If A is noetherian, then Spec(A) is noetherian

Let $A$ be a Noetherian ring. Let $I$ be an ideal of $A$. We denote by $V(I)$ the set {$P \in$ Spec($A$); $I \subset P$}. Let $N$ be the nilradical of $A/I$. We denote by rad($I$) the inverse image of $N$ by the canonical homomorophism $A \rightarrow A/I$, i.e. rad($I$) = {$x \in A$; $x^n \in I$ for some $n$ ($n$ depends on $x$)}. Since $N$ is an ideal(Atiyah-MacDonald Proposition 1.7), so is rad($I$). Clearly $V(I)$ = $V$(rad($I$)).

Since the nilradical of $A/I$ is the intersection of all the prime ideals of $A/I$(Atiyah-MacDonald Proposition 1.8), rad($I$) = $\cap$ {$P \in V(I)$}.

Suppose $V(I_1) \supset V(I_2) \supset \dots$ is a descending sequence of closed subsets of Spec($A$). Then rad($I_1$) $\subset$ rad($I_2$) $\subset \dots$ by the above claim. Since $A$ is Noetherian, there exists $n$ such that rad($I_n$) = rad($I_{n+1}$) = $\dots$

Hence $V(I_n) = V(I_{n+1}) = \dots$

Hence Spec($A$) is Noetherian and we are done.

I find it helpful to think in terms of two somewhat more precise results.

Step 1: The mappings

$J \mapsto V(J) = \{\mathfrak{p} \in \operatorname{Spec}(R) \ | \ J \subset \mathfrak{p} \}$ from ideals of $R$ to subsets of $\operatorname{Spec}(R)$

and

$S \mapsto I(S) = \bigcap_{\mathfrak{p} \in S} \mathfrak{p}$ from subsets of $\operatorname{Spec}(R)$ to ideals of $R$

induce mutually inverse bijections from the set of radical ideals in $R$ to the family of Zariski-closed subsets of $\operatorname{Spec}(R)$.

(If I remember correctly, what is given in Atiyah-Macdonald before the exercise in question is sufficient to establish this without much trouble.)

Step 2: Therefore for any commutative ring $R$, $\operatorname{Spec} R$ is Noetherian -- i.e., satisfies DCC on Zariski-closed subsets -- iff $R$ satisfies ACC on radical ideals.

The details can be found in $\S 13.5$ of my commutative algebra notes.