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Math Help - Isomorphic

  1. #1
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    Isomorphic

    Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)
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  2. #2
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    I'll give it a shot. Assume \phi: \mathbb{R} \rightarrow \mathbb{Q} is such an isomorphism. Say \phi(1)=k where k \in \mathbb{Q}. \phi must satisfy
    \phi(ab) = \phi(a)\phi(b)
    so
    kab = k^2 ab \Rightarrow k = 1

    Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist.
    Last edited by eeyore; October 13th 2009 at 11:48 AM. Reason: typo
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  3. #3
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    Hi,
    here's my try:
    sentence
    \forall x \exists y \; x=y\cdot y\cdot y
    holds in \mathbb{R} but not in \mathbb{Q}. So they are not elementarily equivalent, therefore they can't be isomorphic.
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  4. #4
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    Another one: assume f:\mathbb{R}\rightarrow\mathbb{Q} is a ring isomorphism, then f(\sqrt{2})^2=f(\sqrt{2}^2)=f(2)=f(1+1)=f(1)+f(1)=  1+1=2.

    But no rational is a squared root of 2, so f cannot exist.

    (assume p,q belong to \mathbb{N}^*, gcd(p,q)=1 and (\frac{p}{q})^2=2 then p^2=2q^2.
    But this is impossible since \nu_2(p^2) is even and \nu_2(2q^2) is odd (where \nu_2(n) is the greatest k such that 2^k divides n))
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