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.