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Let $X$ be a scheme and $f : X \rightarrow \mathrm{Spec}\, A$ a quasicompact morphism. Are there any easy conditions on $A$ under which we can say that $X$ is quasicompact?

Quasicompact morphism means only that there is an affine cover $\cup_{i \in I} \mathrm{Spec}\, A_i$ where $f^{-1}(\mathrm{Spec}\, A_i)$ is quasicompact. It doesn't seem to be enough.

NotfromBrazil
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    No, it doesn't "only" mean what you say! The correct definition of quasi compact is that the inverse image of every open quasi-compact subset (affine or not) is quasi-compact. Correct means as decreed by Grothendieck in His infinite wisdom: EGA I Chap.I, Définition 6.6.1, page 152. That heresiarchs, dissenters, renegades and apostates replace that definition by some sufficient condition like the one you quote will in no way attenuate the torments they will suffer in the Seventh Circle of the Hell of Algebraic Geometry. – Georges Elencwajg May 25 '13 at 11:20
  • @GeorgesElencwajg Sorry, I didn't know - my definition is from Hartshorne's Algebraic Geometry – NotfromBrazil May 25 '13 at 12:12
  • Dear NotfromBrazil: don't worry , my comment was just a joke: no algebraic geometer was tormented while I made it! Just below the official definition, Grothendieck gives a criterion in the spirit of Hartsorne's definition for checking quasi-compacness. You can also look at Görtz-Wedhorn's Algebraic Geometry I Prop. 10.1, page 242 and on the same page at Remark 10.2 (1), which answers your question. – Georges Elencwajg May 25 '13 at 13:28
  • Hartshorne's book is full of ad hoc definitions and constructions, which are - in most cases - not the best and most transparent ones. They only work somehow within the context he developes. The chapters about sheaves and divisors are just a mess. Nowadays there are a lot of better introductory books to algebraic geometry. Using and quoting Hartshorne is just a habit, because decades ago this was the best book around. – Martin Brandenburg May 26 '13 at 00:09

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$X$ is always quasi-compact. This is a standard result which can be found in any good and complete introduction to algebraic geometry. But you can also prove it yourself. In the end, it is just an exercise in general topology. Hint: Affines schemes are quasi-compact. So choose a finite subcover of $\{\mathrm{Spec} A_i \to \mathrm{Spec} A\}$ and observe that $X$ is a finite union of quasi-compact open subspaces.

Martin Brandenburg
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