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
iii) If $\mathfrak b$ is an ideal of $B$, then $\overline{\phi^\ast (V (\mathfrak b))} = V (\mathfrak b^c)$.

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.

  • $\begingroup$ What is $r(V(\mathfrak{b})^c)$ supposed to mean? In particular, I don't know what the radical of a closed subset in the Zariski topology is, nor what the contraction of such a set should be. $\endgroup$ May 17, 2015 at 23:18
  • $\begingroup$ @TakumiMurayama That is where my confusion lies. Maybe it is a shorthand for the set of all the elements in the contractions of the prime ideals containing $\mathfrak b$ (or a shorthand for the ideal generated by all these elements). The first equality in the last chain of equalities says $r (E^c) = r (E)^c$, which I only know to hold when $E$ is an ideal. I'm asking if there is a way to make sense of this "proof", because if the answer is yes, it is shorter than the other proof I found. But if the answer is no, then I will simply ignore it. $\endgroup$
    – Earthliŋ
    May 17, 2015 at 23:29

2 Answers 2


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 $J^c := \phi^{-1}(J)$.

  1. $\phi^{-1}(r(J)) = r(\phi^{-1}(J))$, that is, the radical and pre-image commute. This is a pretty straightforward chain of if and only if statements: $$a \in \phi^{-1}(r(J)) \iff \phi(a) \in r(J) \iff \exists n \in \mathbb{N} : \phi(a)^n = \phi(a^n) \in J \iff \exists n \in \mathbb{N} : a^n \in \phi^{-1}(J) \iff a \in r(\phi^{-1}(J))$$
  2. $\overline{\phi^*(V(J))} = V(r(\phi^{-1}(J)))$. 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!


I'm just going to talk.

The closure of a set is the intersection of all closed sets containing it. The radical of an ideal is the intersection of all prime ideals containing it (since if the nilradical is the intersection of all prime ideals containing $0$, then the nilradical of $R/I$ is the radical of $I$ in $R$ and the fourth isomorphism theorem gives the proof).

If you think about the closed set $V(E)$ for some set $E\subset A$, then this is the same as the set of ideals which occur in the intersection of $r((E))$.

The notation $r(V(\mathfrak{b})^c)$ should be avoided as it is nonsensical via the definitions. I would assume what they mean is the radical of the ideal generated by the set of the contractions of prime ideals in $V(\mathfrak{b})$. Or possibly the intersection of all contractions of prime ideals appearing in the set $V(\mathfrak{b})$.

I would suggest not using the solutions for these reasons. I don't immediately see why the first claim is true (which, admittedly, could be lack of knowledge on my own part). But the second equality, I would think, is to say that $r(\mathfrak{b})$ is the intersection of all prime ideals containing $\mathfrak{b}$ and $V(\mathfrak{b})$ is the set of all prime ideals containing $\mathfrak{b}$ so I imagine they are using the radical to denote their intersection.


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