I have learned from this question that, in spite of the gap in the proof of 17.1.6 (i) in EGA IV, we can still verify that a morphism of schemes is formally smooth locally on the source. But, even assuming the results of Raynaud-Gruson, I did not get the argument in "Catégories Cofibrées Additives et Complexe Cotangent Relatif" given by Grothendieck, possibly because I do not know the theory of the cotagent complex. For me it seems that he is using some global version of the critère jacobien de lissité formelle [EGA $0_{\text{IV}}$, 22.6.3], but I could not work out the details. Could someone explain me how can we justify this assuming only the results from EGA $0_{\text{IV}}$ and that the projectivity of modules descends along faithfully flat ring maps?
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2This question really belongs to mathoverflow (especially after reading the answer). – Martin Brandenburg Aug 19 '14 at 18:51
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@Martin: I do agree with you, especially after knowing the answer, but sometimes it's hard to tell whether what we're asking is trivial or not, which is even more true when we don't know very well what we are talking about. – Nuno Aug 19 '14 at 22:19
1 Answers
To prove this we can just adapt the proof in [EGA IV$_4$, 17.1.6]. The argument I will give below is taken from [Stacks, Tag 061K]. First of all we need the following observation: given a commutative diagram of morphisms of schemes $$\require{AMScd} \begin{CD} T_0 @>{g_0}>> X\\ @V{i}VV @VV{f}V \\ T @>>> Y, \end{CD}$$
where $i$ is a thickening of order $1$ (i.e. a closed immersion defined by an ideal $\mathscr{I}$ of $\mathscr{O}_T$ of square zero), we can consider the sheaf of sets $\mathscr{P}$ in $T_0$ such that $\mathscr{P}(U_0)$ consists of the $Y$-morphisms $g : U \to X$ that satisfy $g \circ i_{U_0} = g_0|_{U_0}$, where $U$ is the open subscheme of $T$ corresponding to $U_0$ and $i_{U_0} : U_0 \to U$ is the thickening of order $1$ induced by $i$. It is not hard to show that $\mathscr{P}$ is a pseudotorsor under $\mathscr{G} = \mathscr{H\!om}_{\mathscr{O}_{T_0}}(g_0^*(\Omega^1_{X/Y}), \mathscr{I})$ (here we are considering $\mathscr{I}$ as a quasi-coherent $\mathscr{O}_{T_0}$-module, which is possible since $\mathscr{I}^2=0$), see for example [EGA IV$_4$, 16.5.17] or [SGA 1, Exposé III, 5.1].
Let $(U_{\alpha})_{\alpha \in I}$ be an open cover of $X$ and suppose that each $f|_{U_{\alpha}}$ is formally smooth. In this case $\mathscr{P}$ is actually a torsor under $\mathscr{G}$ and it is trivial (i.e. $\Gamma(T_0,\mathscr{P}) \neq \varnothing$) if and only if its class $o(g_0,i)$ in $H^1(T_0, \mathscr{G})$ is zero. To prove that $f$ is formally smooth we need to show that this is the case if $T_0,T$ are affine schemes. We will even prove that $H^1(T_0, \mathscr{G}) = 0$. Note that $\mathscr{G}$ is not necessarily a quasi-coherent $\mathscr{O}_{T_0}$-module, but $g_0^*(\Omega^1_{X/Y})$ is a locally projective one [Stacks, Tag 06B5]. Raynaud-Gruson's theorem guarantees that it is of the form $\widetilde{P}$, where $P$ is a projective $\Gamma(T_0, \mathscr{O}_{T_0})$-module. In order to show that $H^1(T_0, \mathscr{G})$ is zero, we can suppose that $P$ is actually a free module. In this case $\mathscr{G}$ is a product of quasi-coherent modules and the result follows from [Stacks, Tag 060L].
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