# Stein factorization and thm. of formal functions

If $f: X \rightarrow Y$ is a proper morphism of locally noetherian schemes with $f_* \mathcal{O}_X = \mathcal{O}_Y$ then the thm. of formal functions tells us that $f$ has connected fibers, since there is an isomorphism $$\widehat{ (f_* \mathcal{O}_{X,y})} \longrightarrow \lim_{\leftarrow_n} H^0(X_n, \mathcal{O}_{X_n})$$ 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.

$\textbf{Question:}$ 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 $X$.

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Think about the definition of the map in the theorem on formal functions (i.e., don't treat the existence of that isomorphism as a black box: the map is defined in a specific way prior to the statement of the result), and then the affirmative answer should become clear to you. Hint: enough to check the ring property after composing with the map from the inverse limit to each of the constituents of the inverse system. – user27056 Nov 3 '12 at 17:45
Great, thanks xbnv! So the point is that we can get our hands on the base change maps $g^*f_* \rightarrow f'_*g'^*$on cohomology explicitly by Cech cohomology. Is this what you had in mind? – LMN Nov 3 '12 at 19:16
@LMN: Cech theory isn't relevant. You're just using that the base change map on global sections for a quasi-coherent sheaf of algebras is a map of rings, which is clear by inspection. The relevance of Cech theory is in showing that the formation of quasi-coherent cohomology modules (for qcqs schemes over an affine base) commutes with flat base change (in the sense that the base change morphisms are isomorphisms in such cases). – user27056 Nov 4 '12 at 1:04
Dear LMN, A an element of $f_*\mathcal O_{X,y}$ is, by definition, the germ of a section of $f_*\mathcal O_X$ over a n.h. $V$ of $y$, which by definition of $f_*$ is a section of $\mathcal O_X$ over $f^{-1}(V)$. Now $f^{-1}(V)$ is an open set containing $f^{-1}(y)$, so it contains all the $n$th order n.h.s of the fibre $f^{-1}(y)$. Any section of $\mathcal O_X$ over $f^{-1}(V)$ can thus be restricted to a section of the structure sheaf of $X_n$ for any $n$, so we get a canonical map $H^0(f^{-1}(V),\mathcal O_X) \to \varprojlim{n} H^0(X_n,\mathcal O_{X_n})$, hence a canonical map ... – Emerton Nov 4 '12 at 6:08
... $f_*\mathcal O_{X,y} \to \varprojlim H^0(X_n\mathcal O_{X_n})$. Since the target is $\matfrak m_y$-adically complete, this in turn induces a map $\widehat{f_*\mathcal O_{X,y}} \to \varprojlim H^0(X_n,\mathcal O_{X_n})$, and it is this map that the theorem on formal functions addresses. As this discussion shows, it is ultimately induced by a simple restriction, and hence is certainly an algebra map when applied to a sheaf of algebras. Regards, – Emerton Nov 4 '12 at 6:10