# Why is the ring of local-ring valued points of a ring scheme a local ring.

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?

• 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. – JWL May 26 at 14:50
• Oh yeah, of course. Thank you – edgarlorp May 26 at 16:56