Stein factorization and thm. of formal functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:23:11Z http://mathoverflow.net/feeds/question/111393 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111393/stein-factorization-and-thm-of-formal-functions Stein factorization and thm. of formal functions LMN 2012-11-03T16:55:46Z 2012-11-03T16:55:46Z <p>If $f: X \rightarrow Y$ is a proper morphism of locally noetherian schemes with <code>$f_* \mathcal{O}_X = \mathcal{O}_Y$</code> then the thm. of formal functions tells us that $f$ has connected fibers, since there is an isomorphism <code>$$\widehat{ (f_* \mathcal{O}_{X,y})} \longrightarrow \lim_{\leftarrow_n} H^0(X_n, \mathcal{O}_{X_n})$$</code> for any $y \in Y$, where LHS is the stalk of the pushforward completed as an $O_y$ module and $X_n$ denotes the $n$th thickened fiber as usual.</p> <p>$\textbf{Question:}$ In general, for coherent sheaves on $X$ the theorem of formal functions only gives an isomorphism of $\widehat{\mathcal{O}_y}$ modules. In the context of Stein factorization above, since both LHS and RHS are in fact rings, is the isomorphism of the thm. of formal functions "sufficiently functorial" so that it is in fact an isomorphism of rings? Then we know that the spectrum of RHS is connected because LHS is a local ring. This seems to be the argument in Illusie's article "Grothendieck existence thm...", but I can't help but think that the thm of formal functions only gives an isomorphism of $\widehat{\mathcal{O}_y}$ modules, so to be completely precise we need a slightly messier argument. Can you make this a general statement, eg, the thm. of formal functions always gives a map of rings we start with a coherent sheaf of rings on $X$.</p>