This is a followup to an earlier question I asked: 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. 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.


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.

  • $\begingroup$ Thanks Brian, that was exactly the sort of answer I was looking for. $\endgroup$ – mabli Feb 13 '10 at 20:59
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    $\begingroup$ My comment about Matsumura's book is slightly incorrect: this result is stated there, but I'm pretty sure it isn't proved there; he references EGA for the proof (which is too long to fit into Matsumura's book). $\endgroup$ – BCnrd Feb 14 '10 at 6:12

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