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Math Help - Isomorphic

  1. #1
    ux0
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    Isomorphic

    Let G be the additive group if all polynomials in x with coefficients in \mathbb{Z}, and let H be the multiplicative group of all positive rationals. Prove that G \cong H


    I'm think if I list the prime numbers p_0=2, p_1=3, p_2=5,..., and define

    \varphi(e_0+e_1x+e_2x^2+...+e_nx^n)=p_0^{e_0}...p_  n^{e_n}

    But I'm really not sure where to go from here..
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  2. #2
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    Quote Originally Posted by ux0 View Post
    Let G be the additive group if all polynomials in x with coefficients in \mathbb{Z}, and let H be the multiplicative group of all positive rationals. Prove that G \cong H


    I'm think if I list the prime numbers p_0=2, p_1=3, p_2=5,..., and define

    \varphi(e_0+e_1x+e_2x^2+...+e_nx^n)=p_0^{e_0}...p_  n^{e_n}

    But I'm really not sure where to go from here..
    you've done the hard part, which is finding the map. \varphi is obviously well-defined because \{x^n \}_{n \geq 0} is a basis for the free abelian group \mathbb{Z}[x]. to prove that \varphi is a homomorphism,

    let f=\sum_{j=0}^n a_jx^j, \ g=\sum_{j=0}^n b_jx^j, where n=\max \{\deg f, \deg g \}. then f+g=\sum_{j=0}^n (a_j+b_j)x^j and thus \varphi(f + g)=\prod_{j=0}^n p_j^{a_j+b_j}=\prod_{j=0}^n p_j^{a_j} \prod_{j=0}^n p_j^{b_j}=\varphi(f) \varphi(g).

    injectivity and surjectivity of \varphi follows from the unique prime factorization that we have in \mathbb{Z}.
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