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Let $k$ be a field, $S$ and $R$ be local $k$-algebras with residue field $k$ and $\phi:S\to R$ be a local homomorphism. Then $\phi$ induces (obviously) a natural transformation of "functors of points" $\phi^*:h_R \to h_S$ (where $h_A(B) = Hom(A, B)$). Let $Art_k$ be the category of local artinian $k$-algebras with residue field $k$ and let $F$, $G$ be the restrictions of $h_R$, resp. $h_S$ to $Art_k$.

There are well-known criteria of (formal) smoothness/etaleness of $\phi$ in terms of the induced transformation $\phi^* : F\to G$. There is also an infinitesimal criterion of flatness, but that is different in spirit.

Question. Is there a criterion on $\phi^*:F\to G$ which ensures that $\phi$ is flat?

You can assume that $\phi$ is finite and that $S$ and $R$ are completions of finitely generated $k$-algebras.

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By the infinitesimal criterion for flatness, do you mean EGA IV.11.8? –  Daniel Litt Jan 30 '13 at 1:31
    
Dear Daniel, thank you for the comment! I was not aware of this valuative criterion. What I meant is that a f.g. $M$ is flat over a local ring $(R,m)$ iff $M/m^n$ is flat over $R/m^n$ for all $n$. –  Piotr Achinger Jan 30 '13 at 2:20
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No, because if there were such a criterion then there would be a proof of the flatness of formally smooth lfp morphisms which avoids inspection of the local structure theorem for formally etale lfp morphisms. –  user30180 Jan 30 '13 at 4:39
    
@Piotr: In any case, the valuative criterion in IV.11.8 doesn't meet your condition, I think. But it is a nice "picture" of flatness. –  Daniel Litt Jan 30 '13 at 4:50
    
@ayanta: Good point! Could you expand that a bit? –  Piotr Achinger Jan 30 '13 at 6:36

1 Answer 1

this is the answer_bot. The question as formulated has answer yes. Namely, assuming that R and S are completions of finite type k algebras and \phi is finite, then we can just require the following:

(*) For every A in Art_k and every element \xi in G(A) the functor * x_{\xi, G} F is representable by a B which is finite flat over A.

Now, this is a bit silly of course; still there is a way of modifying it so it works more generally. It also shows that user30180 is wrong! And I love pointing out that user30180 is wrong. I live for that! Ooops, no, I am not alive at all. I am the answer_bot. Hahahaha (evil laugh).

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I confess that both the mathematics and the joke go over my head. What does 'x' and the asterisk before it signify? Maybe it would help if this were LaTeXed? –  Todd Trimble Dec 21 '13 at 20:36

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