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

Given a unital commutative ring $A$ (not necessarily noetherian) and a formally smooth morphism of rings $f:A \to B$, where $B$ is not necessarily 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

We give a counterexample to the following statement that came up in the question/comments:

Any formally smooth flat map $A \to B$ is a filtered colimit of smooth maps.

In the example, $A$ is noetherian (even a field), but $B$ is not. The example takes place in characteristic $p > 0$.

Say $A = \mathbb{F}_p$ is a finite field with $p$ elements. Let $B_0 = A[x,y]/(xy)$ be the co-ordinate ring of the union of the two axes in the plane, and let $B = B_{0,\mathrm{perf}}$ be the perfection of $B_0$, so $B = \mathrm{colim}_i B_i$ is a filtered colimit of copies of $B_i := B_0$ along Frobenius maps. Note that $B_0$ is not a domain. Now:

Claim: The map $A \to B$ is formally smooth and flat, but not a filtered colimit of smooth $A$-algebras.

The flatness is clear as $A$ is a field. Also, any map between perfect rings is formally smooth by an elementary lifting argument. (In fact, $L_{B/A} \simeq 0$.) Now assume that we can write $B = \mathrm{colim}_j C_j$ as a filtered colimit of smooth $A$-algebras $C_j$; this will lead to a contradiction.

As each $C_j$ and $B_i$ is finitely presented, by comparing the two filtered inductive systems $\{B_i\}$ and $\{C_j\}$, one finds a factorisation

$$B_i \to C_j \to B_{i'}$$

of the structure map $B_i \to B_{i'}$ for $i' \gg i$ and suitable $j$. As the transition maps in $\{B_i\}$ are Frobenius, we conclude: there is a smooth $A$-algebra $C$ and a factorisation

$$B_0 \stackrel{a}{\to} C \stackrel{b}{\to} B_0$$

with $b \circ a = \mathrm{Frob}_{B_0}^n$ for some $n \geq 0$. As the target $B_0$ is connected, after replacing $C$ by a connected component, we may assume $C$ is also connected, and thus a domain (by smoothness). The existence of the above factorisation shows that $a$ is injective (as the composite $b \circ a$ is injective), so $B_0$ is also a domain, which is nonsense.

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