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I was able to recover the original statement, which is: Let A be a local ring (for Zariski, henselian for Nisnevic and strictly henselian for étale) and $\{U_i\rightarrow Spec A\}$ a covering. Then there exists an $i$ and a morphism $Spec A\rightarrow U_i$ (open/étale) which admits a lift of $U_i\rightarrow Spec A$. (For Zariski this is just the statement below, that each covering of $Spec A$ already includes the identity.) Thus the global sections are unchanged for a presheaf and its associated sheaf.

It remains to ask why this implies that the suitable notion of points in the étale site are the geometric points, but I guess this is too fuzzy to be an actual question.

As seen in the comments the original question didn't really make sense. Sorry for any inconvenience.


The question:

Is any presheaf (of sets, maybe more structure) on the spectrum of a (edit: reduced) local ring A (with the Zariski topology) automatically a sheaf?

The motivation:

For the Zariski and Nisnevich topology, stalks are defined for every "classical point" $Spec k\rightarrow Spec A$, whereas for the étale topolgy we take geometric points $Spec k^s\rightarrow Spec A$, where $k^s$ is the separable closure. The reason is somewhat that we want sheafification (and other constructions) to work and hence want the small bits from which we build the sheaf associated to a presheaf (i.e stalks) to fulfill the sheaf properties to begin with. The above question is definitely false for the étale topology, but we should be able to fix it, if we require A to be strictly henselian, which then inevitably leads to the definition of geometric points.

How to tackle the question:

It would be nice to have a proof which refrains from using stalks itself, as this would appear a bit circular.

What I have so far:

For any covering of the whole space $Spec A$ by standard open sets $D(f_i)$, we know that the maximal ideal $m$ must be contained in at least one of the $D(f_i)$, which means $f_i$ is not an element of $m$, which implies $f_i$ is a unit and hence $D(f_i)=Spec A$. This reasoning fails for open subsets $D(f)$ of $Spec A$, since $A_f$ doesn't need to be local (or does it?).

Additional requirements:

I have heard the statement somewhere and have failed proving it. Maybe there are some extra conditions which are "standard" in the setting people usually work with. However I guess it should hold in this full generality.

edit: One extre condition should be $F(\emptyset)$ is the final object. Hopefully there aren't too many similar conditions I forgot.

Thanks for your help!

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2  
This already fails for the spectrum $X$ of a field whenever $F(X)=F(\varnothing)$ is not the final object. –  user2035 Nov 24 '11 at 11:32
    
Valid point a-fortiori. As I said maybe there are some minor extra conditions. Let us just assume $F(\emptyset)$ is the final object. –  Simon Markett Nov 24 '11 at 11:48
    
Doesn't the etale topology usually include open sets which are $n$ copies of the base space, for instance? You need to account for those. Accounting for that, I'm not sure that it's even true for the spec of a field in general, because the sheaf condition says something about non-Galois extensions. There tend to be a LOT of presheafs you can construct on any given site. This should only work if you have very few open sets. –  Will Sawin Nov 24 '11 at 11:58
5  
Let $X$ be the spectrum of a one-dimensional local ring having two minimal prime ideals $x_1,x_2$. Now choose $F(X)=F(\{x_1,x_2\})\ne F(\{x_1\})\times F(\{x_2\})$. –  user2035 Nov 24 '11 at 11:59
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There are lots of one-dimensional reduced local rings having more than one minimal prime. You could take $\mathbb{Z}[x]_{(2,x)} / (2x)$ or $k[[x, y]]/ (xy)$ ($k$ any field), for instance. –  Neil Epstein Nov 24 '11 at 14:41
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