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Recall that a morphism of rings $R\to S$ is called (essentially) smooth if it is formally smooth and (essentially) finitely presented.

(Note: $R\to S$ is essentially finitely presented provided that $S$ is the localization of some finitely
presented $R$-algebra $T$ at some multiplicative system $A \subset T$, that is, $S=A^{-1}T$.)

In class, our professor said that working with smooth or essentially smooth morphisms yields an effectively equivalent theory. This motivates my question: Is there a general technique to lift results from the smooth case to the essentially smooth case?

Edit: According to Mel, every essentially smooth morphism is a localization of a smooth morphism. However, this direction is much more involved than the other direction, which is immediate from the definitions. Anyway, this would be the answer to the question.

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  • $\begingroup$ Harry, you should get Mel to register on MO. $\endgroup$
    – Hailong Dao
    Apr 11, 2010 at 4:16
  • $\begingroup$ I feel like you'd have a better chance than I would. =) $\endgroup$
    – Harry Gindi
    Apr 11, 2010 at 4:19
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    $\begingroup$ EGA IV$_4$, 17.5.1(a),(c) $\endgroup$
    – BCnrd
    Apr 11, 2010 at 16:10
  • $\begingroup$ Added a bounty because I couldn't really see how that was at all related to being essentially of finite presentatation. $\endgroup$
    – Harry Gindi
    Apr 12, 2010 at 3:22
  • $\begingroup$ Not saying that it's not, just that I couldn't figure out how to show that they were equivalent, having not read any other parts of EGA 4. $\endgroup$
    – Harry Gindi
    Apr 12, 2010 at 3:23

1 Answer 1

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It seems that what you are looking for is theorem 5.11 here. See also example (e) on section 5.12. Also if you don't feel like reviewing from EGA you can look at section 1.5 of "Introduction to algebraic stacks" by A. Canonaco which I think covers the relevant facts (including 17.5.1)

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  • $\begingroup$ The problem I had with EGA was that I have all of my definitions set up in "commutative algebra form", as well as all of these theorems (the generalization to schemes is straighforward, so it doesn't really matter "where you work"). The problem I was having with EGA was that everything is phrased in terms of schemes, so I would have to go back and check all of the theorems that say "oh yes, our definition of smooth at a point means blah and our definition of smooth in an open affine neighborhood means blah". $\endgroup$
    – Harry Gindi
    Apr 15, 2010 at 9:52
  • $\begingroup$ Also, the statement in EGA that smoothness is a "local property" is given without proof, since Grothendieck mistakenly thought that it was straightforward. However, as noted in a post (somewhere on here) of Jim Borger, this is actually a somewhat deep result that wasn't published until a while after EGA. I haven't checked out the second source you gave yet, but I'll give you +1 anyway. $\endgroup$
    – Harry Gindi
    Apr 15, 2010 at 9:55
  • $\begingroup$ Yeah, it is also mentioned in remark 1.5.12 of "Introduction to algebraic stacks". $\endgroup$
    – Gjergji Zaimi
    Apr 15, 2010 at 23:38
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    $\begingroup$ Harry, you need to learn some humility. Borger's comment was about formal smoothness, which is a different concept from smoothness. It is obvious that smoothness is a local property: at that point EGA did the hard work to make it clear. Takes a lot of chutzpah to pass judgement on EGA when you barely know how to work with non-affine schemes. (I will also remind you that the proof in which the EGA glitch occurred was trivially fixed by imposing a mild extra hypothesis satisfied in pretty much all interesting cases. When you wrote a multi-thousand page work with no glitches, let me know.) $\endgroup$
    – BCnrd
    Apr 18, 2010 at 19:30

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