Let $X$ be a scheme and $\operatorname{Spec} A\subset X$ be an affine with $x\in\operatorname{Spec} A$,
Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$?
I ask this question because I feel that many "obvious" remarks made in Gortz-Wedhorn and Vakil FOAG seem to rely on this or a similar fact: such as in Vakil
"We say a ringed space is a locally ringed space if its stalks
are local rings. Thus Exercise 4.3.F shows that schemes are locally ringed spaces." (Where 4.3.F is Show that the stalk of $\mathscr{O}_{\operatorname{Spec} A}$ at the point $[\mathfrak{p}]$ is the local ring $A_{\mathfrak{p}}$).
Is this the case? If not, why is the above statement immediate from 4.3.F?
Best Answer
Yes that's true, the statement you want to prove is that if $U$ is any open subset of any ringed space $X$ then for $x\in U$ the natural map $\mathcal O_{U,x}\to\mathcal O_{X,x}$ is an isomorphism. It's pretty straightforward, for instance for surjectivity if you have an element $f_x\in\mathcal O_{X,x}$ then it is represented by a function $f\in\mathcal O_X(V)$ for some $V$ containing $x$, and then the stalk of $f|_{U\cap V}$ in $\mathcal O_{U,x}$ is the element you want mapping to $f_x$.