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I was surprised to find out that, even if the normalization $X^\nu$ of a scheme $X$ is affine, $X$ may not be affine (remove the line $x=y$ from their example to make the source affine). In the example I'm working on, both $X$ and $X^\nu$ are quasi-affine. In fact $X \to \text{Spec }A$ is a quasi-affine open of affine space $\mathbb{A}^k_A \to \text{Spec }A$ and so $X^\nu$ is the pullback of the normalization of $\text{Spec }A$. Does it follow then that if $X^\nu$ is affine, so is $X$?

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    $\begingroup$ Have you seen Tag 01YQ? (I apologize that I haven't read through Tag 0271.) $\endgroup$ Dec 1 '20 at 0:06
  • $\begingroup$ Thank you so much! I was trying to do a Leray Spectral Sequence argument that was a waste of time, and the stacks project tag I found made me nervous it was false. $\endgroup$
    – Leo Herr
    Dec 1 '20 at 1:21
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    $\begingroup$ For the record, the Tag 01YQ result is also EGA II, (6.7.1) and is due to Chevalley in the case of algebraic schemes. For a non-noetherian generalization see Tag 05YQ. I believe I have seen a generalization to algebraic spaces but can't remember where. $\endgroup$ Dec 1 '20 at 14:05
  • $\begingroup$ Correction to previous comment: the non-noetherian case is Tag 05YU, not 05YQ. $\endgroup$ Dec 6 '20 at 15:25

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