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Thread: Help Proving a Theorem (Ring Isomorphisms)

  1. #1
    Junior Member ginafara's Avatar
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    Help Proving a Theorem (Ring Isomorphisms)

    Can someone help me get started. I start by assuming 1-1 and know that i go the other direction, but can't seem to see how to use 1-1 to show anything...Thanks in advance.

    Theorem:
    Let $\displaystyle \Phi :R \rightarrow S $ be a ring homomorphism. Show that $\displaystyle \Phi $ is 1-1 iff $\displaystyle ker(\Phi) $ is trivial
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by ginafara View Post
    Can someone help me get started. I start by assuming 1-1 and know that i go the other direction, but can't seem to see how to use 1-1 to show anything...Thanks in advance.

    Theorem:
    Let $\displaystyle \phi :R \rightarrow S $ be a ring homomorphism. Show that $\displaystyle \phi $ is 1-1 iff $\displaystyle ker(\phi) $ is trivial
    1-1 $\displaystyle \Rightarrow ker(\phi)$ is trivial

    Since for any ring,$\displaystyle \phi(0) = 0$, 1-1 nature of $\displaystyle \phi$ forces 0 to be the only number of R to map to the 0 of S. Thus $\displaystyle \phi(x) = 0$ only for x=0. Hence $\displaystyle ker(\phi)$ is trivial

    $\displaystyle ker(\phi) = \{0\} \Rightarrow \phi$ is 1-1

    For any ring,$\displaystyle \phi(0) = 0$. If $\displaystyle \phi(x) = \phi(y)$, then $\displaystyle \phi(x-y) = 0$. But since $\displaystyle ker(\phi) = \{0\}, x=y$
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