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Math Help - [solved] ss

  1. #1
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    [solved] ss

    Hi:
    Let K be a field and, for a, b in K \ {0}, write a ~ b if ab is a sum of two squares in K. The author writes: why is ~ an equivalence relation? I see that for any a, b in K \ {0}, ab= ab + 0, where 0 is a square and ab is the square of an element in some extension of K and ab , 0 are in K. So a ~ b for every a, b in K \ {0}.

    However the author proceeds: Does the analogous statement hold if one replaces 2 by some arbitrary power 2^n? (supposedly the first statement was ~ is an equivalence relation). And here I could write ab= ab + 0 + 0 + 0 (for n= 2). But this seems too obvious. There must be something I don't understand well in the statement of the problem.

    Any hint will be welcome. Regards.
    Last edited by ENRIQUESTEFANINI; November 23rd 2009 at 04:48 PM.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:
    Let K be a field and, for a, b in K \ {0}, write a ~ b if ab is a sum of two squares in K. The author writes: why is ~ an equivalence relation? I see that for any a, b in K \ {0}, ab= ab + 0, where 0 is a square and ab is the square of an element in some extension of K and ab , 0 are in K. So a ~ b for every a, b in K \ {0}.
    well, you have to prove three conditions as you know: that the relation you've defined is reflexive, symmetric and transitive. the reflexive and symmetric is trivial. for the transitive, suppose

    ab=x^2+y^2 and bc=z^2+t^2, for some x,y,z,t \in K. then ac=(x^2+y^2)(z^2+t^2)b^{-2}=(xzb^{-1} + ytb^{-1})^2 + (xtb^{-1} - yzb^{-1})^2.


    However the author proceeds: Does the analogous statement hold if one replaces 2 by some arbitrary power 2^n? (supposedly the first statement was ~ is an equivalence relation).
    are you assuming that n is fixed? (i'm quite sure the answer is no because otherwise the claim would be trivially false in general!)
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post

    are you assuming that n is fixed? (i'm quite sure the answer is no because otherwise the claim would be trivially false in general!)
    I think the author wants me to prove the following: let a ~ b iff ab is the sum of 2^n squares. Then there exists n > 1 such that ~ is not an equivalence relation.

    However of much more interest to me is what's wrong in my line of reasoning that a ~ b for every a, b in K\{0}. That is, the partition of K\{0} induced by ~ consists of only one equivalence class. Very kind of you to have answered my post.

    P.S.: I tried to prove transitivity but, unfortunately, I came to think that all reduced to prove that a sum of four squares can always be written as a sum of two squares. And the proof was so simple!
    Last edited by ENRIQUESTEFANINI; November 24th 2009 at 03:46 AM.
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