2
$\begingroup$

A morphism of set-valued functors $\eta: F \to G$ on $\mathcal{C}$ is called smooth if for all epimorphisms $B \to A$, the natural morphism $F(B) \to F(A) \times_{G(A)} G(B)$ is surjective.

Obviusly "smooth => formally" smooth for $\mathcal{C} = \mathrm{Sch}$.

Now my question: Does the converse hold?

My thoughts: 1. Assume the morphism is of finite presentation.

  1. Assume it is in the local standard form: an étale morphism followed by an affine projection

  2. It is clear that an affine projection is smooth in the above sense, so we have reduced the problem to étale morphisms.

$\endgroup$
4
  • $\begingroup$ Possible duplicate: mathoverflow.net/questions/195/… $\endgroup$ Sep 5, 2011 at 10:42
  • 2
    $\begingroup$ I think my definition of "smooth" is a priori different from this. $\endgroup$
    – user12832
    Sep 5, 2011 at 11:08
  • 1
    $\begingroup$ I don't have the book in front of me, but there is an appendix to Loday's "Cyclic Homology" where various definitions of 'smooth' are compared quite carefully. $\endgroup$ Sep 5, 2011 at 13:08
  • 2
    $\begingroup$ Your definition of "smooth" is for me the definition of "formally smooth", and "smooth" would be "formally smooth"+"locally of finite presentation". What in fact is your definition of formally smooth? $\endgroup$ Sep 5, 2011 at 17:40

1 Answer 1

5
$\begingroup$

In the case where $F\to G$ is representable by an epimorphism $C\to D$ of rings, then your smoothness condition implies that $C\to D$ has a section. (Take $A\to B$ to be the given map $C\to D$.) But it is not hard to find a formally smooth epimorphism that does not admit a section. Any localization will do. For instance, you can take $D$ to be the zero ring and $C$ any ring but the zero ring.

If you visualize the geometry here, it's pretty clear that your smoothness condition is much stronger than anything normally called smoothness.

$\endgroup$
1
  • $\begingroup$ Ah, I missed the part where the OP allowed any epimorphism. $\endgroup$ Sep 5, 2011 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.