Closed points of $\text{Proj}(\mathbb R[x,y,z]/(x^2+y^2+z^2))$ Consider $X=\text{Proj}(\mathbb R[x,y,z]/(x^2+y^2+z^2))$, I want to show that $X(\mathbb R)=\emptyset$.
I have two problems with this. The first one is that I'm not sure of the definition of $X(\mathbb R)$. Is it the set of closed points of $X$ such that $\exists K(x)\hookrightarrow \mathbb R$ an injection ? Do we require that this extension has something more, to be algebraic for example ? Maybe there is a reference I could use.
Since I'm quite sure about the closed points part, we can try to find closed points. Closed points of $X$ are the homogeneous maximal ideals of $\mathbb R[x,y,z]$ containing $(x^2+y^2+z^2)$, which is the set $\{(x+y+z)\}$.
We write $x=(x+y+z)$.
My second problem is the computation of $K(x)$.
We get $\mathcal O_{X,x}=T^{-1}(\mathbb R[x,y,z]/(x^2+y^2+z^2))$ where $T$ is the set of homogenous elements of $R[x,y,z]/(x^2+y^2+z^2)$ not contained in $\overline{(x+y+z)}$. Since I don't get what this thing is I tried to consider the open subset $D(x)$ and compute the stalk on this subset but I'm struggling. Is there a better approach ?
 A: By definition, for any scheme $X$ and any commutative ring $R$,
$$X(R) = \operatorname{Hom}_{\mathsf{Sch}}(\operatorname{Spec}(R),X).$$
This definition makes sense for schemes in general, although if $X$ is an $R$-scheme you can also take the hom in the category of $R$-schemes, which in some cases can be different. In this problem, $X$ is an $\mathbb{R}$-scheme, but it doesn't matter which interpretation we use: it turns out that $\operatorname{Hom}_{\mathsf{Sch}}(\operatorname{Spec}(\mathbb{R}),X) = \operatorname{Hom}_{\mathsf{Sch}_\mathbb{R}}(\operatorname{Spec}(\mathbb{R}),X)$ for any $\mathbb{R}$-scheme $X$. This is something a bit special about the field $\mathbb{R}$.
Anyway, this just means that an element of $X(\mathbb{R})$ (aka an "$\mathbb{R}$-point of $X$") is just a scheme morphism $\operatorname{Spec}(\mathbb{R}) \to X$. Unwinding definitions, such a morphism is given by a pair $(x,\varphi)$, where $x$ is a point of $X$ and $\varphi : k(x) \to \mathbb{R}$ is a homomorphism, where $k(x)$ is the residue field of $x$. This is close to what you said, but a bit different: the map $\varphi$ is part of the data of the $\mathbb{R}$-point! The same point $x \in X$ may give rise to many different $\mathbb{R}$-points if there are many different homomorphisms $k(x) \to \mathbb{R}$.
Now let's think about the $\mathbb{R}$-scheme structure. Since $\mathbb{R}[x,y,z]/(x^2+y^2+z^2)$ is an $\mathbb{R}$-algebra, every ring of local sections of the structure sheaf of $X$ is naturally an $\mathbb{R}$-algebra, and thus so are the local rings $\mathcal{O}_x$ and the residue fields $k(x)$ for all points $x \in X$. Now if $(x,\varphi)$ is an $\mathbb{R}$-point of $X$, the composite $\mathbb{R} \to k(x) \xrightarrow{\varphi} \mathbb{R}$ must be the identity, because $\mathbb{R}$ has no nontrivial endomorphisms as a field, and therefore $\varphi$ is an isomorphism. Moreover, since there are no nontrivial endomorphisms of $\mathbb{R}$, whenever $k(x)$ is isomorphic to $\mathbb{R}$, the isomorphism is unique! Thus, an $\mathbb{R}$-point of $X$ is simply given by a point $x \in X$ such that $k(x) \cong \mathbb{R}$.
Such points are necessarily closed, but this takes a little more working out. It's probably more effective here to just think directly about all of the points of this scheme and work out that none of the residue fields can be isomorphic to $\mathbb{R}$. The following lemma might help:
Lemma Let $R$ be a commutative graded ring and let $X = \operatorname{Proj}(R)$. Then for each point $\mathfrak{p} \in X$, the local ring $\mathcal{O}_{X,\mathfrak{p}}$ is isomorphic to the $0$th graded piece of $R_{\mathfrak{p}}$.
