The localization of a regular local ring is regular I've heard, as I'm sure many have, that the theorem that the localization of a regular local ring at any prime ideal is regular is one of the first major applications of homological methods to pure algebra.
Alas, I fail to see a geometric picture attached to this theorem. To quote wikipedia: "Geometrically, this corresponds to the intuition that if a surface contains a curve, and that curve is smooth, then the surface is smooth near the curve."
When I imagine local rings, I always prefer complete local rings. $k[[t]]$ is much nicer than $k[t]_{(t)}$ to imagine.
So let's say we have $R=k[[x,y]]$. The theorem would then tell us that because $R$ is regular, $k[[x,y]]_{(x)}=k[[x,y]][y^{-1}]$ $=k[[x]]$((y)) (and unless I'm very much mistaken this also equals $k((y))[[x]]$) is regular.
How does this correspond to wikipedia's intuition? Let's say we have a surface over $k$, and there's a smooth curve going through it. Let's say the point $P$ is on it. Look formally locally near $P$ to get $R$. Why does this tell us that $R$ is regular? If anything the theorem seems to start from the assumption that $R$ is regular.
As you can see, my intuition is out of kilter. 
 A: The Wikipedia "intuition" you mention above should actually be about the (nearly trivial) fact that if $R$ is a local ring, $r\in R$ a regular element and $R/(r)$ is regular, then $R$ is regular. In other words, if a scheme $X$ contains a Cartier divisor which is smooth at a point, then the scheme itself is smooth at that point (hence in a neighbourhood).
And, of course, as Laurent Moret-Bailly points out, the statement "as is" is incorrect. It should say that "if a surface contains a curve, which is locally defined by a single equation, and that curve is smooth, then the surface is smooth near the curve".
A: I would say that Wikipedia (as it is sometimes) wrong in this "intuition". I think that the geometric meaning of this localization theorem is more in the line of: if a variety is smooth at a point then it is smooth in a neighbourhood, or more generally, if a variety is smooth at a point, then any irreducible subvariety through that point is generically smooth (in particular the ambient variety) near that point. An argument to support this could be that localization of a local ring means something like going from a point to an irreducible subvariety going through that point. In other words, if that is a smooth (say closed) point, then the general point of any irreducible subvariety containing that point is also smooth, in other words, it is generically smooth. 
I guess an even better way to think about this is to consider a local ring which is localized not at a prime ideal, but at an element, so what's happening is exactly going from a point to an open set. So it is saying "exactly" that if a variety is smooth at a point, then it is smooth in an open set near that point.
A: Besides the geometric intuition thanks to Wikipedia wisdom, there is also a simple algebraic intuition: localization functor is exact! Notice that for a $R_{(I)}$-module $M$ you have $R_{(I)}\otimes_R M \cong M$, hence you pick a finite free resolution of the $R$-module $M$, and localize it. Bingo!
