A proof for Atiyah-Macdonald Exercise I.21.iii
The following is exercise I.21.(iii) of Atiyah-Macdonald:
Let $\phi \colon A \to B$ be a ring homomorphisms. Let $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$ [and let $\phi^\ast \colon Y \to X$ be the induced mapping, $\phi^\ast (\mathfrak q) = \phi^{-1} (\mathfrak q)$]. Show that
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iii) If $\mathfrak b$ is an ideal of $B$, then $\overline{\phi^\ast (V (\mathfrak b))} = V (\mathfrak b^c)$.
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After trying myself for a while, I looked at solutions posted on the internet, one of which presented the following solution.
It is clear that $\mathfrak p \in \overline{\phi^\ast (V (\mathfrak b))} \color{blue} \Leftrightarrow r (V (\mathfrak b)^c) \subset \mathfrak p$ and $\mathfrak p \in V (\mathfrak b^c) \Leftrightarrow \mathfrak b^c \subset \mathfrak p$. Then it suffices to show that $\mathfrak b^c \subset \mathfrak p$ if and only if $r(V (\mathfrak b)^c) \subset \mathfrak p$, which is true because $r(V (\mathfrak b)^c) = r(V (\mathfrak b))^c \color{blue} = r(\mathfrak b)^c = r(\mathfrak b^c)$.
where I have marked the things I don't understand in blue.
In the meanwhile I have found a different proof for the exercise, but I still don't understand this proof. In particular
I don't understand the connection between taking the closure and taking the radical.
I don't know if $r(V (\mathfrak b)^c)$ is a shorthand for the set of elements contained in the contractions of prime ideals contained $\mathfrak b$ or for the ideal generated by these elements or something else.
I don't see why the marked equality holds.
I know this question is old, but I stumbled across it because I had the same questions as the OP. There seem to be two issues the OP was struggling with. I am going to avoid Atiyah-MacDonald's $J^c$ notation because it is nonstandard, but if you have a (commutative) ring homomorphism $\phi \colon A \to B$ between commutative rings $A$ and $B$, then, for $J$ an ideal of $B$, we have $J^c := \phi^{-1}(J)$.
- $\phi^{-1}(r(J)) = r(\phi^{-1}(J))$, that is, the radical and pre-image commute. This is a pretty straightforward chain of if and only if statements: $$a \in \phi^{-1}(r(J)) \iff \phi(a) \in r(J) \iff \exists n \in \mathbb{N} : \phi(a)^n = \phi(a^n) \in J \iff \exists n \in \mathbb{N} : a^n \in \phi^{-1}(J) \iff a \in r(\phi^{-1}(J))$$
- $\overline{\phi^*(V(J))} = V(r(\phi^{-1}(J)))$. We know that $\overline{\phi^*(V(J))}$ must be of the form $V(I)$ for some $I$ an ideal of $A$ by definition of the Zariski topology on $Spec(A)$ (which is covered in a prior exercise in AM). Let $Y := Spec(B)$. Such an ideal $I$ must have $$I = \bigcap_{Q \supset J, Q \in Y} \phi^{-1}(Q) = \phi^{-1} \left( \bigcap_{Q \supset J, Q \in Y} Q \right) = \phi^{-1}(r(J)) = r(\phi^{-1}(J))$$ where the preimage/intersection commute by an earlier exercise, we use a characterization of the radical as an intersection of prime ideals containing $J$, and the previous item above to get the equalities.
I hope that helps the OP or any future readers!