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Let $X$ be a scheme. An open atlas for $X$ is a jointly epimorphic family of Zariski-open immersions $\{X_i\to X\}$ where each $X_i$ is an affine scheme.

A morphism $X\to S$ of schemes is called representable by an affine if for any map $Y\to S$ where $Y$ is affine, the pullback $X\times_S Y$ is itself affine.

Then the question:

Given a scheme $S$, does there exist an open atlas for $S$ consisting only of morphisms representable by an affine?

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    $\begingroup$ Try the canonical non-quasi-separated scheme (two copies of the affine plane glued together everywhere other than at their origins) for a counterexample, if I've got it right. The problem is that no affine will contain both origins, and for any affine containing one, the pullback via an affine containing the other will give you problems. $\endgroup$ – Kevin Buzzard Oct 11 '10 at 19:44
  • $\begingroup$ @Kevin: Thanks. I got it now. I was trying to make sense of Toën-Vezzosi's inductive definition for algebraic n-stacks, but I misread the lemma where they show that a scheme is a 1-geometric sheaf, while a scheme with affine diagonal is a 0-geometric sheaf (this is defined as a sheaf admitting an open atlas consisting of affine morphisms (as above)). A scheme in general is 1-geometric (the same conditions as above, but change "representable by an affine" to "representable by a scheme with affine diagonal). $\endgroup$ – Harry Gindi Oct 11 '10 at 20:21
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    $\begingroup$ [I was always told by my logic lecturer that there's no such thing as an inductive definition ;-) You prove things by induction, and define them by recursion. I've always assumed he was right...] $\endgroup$ – Kevin Buzzard Oct 11 '10 at 20:27
  • $\begingroup$ Good point. Recursive definition it is! $\endgroup$ – Harry Gindi Oct 11 '10 at 20:36
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A morphism $X \to Y$ is representable by an affine iff it is an affine morphism (preimages of open affines are open affine), since the latter is stable under base change. Now an open immersion $U \to X$ is affine iff $U \cap V$ is open affine in $X$ for every open affine $V$ in $X$. Thus $X$ has an atlas consisting of such maps iff every two open affines intersect in an open affine, i.e. iff the diagonal $\Delta : X \to X \times_{\mathbb{Z}} X$ is affine (which is the case when $X$ is separated).

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    $\begingroup$ This is not quite right: the condition that the intersection of two affines is affine is equivalent to saying that the diagonal is affine, but this does not imply that $X$ is separated. $\endgroup$ – Angelo Oct 11 '10 at 19:02

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