21
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
41
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks Brian, that was exactly the sort of answer I was looking for. $\endgroup$
    – mabli
    Commented Feb 13, 2010 at 20:59
  • 3
    $\begingroup$ 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
    Commented Feb 14, 2010 at 6:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .