If $Z:=\bigcap_{x\in U}U$, prove that $(Z,\mathcal{O}_X\big|_Z)\simeq (\text{Spec}(\mathcal{O}_{X,x}),\mathcal{O}_{\text{Spec}(\mathcal{O}_{X,x})})$

Question

Let $(X,\mathcal{O}_X)$ be a scheme. For a fixed $x\in X$, let $Z:=\bigcap_{U\ni x}U$, where each $U$ is open in $X$. Prove that the scheme $(Z,\mathcal{O}_X\big|_Z)$ is isomorphic to $(\text{Spec}(\mathcal{O}_{X,x}),\mathcal{O}_{\text{Spec}(\mathcal{O}_{X,x})})$.

Here's where I'm at:

Because of the local nature of the problem, we may assume $X$ to be affine, say $X=\text{Spec}(A)$. Since the principal open sets are a basis of the topology, we may say $Z=\bigcap_{x\in X_f}X_f$, so $\mathfrak{p}\in Z\Leftrightarrow f\notin\mathfrak{p}$ for all $f\notin x\Leftrightarrow \mathfrak{p}\subset x$ and $Z=\{\mathfrak{p}\in X \mid \mathfrak{p}\subset x\}$. Then it's natural to define \begin{align*} \varphi:\text{Spec}(\mathcal{O}_{X,x})&\to Z\\ \mathfrak{p}\cdot \mathcal{O}_{X,x}&\mapsto \mathfrak{p} \end{align*}

which is easily seen to be a homeomorphism.

My difficulty is how to define the morphism of sheaves $\varphi^\#:\mathcal{O}_X\big|_Z\to \varphi_*\mathcal{O}_Y$, where $Y:=\text{Spec}(\mathcal{O}_{X,x})$. Since $\mathcal{O}_X\big|_Z$ is the sheafification of the pre-sheaf $\mathcal{F}$ on $Z$ given by $U\subset Z\mapsto \varinjlim_{V\supset U}\mathcal{O}_X(V)$, then I should define a morphism $\psi:\mathcal{F}\to \varphi_*\mathcal{O}_Y$ in order to obtain a morphism $\mathcal{O}_X\big|_Z\to \varphi_*\mathcal{O}_Y$ by the universal property.

How am I supposed to define $\psi$?

Answer 1

Somehow this is all just definition-pushing and there's only ever one thing you could do, but it can be kind of confusing. Let me first offer you a hint: how do you construct a map out of the directed limit? Do you have anything around which might look like that?

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Now let us explain. Recall how we get a map out of a directed limit: if we have a limit $A=\lim_{\rightarrow} A_i$, then in order to get a map $A\to B$, we can specify a compatible system of maps $A_i\to B$ as $i$ ranges over our index category.

By the fact that $i:\operatorname{Spec} \mathcal{O}_{X,x}\to X$ is a morphism of schemes, we get a morphism of sheaves $\mathcal{O}_X \to i_*\mathcal{O}_{\operatorname{Spec} \mathcal{O}_{X,x}}$, or equivalently for every open subset $W\subset X$, we get a map $$\mathcal{O}_X(W)\to (i_*\mathcal{O}_{\operatorname{Spec} \mathcal{O}_{X,x}})(W)=\mathcal{O}_{\operatorname{Spec} \mathcal{O}_{X,x}}(W\cap \operatorname{Spec}\mathcal{O}_{X,x})$$ which is compatible with restrictions from $W$ to any smaller open subset $W'$. This means that if we fix the open set $U=W\cap\operatorname{Spec}\mathcal{O}_{X,x}$, we get a compatible system of maps from $\mathcal{O}_X(V)\to \mathcal{O}_{\operatorname{Spec}\mathcal{O}_{X,x}}(U)$ where $V$ ranges over all open sets of $X$ which contain $U$. But by the definition of the directed limit, this means we have a map from the directed limit $\lim_{\rightarrow V\supset U}\mathcal{O}_X(V)$ to $\mathcal{O}_Y(U)$, where I commit the sin of identifying $U$ with it's image under the homeomorphism between $Z$ and $\operatorname{Spec} \mathcal{O}_{X,x}$. Compatibility with restriction is clear from the construction above, so this is exactly the data we need to define the morphism $\mathcal{O}_X|_Z \to \varphi_*\mathcal{O}_Y$ and we are done.