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Show R is not a domain

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that R = {0.5(a+b(root2)) : a,b are integers} is not a domain

    2. Relevant equations
    A commutative ring (non trivial) is a domain iff (ab=0 => a=0 or b=0)

    3. The attempt at a solution
    I've try to construct a counterexample where x, y in R and xy=0 but x,y=/=0 but this has been hard to do since i realize that x and y mostly like have to be "conjugates of each other" and xy is of some form 0.25(a^2 - 2b^2) and (a^2 - 2b^2) = 0 and its impossible to do this just by using integers for a and b.
  2. jcsd
  3. Oct 21, 2008 #2


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    Staff Emeritus
    Science Advisor

    If x= 0.5(a+ bsqrt(2)) and y= 0.5(c+ dsqrt(2)) then xy= 0.25(ac+ (ad+bc)sqrt(2)+ 2cd)

    Obviously you can ignore the "0.5". You need to find a, b, c, and d, non-zero, so that ad+ bc= 0 and ac+ 2cd= 0.

    In your equations, you seem to be assuming x=y.
  4. Oct 21, 2008 #3
    doesn't xy=0.25(ac+2bd+(ad+bc)sqrt(2))=0.5 ( (ac+2bd)/2 + sqrt(2) (ad+bc)/2) imply that R is not closed under multiplication? since (ac+2bd)/2 and (ad+bc)/2 are not necessarily integers. R is therefore not even a ring.
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