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 $\hat{\mathcal{F}} = \mathfrak{F}$?

$\begingroup$ Let me just make the orthogonal comment that this is true if $X$ is projective, by formal GAGA. Algebraicity in the affine case is pretty rare as you can see from ts3uji's example below. $\endgroup$ – Keerthi Madapusi Pera Mar 18 '11 at 0:59
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.

3$\begingroup$ But does this answer the original question? You show there is a f.g. $\hat A$module, but is it induced by a f.g. $A$module. $\endgroup$ – t3suji Mar 17 '11 at 19:49

2$\begingroup$ I think this is a problem of definitions. Usually an adic ring $A$ is already defined to be complete and thus $A = \hat A$. Then coherent modules over $O_{{\rm Spec}(A)}$ and $O_{{\rm Spf}(A)}$ are defined in a different way but EGAI tells you that there is not really a difference. If you define it to be only a topological ring endowed with some $I$adic topology for some ideal $I$  and thus $A \ne \hat A$ in general, then the answer is No as you explained. $\endgroup$ – Torsten Wedhorn Mar 18 '11 at 8:51

2$\begingroup$ @Wedhorn: I completely agree that there is a potential confusion. My interpretation was based on the fact that the OP uses two different letters $A$ and $\hat A$ for the rings. $\endgroup$ – t3suji Mar 18 '11 at 11:41
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, $f=yexp(x)$, 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 torsionfree. 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.)