# If the normalization is affine, is it affine? (if quasiaffine)

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 $$X$$?

• Have you seen Tag 01YQ? (I apologize that I haven't read through Tag 0271.) Dec 1 '20 at 0:06
• 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. Dec 1 '20 at 1:21
• 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. Dec 1 '20 at 14:05
• Correction to previous comment: the non-noetherian case is Tag 05YU, not 05YQ. Dec 6 '20 at 15:25