**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.16**

Solution: Suppose $r \in \varphi[Z(R)]$. Then $r = \varphi(z)$ for some $z \in Z(R)$. Now let $x \in S$. Since $\varphi$ is surjective, we have $x = \varphi y$ for some $y \in R$. Now $$xr = \varphi(y)\varphi(z) = \varphi(yz) = \varphi(zy) = \varphi(z)\varphi(y) = rx.$$ Thus $r \in Z(S)$.