Lifting idempotents à la Grothendieck

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 1 Let ${A}$ be a ring complete with respect to the ${I}$ -adic filtration. Then if ${\bar{e} \in A/I}$ is idempotent (i.e. ${\bar{e}^2=\bar{e}}$ ) then there is an idempotent ${ e \in A}$ such that ${e}$ reduces to ${\bar{e}}$ .

More elegantly, if ${Idem(R)}$ for a ring ${R}$ denotes the set of idempotents, we have ${Idem(A) \rightarrow Idem(A/I)}$ 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 2 Let ${A}$ be a ring and ${I}$ a nilpotent ideal. Then ${Idem(A) \rightarrow Idem(A/I)}$ is surjective.

Indeed, the topological spaces of ${\mathrm{Spec} \ A}$ and ${\mathrm{Spec} \ A/I}$ are the same. The result then follows from the next section.

Idempotents and Connectedness

Idempotents measure the disconnectedness of ${\mathrm{Spec} \ A}$ for a ring ${A}$ :

Proposition 3 If ${X = \mathrm{Spec} \ A}$ , then there is a one-to-one correspondence between ${Idem(A)}$ and the open and closed subsets of ${X}$ .

Given an open and closed ${Y \subset X}$ , we have ${X = Y \cup (X-Y)}$ . Since ${Y \cap (X-Y) = \emptyset}$ , we can define a global section ${e \in \Gamma(X, O_X)=A}$ of the structure sheaf by ${e :=1}$ on ${Y}$ , ${e := 0}$ on ${X-Y}$ .

Similarly, if ${e \in \Gamma(X, O_X)}$ is an idempotent, then ${e_x \in O_x}$ must be either ${0}$ or ${1}$ for ${x \in X}$ , because there are no nontrivial idempotents in a local ring. (If ${f}$ were a nontrivial idempotent in a local ring, then ${f(1-f)=0}$ , and either ${f}$ or ${1-f}$ is necessarily invertible.) So we can set ${Y := \{ x: e_x = 1 \ \mathrm{in} \ O_x \}}$ . It can be checked that ${Y}$ is open, and so is ${X-Y}$ by symmetry. This establishes the bijection I claimed. (Another approach here is to note that idempotents decompose ${A}$ as a product ${A_1 \times A_2}$ in the category of rings, which corresponds by fully faithful contravariantness the contravariant equivalence of ${ \mathrm{Spec}: \mathbf{Rings} \rightarrow \mathbf{Affine \ schemes}}$ 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 ${\bar{e} \in A/I}$ . Lift it to ${\bar{e}' \in A/I^2}$ , and inductively to ${\bar{e}^{(n)} \in A/I^{n+1}}$ . In the inverse limit, we get an idempotent ${e \in A}$ reducing to ${\bar{e}}$ .

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.

Climbing Mount Bourbaki