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To be or not to be (an ideal)

  1. Sep 24, 2006 #1
    let F be a field.x^2/y^2 is not an element of F[x,y](is it?)
    (x^2/y^2) can or can not be ideal in F[x,y] ?
     
  2. jcsd
  3. Sep 24, 2006 #2

    Hurkyl

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    Strictly speaking, the expression

    x²/y²

    is undefined, because one cannot divide in the ring F[x, y]. And because x²/y² is undefined, so is the expression (x²/y²).


    We're usually more generous with notation, though; rather than leave x²/y² undefined, we implicitly shift our attention to the field F(x, y), which does contain an element by that name.

    Where did this come from?
     
  4. Sep 24, 2006 #3
    the origin of my question:i have to prove F[x,y]/(x^2/y^2) is a vector space it seemed a bit meaningless anddid not remember fraction fields
    probably it was F(x,y)/(x^2/y^2) and i misread it
    sorry:uhh:
     
  5. Sep 24, 2006 #4

    Hurkyl

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    F(x, y) / (x² / y²) doesn't make much sense either; the only ideals of a field are the zero ideal and the whole field itself.

    It probably said F[x, y] / (x², y²)
     
  6. Sep 24, 2006 #5
    now it is clear thank you very much
     
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