Recall that a morphism of schemes $X\rightarrow Y$ is regular if it's flat with geometric fibres that are regular schemes.

Fix a field $k$ and consider a morphism $f:X\rightarrow Y$ of noetherian $k$-schemes, do we have an equivalence of:

1/ the cotangent complex $L_{X/Y}$ is concentrated in degree zero and is a flat module.

2/ $f$ is a regular morphism.

(2 implies 1 follows from Popescu's theorem).

  • $\begingroup$ It this known when $Y = \operatorname{Spec} k$ with $k$ algebraically closed? It might in principle be possible to construct some tower $X_i$ over $Y = \operatorname{Spec} k$ such that the image of $H_j(L_{X_i/Y})$ dies in $H_j(L_{X_{i+1}/Y})$ for example, yet $X = \lim X_i$ is not regular. $\endgroup$ Jun 15, 2017 at 17:14

1 Answer 1


Yes, these are equivalent. See Srikanth Iyengar's write-up on Andr\'e-Quillen homology. Specifically, Theorem 9.5 (together with Proposition 5.9) proves exactly what you want, and a little more. The theorem is actually true for morphisms between Noetherian schemes, no base field required.

As you probably expect, this paper makes heavy use of simplicial module categories. It does, however, contain a lovely introduction to them.


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