I am going to get back shortly to discussing algebraic number theory and discrete valuation rings. But this tidbit from EGA 1 that I just learned today was too much fun to resist. Besides, it puts the material on completions in more context, so I think the digression is justified.

**Lifting Idempotents **

The theorem says we can lift “approximate idempotents” in complete rings to actual ones. In detail:

Theorem 1Let be a ring complete with respect to the -adic filtration. Then if is idempotent (i.e. ) then there is an idempotent such that reduces to .

More elegantly, if for a ring denotes the set of idempotents, we have surjective.

The proof I knew of earlier is a fairly straightforward application of Hensel’s lemma, that more general result which I plan to cover in the future. But there is a proof using (a very little bit of) algebraic geometry.

The first step is to prove:

Proposition 2Let be a ring and a nilpotent ideal. Then is surjective.

Indeed, the topological spaces of and are the same. The result then follows from the next section.

**Idempotents and Connectedness **

Idempotents measure the disconnectedness of for a ring :

Proposition 3If , then there is a one-to-one correspondence between and the open and closed subsets of .

Given an open and closed , we have . Since , we can define a global section of the structure sheaf by on , on .

Similarly, if is an idempotent, then must be either or for , because there are no nontrivial idempotents in a local ring. (If were a nontrivial idempotent in a local ring, then , and either or is necessarily invertible.) So we can set . It can be checked that is open, and so is by symmetry. This establishes the bijection I claimed. (Another approach here is to note that idempotents decompose as a product in the category of rings, which corresponds by fully faithful contravariantness the contravariant equivalence of to a coproduct in the category of affine schemes. Except one has to check that an open and closed subscheme of an affine scheme is itself affine. Perhaps this is easy, but a quick proof wasn’t obvious to me at least.)

So the first proposition is now proved. Finally, we need to prove the theorem. Well, choose an idempotent . Lift it to , and inductively to . In the inverse limit, we get an idempotent reducing to .

There are some useful applications of this in representation theory, because one can look for idempotents in endomorphism rings; these tell you whether a module can be decomposed as a direct sum into smaller parts. Except, of course, that endomorphism rings aren’t necessarily commutative and this proof breaks down.

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