Assume you have a smooth quasi-projective scheme $X$ (you can actually assume $X$ is projective over an affine scheme of finite type) defined over $\mathbb Z$ (or if you prefer, a discrete valuation ring), and let $F$ be a locally free tilting sheaf on $X$, that is, a locally free sheaf such that $\mathrm{Ext}^i(F,F)=0$ for all $i>0$. Assume that this sheaf is a generator of the derived category of coherent sheaves after base change to $\mathbb{Q}$ (or basically equivalently, to $\mathbb{F}_p$ for all sufficient large primes); does it follow that it is a generator over $\mathbb{Z}$ or modulo all primes?

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    $\begingroup$ Why is generating over $\mathbb Q$ equivalent to generating over $\mathbb F_p$ for large primes $p$? $\endgroup$ – Will Sawin Aug 15 '15 at 14:31
  • $\begingroup$ @WillSawin I did say "basically", but I believe this is true. By 3.3 in Kaledin, $F$ is a generator if a particular object in the derived category of coherent sheaves is trivial; generation over $\mathbb{Q}$ is that this sheaf is torsion, which is the same as saying that it is trivial after base change to $\mathbb{F}_p$ for sufficiently large $p$. $\endgroup$ – Ben Webster Aug 15 '15 at 15:39
  • $\begingroup$ Can you clarify what you mean by "variety defined over $\mathbf{Z}$"? $\endgroup$ – Dracula Aug 16 '15 at 11:51
  • $\begingroup$ @Dracula Clarified in the question. $\endgroup$ – Ben Webster Aug 16 '15 at 14:57

Let $X = \lim X_i$ be a scheme written as a cofiltered limit of qcqs schemes with affine transition morphisms. Then the category of perfect objects in $D(O_X)$ is the colimit of the categories of perfect objects in $D(O_{X_i})$. Moreover, if $E$ is a perfect object and generator for $D_{QCoh}(O_X)$, then for some $i$ there exists an $E_i$ over $X_i$ which is perfect, a generator, and pulls back to $E$ (we will see an argument in a special case for this below --- but the argument in general is the same).

More specifically, suppose that we have a ring $A = \text{colim}_{i \in I} A_i$, we have a smallest element $0 \in I$ we have a projective morpihsm $X_0 \to \text{Spec}(A_0)$ projective and $X_i$, resp. $X \to \text{Spec}(A)$ is the base change of $X_0$ via the ring maps $A_0 \to A_i$, resp. $A_0 \to A$. Then we know there is a generator of the form $F_0 = O_{X_0} \oplus O_{X_0}(1) \oplus \ldots \oplus O_{X_0}(n)$ for a suitable $n \geq 0$ which also is a generator after restriction to the fibres of $X_0 \to \text{Spec}(A_0)$. Moreover the same thing holds for the pullbacks $F_i$ and $F$ of $F_0$ to $X_i$ and $X$.

Going back to our $E$, we can construct $F$ out of $E$ using a finite number of cones and direct summands. Hence we find that for sufficiently large $i$ we can construct $F_i$ out of $E_i$ by a finite number of cones and direct summands. Since taking cones and direct summands commute with restriction to fibres we see that, for some $i$, the restriction of $E_i$ to the fibres of $X_i \to \text{Spec}(A_i)$ generates.

Apply with $A = \mathbf{Q}$ and $A_i = \mathbf{Z}[1/i]$ and $0 = 1$ to get what you want in case $X$ is projective over $\mathbf{Q}$. I think the same argument works if $X$ is just quasi-projective over $\mathbf{Q}$. For general varieties you first have to show a priori that there is some generator which generates on all the fibres. Presumably you can do this with some argument using Noetherian induction, but I didn't try.

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    $\begingroup$ Sorry, I don't really follow. It seems like your argument gets a generator after inverting finitely many primes; my question was precisely whether it's really necessary to actually invert those primes. Did I miss something? $\endgroup$ – Ben Webster Aug 15 '15 at 16:08
  • $\begingroup$ OK, no, I think you followed it fine. I was answering Sawin's implicit question. $\endgroup$ – Dracula Aug 16 '15 at 11:50

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