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Let $f \colon X \to Y$ be a morphism of schemes, where $X$ and $Y$ are separated integral Noetherian schemes. Does there necessarily exist a nonempty open affine $U \subset Y$ such that $f^{-1}(U)$ is affine? If $X \to Y$ is not dominant, then the answer is clearly yes (take $U$ to have empty inverse image). Even for dominant morphisms, I feel like this should be related to Chevalley's theorem that the image of a constructible set is constructible, but I can't put it together into a proof.

Note: this question is silly (see comments/answer below). The specific case I had in mind was when $f$ is unramified over the generic point of $Y$; however, if I decide I want the answer to this, I'll ask in a separate question.

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I do not think so. What if you take $X=Y \times \mathbb{P}^1$ and $f$ the projection onto the first factor? – Francesco Polizzi Jul 24 '10 at 17:47
up vote 8 down vote accepted

I think the answer is no. Take any field $k$, and consider the natural morphism $\mathbb P^n_k \to spec(k)$, where $\mathbb P^n_k$ is the $n$-dimensional projective space. Since $spec(k)$ is, as topological space, a single point and $\mathbb P^n_k$ is not affine, your open subset cannot exist.

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Thanks for the answer. I just realized this also, so I'm modifying the question to make it (hopefully) more subtle. – Charles Staats Jul 24 '10 at 17:57
Charles, instead of trying to make it more subtle and re-posting, why not remove the question and think about the more subtle one on your own for a few days (or more) and discuss with classmates? – BCnrd Jul 24 '10 at 17:59
I'm not sure about removing the question (I prefer not to delete a question that someone has already answered), but for the rest, if you'll notice, I came to a similar conclusion myself. – Charles Staats Jul 24 '10 at 18:07

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