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Math Help - Hom

  1. #1
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    Hom

    Prove that:

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


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

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

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

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

    c) Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0
    let f \in \text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}). then for any integer n > 0: 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: f=0. why?
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