If A is noetherian, then Spec(A) is noetherian Let A be a noetherian ring. How can I show that Spec(A) is noetherian?
Also, is there a way to show this by showing directly that the closed sets in Spec(A) satisfy the descending chain condition?
(This is exercise 6.8 from Atiyah and Macdonald.)
 A: 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.
A: 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.
