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Let me state the question for rings (rather than schemes) for simplicity. Let $R$ be a commutative ring with unit and $A$ an $R$-algebra of finite presentation. Recall that $R\to A$ is called a syntomic morphism if it is flat and for each morphism $R\to k$ to a field, the ring $A\otimes_R k$ is a local complete intersection in the sense of commutative algebra.

Do you know a simple functorial criterion for $R\to A$ to be syntomic, in the spirit of the functorial criterion for smoothness saying that $R$-morphisms $A\to B/I$ lift to $A\to B$ whenever $I\subset B$ is a square-zero ideal?

Variants of such criteria in this functorial flavour, or examples in the literature, are welcome. Variants where $R\to A$ is already known to be flat are also welcome (for smooth morphisms, flatness somehow comes for free but it is sort of a small miracle).

My motivation to ask this question is that I have a morphism which I can prove is syntomic, but only indirectly, using smoothness in an indirect way. I'm wondering whether a direct attack would be possible.

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  • $\begingroup$ Unlike syntomic, the notion of lci morphism usually does not include flatness, as far as I'm aware. See for instance Tags 069F and 069K. $\endgroup$ Commented Sep 7, 2022 at 9:49
  • $\begingroup$ It does according to EGA (IV, Def. 19.3.6), and also SGA 7, Exp. VIII, Def. 1.1. $\endgroup$
    – abx
    Commented Sep 7, 2022 at 9:58
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    $\begingroup$ @abx Not so, EGA only defines the lci condition for flat morphisms, but does not say that lci morphisms should be flat, and SGA has the general definition. $\endgroup$ Commented Sep 7, 2022 at 11:14
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    $\begingroup$ The OP states that his morphism is flat. There is a valuative criterion for flatness. For a flat morphism, the morphism is LCI if and only if the cotangent complex is bounded. At least in the category of dg schemes / rings, there is a universal property of the cotangent complex that can be used to write a functorial property for boundedness of the cotangent complex. $\endgroup$ Commented Sep 7, 2022 at 11:37
  • $\begingroup$ Thanks for these clarifications, I'm editing the question to focus on syntomic morphisms then. $\endgroup$ Commented Sep 7, 2022 at 14:55

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