For a commutative ring $A$ and $f\in A$ a non-zero divisor, the Beauville-Laszlo theorem gives a gluing statement for vector bundles on $A$ in terms of a vector bundle on $A\big[\frac{1}{f}\big]$, a vector bundle on $\hat{A}$ the ($f$)-adic completion of $A$ and an isomorphism on $\hat{A}\big[\frac{1}{f}\big]$.

Is there a similar statement for a scheme $X$ and an effective Cartier divisor $D\subset X$?

  • $\begingroup$ For the affine case, see also Theorem 1.7 of Landsburg, Patching theorems of projective modules (link). (He says in the second half the introduction that it was also known to Karoubi and Swan.) $\endgroup$ – Minseon Shin Sep 16 '19 at 9:07

Completing the discussion under Will Sawin's answer. The question has been answered completely and affirmatively by Ben-Bassat and Temkin in their paper "Berkovich spaces and tubular descent" (Adv. Math. 2013), but one has to be careful when globalizing $\hat A$ and (especially) $\hat A[1/f]$.

(1) Beauville and Laszlo treat the non-affine case in the Corollary on p. 8 of their paper. However, what they do to define $\hat X$ (by taking the relative spec over $X$ of $\varprojlim \mathcal{O}_X/I_Z^n$) seems flawed for the reason that completion does not commute with localization, e.g. $$ (\varprojlim k[[t]][x]/(t^n))[1/x] \neq \varprojlim k[[t]][x, x^{-1}]/(t^n). $$ Therefore there does not exist a quasi-coherent $\mathcal{O}_X$-algebra whose value on an affine open $U= \operatorname{Spec} A$ is $\hat A$ (completion w.r.t. the ideal of $U\cap Z$ in $U$). See also paragraph 1.3 (p. 219) in Ben-Bassat and Temkin.

(2) Clearly, the global version of $\hat A$ should be $\hat X$, the formal completion of $X$ along $Z$, which is a formal scheme but not a scheme. It is more delicate to define the "tubular neighborhood" $W = "\hat X \setminus Z"$ (quotes because, taken literally, this is the empty formal scheme). The answer is that $W$ is an object of rigid analytic geometry, to which there are several approaches. Ben-Bassat and Temkin choose to work in Berkovich spaces, but discuss the other approaches as well in Section 4.6. They remark that whatever framework we choose, the category of coherent sheaves or vector bundles will be the same.

(3) Their main result is:

Theorem (2.4.8). Let $X$ be a scheme of finite type over a field $k$, and let $Z$ be a closed subscheme of $X$. We denote by $\hat X$ the formal completion of $X$ along $Z$, and by $W$ the corresponding Berkovich space (which the authors define). Then $$ \operatorname{Coh}(X) \stackrel{\sim}{\longrightarrow} \operatorname{Coh}(X\setminus Z) \times_{\operatorname{Coh}(W)} \operatorname{Coh}(\hat X). $$

| cite | improve this answer | |

This is stated (in French) as a corollary in page 8 of Beauville-Laszlo.

| cite | improve this answer | |
  • $\begingroup$ I am baffled by their definition of $\widehat{X}$ as a scheme, as completion does not commute with localization. After all, this is why formal schemes were invented. In any case it seems more natural to replace $\widehat{X}$ with the respective formal scheme, but then $\widehat{X}^*$ should be a rigid space a'la Raynaud... $\endgroup$ – Piotr Achinger Sep 15 '19 at 16:00
  • $\begingroup$ My issue is the same as Achinger, their definition does not make sense already for $\widehat{X}$, because $\mathcal{O}_{S}[[t]]$ is not quasi-coherent. Does it hold if we go in the world of rigid spaces? $\endgroup$ – prochet Sep 15 '19 at 18:03
  • $\begingroup$ The paper "Berkovich spaces and tubular descent" by Ben-Bassat and Temkin seems relevant, though they work in the Berkovich setup. From the abstract: "We consider an algebraic variety $X$ together with the choice of a subvariety $Z$. We show that any coherent sheaf on $X$ can be constructed out of a coherent sheaf on the formal neighborhood of $Z$, a coherent sheaf on the complement of $Z$ , and an isomorphism between certain representative images of these two sheaves in the category of coherent sheaves on a Berkovich analytic space $W$ which we define." $\endgroup$ – Piotr Achinger Sep 15 '19 at 18:11
  • $\begingroup$ @PiotrAchinger Since the idea of working in the Raynaud setup and not another one was yours, that might be a complete answer to the question... $\endgroup$ – Will Sawin Sep 15 '19 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.