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  • Which morphisms of schemes (or varieties, if you prefer) $\pi: X \rightarrow Y$ are quotient morphisms, i.e. satisfy the following universal property (*)?

(*) For any morphism $f:X \rightarrow Z$, such that its restriction to the fibers of $\pi$ is constant, there is an $\bar{f}: Y \rightarrow Z$ with $f= \bar{f} \circ \pi$.

[btw: is it the right definition of quotient morphism? Is it ok if we consider only fibers over closed points?]

Categorical quotients by group actions $X \rightarrow X/G$ have this property.

  • Do flat surjective morphisms have this property?
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The constancy condition on fibers is too weak; just consider the finite flat bijective Frobenius morphism for a smooth scheme over a field of positive characteristic. For (*) to be reasonable, the fibers cannot be too bad, or should assume $f_{\ast}(O_X) = O_Z$ with $f$ proper surjective, or something like that. This is too special to be regarded as a good "general" notion of quotient morphism. –  BCnrd Apr 23 '10 at 16:26
    
I took one defn. that just parallels the definition in the topological category. So, maybe it's the wrong one in the algebraic category. By the way, I was mainly thinking of the case of varieties over the complex numbers. What do you mean by "too special"? In char zero a generic fibre is reduced (in the variety case): can it still be "too bad" a situation? Any example (of (*) not being a sensible definition) in char zero? –  Qfwfq Apr 23 '10 at 18:51
    
The manifold category would be a better guide than the topological category. Even in char 0 there are finite bijective non-isoms between varieties, such as normalization map of cuspidal cubic (i.e., $t \mapsto (t^2, t^3)$). Such maps do not deserve to be regarded as "quotient maps" in any general sense. I suggested one way out in the proper case, but such hypotheses are too special to regard that as a general notion of quotient either. I recommend learning faithfully flat descent theory (e.g., 6.1--6.2 of "Neron Models"), and then you'll have better intuition for quotients in alg. geom. –  BCnrd Apr 23 '10 at 19:46
    
I'm a bit confused: does the normalization map $\pi$ of the cusp verify the above universal property? Take $f$ to be the identity map of $\mathbb{A}^1$; being constant along the fibers of $\pi$, it should induce an injective map $g$ from the cusp to $\mathbb{A}^1$. Is it possible? –  Qfwfq Apr 24 '10 at 7:59
    
Ops: of course it's possible! :) So, in the end it really seems that the universal property as stated in my question is not well suited for algebraic varieties and schemes in general. Nevertheless, it has sense to ask for a characterization of the morphisms for which that property holds. –  Qfwfq Apr 24 '10 at 9:02

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