wondering, this came up because I was interested in generalizations of metrics where the metric function is into any ordered ring, instead of the reals. I think that nonstandard models of the reals might make a nice class of alternatives, and I think that any ordered ring can be embedded into one such model (and this is what I'm really trying to prove)
  2. jcsd
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