I'm trying to prove or find a counterexample to the following:
Let $R$ be a ring and $z \in Z(R)$. Do we have an isomorphism $$ Z(R/zR) \cong Z(R)/zZ(R) \hspace{2pt}?$$
When I've been trying to prove this, a sensible thing to do was to consider the map $Z(R) \to R/zR, \hspace{2pt} x \mapsto x + zR$ and try to apply the first isomorphism theorem. I ran into trouble when trying to show that it's image was $Z(R/zR)$. I'm happy to assume some stronger hypotheses on $R$, such as it being graded (with $z$ homogeneous, say) or $R$ and/or $R/zR$ being a domain (where the latter implies the former in the graded case if we also assume that $z$ is a nonzerodivisor).
It's been true for the examples I've tried, but it might be the case that I've been picking examples which are "too nice".
Consider the ring $R$ freely generated by by three letters $x$, $y$ and $z$ subject to the relations $$xz=zx, \qquad yz=zy, \qquad xy-yx=z.$$ The center of this $R$ is $Z(R)=k[z]$, the subalgebra generated by $z$, and then $Z(R)/zZ(R)\cong k$. On the other hand, $R/zR$ is a commutative algebra isomorphic to thepolynomial algebra on two generators.