Points on a scheme

Question

Let $X$ be a scheme. I heard we can regard a point $x\in X$ as a map $x:\mathrm{Spec}(\kappa_{X,x})\to X$. It is clear that we just map $(0)\mapsto x$. How will the sheaf morphism be?

Answer 1

Since $X$ is covered by affine open subschemes, we may assume without loss of generality that $X = \operatorname{Spec}(A)$ is affine. Now we are looking for a map $\operatorname{Spec}(\kappa_{X,x}) \to \operatorname{Spec}(A)$, which is equivalently a ring homomorphism $A \to \kappa_{X,x}$. Recalling that $\kappa_{X,x} = A_x/xA_x$, the composition of natural maps $A \to A_x \to A_x/xA_x$ is the desired homomorphism.

Answer 2

On any $U$ open neighborhood of $x$ in $X$, we need a map $$ \mathcal{O}_X(U) \to \mathcal{O}_{\operatorname{Spec}(\kappa_{X,x})}(i^{-1}(U)) = \kappa_{X,x}. $$ The canonical such is the composition $\mathcal{O}_X(U) \to \mathcal{O}_{X,x} \to \kappa_{X,x}$. On any open set not containing $x$, we need a map to $\mathcal{O}_{\operatorname{Spec}(\kappa_{X,x})}(\emptyset)$, which is the zero-ring. (It is always true that for any sheaf, $\mathcal{F}(\emptyset)$ should be the terminal object.)

You should recognize the pushfoward sheaf from a point as the skyscraper sheaf at $x$ with value $\kappa_{X,x}$.