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Let $X,Y$ be integral Noetherian schemes. Let $f:X\to Y$ be a finite map of schemes. I recently had to show that the set of points $V\subset Y$ over which $f$ is flat is open, as is for instance proven in this stackexchange post.

The exercise asked for a counterexample when we drop the finiteness assumption on $f$. The counterexample I came up with is the following, let $p:Z\to \mathbb{A}^3$ be the blowup of $\mathbb{A}^3$ along $\mathbb{A}^1= V(x,y)$ and let $X= Z\setminus p^{-1}(0)$, then the flat locus of $p:X\to \mathbb{A}^3$ is $D(x,y)\cup V(x,y,z)$ which is not open.

There are two things I dislike about this example

  • $X$ is not affine, are there counterexamples where the domain and codomain are affine?
  • the map $p:X\to \mathbb{A}^3$ is not surjective.

Are there counterexamples under these additional assumptions?

David Wiedemann
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  • This is a little confusing - normally when people talk about the flat locus of a morphism $f:X\to S$, they're talking about the set ${x\in X\mid \mathcal{O}_X \text{ is flat over } S \text{ at } x}$. But you're talking about the image of this set. Could you [edit] your question to clarify? – KReiser Jun 10 '23 at 15:31
  • @KReiser, I edited it, I'm not quite talking about the image of this set as the map $f$ might not be surjective, I hope it is clearer like this – David Wiedemann Jun 10 '23 at 16:01
  • Haha, I guess I misinterpreted it slightly too. Thanks for the clarification. – KReiser Jun 10 '23 at 16:39

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