# Principal opens of spectrum of a ring are (quasi-)compact

I am doing exercise 17 in Atiyah-MacDonald, in particular I am not confident with my solution for part vi):

[Let $$A$$ be a ring.] For each $$f \in A$$, let $$X_f$$ denote the complement of $$V(f) [= \{\text{prime ideals containing }f\}]$$ in $$\mathrm{Spec}(A)$$. The sets $$X_f$$ are open. Show that they form a basis for the Zariski topology, and that

• [...]
• v) $$X[=\mathrm{Spec}(A)]$$ is quasi-compact (that is, every open covering of $$X$$ has a finite subcovering).
• vi) More generally, each $$X_f$$ is quasi-compact.
• [...]

## My solution for (v) (to give context)

I've proven (v) as follows. It suffices to consider open covers with principal open sets. Let $$(X_{f_i})_{i \in I}$$ be one such cover. Then for any ideal $$\mathfrak{a} \triangleleft A$$, if $$\mathfrak{a}$$ is proper then we can extend it to a prime (maximal) ideal $$\mathfrak{p}$$ and then, due to the open cover there is an $$i \in I$$ such that $$f_i \notin \mathfrak{p} \supseteq \mathfrak{a}$$, so $$f_i \notin \mathfrak{a}$$. Taking the contrapositive, any ideal containing all of the $$f_i$$ will be the whole ring $$A$$. Therefore the ideal generated by all the $$f_i$$ is $$(1)$$, so there is a finitely supported $$A$$-linear combination of the $$f_i$$ that equals $$1$$: $$\sum_{i \in J} g_i f_i = 1 \qquad J \subseteq I,\ J\text{ is finite}$$ Therefore, a prime ideal cannot contain $$f_{i}$$ for all $$i \in J$$, as otherwise it would contain 1. Hence we obtain the finite subcover $$(X_{f_i})_{i \in J}$$.

## First (failed) attempt for (vi)

For part (vi), first I noticed that any open set $$V$$ in $$X_f$$ with the subspace topology can be written as $$X_f \cap U$$ where $$U$$ is open in $$X$$, which in turn can be written as a union of principal open sets $$U = \bigcup_{i \in I} X_{f_i}$$; so we can write $$V = X_f \cap U = X_f \cap \bigcup_{i \in I} X_{f_i} = \bigcup_{i \in I} X_f \cap X_{f_i} = \bigcup_{i \in I} X_{f \cdot f_i}$$ So I tried using the same reasoning as before. The claim I was trying to prove is that if an ideal does not contain $$f$$ then it does not contain all of the $$f \cdot f_i$$, but this is not necessarily true as when we extend an arbitrary proper ideal to a prime ideal, we might be adding $$f$$ in the process.

## Second attempt for (vi)

So I tried this other approach. I claim that $$X_f$$ with the subspace topology is homeomorphic to $$\mathrm{Spec}(A_f)$$ with the Zariski topology, where $$A_f = S^{-1}A$$ denotes the localization of $$A$$ at $$S = \{f^n\}_{n \ge 0}$$. If this claim is true, then I can use part v) to argue that $$\mathrm{Spec}(A_f)$$ is compact, hence so is $$X_f$$.

To prove the claim, I argue that localization gives us a bijection $$\mathfrak{p} \mapsto S^{-1}\mathfrak{p}$$ between the prime ideals of $$A$$ that do not meet $$S$$ and the prime ideals of $$S^{-1}A$$; but the prime ideals of $$A$$ that do not meet $$S$$ (i.e. that do not contain any power of $$f$$) are, by primality, the prime ideals that do not contain $$f$$ itself; so this is a bijection $$X_f \to \mathrm{Spec}(A_f)$$. Moreover, both this map and its inverse are continuous because the image of a closed set $$X_f \cap V(E)$$ is $$V(i(E))$$ (which is closed) and the preimage of a closed set $$V(F)$$ is $$X_f \cap V(i^{-1}(F))$$ (which is closed), where $$i \colon A \to A_f$$ is the natural map, because extension and contraction of ideals preserves inclusions and $$V(E)$$ is defined as the set of prime ideals containing the set $$E \subseteq A$$.

I am not very confident about the continuity argument when showing that this bijection is a homeomorphism. Is it correct? Alternatively, can the first, more elementary approach also be continued?

Let $$\{X_{f_i}\}_{i\in I}$$ be a cover of $$X_f$$, i.e., $$X_f\subset\bigcup_{i\in I}X_{f_i}$$. Taking the compliment, we have $$V(f)\supset \bigcap_{i\in I}V(f_i)=V\big((f_i)_{i\in I}\big),$$ i.e., $$(f^m)\subset (f_i)_{i\in I}$$ for some $$m\ge1$$. Thus, $$f^m=\sum_{i\in J}a_if_i$$ for some finite subset $$J\subset I$$ and $$a_i\in A$$. Now $$X_f\subset \bigcup_{i\in J}X_{f_i}$$.
• How can we deduce $(f) \subseteq (f_i)_{i \in I}$ from $V(f) \supseteq V((f_i)_{i \in i})$? The latter inclusion tells us that every prime ideal containing all of the $f_i$ also contains $f$. This just tells us that $(f) \subseteq \sqrt{(f)} \subseteq \sqrt{(f_i)_{i \in I}}$, since the radical of an ideal is the intersection of all prime ideals that contain it. But the ideal generated by the $f_i$ could be non-radical. Commented Jan 9 at 9:21