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Valid argument ?

  1. Mar 19, 2008 #1
    I was proving 2Z is non-isomorphic to 3Z
    I tried it by contradiction, of course.
    If possible there exists a ring homomorphism f : 2Z ---> 3Z
    Then,
    f (x.y) = f(x).f(y) must hold
    x,y belong to 2Z
    So x = 2a, f.s. a belonging to Z
    y = 2b, f.s. b belonging to Z
    so x.y = 2z, f.s. z belonging to Z
    Also, f(x) belongs to 3Z, and so does f(y)
    => f(x).f(y) = 3n , f.s. n belonging to Z
    So,
    f(2z) = 3n
    f(2).f(z) = 3n, (since f is a ring homomorphism)
    which is a meaningless statement, since f is not defined for all z belonging to Z, but only for elements of the form 2z.

    Is it correct ? Someone got something neater ?
     
  2. jcsd
  3. Mar 20, 2008 #2
    no

    f(2z) = 3n, then you said f(2)f(z) = 3n, you can't do this because you don't where z lives and f is a ring homo on 2Z, what if z is in 5Z, etc


    you need to use the fact you have an isomorphism, in your proof you are only using the fact you have a homomorphism but those do exist, so what you are trying to do is not going to be enough

    edit: also you should be using additive notation here, ie, f(x) + f(y) = f(x + y)
     
    Last edited: Mar 20, 2008
  4. Mar 20, 2008 #3
    let f(2) = 3n , f.s. n in Z
    f(2 + 2) = f(4) , and f(2.2) = f(4)

    f(2 + 2) = f(2) + f(2) = 6n
    f(2.2) = f(2)f(2) = 9n^2

    => 6n = 9n^2
    => n = 0 (contradiction as f is an isomorphism, so must be injective, so only 0 maps to 0)
    or, n = 2/3 (contradiction as n belongs to Z)

    Q.E.D.

    thanks for the help
     
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