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After studying about the completion of a module $M$ over a ring $A$ (e.g. $I$-adic completion), I am left with the following questions:

(i) What is the usefulness of the concept of completion in commutative algebra or algebraic geometry, apart from the philosophical point of every Cauchy sequence being convergent?

(ii) What does it allow us to achieve on the technical level and what type of geometric insights does it lead to?

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There are several applications in number theory: look up "Hasse principle" for example – Geoff Robinson Mar 30 '13 at 16:20
A completion is a lot like the localization. For instance $\mathbb{Z}_p$ is obtained by completing the ring $\mathbb{Z}$. So we get a nice ring with nice topological properties that we can use to analyse a ring closer. For instance we can apply Hensel lifting to $\mathbb{Z}_p$ – Brent J Mar 30 '13 at 16:20
To add to @Brent’s comment, it often happens that the algebra in the completed object is simpler than in the original. This may help you with the properties of the original object. – Lubin Mar 30 '13 at 16:23
Maybe I should change my comment to an answer... – Brent J Mar 30 '13 at 16:45
@BrentJ: I am more interested in algebraic geometric examples. Lubin's comment is very interesting as well. – Manos Mar 30 '13 at 16:51
up vote 11 down vote accepted

Matt's answer gives some justification from a geometric perspective. However, I thought you might benefit from an algebraic point of view as well. From this perspective, the main point is something Matt mentioned in passing -- the Cohen structure theorem.

Suppose your domain of inquiry is Noetherian rings, and that you are interested in using homological tools. An arbitrary Noetherian ring can have very strange structural properties. However, Cohen's structure theorem says that if $R$ is a complete Noetherian local ring, then

  1. $R$ is isomorphic to a quotient (factor ring) of a power series ring in finitely many variables over a field (or over a complete DVR, in the mixed characteristic case), and

  2. If $R$ contains a field, then $R$ is module-finite over a regular local subring. Indeed, if $x_1, \ldots, x_d$ is any system of parameters for $R$, and $k$ is the coefficient field for $R$ (i.e. a subfield of $R$ that is isomorphic to its residue field, which always exists, again by Cohen's results), then the subring $A=k[[x_1, \ldots, x_d]]$ of $R$ is isomorphic to a power series ring in $d$ variables, and $R$ is finite as an $A$-module.

These two facts are incredibly useful if you happen to be working with a complete ring. But your question was about the utility of the completion process. The point here is that when $R$ is a local Noetherian ring, the morphism $R \rightarrow \hat R$ is faithfully flat, and for any finite $R$-module $M$ we have $\hat M \cong M \otimes_R \hat R$. This means that to find homological (and other) information about $R$, its modules, and its ideals, one can often pass to the completion, which (as outlined above) is typically a much easier ring to work with than $R$ itself.

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Thank you, that's a useful answer +1 – Manos Apr 1 '13 at 16:06
You're welcome. – neilme Apr 3 '13 at 18:17

If you haven't gone very far in algebraic geometry, then this example is probably off your radar. Suppose you have a "plane curve" in $\mathbb{P}^2$ given by $f(x,y)=0$. We can study it "analytically locally" (which is usually interpreted as locally in the analytic topology as opposed to Zariski topology).

Suppose we want to study a neighborhood of a closed point defined by $\frak{m}$. To get a Zariski neighborhood we just localize the ring at $\frak{m}$. But now we have a local ring and we could take the $\frak{m}$-adic completion. This "zooms in" to an analytic neighborhood of the point.

If the point is regular/smooth, then it will look "flat" since the ring will just be isomorphic to $k[[s,t]]$ (by Cohen structure theorem or probably something weaker). So analytic locally all smooth points of plane curves look the same (which we'd expect of course).

The interesting thing is with singularities. It would be odd to try to classify singularities Zariski locally. To see this draw (over $\mathbb{R}$) the curve $y^2=x^2$ which is just a pair of lines and draw $y^2=x^2-x^3$ the nodal curve. The singularity at $(0,0)$ "ought" to be the same because locally they are two lines intersecting even with the same slopes!!

By passing to the completion we remember this information. Try completing the rings at $(x,y)$ and doing a change of coordinates to show they are isomorphic. There is a vast literature out there on this topic so I won't go any further. The idea of classifying singularities analytic locally has had great success and the theory of just plane curve singularities is already interesting and should be approachable for you if you have a handle on completions.

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You are right, singularities are outside my radar for now, but i think i get the big picture +1 – Manos Apr 1 '13 at 16:08
:( You said in a comment "I am more interested in algebraic geometric examples." I give an algebraic geometry example and the other poster gives a purely algebraic answer and you accept the other one? – Matt Apr 1 '13 at 23:11
Dear @Matt, how can I see that the local ring of $k[x,y]/(y^2-x^2)$ at the origin is different from the local ring of $k[x,y]/(y^2-x^2+x^3)$ at the origin? – Ronald Bernard Apr 5 '13 at 6:54
Oh no. You are totally right. How embarrassing. I've said essentially the exact opposite of what is true. Of course, Zariski locally these two singularities are the same because after localizing $(1-x)$ is just a unit, so $y^2-x^2+x^3=u(y^2/u - x^2)$ so there is a clear change of coordinates $x\mapsto x$ and $z\mapsto y^2/u$ to get the isomorphism. – Matt Apr 5 '13 at 16:47
I guess I didn't actually claim this, but merely implied it in the post on re-reading. The point is that Zariski locally is really weak, so if we took lines intersecting transversally of different slope, then in the Zariski topology they would still be a the same singularity, but analytically locally you remember the "slope information" and you'll be able to distinguish them. – Matt Apr 5 '13 at 16:50

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