Atiyah MacDonald's exercise 1.21 on Zariski topology The exercise is the following:

Let $\phi:A\longrightarrow B$ be a ring homomorphism. Let $X = Spec(A)$ and $Y = Spec(B)$. If $\mathfrak{q} \in Y$, then $\phi^{-1}(\mathfrak{q})$ is a prime ideal of $A$. Hence $\phi$ induces a mapping $\phi^*:Y \longrightarrow X$. Show that:
  [...]
  iii) If $\mathfrak{b}$ is an ideal of $B$, then $\overline{\phi^*(V(\mathfrak{b}))} = V(\mathfrak{b}^c)$

So, i know that  $\overline{\phi^*(V(\mathfrak{b}))} = V(\bigcap \phi^*(V(\mathfrak{b})))$, i.e. the closure of $\phi^*(V(\mathfrak{b}))$ is the set of all prime ideals in $A$ which contain the intersection of all the contracted (prime) ideals containing $\mathfrak{b}$. 
I'm having some problems in showing that every prime ideal in $A$ containing $\mathfrak{b}^c$ also contains the intersection defined above. Could you help me? Thanks 
 A: First, I will make some statements without proof. For an ideal $\mathfrak{a} \unlhd A$, $\sqrt{\mathfrak{a}} = \cap_{\substack{p \in Spec(A) \\ p \supset a}} p$. Also, $V(\mathfrak{a}) \subset Spec(A)$ is a closed set for any $\mathfrak{a} \unlhd A$. Lastly, for $\phi : A \rightarrow B$ a ring homomorphism, $\mathfrak{a} \unlhd A$, we have $(\phi^{*})^{-1}(V(\mathfrak{a})) = V(\mathfrak{a}^{e})$, where $\mathfrak{a}^{e} \unlhd B$ is the ideal generated by the image of $\mathfrak{a}$. We want to show $(\overline{\phi^{*}(V(\mathfrak{b}))}) = V(\mathfrak{b}^{c})$.


$\boxed{ \subseteq }$ Suppose that $x \in \phi^{*}(V(\mathfrak{b}))$. Then there is $y \in V(\mathfrak{b})$, $\phi^{*}(y) = x \implies x = y^{c}$. But $y \supseteq \mathfrak{b} \implies x = y^{c} \supseteq \mathfrak{b}^{c} \implies x \in V(\mathfrak{b}^{c})$. Since $\mathfrak{b}^{c} \unlhd A$, $V(\mathfrak{b}^{c})$ is closed, which implies $\overline{\phi^{*}(V(\mathfrak{b}))} \subseteq V(\mathfrak{b}^{c})$

$\boxed{ \supseteq }$
Observe that $\overline{\phi^{*}(V(\mathfrak{b}))} \subseteq Spec(A)$ is closed, so $\overline{\phi^{*}(V(\mathfrak{b}))} = V(\mathfrak{a})$, for some $\mathfrak{a} \unlhd A$. Then:
$$
V(\mathfrak{a}^{e}) = (\phi^{*})^{-1}(V(\mathfrak{a})) = (\phi^{*})^{-1}(\overline{\phi^{*}(V(\mathfrak{b}))}) \supseteq V(\mathfrak{b})
$$
which implies that $\mathfrak{a}^{e} \subseteq \sqrt{\mathfrak{b}}$. Let $t \in \mathfrak{a}$, then $\phi(t) \in \mathfrak{a}^{e} \subseteq \sqrt{\mathfrak{b}}$. Then $\phi(t^{n}) = \phi(t)^{n} \in \mathfrak{b}$ for some $n \in \mathbb{N}$. So $t^{n} \in \mathfrak{b}^{c} \implies t \in \sqrt{\mathfrak{b}^{c}} \implies \mathfrak{a} \subseteq \sqrt{\mathfrak{b}^{c}}$. So:
$$
\overline{\phi^{*}(V(\mathfrak{b}))} = V(\mathfrak{a}) \supseteq V(\sqrt{\mathfrak{b}^{c}}) = V(\mathfrak{b}^{c})
$$
which completes the proof. Let me know if you have questions.
A: $$
\begin{align*}
\overline{\phi^*(V(\mathfrak{b}))}
&= V\left(\bigcap \phi^*(V(\mathfrak{b}))\right) && \text{as you observed,} \\
&= V\left(\bigcap_{\mathfrak{q} \in V(\mathfrak{b})} \mathfrak{q}^c\right) && \text{because $\phi^*(\mathfrak{q}) = \mathfrak{q}^c$ by definition,} \\
&= V\left(\left(\bigcap_{\mathfrak{q} \in V(\mathfrak{b})} \mathfrak{q}\right)^c\right) && \text{because preimages commute with intersections,} \\
&= V(\sqrt{\mathfrak{b}}^c) && \text{by Proposition 1.14,} \\
&= V(\sqrt{\mathfrak{b}^c}) && \text{because contraction/radicals commute,} \\
&= V(\mathfrak{b}^c).
\end{align*}
$$
