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Why isn't this a domain?

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data


    If we have [tex]R = \{ \frac{1}{2}(a+b\sqrt{2}) \}[/tex].

    2. Relevant equations



    3. The attempt at a solution

    For the identity, we have a=2 and b=0. So the identity is not equal to zero. Also, there can't be any zero divisors, because...
    [tex][\frac{1}{2}(a+b\sqrt{2})][\frac{1}{2}(a'+b'\sqrt{2})] [/tex] = [tex]\frac{1}{4}(aa' + ab'\sqrt{2} + a'b\sqrt{2} + 2bb')[/tex]. If a,a',b, b' are not zero, then this can't be zero.

    So why isn't it a domain?

    Thanks in advance
     
    Last edited by a moderator: Dec 27, 2012
  2. jcsd
  3. Dec 27, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You haven't explained this very well! I take it that you do NOT mean "[itex]R= (1/2)(a+ b\sqrt{2})[itex]", a single number, but rather "[itex]R= \{(1/2)(a+ b\sqrt{2})\}[itex]", the set of all such numbers with a and b integers. Yes, if a= 2 and b= 0, we have [itex](1/2)(2+ 0\sqrt{2})= 1[/itex] as multiplicative identity which is not 0, the additive indentity, so this is a ring with more than one member.

    Yes, in order that [itex][(1/2)(a+ b\sqrt{2}][(1/2)(a'+ b'\sqrt{2})]= (1/4)(aa'+ ab'\sqrt{2}+ a'b\sqrt{2}+ 2bb')[/itex] be 0, we must have aa'+ 2bb'= 0 and ab'+ a'b= 0. The second gives ab'= -a'b or a/b= -a'/b'. The first gives a/b= -2b'/a'. that is, -a'/b'= -2b'/a' or [itex]a'^2= 2b'^2[/itex] which is impossible.


    Now, what makes you believe this is not an integral domain?
     
  4. Dec 27, 2012 #3
    Thanks. I actually did write R={...}, but the brackets automatically got deleted. This is a question in my textbook. It tells us to prove that this set is not a domain, and I'm confused about that...
     
  5. Dec 27, 2012 #4

    Borek

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    Staff: Mentor

    You may want to learn about the difference between {} and \{\} in LaTeX :wink:
     
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