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Lifting idempotents à la Grothendieck August 29, 2009

Posted by Akhil Mathew in algebra, algebraic geometry, commutative algebra.
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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.

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