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Homework Help: Uniqueness of ordering of a ordered field.

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Is the ordering of a ordered field unique? That is, is it possible to have different ordering set(the order), we call P1 and P2, both able to make a field F into a ordered field?

    2. Relevant equations

    no.

    3. The attempt at a solution

    First I tried to assume now there's a X in P1, but not in P2(because P1 != P2), and therefore by definition -X must be in P2. After that, I tried to get to a point where I can make X also in P2 to have contradiction. But I can't.

    Then I tried to go from ration numbers. With some existing theorem, Q(rational number set) is the smallest ordered field, and a subfield of any ordered field. Therefore if there're such P1 and P2 in some ordered field, they will both have Q, thus the element that is not shared can not be a rational number. After that, I stopped without any idea how i could progress.

    Thank you very much for viewing, and lemme know anything I unclearly stated.
     
  2. jcsd
  3. Sep 29, 2010 #2

    Dick

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    Think about extension fields of Q. Like Q[sqrt(2)]. Is there anything in the ordered field properties that forces sqrt(2) to be in P?
     
    Last edited: Sep 29, 2010
  4. Sep 29, 2010 #3
    By applying taylor series to express sqrt(2) as result of a series of rational numbers I guess?

    Therefore any irrational number must be in P...and then we could get to real number field. But it's still not a general field, and I can't prove the uniqueness of its order.

    It's a real analysis class and I'm so not sure of what I can use and what I can't, because everything I know for sure is getting proved.
     
  5. Sep 29, 2010 #4

    Dick

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    Any irrational number must be in P? Now that's gibberish. I'm suggesting you use Q[sqrt(2)] to construct a counterexample. There are two field orderings of Q[sqrt(2)].
     
  6. Sep 30, 2010 #5
    Ehh sorry my bad understanding and stupidity.

    So for Q[sqrt(2)], either sqrt(2) is in P or -sqrt(2) is in P. And we know all a+bsqrt(2) will be in Q.

    by constructing two field ordering of Q[sqrt(2)], and by proving that sqrt(2) is always in P, we have the contradiction, therefore the uniqueness, right?

    I come up with something like a+bsqrt(2) where b is postive is in P, and a+bsqrt(2) where b is negative is in P, thus two different ordering but both satisfying the requirement of ordering. And by fact that sqrt(2) is always there, we have the contradiction?

    My approach still sounds not flawless. Am I wrong at some steps this time?
     
  7. Sep 30, 2010 #6

    Dick

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    That's not really it. Think about this. Q[sqrt(2)] is the set of all numbers a+sqrt(2)*b with a and b rational. There is a field automorphism that exchanges sqrt(2) and -sqrt(2) sort of like complex conjugation, right? If you decide to put sqrt(2) in P, then you don't have much choice where to put it in the ordering of Q. It has to go between 1 and 2 in the usual 1.414... spot. But you could also have put -sqrt(2) there. You can't tell the difference between the two using field operations.
     
  8. Sep 30, 2010 #7
    Oh good morning, wish you a nice dream.

    yes I do have no way of telling the difference between the two using field operations, just like that I can't tell the difference between P1 with Z in it and P2 with -Z in it. Is that breaking the uniqueness? I thought I could find a contradiction and prove the uniqueness. Ahh, I can't even understand how you go from there to the proof.

    Sorry I'm just starting set theory and abstract math this month. There's too much I don't know and learning day by day. That must be causing toughness for you to guide me through it, though I see you're trying to teach me the way not the answer. PF is full of good guys woot.
     
  9. Sep 30, 2010 #8
    Are you still there, Dick........m(-_-)m
     
  10. Sep 30, 2010 #9

    Dick

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    Sure. I wasn't sure if you still had a question. All the field Q knows about sqrt(2) or -sqrt(2) is that they satisfy x^2=2. That's all. You can either make the standard order where you put sqrt(2) in between 1 and 2 where it belongs, or you can make a nonstandard order where you put -sqrt(2) there instead. The rationals don't care.
     
  11. Sep 30, 2010 #10
    Ouch, so the uniqueness I'm trying to find is actually not there? Ahh the theorem of real closed field with unique ordering must be understood terribly be me.

    I think for one day I totally complicated the problem too far, while at first guess that P1,P2,+Z,-Z(which I can't prove to be wrong, and actually correct if I make Z a unrelated thing to all members left). Thx bro, you helped a lot.

    Time for consuming your replies and get every new words there wikified!

    Good night.
     
  12. Oct 1, 2010 #11

    Dick

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    Show this, ok? If F is a field, f:F->F is a field isomorphism and P is a field ordering then f(P) is ALSO a field ordering. It's pretty easy. That might make what's going on clearer.
     
  13. Oct 1, 2010 #12
    The problem is I don't know what fied isomorphism is (maybe for the whole semester) and when I asked professor if I needed to know about field isomorphism to solve this question he anwsered no. Only thing I know and related to isomorphism is Peano isomorphism I learned in number system.

    I totally get what you're saying, and keep reading stuff about isomorphism and automorphism, trying to get a automorphism from Q[sqrt(2)] to Q[sqrt(2)]. But right now I'm still not capable of doing that, though I know it could be simple as mapping but on field level.
     
  14. Oct 1, 2010 #13

    Dick

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    Take f(a+b*sqrt(2))=a-b*sqrt(2). It just exchanges sqrt(2) and -sqrt(2).
     
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