most recent 30 from http://mathoverflow.net 2013-05-22T01:24:58Z http://mathoverflow.net/feeds/question/58761 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58761/coherent-sheaves-on-affine-formal-schemes lorca 2011-03-17T15:03:12Z 2011-03-17T19:48:33Z

<p>Let $\hat{X} = \text{Spf} \hat{A}$ be obtained as the formal completion of an affine scheme $X = \text{Spec} A$ where $A$ is an adic noetherian ring. Given a coherent sheaf $\mathfrak{F}$ on $\hat{X}$, is it always possible to find a coherent sheaf $\mathcal{F}$ on $X$ such that $\hat{\mathcal{F}} = \mathfrak{F}$? </p>

http://mathoverflow.net/questions/58761/coherent-sheaves-on-affine-formal-schemes/58770#58770 Torsten Wedhorn 2011-03-17T17:29:47Z 2011-03-17T17:29:47Z

<p>According to EGA I (Springer edition), Theoreme (10.10.2), there is an equivalence of categories between the category of finitely generated $A$-modules and the category of coherent $O_{{\rm Spf}(A)}$-modules. In particular, the answer to your question is Yes.</p>

http://mathoverflow.net/questions/58761/coherent-sheaves-on-affine-formal-schemes/58780#58780 t3suji 2011-03-17T19:48:33Z 2011-03-17T19:48:33Z

<p>Take $A=k[x,y]$, $\hat A=k[[x,y]]$, and suppose coherent sheaf corresponds to the $\hat A$-module $N=\hat A/(f)$, where $f\in k[[x,y]]$ is not "algebraic": say, $f=y-exp(x)$, assuming $k$ has characteristic zero. It is clear that $N$ does not come from completion of any f.generated $A$-module $M$. (Proof: Suppose $N=\hat M$. Since $N$ has no $A$-torsion, any torsion of $M$ maps to zero, so we can replace $M$ with $M/(torsion)$ and assume $M$ is torsion-free. Then $M$ embeds in a locally free module (its second dual), but that would make $N$ embed in the free $\hat A$-module, which is false.)</p>