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Lemma 1. All preimages of affine open subschemes are affine (Inverse of open affine subscheme is affine).

Lemma 2. All affine schemes are quasi-compact (Proof that an affine scheme is quasi compact).

The definition of a quasi-compact morphism between schemes is the following (Hartshorne, excercise II.3.2): A morphism $f: X \rightarrow Y$ of schemes is quasi-compact if there is a cover of $Y$ by open affines $V_i$ such that $f^{-1}(V_i)$ is quasi-compact for each $i$.

By the two lemmas above, however, the preimages $f^{-1}(V_i)$ are all affine, hence quasi-compact.

What am I missing? Which definition am I misunderstanding?

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  • $\begingroup$ I fixed the title of the first question so it's more obvious what's going on here and hopefully in the future folks don't get tripped up in the same way as you did. $\endgroup$
    – KReiser
    Apr 11, 2022 at 8:36

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Lemma 2 is fine, but lemma 1 is inaccurately stated since the morphism is assumed to be between two affine schemes.

For a counterexample to lemma 1, take any non-affine scheme $X$ and consider $X\to \operatorname{Spec}\mathbb{Z}$. The preimage of $\operatorname{Spec}\mathbb{Z}$ is $X$.

For a counterexample to your original claim, take $X$ to be a non-quasi-compact scheme and do the same thing.

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