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Is there a known characterization of epimorphisms in the category of schemes?

It is easy to see that a morphism $f : X \to Y$ such that the underlying map $|f|$ is surjective and the homomorphism $f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$ is injective, is an epimorphism. But there are other examples, too, see for example this problem sheet.

If this is is not possible, what about regular, extremal, or effective epimorphisms? Here, again, I know only some examples.

My background is that I want to know if there is a categorical characterization of the spectra of fields in the category of schemes. In the full subcategory of affine schemes, they are characterized by the property: $X$ is non-initial and every morphism from a non-initial object to $X$ is an epimorphism. But I doubt that this characterization takes over to the category of schemes. EDIT: Kevin Ventullo has shown below that the characterization takes over. Thus my original question has been answered (and I wonder if its appropriate to accept it as an answer). But of course every other hint about the characterization of epimorphisms of schemes is appreciated.

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One thing to keep in mind, although this is not your original purpose, is that epimorphisms in $Sch$ might in general be 'badly behaved'. In $Aff$ the 'real' notion of epimorphism you need (regular, effective or what-have-you) might coincide with ordinary epis. –  David Roberts Feb 24 '11 at 23:09

1 Answer 1

Actually, your suggested categorical characterization of spectra of fields does work.

Edit: (I had written something incorrect here)

By Martin's comment below, we just have to show that maps from affines into $\text{Spec}(k)$ are epis in the full category. But if we had two maps $f,g: \text{Spec}(k) \rightarrow Y$ which agreed on some affine mapping into Spec(k), then first of all $f$ and $g$ would have to be the same topological map. Then both would land inside some affine $\text{Spec}(R)\subset Y$, and now we're reduced to the affine situation where we know it holds.

Conversely, suppose X is not the spectrum of a field. If every point is dense, X is affine and we are done by what we know about the affine subcategory. Otherwise, we can find an open subscheme $U\subsetneq X$. Then the inclusion of $U$ into $X$ is not an epi as is witnessed by the two inclusions

$X \rightrightarrows X\coprod_U X$,

where this last object is $X$ glued to itself along $U$.

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The observation that maps from $Spec(k)$ to schemes are "completely determined" by maps to affine schemes is neat. It also works for $Spec(R)$, where $R$ is a local ring. This answer (mathoverflow.net/questions/63/…) suggests this approach might be able to produce an example of a coequalizer of schemes which is not surjective, but I wasn't ever able to get it to work. –  Anton Geraschenko Feb 25 '11 at 2:40
    
I don't think that the use of $Spec(O_X(X))$ is correct here. This would only work if $X \to Spec(O_X(X))$ is an epi. Rather, we compose $X \to Spec(k)$ with the inclusion of a nonempty open affine part of $X$ and it suffices to prove that this composition is epi. –  Martin Brandenburg Feb 25 '11 at 7:34
    
... but I didn't manage to prove the other direction, which is now easier than I thought. Thank you! @Anton: I also tried to produce an example some time ago, but it is quite hard. It resulted in mathoverflow.net/questions/24066/…. –  Martin Brandenburg Feb 25 '11 at 8:09
    
@Martin: Good call; affinification need not be an epi. My mistake. –  Kevin Ventullo Feb 25 '11 at 9:35

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