# Is a formally smooth morphism a filtered inductive limit of smooth algebras?

Given a unital commutative ring A (not nec. noetherian) and a formally smooth morphism of rings f:A --> B, where B is not nec. noetherian, is (or when is) B a filtered inductive limit of smooth A-algebras?

There is the partial result of D. Popescu, Thm. 1.1, in

MARK SPIVAKOVSKY, A NEW PROOF OF D. POPESCU'S THEOREM ON SMOOTHING OF RING HOMOMORPHISMS, JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY, Volume 12, Number 2, April 1999, Pages 381-444,

where the rings are assumed to be noetherian.

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A weaker question is: does the property f is formally smooth imply $H_1(L_{B/A})=0$, where $L_{B/A}$ is the cotangent complex of B/A? –  Lutz Geissler Jul 17 '10 at 7:41
I remember that when B. Teissier gave a Bourbaki talk (1995) on Popescu's theorem, both J.-P. Serre and O. Gabber insisted in the noetherian hypothesis. –  Qing Liu Jul 17 '10 at 13:53
A necessary condition is flatness, and so a counterexample is the non-flat quotient map $A \rightarrow A/J$ for a local ring $A$ and nonzero proper ideal $J$ such that $J = J^2$ (of which there are many examples, such as by using suitably crazy valuation rings). The condition $J = J^2$ ensures it is formally etale. These are also counterexamples to EGA 0$_{\rm{IV}}$, 19.10.3(i), and also counterexamples to EGA IV$_4$, 18.4.6(i) (whose proof uses 19.10.3(i) right at the end). I expect there should be flat counterexamples also, which would doom any reasonable sufficient criterion. –  BCnrd Jul 18 '10 at 18:39
(i) Of course flatness is a necessary condition. Thanks for adding it. (The initial question arose out of trying to examine whether H_1(L_{B/A})=0 for f:A --> B formally smooth without any flatness assumption, cf. Illusie's Complexe cotangent et déformations I, esp. Chp. III.3.1.2.) –  Lutz Geissler Jul 19 '10 at 2:41
(ii) There is a characterisation of flat quotient maps A \to A/I (by any ideal I): the following are equiv. (ii.1) A \to A/I is flat; (ii.2) A \to A/I is a localisation via a multiplicative subset of A; (ii.3) Every prime p containing I satisfies I_p = 0 (I localised at p). A nondiscrete valuation ring of rank one A with maximal ideal J gives a ring with idempotent ideal (as in p-adic HT), but such examples do not satisfy (ii.3) and so are not flat. –  Lutz Geissler Jul 19 '10 at 2:41