In which precise geometric sense is localizing at a prime ideal $P$ like looking "near" $P$. I have heard this statement various times in my life before, but never knew what was the precise geometric statement behind it.
Let us start with an example. Let $R = \mathbb{C}[x]$, which corresponds to the complex affine line. The ideal $(x)$ of $R$ corresponds to the origin on that line. The localization $R_{(x)}$ of $R$ at the prime ideal $(x)$ has only 2 prime ideals, namely $0$ and $(x/1)$. These are in 1-1 correspondence with the prime ideals $0$ and $(x)$ in $R$. $(x)$ corresponds to the origin, and $0$ corresponds to a generic point on the complex affine line.
Thus, by localizing at $(x)$, we only "picked up" the "point" $(x)$ itself, as well as the generic point on the complex affine line. I can see why this is in some sense like looking near $(x)$, but can someone make this more precise please? In general, can one say perhaps that if one localizes at a prime ideal $P$, then one picks up only the prime ideals $Q$ whose closure in the Zariski sense contains the "point" $P$? Can someone perhaps comment on that?
 A: I sorted it out. While I am mostly interested in the geometric setting, here is the answer for arbitrary commutative rings with 1. Let $R$ be a commutative ring with 1, and let $P$ be a prime ideal of $R$. Just to distinguish, we will denote the point in $\operatorname{Spec}(R)$ corresponding to a prime ideal $Q$ of $R$ by $q$, with lowercase. We denote by $\bar{q}$ the closure of $q$ in $\operatorname{Spec}(R)$ with respect to the Zariski topology on $\operatorname{Spec}(R)$.
Then the elements of $\operatorname{Spec}(R_P)$ are in one-to-one correspondence with the set of all $q$'s in $\operatorname{Spec}(R)$ whose closure $\bar{q}$ contains $\bar{p}$ or, equivalently, whose closure $\bar{q}$ contains $p$. So in some sense, what we have is $P \subseteq Q$, but we want to translate that geometrically, in the Zariski topology sense, and this is how one does it.
The question turns out to be quite easy to answer. Somehow, if someone else hears that sentence mentioned in the post, this is the precise meaning of it. 
