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Thread: show ring isomorphism

  1. #1
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    show ring isomorphism

    Let $\displaystyle R$ be a ring and $\displaystyle M$ an $\displaystyle R$-module.
    $\displaystyle Hom_R(R,M)$ is the set of $\displaystyle R$-homomorphisms from $\displaystyle R$ to $\displaystyle M$.
    Define $\displaystyle \phi: Hom_R(R,M) \rightarrow M$ by $\displaystyle \phi(f)=f(1)$. Show $\displaystyle \phi$ 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 $\displaystyle R$ be a ring and $\displaystyle M$ an $\displaystyle R$-module.
    $\displaystyle Hom_R(R,M)$ is the set of $\displaystyle R$-homomorphisms from $\displaystyle R$ to $\displaystyle M$.
    Define $\displaystyle \phi: Hom_R(R,M) \rightarrow M$ by $\displaystyle \phi(f)=f(1)$. Show $\displaystyle \phi$ 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 $\displaystyle \phi$ is onto by using a $\displaystyle \phi$ alone.

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

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

    Then $\displaystyle \phi\psi=1_M$ and $\displaystyle \psi\phi=1_{Hom_{R}(R,M)}$ (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 $\displaystyle f_m$ function that aliceinwonderland constructed is exactly demonstrating this idea.
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