Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
It seems to me $\sum_i x^n/y^{n^2}$ is an element of $k((y))[[x]]$ that will not be in $k[[x]]((y))$. In the latter ring any element can be brought into $k[[x,y]]$ by multiplying by a finite power of $y$. See mathoverflow.net/questions/34010/… for related differences. –  solbap Mar 5 at 8:46
add comment

3 Answers

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.

share|improve this answer
    
Interesting. So the theorem is about localization by any set, not nec. by a prime ideal? –  James D. Taylor Oct 4 '10 at 0:41
3  
I guess I was a little sloppy on this one. If you don't localize at a prime then you may not get a local ring, so one needs to adjust the definition of "regular". I suppose one could say that a ring is regular if all of its localizations at primes are regular local rings (is there a notion like this?). In that case the theorem should be OK, because the primes of the localized (but not necessarily local) ring come from the original, so localizing it at a prime is the same as localizing the original at a prime. Now that we're here, there is another slight problem with the analogy. (cont.) –  Sándor Kovács Oct 4 '10 at 0:56
3  
(cont.) Since we're starting with a local ring, we don't really get a ring that corresponds to a neighbourhood, because we will never get any closed points back that are different from the original (say we start with a closed point). But I think this is OK, intuitively, since we cannot expect to say anything about any particular closed points explicitly, only about their collection generically. And those points we do have, they are the general points of the subvarieties through our original closed point. I guess we're back to the first paragraph and perhaps I should erase the second.... –  Sándor Kovács Oct 4 '10 at 1:00
    
I don't think you should. The stream-of-thought was quite nice. There certainly is a notion of a regular (not nec. local) ring with exactly the meaning you wrote. I'm actually quite pleased with the interpretation of the second paragraph. The intuition of switching from k[[x,y]] to k[[x]]((y)) being analogous to "something like going from a point to an irreducible subvariety going through that point" sits much less well with me. –  James D. Taylor Oct 4 '10 at 1:06
add comment

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).

share|improve this answer
1  
In fact, Wikipedia's statement is quite misleading. If you take a cone in 3-space, then every line through the vertex is a smooth curve although the surface is not smooth there. What is misleading is the phrase "near the curve". –  Laurent Moret-Bailly Oct 5 '10 at 13:24
add comment

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!

share|improve this answer
4  
This is not an intuition, but rather the proof using homological algebra (namely the homological characterization of regular rings). –  Martin Brandenburg Oct 4 '10 at 7:00
    
LOL, if the proof takes one line, the intuition must be somewhere in the bushes... –  Bugs Bunny Oct 5 '10 at 18:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.