Let $R$ be a commutative ring with a unit and let $J\subseteq R$ be an ideal. Let $\psi:\operatorname{spec}(R/J)\to \operatorname{spec} R$ be the natural mapping (defined by $Z\mapsto\phi^{-1}(Z)$ where $\phi: R\to R/J,\ r\mapsto r+J$).
Let $Z\subseteq \operatorname{spec}(R/J)$ such that $\psi(Z)\subseteq\operatorname{spec}(R)$ is Zariski closed.
Show that $Z$ is Zariski closed.
Attempt:
$\psi(Z)$ is Zarisky closed so $\exists M\in\text{spec}(R)$ s.t. $$ \psi(Z)=\phi^{-1}(Z)=\{p\in\text{spec}(R):M\subseteq p\} \\ \Rightarrow \phi(\phi^{-1}(Z))=\phi(\{p\in\text{spec}(R):M\subseteq p\}) \\ =\{\phi(p):M\subseteq p\in\text{spec}(R)\} \\ =\{\{t+J:t\in p\}:M\subseteq M\in \text{spec}(R)\} $$ But here I stuck (also because $\phi(p)$ is not necessarily an element of $\text{spec}(R/J)$ for some $p\in\text{spec}(R)$).
If $\psi(Z)$ is a closed subset of $\operatorname{spec}(R)$, then there is an ideal $I$ of $R$ such that $\psi(Z)=V_R(I):=\{p\in \operatorname{spec}(R)\mid I\subseteq p\}$. Set $K:= \frac{I+J}{J}$ which is an ideal of $\frac{R}{J}$. Thus we have $\frac{p}{J}\in V_{\frac{R}{J}}(K)$ if and only if $\frac{I+J}{J}=K\subseteq \frac{p}{J}$ if and only if $I+J\subseteq p$ if and only if $I\subseteq p$ and $J\subseteq p$ if and only if $p\in V_R(I)=\psi(Z)$ and $\frac{p}{J}\in \operatorname{spec}(\frac{R}{J})$ if and only if $\frac{p}{J}\in Z$. Hence, $Z=V_{\frac{R}{J}}(K)$ .