Let $X$ be a scheme and $x \in X$. Then there is a canonical scheme morphism $\operatorname{Spec} \mathcal O_{X,x} \to X$ (Vakil’s Exercise 6.3.J(a)) and the quotient map $\mathcal O_{X,x} \to \kappa(x)$ induces a map between their spectra. Then composition gives a canonical scheme morphism $\operatorname{Spec} \kappa(x) \to X$.
However, there is also a more explicit way for a canonical morphism. Namely, we define $(\pi,\alpha): \operatorname{Spec} \kappa(x) \to X$ as follows. As a continuous function, $\pi$ sends the unique point of the spectrum to the point $x$. (I mean, what else can you do?) Then for every open subset $U$ of $X$, if $x \not\in U$, then $\alpha_U$ is the unique map to the zero ring. If $x \in U$, then $\alpha_U$ is the composition
$$\mathcal O_X(U) \to \mathcal O_{X,x} \to \kappa(x)$$
where the second map is the quotient map. So there is a few naturally-arising questions.
- $(\pi,\alpha)$ defines a morphism of ringed spaces, but does it defines a scheme morphism? (I honestly have no idea on showing the induced map on stalks is local.)
- If it is a scheme morphism, then is it the same as the one obtained in the first paragraph? (It should, as they are both “canonical”)
- Assuming question 2 (i.e. $\operatorname{Spec} \kappa(x) \to X$ factors through $\operatorname{Spec} \mathcal O_{X,x} \to X$), is there a choice-free description of the scheme morphism $\operatorname{Spec} \mathcal O_{X,x} \to X$. (I mean, the map is choice-free, but can we define it without choosing an open affine in the first place?)
Any help is appreciated.
Best Answer
is easy since you only need to check the condition for the unique point in $\operatorname{Spec}\kappa(x)$, in which case the map on local rings is given by the quotient map $\mathcal O_{X,x}\to \kappa(x)$.
yes, both maps are the same.
We can write $\operatorname{Spec} \mathcal O_{X,x}$ as the limit over all open subschemes of $X$ that contain $x$. Then the map into $X$ is just the one induced by the inclusions of the open subschemes into $X$.