Nice fact: Suppose f:X->Y is a map of schemes and Z⊆Y is a subscheme (locally closed immersion) containing the set-image of X. If X and Z are reduced, then it follows that f factors through Z. This is nice because it makes "factoring through" purely a consideration of the underlying topological spaces.

So now I'm wondering, to what extent does "reduced" allow us to think only terms of topological spaces? Suppose we weaken the assumption that Z→Y is an inclusion. When can we say f factors through Z? More precisely:

Suppose X,Z are reduced schemes, f:X→Y and g:Z→Y are scheme morphisms such that f factors through g in Top. When does f factor through g in Sch?

I know the answer is "not always", for example if Y is a field and X,Z are incomparable field extensions of Y (in Ringop). But does anyone know any positive results we can state here?

  • $\begingroup$ Given that no-one has bitten yet, can I explicitly ask whether people think it might hold if X and K are reduced varieties over a field? Maybe one needs X to be smooth? I'm not sure. I'm a bit concerned about the resolution of the singularity of a cuspidal cubic being a homeomorphism on the top spaces (and the map only existing in one direction in alg geo) but can't make an explicit counterexample. $\endgroup$ – Kevin Buzzard Nov 4 '09 at 23:25
  • $\begingroup$ @KMB: Let f be the identity on the cuspidal cubic X=Y=Spec k[x,y]/(y^2-x^3), and let g be the normalization from A^1. Then there is no factorization, since k[x] doesn't map to k[x^2,x^3] in a way that commutes with the reverse inclusion. $\endgroup$ – S. Carnahan Nov 5 '09 at 2:01
  • $\begingroup$ Ah, can you use Z or something instead of K? When I see K, I can only think "field". I suppose this is not very mathematicianly of me. It's kind of physicistly, maybe -- well, it's "field theory" after all ;-) $\endgroup$ – Kevin H. Lin Nov 5 '09 at 2:42
  • $\begingroup$ @Kevin: replaced K by Z. $\endgroup$ – Andrew Critch Nov 5 '09 at 4:47
  • $\begingroup$ What you are asking will (in my opinion) always require a scheme theoretic condition.. For instance, your example Z⊆Y being a subscheme is already a scheme theretic condition, you require something about the rings involved in the structure sheaf. $\endgroup$ – Jose Capco Nov 5 '09 at 10:17

Here's an example that is not completely silly. I think you get scheme-theoretic factorization if g is etale, and X is simply connected.


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