Given $\varphi: R \rightarrow S$ be a surjective ring homomorphism. Prove that the image of the center of $R$ is
contained in the center of $S$.
$\textit{Proof}$ : Denote the center of $R$ as $Z(R)$ and the center of $S$ as $Z(S)$. Let $x \in{Z(R)}$. Then, $xr=rx \quad \forall r\in{Z(R)}$.
We want to show that $\varphi(x)s=s\varphi(x) \quad \forall s\in{S}$.
With $s\in{S}$, then since $\varphi$ is surjective, $\exists a\in{R}$ s.t. $\varphi(a)=s$. Then,
$$s\varphi(x)=\varphi(a)\varphi(x)=\varphi(ax)=\varphi(xa)$$
since $x\in{Z(R)}$.
$$\varphi(xa)=\varphi(x)\varphi(a)=\varphi(x)s \Longrightarrow \varphi(x)\in{Z(S)}$$
We have shown that the image of an arbitrary element in $Z(R)$ commutes with any element in $S$.
$\therefore \varphi(Z(R)) \subseteq Z(S) \quad \blacksquare$
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