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R^2 a field?

  1. Sep 18, 2010 #1

    quasar987

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    Hello,

    Am I missing something or is R^2 a field with the obvious component wise addition and multiplication (a,b)*(c,d)=(ac,bd)?
     
  2. jcsd
  3. Sep 18, 2010 #2
    EDIT: Completely ignore this. Didn't think it through. For an example of why it's false see Office Shredder's reply.

    Yes if R is a field, then R^2 is a field (clearly commutative, and (a,b) has inverse (1/a,1/b) ).
     
    Last edited: Sep 18, 2010
  4. Sep 18, 2010 #3

    Office_Shredder

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    What's the inverse of (3,0)?
     
  5. Sep 19, 2010 #4

    quasar987

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    Ok, that's what I was missing :)
     
  6. Sep 19, 2010 #5

    Landau

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    If R, S are non-trivial integral domains, then their direct product [itex]R\times S[/itex] is never an integral domain, because it always has zero-divisors: (a,0).(0,b)=0. In particular, this holds for fields.
     
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