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A (commutative) algebra $A$ is called formally smooth if for any (commutative) algebra $R$ and an ideal $I\subset R$ such that $I^2=0$, any morphism $A\to R/I$ lifts to a morphism $A\to R$.

Suppose now that $X$ is a variety and $A$ is the algebra of regular functions on it.

How are the definitions of formal smoothness of $A$ and smoothness of the variety $X$ related?

Are the two notions equivalent? It doesn't seem to be the case, but I don't know how to prove it. I might be very silly here. In this case I would be glad if you explain that to me.

Thank you very much!

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1 Answer 1

up vote 7 down vote accepted

You haven't introduced a base-ring for the various algebras in play; let me denote it by $A_0$. (In the case of varieties, we would take $A_0$ to be a field, but that doesn't affect anything.)

For a finitely presented $A_0$-algebra $A$, formal smoothness is equivalent to smoothness. (And if $A_0$ is Noetherian, e.g. a field, then f.p. is equivalent to finite type.)

In particular, for the affine ring of an affine variety, formal smoothness over the ground field $k$ is equivalent to the variety being smooth.

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Dear Matthew, can you, please, give me a reference? Or is it supposed to be obvious and I am just being stupid? –  Sasha Patotski Mar 10 at 0:39
    
Dear Sasha, It's a theorem of Grothendieck, originally proved somewhere in SGA IV, I think. The reference in the stacks project is here. There is also an exercise in Hartshorne where he sketches how to deduce one direction (smoothness implies formal smoothness), probably in the section that discusses smooth morphisms. Regards, –  Matt E Mar 10 at 0:45
    
@SashaPatotski: Dear Sasha, Sorry, SGA IV should have read EGA IV. Cheers, –  Matt E Mar 10 at 2:34
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@SashaPatotski It is probably worth mentioning that you're not stupid, that this theorem, unlike the analogue for, say, unramifiedness, is difficult. –  Alex Youcis Mar 10 at 5:37

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