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R^n and field

  1. May 6, 2008 #1
    Is there a proof that shows if R^n can be turned into a field, for specific n?

    Obviously, n=1 is a field, and n=2 can be made into a field (which is just the complex plane.)

    So what about n>2?
     
  2. jcsd
  3. May 6, 2008 #2
    What invertable, closed, multiplication and division operation are you going to define for n>2?

    Even for R^2, your complex numbers, that's a space that's isomorphic with R^2, and is not R^2 itself. You can introduce invertable vector product for R^n (I like the geometric/clifford algebra product for this). But to get invertable and closed with that product you have to combine components of such vector products (ie: complex numbers, and quaternions, or other generalizations of these get by adding grade 0 and 2 components from this larger algebraic space).
     
  4. Jun 5, 2008 #3
    it is a theorem that for dimensions three or higher euclidean space cannot be turned into a field, or if we want to be pedantic: for n >/= 3 there is no field isomorphic to R^n.

    I saw a proof of the fact in a complex analysis class sometime ago so I don't remember it but it uses (very) elementary methods and as such should be found easily.
     
  5. Jun 5, 2008 #4

    matt grime

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    I have a small technical problem with the explanation given in 3: in what sense are you asserting that there are no fields isomorphic to R^n for n>=3? Or more accurately, in what category?

    I think that it might be better to say - if F is a field, and F is in VECT(R) - cat of real vector spaces - then F has dimension 1 or 2.
     
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