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Thread: Hom

  1. #1
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    Hom

    Prove that:

    a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$
    b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
    c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$


    thanks!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by roporte View Post
    Prove that:

    a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$
    b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
    c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$


    thanks!!!
    what does $\displaystyle \text{Hom}_{\mathbb{Z}} (A,B)$ mean?
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  3. #3
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    $\displaystyle Hom_{R}(A,B)$ are all R-module homomorphisms from A to B.

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  4. #4
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    Quote Originally Posted by roporte View Post
    Prove that:

    a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$
    the natural $\displaystyle \mathbb{Z}-$homomorphism $\displaystyle f: \mathbb{Z} \longrightarrow \mathbb{Z}_n$ defined by $\displaystyle f(k)=[k]_n$ is obviously nonzero.

    b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
    let $\displaystyle f \in \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_5, \mathbb{Z}_7).$ then: $\displaystyle [0]_7=f([0]_5)=f([5]_5)=5f([1]_5).$ thus: $\displaystyle f([1]_5)=[0]_7.$ why? hence: $\displaystyle f=0.$

    c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$
    let $\displaystyle f \in \text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}).$ then for any integer n > 0: $\displaystyle nf(1/n)=f(1).$ so all (positive) integers are divisors of the integer f(1), which is possible only if f(1) = 0. therefore: $\displaystyle f=0.$ why?
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