Is formal smoothness a local property? Is the following statement true?


Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps $R\to S_{\mathfrak{p}}$ are formally smooth for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally smooth.  


I've looked all over, and I have not been able to find a proof or counterexample for this claim.  It might even be an open problem.  This statement is true for formally unramified and formally étale morphisms, but the proof for formally étale morphisms uses the fact that the module of Kähler differentials is $0$, so it doesn't seem like the same approach will work for the formally smooth case. 
 A: I think this works. Suppose we have a ring $R$ and an $R$-module $M$ all of
whose localisations are projective and consider $S=S^\ast_RM$, the symmetric algebra
on $M$. Then $R \rightarrow S$ is formally smooth precisely when $M$ is
projective. Let $\mathfrak p$ be a prime ideal of $S$ and $\mathfrak q$ the
prime ideal of $R$ lying below $\mathfrak p$. Then $S_{\mathfrak p}$ is a
localisation of $S_{\mathfrak q}=S^\ast_{R_{\mathfrak q}}M_{\mathfrak q}$ which is a
formally smooth $R_{\mathfrak q}$-algebra and hence $S_{\mathfrak p}$ is a
formally smooth $R$-algebra (localisations being formally étale). To show that
the statement is false it thus is enough to find a non-projective $R$-module all
of localisation are projective.
If $R$ is a boolean ring (i.e., $r^2=r$ for all $r\in R$) then all its
localisations are isomorphic to $\mathbb Z/2$ all of whose modules are
projective. Hence it suffices to find a boolean ring $R$ and a non-projective
$R$-module. This is easy: Let $S$ be a totally disconnected compact topological
space and $s\in S$ a non-isolated point and put $R$ equal to the boolean ring of
continuous maps $S \rightarrow \mathbb Z/2$. Evaluation at $s$ gives a ring
homomorphism $R \rightarrow \mathbb Z/2$ making $\mathbb Z/2$ an $R$-module. If
it were projective there would be a module section $\mathbb Z/2 \rightarrow
R$. Let $e$ be the image of $1$. Then for any $f\in R$ we have $fe=f(s)e$. This
means that $e$ must be the characteristic function of $s$. Indeed, as $e \mapsto
1$ under evaluation at $s$ we have $e(s)=1$. On the other hand, if $t\neq s$
there is an $f\in R$ with $f(t)=1$ and $f(s)=0$ giving $e(t)=f(t)e(t)=f(s)e(t)=0$.
Hence $\{s\}=\{t\in S:e(t)=1\}$ is open so that $s$ is isolated.
Note that we may take for $S$ any infinite compact totally disconnected set as
some point of it must be non-isolated. Nice examples are the Cantor set (all of whose points are non-isolated) and the
spectrum of $\prod_T\mathbb Z/2$ for any infinite set $T$ (which is the
ultrafilter compactification of $T$, any non-principal ultrafilter is
non-isolated I think).
