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Math Help - show ring isomorphism

  1. #1
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    show ring isomorphism

    Let R be a ring and M an R-module.
    Hom_R(R,M) is the set of R-homomorphisms from R to M.
    Define \phi: Hom_R(R,M) \rightarrow M by \phi(f)=f(1). Show \phi is a ring isomorphism.
    I know how to show it is a homomorphism and one-to-one, but having trouble with onto. Can I get some help?
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Let R be a ring and M an R-module.
    Hom_R(R,M) is the set of R-homomorphisms from R to M.
    Define \phi: Hom_R(R,M) \rightarrow M by \phi(f)=f(1). Show \phi is a ring isomorphism (R-module isomorphism ?).
    I know how to show it is a homomorphism and one-to-one, but having trouble with onto. Can I get some help?
    Your M seems to be a unitary R-module from your question because f(1) is defined.

    I think it is difficult to deduce that \phi is onto by using a \phi alone.

    Rather, define a map \psi:M \rightarrow Hom_R(R,M) given by m \mapsto f_m, where f_m(r) = rm.

    Suppose m_1 = m_2 ( m_1, m_2 \in M). Then, \psi(m_1) = f_{m_{1}}=f_{m_{2}}=\psi(m_2), because f_{m_{1}}(r)=rm_1=rm_2=f_{m_{2}}(r). Thus \psi is well-defined. I'll leave it to you to show that \psi is an R-module homomorphism.

    Then \phi\psi=1_M and \psi\phi=1_{Hom_{R}(R,M)} (verify this).
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  3. #3
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    Just to show that given any element x in M, you can construct a well-defined R-homomorphism f from R to M such that f(1)=x. The f_m function that aliceinwonderland constructed is exactly demonstrating this idea.
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