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of course, one should expect that the concept of ring-valued points is not well-behaved for locally ringed spaces (LRS). I want to see examples for this.

so consider $LRS \to Set^{Ring}, X \mapsto X(-)=Hom(Spec - , X)$. if $A$ is a local ring, whose maximal ideal is principal, and $\hat{A}$ its completion, and we regard local rings as locally ringed spaces whose underlying set is just one point, then $A \to \hat{A}$ induces a bijection $Hom(Spec R,A) \to Hom(Spec R,\hat{A})$ (I'll add the proof if you want). this shows that the functor is not full. but how can we see that it is not faithful?

For example, for local rings $A$, we have

$Hom_{LRS}(Spec R,A)=\{\phi \in Hom_{Ring}(A,R) : \phi(\mathfrak{m}_A) \subseteq rad(R)\}$.

If $f,g$ are local homomorphisms inducing the same maps $Hom_{LRS}(Spec -,B) \to Hom_{LRS}(Spec -,A)$, it seems that they don't have to be identical ...

edited Jan 23 2010 at 20:25

asked Dec 30 2009 at 11:38

Martin Brandenburg
29.4k●2●47●133

There's something funny here in your set up, I think. A morphism of LRS is a continuous map and a map of sheaves in the other direction, so that natural map $A\to \hat{A}$ would give a map $|\hat{A}|\to|A|$ (using $|A|$ to denote the LRS you defined), so if it induces a bijection, then should be $Hom(Spec R,\hat{A})\to Hom(Spec R,A)$, but I still don't think I believe your statement. My proposed counterexample is in an answer below. – Charles Siegel Jan 7 2010 at 14:50

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There's an additional restriction on morphisms of locally ringed spaces: the maps on stalks must be local maps. (This is not automatic; it must be imposed. Thus locally ringed spaces are not a full subcategory of ringed spaces.) This means that the map $A \to R$ must take the maximal ideal of $A$ into the nilradical of $R$, and hence if that maximal ideal is f.g. (e.g. principal), to a nilpotent ideal. Thus Martin's statement is correct (even for all Noetherian local rings). – Emerton Jan 7 2010 at 15:46

Ahh, of course. Don't know what I was thinking. I'm withdrawing my objection. – Charles Siegel Jan 7 2010 at 16:08

This is of course why we need to impose the locality condition for covering functors of points. – Harry Gindi Jan 23 2010 at 20:32

ring-valued points of locally ringed spaces

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