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I will start with a motivation. Let $X$ be a compat hausdorff space and let $A$ be the ring ($\mathbb{R}$-algebra) of continuous functions $X\to \mathbb{R}$. We define $Y=MaxSpec(A)$ to be the set of maximal ideals of $A$ and we can endow it with the subspace topology induced from $Spec(A)$ with the usual Zariski topology. Now, for each point $x\in X$ the subset $\mathfrak{m}_x\subseteq A$ of functions vanishing at $x$ is a maximal ideal and therefore we can define a function $\mu:X\to Y$ by $\mu(x)=\mathfrak{m}_x$. It is known that $\mu$ is a homeomorphism (so far, this is an exercise in Atiyah- Macdonald).

Now, $X$ can be turned into a locally-ringed space by taking the structure sheaf $\mathcal{O}_X$ to be the sheaf of continuous real functions. For each point $x\in X$ we can look at the stalk $\mathcal{O}_{X,x}$ at $x$ and it turns out (if I am not mistaken) that it is exactly $A_{\mathfrak{m}_x}$ (there is a canonical isomorphism). So, is it possible to turn $Y$ into a locally ringed space such that the stalk at $\mathfrak{m}$ will be exactly $A_\mathfrak{m}$ (like with the usual $spec$)? and if it is, $\mu$ will be an isomorphism of locally-ringed spaces, right? what will $MaxSpec(A)$ be as a locally-ringed space for general rings?

I am sorry that the formulation of the question is not very sharp, its just an idea I am trying to make sense of.

KotelKanim
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1 Answers1

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If $X=(|X|,\mathcal{O}_X)$ is a ringed space, $|Z|$ is a topological space, and $i : |Z| \to |X|$ is a continuous map, then $\mathcal{O}_Z := i^{-1} \mathcal{O}_X$ is a sheaf of rings on $|Z|$, hence $Z:= ( |Z|,\mathcal{O}_Z)$ is a ringed space. We have $\mathcal{O}_{Z,z}=\mathcal{O}_{X,i(z)}$ for all $z \in |Z|$. From this we see that $Z$ is a locally ringed space if $X$ is a locally ringed space, and that $i$ becomes a morphism of locally ringed spaces $Z \to X$ (in fact on stalks we have isomorphisms). In fact $Z \to X$ is the terminal morphism of locally ringed spaces with underlying map $i$. This can be applied in particular to a subspace $|Z|$ of $|X|$, where $i$ is the inclusion. For example, $|Z|$ may be the set of closed points. For $X=\mathrm{Spec}(A)$ we get $Z=\mathrm{MaxSpec}(A)$.

Martin Brandenburg
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    This is exactly what I was looking for. I want to accept this answer. The only thing I would still like is a reassurance that with this definition, the map $X\to MaxSpec(A)$ in the question is an isomorphism of locally-ringed spaces (i.e. that I calculated the localization correctly and didn't miss anything else in the argument) – KotelKanim Mar 19 '13 at 14:58
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    Yes I think so. There is a canonical homomorphism $C(X){\mathfrak{m}_x} \longrightarrow \mathcal{O}{X,x}$. If I remember correctly, it is not completely trivial that this is an isomorphism, for example one has to use Urysohn's Lemma, bump functions etc. – Martin Brandenburg Mar 19 '13 at 15:28
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    Yes martin, my argument uses normality and Tietze extension theorem. I checked the details myself and yet I was a bit unsure, since this conflicts with my intuition that in honest geometrical situation open sets support more regular functions than just quotients of global regular functions (like functions with essential singularity in the complex case), but I guess that since there are a lot of continuous functions it somehow works. – KotelKanim Mar 19 '13 at 16:07