Let $\phi:A\rightarrow B$ be a ring homomorphism. Let $X＝ \operatorname{Spec}(A)$, $Y= \operatorname{Spec}(B)$. If $q\in Y$, then $\phi^{-1}(q)$ is a prime ideal of $A$. Hence $\phi$ induces a mapping $\phi^* :Y\rightarrow X$. Show that:
If $b$ is an ideal of $B$, then $$\overline{\phi^*(V(b))}＝V(b^c).$$
Here $b^c$ is the ideal $\phi^{-1}(b)$.
I can prove that the left side is contained in the right side. But I can't prove the right side is contained in the left side.