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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?

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1 Answer 1

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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 $X_f\subset \bigcup_{i\in J}X_{f_i}$.

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  • $\begingroup$ 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. $\endgroup$
    – Anakhand
    Commented Jan 9 at 9:21
  • $\begingroup$ @Anakhand Sorry yes you are correct, I have fixed the claim. $\endgroup$
    – Kenta S
    Commented Jan 9 at 9:42

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