# Thread: center of a ring

1. ## 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 $\displaystyle S$.

2. Originally Posted by dori1123
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 $\displaystyle S$.

3. 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.