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I'm confused on a supposedly easy claim:

Let $T$ be a base scheme, and let $\mathbf{R}$ be a ring scheme over $T$, i.e. a scheme $\mathbf{R} \to T$ such that for all $E \in \operatorname{Sch}_{/T}$ $\mathbf{R}(E) = \text{Hom}_T(E,R)$ is a ring. I have a certain functor $F$ that takes a scheme $X$ to the ringed space $(X, Hom_T(*,\mathbf{R}))$ where the sheaf is the sheaf of morphisms $U \subset X \mapsto Hom_{T}(U,\mathbf{R}))$. By virtue of $\mathbf{R}$ being a ring-scheme, each $\mathbf{R}(U)$ is a ring, hence is indeed a ringed space.

Suppose further that $\mathbf{R}$ is such that $F(X)$ is a locally ringed space, i.e. that moroever the stalks of $Hom_{T}(*,\mathbf{R}))$ are local rings. Then the following claim is made:

If $S$ is a local ring, then $\mathbf{R}(\operatorname{spec}(S)) = Hom_{T}(S,\mathbf{R}))$ is a local ring.

This should follow by looking at the stalk over the unique closed point in $\operatorname{spec}(S)$, but I just don't see it. Why is this the case?

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  • $\begingroup$ There is a unique open set containing the closed point in $\operatorname{Spec}S$, the whole set. So the computation of the stalk at the closed point becomes trivial. $\endgroup$ – JWL May 26 at 14:50
  • $\begingroup$ Oh yeah, of course. Thank you $\endgroup$ – edgarlorp May 26 at 16:56

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