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Math Help - Group Isomorphism

  1. #1
    Super Member Gamma's Avatar
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    Group Isomorphism

    Prove that (\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot).

    I am not seeing it.
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    Quote Originally Posted by Gamma View Post
    Prove that (\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot).

    I am not seeing it.
    If a\in \mathbb{Q}^{\times} we can write a = \prod_{n=0}^{\infty} p_n^{e_n} where e_n\in \mathbb{Z} and p_1,p_2,p_3,..., are the primes in increasing order.
    Thus, I think the natural map is \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) = \prod_{n=0}^{\infty} p_n^{e_n}.

    Of course, the infinite product is really finite and the infinite sum is really finite because
    almost all terms are 1 in the infinite product and almost all terms are zero in the infinite sum.

    Now of course \phi \left( \sum_{n=0}^{\infty}a_n x^n + \sum_{n=0}^{\infty}b_n x^n \right) = \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) \phi \left( \sum_{n=0}^{\infty} b_n x^n \right) since addition of exponents is adding the exponents.
    This happens to be one-to-one and onto.
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    Quote Originally Posted by ThePerfectHacker View Post

    Thus, I think the natural map is \phi \left( \sum_{n=0}^{\infty} a_n x^n \right) = \prod_{n=0}^{\infty} p_n^{e_n}.
    the power of p_n in the RHS should be a_n. so i think you meant \phi \left( \sum_{n=0}^{m} a_n x^n \right) = \prod_{n=0}^{m} p_n^{a_n}, which is indeed a group isomorphism.
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    Senior Member TheAbstractionist's Avatar
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    Argh! I spent the whole night thinking of the problem, and now that I have found a solution, I see that ThePerfectHacker and NonCommAlg have got here before me and deprived me of the chance to show off.

    Only kidding. I come here to help, to learn from others and to share what I know, not to show off.

    Anyway, this is a very beautiful problem. The fact that every positive rational number can be written uniquely in the form 2^{a_1}3^{a_2}5^{a_3}\cdots where a_i\in\mathbb Z is very beautiful indeed.
    Last edited by TheAbstractionist; May 28th 2009 at 02:52 AM.
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    Quote Originally Posted by NonCommAlg View Post
    the power of p_n in the RHS should be a_n. so i think you meant \phi \left( \sum_{n=0}^{m} a_n x^n \right) = \prod_{n=0}^{m} p_n^{a_n}, which is indeed a group isomorphism.
    Of course that is what I meant. Thank you for the correction.
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  6. #6
    Senior Member TheAbstractionist's Avatar
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    By the way, just realized of something.

    Quote Originally Posted by Gamma View Post
    Prove that (\mathbb{Z}[x],+) \cong (\mathbb{Q}^{\times},\cdot).
    The right-hand group should be (\mathbb Q^{\color{red}+}\color{black},\cdot) that is, positive rationals, not nonzero rationals.
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  7. #7
    Super Member Gamma's Avatar
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    Interesting

    Yeah, as stated in my sheet of group theory questions it is \mathbb{Q}^{\times}.

    It must be incorrect as stated because I am pretty sure (\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot).

    The first has two elements of finite order \pm 1 while the second has only one of finite order 1.

    And we have demonstrated the isomorphism to the latter.
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Gamma View Post
    It must be incorrect as stated because I am pretty sure (\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot).
    No, the two groups are not isomorphic. To see this, let f:\mathbb Q^\times\to\mathbb Q^+ be a homomorphism. Then obviously f(1)=1 and 1=f(1)=f((-1)(-1))=f(-1)f(-1)=f(-1)^2. Since f(-1) must be in \mathbb Q^+, we must have f(-1)=1. Hence f is not injective and so cannot be an isomorphism.

    However \mathbb Q^\times/\{1,-1\}\cong\mathbb Q^+ as the mapping f:\mathbb Q^\times\to\mathbb Q^+ where f(a)=a if a>0 and f(a)=-a if a<0 is a surjective homomorphism with kernel \{1,-1\}.
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    Quote Originally Posted by TheAbstractionist View Post
    By the way, just realized of something.



    The right-hand group should be (\mathbb Q^{\color{red}+}\color{black},\cdot) that is, positive rationals, not nonzero rationals.
    Quote Originally Posted by Gamma View Post
    Yeah, as stated in my sheet of group theory questions it is \mathbb{Q}^{\times}.

    It must be incorrect as stated because I am pretty sure (\mathbb{Q}^{\times}, \cdot )\not \cong (\mathbb{Q}^{+},\cdot).

    The first has two elements of finite order \pm 1 while the second has only one of finite order 1.

    And we have demonstrated the isomorphism to the latter.
    Quote Originally Posted by TheAbstractionist View Post
    No, the two groups are not isomorphic. To see this, let f:\mathbb Q^\times\to\mathbb Q^+ be a homomorphism. Then obviously f(1)=1 and 1=f(1)=f((-1)(-1))=f(-1)f(-1)=f(-1)^2. Since f(-1) must be in \mathbb Q^+, we must have f(-1)=1. Hence f is not injective and so cannot be an isomorphism.

    However \mathbb Q^\times/\{1,-1\}\cong\mathbb Q^+ as the mapping f:\mathbb Q^\times\to\mathbb Q^+ where f(a)=a if a>0 and f(a)=-a if a<0 is a surjective homomorphism with kernel \{1,-1\}.
    Yes I agree, when I typed up the solution I was thinking to myself that they numbers are assumed to be positive rationals.
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