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Suppose $A_p$ is the stalk of a ring $A$ at a prime ideal $p$.

Consider the (opposite) system of those open immersions $\operatorname{Spec}(A)\leftarrow \operatorname{Spec}(B)$ such that the scheme map $\operatorname{Spec}(k(p))\to \operatorname{Spec}(A)$ factorizes over these $\operatorname{Spec}(B)\to \operatorname{Spec}(A)$. Here, $k(p)=A_p/m_p$ denotes the residue field of $p$. Geometrically, these are smaller and smaller opens around the point $p$ of $\operatorname{Spec}(A)$, I think.

Is it true that $\underset{\rightarrow}{\lim} B=A_p$?

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up vote 1 down vote accepted

Yes ! Because $A_p$ is the union (in its total ring of fractions) of all the $A_{(f)}$, for $f$ not in $p$. Each of these $A_{(f)}$ correspond to an open immersion.

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