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Math Help - center of a ring

  1. #1
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    center of a ring

    The center of a ring R is { z \in R | zr = rz \forall r \in R}

    Let \phi : R --> S be a surjective homomorphism of rings. Prove that the image of the center of R is contained in the center of S.
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    The center of a ring R is { z \in R | zr = rz \forall r \in R}

    Let \phi : R --> S be a surjective homomorphism of rings. Prove that the image of the center of R is contained in the center of S.
    Let z be in the center of R then zr=rz for all r\in R. But then it means \phi(zr) = \phi(rz) \implies \phi(z)\phi(r) = \phi (r)\phi (z) for r\in R. But \{ \phi (r) |r\in R \} = \{ s |s\in S\} because \phi is surjective. Thus, \phi(r) runs through all values in S. This means  \phi(z) s = s \phi (z) for each s\in S. We see that \phi( z), the image of the center, is contained in the center of S.
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  3. #3
    Super Member Gamma's Avatar
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    Let z be in the center of R ( Z(R)) \phi(z)\in Im(Z(R)). Then for any s\in S we can find a preimage in R since \phi is surjective. Then consider

    \phi ^{-1}(s)*z=z*\phi ^{-1}(s) \Rightarrow \phi(\phi ^{-1}(s)*z)= \phi(z*\phi ^{-1}(s))
    \Rightarrow \phi(s*z)=\phi(z*s)\Rightarrow\phi(s)\phi(z)=\phi(  z)\phi(s).

    Thus every element of the image of the center of R commutes with any element of S, so it must be contained in the center of S by definition.
    Last edited by Gamma; December 6th 2008 at 08:38 PM. Reason: Um oops, looks like you beat me to the punch, sorry
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