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Thread: Compute Hom(q,q/z)

  1. #1
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    Compute Hom(q,q/z)

    Please compute;
    Hom(z,q/z)=?
    Hom(q,q/z)=?

    and also please prove that Q is an injective Z-module.
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  2. #2
    Super Member
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    Re: Compute Hom(q,q/z)

    $\displaystyle Hom (\mathbb{Z} , \mathbb{Q} / \mathbb{Z} )$ is trivially isomorphic to $\displaystyle \mathbb{Q} / \mathbb{Z}$ since any morphism is determined by where it sends $\displaystyle 1$. To prove that $\displaystyle \mathbb{Q}$ is injective take any morphism $\displaystyle f: n\mathbb{Z} \rightarrow \mathbb{Q}$ with $\displaystyle f(n)=a$ then, since $\displaystyle \mathbb{Q}$ is divisible there exists $\displaystyle b\in \mathbb{Q}$ with $\displaystyle nb=a$ take $\displaystyle g: \mathbb{Z} \rightarrow \mathbb{Q}$ with $\displaystyle g(1)=b$ then $\displaystyle gi=f$ where $\displaystyle i :n\mathbb{Z} \rightarrow \mathbb{Z}$ is the inclusion, conclude by Baer's criterion. I don't have the answer to the second but maybe the fact that $\displaystyle \mathbb{Q} / \mathbb{Z} \cong \oplus_{p \ prime } \mathbb{Z}_{p^\infty}$ is of use.
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  3. #3
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    Re: Compute Hom(q,q/z)

    thank you jose i got it. I will try for the others.
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