7
$\begingroup$

Let $X$ be a quasi-compact scheme. Say that $X$ has property $\mathbf{P}_{n}$ if $X$ admits an open cover $X = \bigcup_{i=1}^{n} U_{i}$ such that each $U_{i}$ is affine and each pairwise intersection $U_{i} \cap U_{j}$ is distinguished affine open (also "principal affine open") in both $U_{i}$ and $U_{j}$.

Does every quasi-compact (semi-)separated scheme $X$ have property $\mathbf{P}_{n}$ for some $n$?

Thoughts:

  1. If $X$ admits an ample line bundle $\mathcal{L}$ then $X$ has property $\mathbf{P}_{n}$ for some $n$. Namely if $s_{1},\dotsc,s_{n} \in \Gamma(X,\mathcal{L})$ are global sections such that each $X_{s_{i}}$ is affine and $X = \bigcup_{i=1}^{n} X_{s_{i}}$, then $X_{s_{i}} \cap X_{s_{j}}$ is the nonvanishing locus of $\varphi_{i}(s_{j}|_{X_{s_{i}}})$ where $\varphi_{i} : \mathcal{L}|_{X_{s_{i}}} \to \mathcal{O}_{X_{s_{i}}}$ is a trivialization.

  2. If $f : X \to Y$ is an affine morphism and $Y$ has $\mathbf{P}_{n}$, then $X$ has $\mathbf{P}_{n}$. Is the converse true if $f$ is also finitely presented and faithfully flat?

  3. Using 2 and Noetherian approximation, we may assume that $X$ is of finite type over $\mathbb{Z}$.

The motivation for my question was a naive/"crank"y attempt to construct ample families of line bundles on a scheme with affine diagonal (c.f. Question 1 in the introduction to Totaro's "The resolution property for schemes and stacks"). Here is a cute fact: If $X$ has $\mathbf{P}_{2}$, then $X$ admits an ample family of line bundles (and hence has the resolution property). Proof: Set $U_{12} := U_{1} \cap U_{2}$. Let $s_{i} \in \Gamma(U_{i},\mathcal{O}_{X})$ such that $(U_{i})_{s_{i}} = U_{12}$. Let $\mathcal{L}_{1}$ be the line bundle on $X$ defined by the transition map $\times s_{1}|_{U_{12}} : \mathcal{O}_{U_{2}}|_{U_{12}} \to \mathcal{O}_{U_{1}}|_{U_{12}}$. Then there is a section $t_{1} \in \Gamma(X,\mathcal{L}_{1})$ which restricts to $s_{1}$ on $U_{1}$ and $1$ on $U_{2}$, so that $X_{t_{1}} = U_{2}$. Similarly, there is a line bundle $\mathcal{L}_{2}$ on $X$ and a section $t_{2} \in \Gamma(X,\mathcal{L}_{2})$ such that $X_{t_{2}} = U_{1}$. Then $\{\mathcal{L}_{1} , \mathcal{L}_{2}\}$ is an ample family of line bundles on $X$. If $X$ has $\mathbf{P}_{n}$ for $n \ge 3$, it seems much more complicated and I am wondering whether this sort of argument can be made to work.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Is there a motivation for asking this other than idle curiosity? $\endgroup$
    – nfdc23
    Commented Jun 3, 2017 at 23:45
  • 1
    $\begingroup$ Regarding your edit: the "non-projective, projective plane" from Exercise III.5.9 of Hartshorne's "Algebraic Geometry" satisfies the property that you call $\mathbf{P}_3$, yet it has only the trivial invertible sheaf. It does not have an ample family of invertible sheaves. $\endgroup$
    – Jason Starr
    Commented Jun 4, 2017 at 11:58
  • 1
    $\begingroup$ The converse of 2 is very much false. For example, for $n = 1$ it would say that $X$ is affine if and only if $Y$ is affine. But Jouanolou's trick shows that the opposite is true: for every quasi-projective $k$-scheme $Y$, there exists an affine morphism $X \to Y$ which is a torsor under a vector bundle such that $X$ is affine. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 4, 2017 at 17:41
  • 2
    $\begingroup$ If $X$ is Noetherian, irreducible, and locally $\mathbb Q$-factorial (and still has affine diagonal), then every affine open covering can be refined to one where the intersections are standard open. Indeed, by Tag 0BCV, the affine opens $U_i \cap U_j \subseteq U_i$ have complement a Weil divisor $Z_j$. Since $X$ is $\mathbb Q$-factorial, this is the support of a Cartier divisor, hence locally principal. Cover $U_i$ by standard affine open subsets $U_{ik}$ such that each $U_{ik} \cap Z_j$ for $j \neq i$ is principal. The $U_{ik}$ will do the trick. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 4, 2017 at 19:15
  • 2
    $\begingroup$ This suggests that we should look at what happens for (singular) proper $k$-varieties $X$ with $\operatorname{Pic}(X) = 0$. These are necessarily non-projective and not locally $\mathbb Q$-factorial, and it would be interesting to see how we're going to deal with this case. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 4, 2017 at 19:21

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .