This is a followup to an earlier question I asked: [Does formally etale imply flat](https://mathoverflow.net/questions/11868/does-formally-etale-imply-flat/)? After some remarks I received on MO I noticed that this was answered to the negative by an answer to an earlier question [Is there an example of a formally smooth morphism which is not smooth](https://mathoverflow.net/questions/195/is-there-an-example-of-a-formally-smooth-morphism-which-is-not-smooth). However, the simple example involves a non-noetherian ring (in fact, a perfect ring; these are seldom noetherian unless they are a field). 


So my challenge is to provide an example of a formally etale map of **noetherian** schemes which is not flat, or otherwise proof that for maps of noetherian schemes formally etale implies flat.



Does formally etale imply flat for noetherian schemes?
Every formally smooth morphism between locally noetherian schemes is flat; this is a deep result of Grothendieck. Indeed, the formal smoothness is preserved by localization on target and then likewise on the source, so we can assume we're deal with a local map between local noetherian rings. By EGA 0$\_{\rm{IV}}$, 19.7.1 (also proved near the end of Matsumura's book on commutative ring theory) a formally smooth local map between local noetherian rings is flat. 


