Viewing a closed point as a scheme

Question

If $P$ is a closed point of a scheme $X$, then pullback of the structure sheaf $\mathscr{O}_X$ is $k(P)$. This really baffles me - by definition, this pullback is just $P \mapsto \lim_{U \supset P}\mathscr{O}_X(U)$, which is the stalk of $\mathscr{O}_X$ at $P$. So why isn't the pullback the local ring $\mathscr{O}_{X,P}$ instead of the residue field?

Answer 1

It looks like you are confusing the functors $i^*$ and $i^{-1}$, where $i$ is the inclusion morphism $\operatorname{Spec} k(P)\to X$. You are right that $i^{-1}(\mathscr{O}_X)$ is the sheaf with value $\mathscr{O}_{X,P}$. However, the definition of $i^*$ is different: for any sheaf $F$ on $X$, $i^*F=i^{-1}F\otimes_{i^{-1}(\mathscr{O}_X)}\mathscr{O}_{\operatorname{Spec} k(P)}$. In particular, this means $i^*\mathscr{O}_X$ is $\mathscr{O}_{\operatorname{Spec} k(P)}$, which has value $k(P)$.

Answer 2

Just see that $\{P\}$ is, as closed subscheme, $\textrm{Spec }k(P)$. Take $U=\textrm{Spec }A$ an affine open containing $P$. Hence $P$ corresponds to a maximal ideal $\mathfrak{m}$ in $A$ and $$A/\mathfrak{m}\cong A_{\mathfrak{m}}/\mathfrak{m}A_{\mathfrak{m}}=k(P).$$