MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


where the rings are assumed to be noetherian.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.