If $R$ is a noetherian ring, prove that every subset of $\mathrm{Spec}(R)$ is quasi-compact.

Question

This is exercise I-17 of The Geometry of Schemes by D. Eisenbud and J. Harris.

I have thought this. For $X\subset\mathrm{Spec}(R),$ an element $U$ of an arbitrary open cover $\mathscr{U}$ of $X$ is given by $$U=\mathrm{Spec}(R)-V(S)=\mathrm{Spec}(R)-\cap_{f\in S}V(f)=\cup_{f\in S}\mathrm{Spec}(R_f), $$ for some $S\subset R\ $ and with $R_f$ being the localization of $R$ at the element $f.$

So we have that $X=\cup_{U\in\mathscr{U}}U=\cup_S(\cup_{f\in S}\mathrm{Spec}(R_f)),$ and the prime ideals in $R_f$ are in one-to-one correspondence with the ideals in $R$ by $\mathfrak{p}\subset R\leftrightarrow \mathfrak{p}R_f\subset R_f.$ Therefore, $\cup_S(\cup_{f\in S}\mathrm{Spec}(R_f))$ can be seen as an ascending chain of ideals in a noetherian ring, so it must stabilize, so we conclude that $\mathscr{U}$ has a finite subcover.

Does this work? Is there any other way to do this?

Answer 1

Not to say that something truly rigorous could not be produced along the line of reasoning you suggested, but allow me to present a somewhat more general perspective (and also to point out that claiming equality between a Zariski-closed set of a ring $A$ and the spectrum of a localization of $A$ is not a rigourous and entirely licit thing to do; the two objects might be canonically isomorphic as topological spaces, however that still leaves you with the issue of explicitly introducing and rigorously putting to use the canonical homeomorphism existent between the two).

We reason as follows. Define first a topological space $(X, \mathscr{T})$ to be noetherian if its topology $\mathscr{T}$ is a noetherian ordered set when ordered by inclusion (we recall that an arbitrary ordered set $(A, R)$ is called noetherian when any nonempty $X \subseteq A$ admits a maximal element with respect to the fixed order $R$). By complementarity-duality (i.e. the dualisation implemented by taking complementaries) the noetherianity of a topological space is equivalent to claiming its collection of closed subsets is an artinian ordered set (this being the notion dual to noetherian ordered sets, namely it refers to an ordered set in which every nonempty subset will admit at least a minimal element).

We have the following results:

Lemma 1. If $(X, \mathscr{T})$ is a noetherian space, then any subspace $M \subseteq X$ is also noetherian.

Proof: the artinianity of an ordered set is equivalent to claiming that any decreasing sequence of elements from that set eventually stabilizes. To this end, let $F$ be a decreasing sequence of sets closed relatively to $M$; by definition of the subspace topology and since $F_n$ is relatively closed we infer: $$\left(\overline{F_n}\right)_{M}=F_n=\overline{F_n} \cap M \tag{clos}$$ where for $T \subseteq M$ the symbol $\overline{T}_M$ denotes the relative closure of $T$ with respect to the subspace topology induced on $M$.

Since the sequence $F$ is decreasing, clearly $\left(\overline{F_n}\right)_{n \in \mathbb{N}}$ will also be a decreasing sequence of closed subsets of $X$ and hence will eventually stabilize by the noetherianity of $X$; the relation (clos) above tells us that $F$ will also stabilize (from any threshhold value of $n \in \mathbb{N}$ at which the sequence of closures in $X$ happens to stabilize). $\Box$

Lemma 2. The space $(X, \mathscr{T})$ is noetherian if and only if any subset of $X$ is quasi-compact.

Proof: Assume first the noetherianity of $(X, \mathscr{T})$ and consider arbitrary $M \subseteq X$ together with $\mathscr{U} \subseteq \mathscr{T}$ such that $M \subseteq \bigcup \mathscr{U}$. By noetherianity it is easy to show (let me know if you need more details) that there exists a finite $\mathscr{P} \subseteq \mathscr{U}$ such that $\bigcup \mathscr{P}=\bigcup \mathscr{U}$ (any arbitrary union of a collection of open sets can be written as the union of a finite subcollection); this in effect establishes the quasi-compactness of $M$.

For the converse implication, it will suffice to exploit the fact that any open subset is quasi-compact: consider an increasing sequence $U \in \mathscr{T}^{\mathbb{N}}$ and the union $$V=\bigcup_{n \in \mathbb{N}}U_n$$ which will of course be also open; as $V$ is quasi-compact, it can be covered by only finitely many of the terms in the sequence $U$, in other words there exists a finite subset $M \subseteq \mathbb{N}$ such that $$V \subseteq \bigcup_{n \in M} U_n$$

It is easy to furthermore see that this $M$ can be taken to be nonempty; hence, as a totally ordered finite nonempty set it will admit a maximum say $m$ and therefore $$\bigcup_{n \in M} U_n=U_m$$ by virtue of the monotony of $U$; from this it follows at once that $V=U_m=U_n$ for any $n \geqslant m$, in other words that the sequence $U$ stabilizes (at $m$). $\Box$

With the previous results in place we can finally establish the result you seek, by stating one more

Proposition: If $A$ is a noetherian commutative ring, then $\mathrm{Spec}(A)$ is a noetherian topological space.

Proof: In order not to prolong the expound anymore I will briefly mention a series of important facts (again, let me know if you are interested in a more detailed argumentation). Define the maps:

$$\varPhi: \mathscr{Id}(A) \to \mathscr{P}(\mathrm{Spec}(A)), \varPhi(I)=\mathscr{V}(I)$$ $$\Psi: \mathscr{P}(\mathrm{Spec}(A)) \to \mathscr{Id}(A),\\ \Psi(\mathscr{X})=\begin{cases} \bigcap \mathscr{X}, \mathscr{X} \neq \varnothing \\ A, \mathscr{X}=\varnothing \end{cases}$$

where $\mathscr{Id}(A)$ denotes the set of all ideals of $A$, including the improper one (the axiomatic system I use and prefer does not handle empty intersections which is why I had to resort to a branched definition for $\Psi$). These maps are decreasing (when ordering both their domains of definition by inclusion) and have the properties that $\Psi(\varPhi(I))=\sqrt{I}$ for any ideal $I$ and respectively $\varPhi(\Psi(\mathscr{X}))=\overline{\mathscr{X}}$ for any subset $\mathscr{X} \subseteq \mathrm{Spec}(A)$.

Hence, they will implement mutually inverse anti-isomorphisms of ordered sets between the set of radical ideals of $A$ and the set of Zariski-closed subsets in the spectrum of $A$; therefore if $A$ is noetherian on radical ideals (in particular if it is noetherian on all ideals), then $\mathrm{Spec}(A)$ is artinian on closed subsets, in other words noetherian as a topological space. $\Box$

Answer 2

A finite union of a quasicompacts is quasi compact. If $U$ is an open, then $U= X\setminus V(I) $ where $I$ is an ideal of $R$, there is $f_1,...f_r$ in$R$ such that $I=(f_1,...f_r)$. Then $U=D(f_1)\cup... \cup D(f_r) $ and quasi compact.