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Thread: Isomorphic

  1. #1
    ux0
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    Isomorphic

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


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

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

    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 $\displaystyle \mathbb{Z}$, and let H be the multiplicative group of all positive rationals. Prove that G $\displaystyle \cong$ H


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

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

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

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

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