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## Finitely-affine morphisms; cohomological dimension of schemes

Let $f\colon X\to U$ be a morphism of Noetherian schemes such that the scheme $U$ is affine and the scheme $X$ is separated and, e.g., quasi-projective over affine. Let $U=\bigcup_\alpha U_\alpha$ be an affine open covering of $U$. Suppose that for each $\alpha$ the full preimage $f^{-1}(U_\alpha)$ can be covered by at most $n$ affine open subschemes. Is there any bound on the number of affine open subschemes covering $X$?

Of course, the bound is supposed to depend only on the number $n$ and not on the number of open subschemes in the covering $U_\alpha$ (or otherwise it is not interesting). It is well-known that if $n=1$ then the morphism $f$ is affine and the scheme $X$ is affine. What happens for $n\ge2$?

One approach to this question would be to notice that one has $R^if_*(\mathcal F)=0$ for any quasi-coherent sheaf $\mathcal F$ on $X$ and all $i\ge n$, hence also $H^i(X,\mathcal F)=0$ for all $i\ge n$. Is there any way to obtain a bound on the number of open affines covering $X$ from this finiteness of cohomological dimension?

The only reference I was able to find is Hartshorne "Cohomological dimension of algebraic varieties", Annals of Math. 88, 1968 (referred to from Exercise III.4.8 in Hartshorne's "Algebraic geometry"), but this doesn't seem to answer my question.

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You might have a look at "The Affine Stratification Number and the Moduli Space of Curves" by Mike Roth and Ravi Vakil. I don't knwon whether it answers your question though. – Qing Liu Nov 11 at 16:54
Thank you for this helpful reference, which explains that a bound on the Krull dimension of a scheme (in fact, of a variety over a field), implying of course a bound on its cohomological dimension, still implies no bound on the number of covering open affines. However, this counterexample (attributed to Starr) is not a quasi-projective variety. The case I seem to be most interested in is when $X$ is quasi-affine, on the other hand. – Leonid Positselski Nov 11 at 21:30
Yes they actually are interested in a different number. It would be interesting to start with a baby case: $X$ is quasi-projective over $U$ of relative dimension $1$. Then I think $n\le 2$. Can $X$ be covered by $2$ affine open subsets ? – Qing Liu Nov 12 at 10:29

"This is too long to be a comment".

A way to construct examples $X\to U$ with a given $n$ is to take a (quasi-)projective scheme $X\to U$ whose fibers have dimension $\le n-1$. Indeed, let $s\in U$. There are $n$ hypersurfaces in $X_s$ with empty intersection, so $X_s$ is covered by the $n$ complements of hypersurfaces (more care are needed if $X_s$ is not projective, anyway, the exercise in Hartshorne you refered to is done this way). Now lift these hypersurfaces to hypersurfaces over $O_{U,s}$, then the same arguments show that $X\times_U \mathrm{Spec}(O_{U,s})$ is covered by $n$ affine open subsets. By standard arguments, this will hold above an affine open neighborhood of $s$. So $X\to U$ satisfies your hypothesis.

Now in a very special case, $X$ is actually covered by $n$ affine open subsets. Namely, if $U$ is the spectrum of a ring of integers or if $U$ is a regular curve over a finite field, then any projective $X\to U$ with fiber dimensions $\le n-1$ admits a finite morphism $\pi: X\to \mathbb P^{n-1}_U$. This is proved independently in a preprint of Chinburg, Moret-Bailly, Pappas, Martin Taylor, and a preprint of Gabber, Lorenzini and myself. As $\mathbb P^{n-1}_U$ is covered by $n$ affine open subsets, the same is true for $X$.

Edit It would be interesting to see whether any projective curve over a Dedekind domain can be covered by two (or more, but absolutely bounded number of) affine open subsets.

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 But is it still possible that $X$ can always be covered with, say, $2n$ open affines? Or some other function of $n$ growing faster than the identity? – Leonid Positselski Nov 14 at 0:59 Sorry, there is a gap in my proof that the projective bundle can't be covered by two affine open subsets. The example I discussed has two obvious disjoint sections. – Qing Liu Nov 14 at 8:41