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I am doing a problem a problem in Hartshorne (2.3.2) which asks to show that a morphism of schemes $f : X \to Y$ is quasi compact iff for every affine open $U \subseteq Y$, $f^{-1}(U)$ is quasi compact. Now one direction is tautological so for the other direction take $U \subseteq Y$ an affine open subset. Let $\{V_i\}$ be a cover of $Y$ by open affines such that for every $i$, $f^{-1}(V_i)$ is quasi - compact. Then for every $i$ we can write

$$U \cap V_i = \bigcup_{j} V_{ij}$$

where the $V_{ij}$ are open and principal in both $U$ and $V_i$. Since $\bigcup_i (U \cap V_i) = \bigcup_i \bigcup_j V_{ij}$ is an open cover of $U$ and since $U$ is quasi compact, this means $U = \bigcup_{k=1}^n V_{i_k j_k}$ for some indices $i_k,j_k$. Thus $f^{-1}(U)$ is a finite union of the open sets $f^{-1}(V_{i_k,j_k})$. Now if each of these is quasi - compact then $f^{-1}(U)$ being a finite union of quasi - compact sets is quasi - compact.

My question is: Why is each $f^{-1}(V_{i_k,j_k})$ quasi - compact? I know that for every $i_k$, $f^{-1}(V_{i_k})$ can be covered by a finite union of open affine sets by assumption. Does this mean then that $f^{-1}(V_{i_k,j_k})$ is principal in each of these open affines (and thus quasi - compact)?

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2 Answers 2

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Notice that you haven't yet used that the $f^{-1}(V_i)$ are quasi-compact. Since $f^{-1}(V_i)$ is quasi-compact, write $f^{-1}(V_i)=\bigcup_{s=1}^t W_s$ with each $W_s$ affine open in $X$. Now let $V^\prime\subseteq V_i$ be a standard open of $V_i$. You have $f^{-1}(V^\prime)=\bigcup_{s=1}^t f^{-1}(V^\prime)\cap W_s$. The key observation is that $f^{-1}(V^\prime)\cap W_s$ is standard open in $W_s$, hence quasi-compact, so $f^{-1}(V^\prime)$ is quasi-compact as well. So in particular all your $f^{-1}(V_{ij})$ are quasi-compact.

As for why $f^{-1}(V^\prime)\cap W_s$ is standard open in $W_s$, if you have a morphism $g:Z\rightarrow\mathrm{Spec}(A)$ (above $V_i$ is playing the role of $\mathrm{Spec}(A)$) then the inverse image of the standard open $D(f)$ is the set $Z_{g^\sharp(f)}=\{z\in Z:g^\sharp(f)_z\notin\mathfrak{m}_z\}$, an open subset of $Z$. It's an exercise that $Z_{g^\sharp(f)}$ intersected with an affine open $U$ of $Z$ is the standard open of $U$ associated to the restriction of $g^\sharp(f)$ to $U$.

In other words, the answer to the boxed question is "yes." Also it is not necessary to assume that the $V_{ij}$ are also standard open in $U$, though the lemma about simultaneous standard opens in intersections of open affines is very handy in proving other results along these lines (for morphisms locally of finite type, finite presentation, etc.).

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    $\begingroup$ Dear Keenan, the second part of your answer was my main obstacle to completing the problem. Thanks for your answer. $\endgroup$
    – user38268
    Aug 9, 2013 at 2:51
  • $\begingroup$ Dear @BenjaLim, I'm glad I added the second part then, and you're welcome! $\endgroup$ Aug 9, 2013 at 2:52
  • $\begingroup$ Dear @Keenan, the exercise you are alluding to in paragraph 2 is exercise 2.2.16 (a)of Hartshorne yes? I will do it then :D $\endgroup$
    – user38268
    Aug 9, 2013 at 2:54
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    $\begingroup$ Dear @BenjaLim, Yes, I personally have found this fact very helpful in thinking "intrinsically" about affine opens of schemes (thinking about the sets $X_f$ for a locally ringed space $X$ and $f\in\mathscr{O}_X(X)$ instead of standard opens of spectra of rings). $\endgroup$ Aug 9, 2013 at 2:55
  • $\begingroup$ @BenjaLim This fact will lead to the affine communication lemma which is very useful in proving equivalence of one definition for every affine open subset and another definition for some affine open cover. $\endgroup$
    – Yuchen Liu
    Aug 9, 2013 at 3:01
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since the preimage of V is quasicompact it can be covered by a finite union of affine opens. And. for a map between two affines the preimage of a principal open is also principal hence quasicompact. A finite unions of quasicompact is quasicompact.

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