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Math Help - Z[x] -> Z_m[x] defined by ... is a homomorphism

  1. #1
    Senior Member jakncoke's Avatar
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    Z[x] -> Z_m[x] defined by ... is a homomorphism

    The question is:

    Let m be a fixed positive int. For any integer a, let
    \bar{a} denote (a mod m) Show that the mapping of \Phi: Z[x] \to Z_{m}[x] given by [Math]\Phi(a_nx^n + a_{n-1}x^{n-1}+...+a_0) = \bar{a_n}x^n + \bar{a_{n-1}}x^{n-1}+...+\bar{a_0}[/tex]
    is a ring homomorphism.

    All i want to know is that is it enough to show that  \phi : Z \to Z_m defined by  \phi(x) =  x mod m is a homomorphism ?
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  2. #2
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    Quote Originally Posted by jakncoke View Post
    The question is:

    Let m be a fixed positive int. For any integer a, let
    \bar{a} denote (a mod m) Show that the mapping of \Phi: Z[x] \to Z_{m}[x] given by [Math]\Phi(a_nx^n + a_{n-1}x^{n-1}+...+a_0) = \bar{a_n}x^n + \bar{a_{n-1}}x^{n-1}+...+\bar{a_0}[/tex]
    is a ring homomorphism.

    All i want to know is that is it enough to show that  \phi : Z \to Z_m defined by  \phi(x) =  x mod m is a homomorphism ?

    Well, yes, but you've to clarify how this is kept by the polynomial ring's operations.

    Tonio
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