Results 1 to 3 of 3

Thread: center of a ring

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    166

    center of a ring

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

    Let $\displaystyle \phi : R$ --> $\displaystyle S$ be a surjective homomorphism of rings. Prove that the image of the center of $\displaystyle R$ is contained in the center of $\displaystyle S$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by dori1123 View Post
    The center of a ring R is {$\displaystyle z \in R | zr = rz \forall r \in R$}

    Let $\displaystyle \phi : R$ --> $\displaystyle S$ be a surjective homomorphism of rings. Prove that the image of the center of $\displaystyle R$ is contained in the center of $\displaystyle S$.
    Let $\displaystyle z$ be in the center of $\displaystyle R$ then $\displaystyle zr=rz$ for all $\displaystyle r\in R$. But then it means $\displaystyle \phi(zr) = \phi(rz) \implies \phi(z)\phi(r) = \phi (r)\phi (z)$ for $\displaystyle r\in R$. But $\displaystyle \{ \phi (r) |r\in R \} = \{ s |s\in S\}$ because $\displaystyle \phi$ is surjective. Thus, $\displaystyle \phi(r)$ runs through all values in $\displaystyle S$. This means $\displaystyle \phi(z) s = s \phi (z)$ for each $\displaystyle s\in S$. We see that $\displaystyle \phi( z)$, the image of the center, is contained in the center of $\displaystyle S$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    Let $\displaystyle z$ be in the center of R ($\displaystyle Z(R)$) $\displaystyle \phi(z)\in Im(Z(R))$. Then for any $\displaystyle s\in S$ we can find a preimage in $\displaystyle R$ since $\displaystyle \phi$ is surjective. Then consider

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

    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; Dec 6th 2008 at 08:38 PM. Reason: Um oops, looks like you beat me to the punch, sorry
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Prove the Artinian ring R is a division ring
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Jun 8th 2011, 03:53 AM
  2. Center of a Ring and an Ideal
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 14th 2011, 02:38 PM
  3. Center of a matrix ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 28th 2011, 11:00 AM
  4. Ideals of ring and isomorphic ring :)
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 24th 2009, 03:23 AM
  5. Rotation Center Vs. Actual Center.
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 7th 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum