Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$ for any $\operatorname{Spec} A$ with $x\in\operatorname{Spec} A$?

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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?

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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 $f_x$.

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Yes, as long as $\operatorname{Spec} A$ is an open affine set in $X$.

More generally if $U$ is an open subset of a ringed space $(X,\mathcal O_X)$, and $\mathcal O_U$ is the sheaf of rings on $U$ defined by $\mathcal O_U(V) = \mathcal O_X(V)$ for all open sets $V \subset U \subset X$, then $\mathcal O_{X,x} = \mathcal O_{U,x}$ for all $x \in U$. This is because

$$\mathcal O_{X,x} = \varinjlim_V \mathcal O_X(V)$$ as $V \subset X$ runs through the open neighborhoods of $x$ ordered by reverse inclusion, and $$\mathcal O_{U,x} = \varinjlim_V \mathcal O_U(V) = \varinjlim_V \mathcal O_X(V)$$

as $V \subset U$ runs through the open neighborhoods of $x$. Although the direct system of rings defining $\mathcal O_{U,x}$ is a subsystem of the one defining $\mathcal O_{X,x}$, it is easy to see that they define the same direct limit. It is the same principle by which the infinite polynomial ring $\mathbb C[x_1, x_2, x_3, ...]$ is the direct limit of the direct system

$$\mathbb C[x_1] \subset \mathbb C[x_1, x_2] \subset \mathbb C[x_1, x_2,x_3] \subset \cdots$$

as well as the subsystem

$$\mathbb C[x_1] \subset \mathbb C[x_1,x_2,x_3] \subset \mathbb C[x_1,x_2,x_3,x_4,x_5] \subset \cdots.$$