Let R be a general ring, not necessarily commutative. An element x∈R is said to be idempotent if x2 = x.
Note
An endomorphism f of an R-module M (i.e. ) is an idempotent if and only if f is a projection, i.e. M = ker(f) ⊕ im(f) and f : M→M projects onto im(f). Indeed ⇐ is obvious, and conversely if f is idempotent, we have:
- Every m∈M is just f(m) + (m – f(m)). The first term is in im(f); the second term lies in ker(f) since f(m – f(m)) = f(m) – f2(m) = f(m) – f(m) = 0. So M = ker(f) + im(f).
- Any element of ker(f) ∩ im(f) can be written as f(m) such that f(f(m)) = 0. But this means f(m) = f2(m) = 0, so ker(f) ∩ im(f) = 0.
Throughout this article, we shall focus on idempotents which commute, i.e. ef = fe. A set of idempotents {ei} is said to be orthogonal if eiej = 0 for all i≠j. The following are easy to prove.
- The sum of two orthogonal idempotents is also an idempotent.
- If e is any idempotent, then e and 1-e are orthogonal idempotents.
The key result is the following.
Theorem. Let R be any ring. There is a 1-1 correspondence between the following:
- a decomposition
as a direct sum of left ideals, and
- orthogonal idempotents
such that
Proof
The correspondence is given as follows: for any , write
with
Then each xi is idempotent since
and each
since Ij is a left module. Since R is the direct sum of Ij‘s we have
for all j≠i and
for all i. Hence the
are orthogonal idempotents.
Conversely given {ei}, let us define Ii := Rei which is a left ideal. Note that since 1 lies in the RHS. On the other hand, an element of
can be written as
Then
since the ei‘s are orthogonal. Similarly,
for all i and thus
It remains to show that the two constructions are mutually inverse.
- Start with
and write
with
We need to show
- Since
we have
- Conversely, if
write
where we have
from what we just proved. Since
we see that
for all j≠i and so
Finally, start with orthogonal ei‘s summing to 1 and let Clearly
with
and this is the unique representation. ♦
Note that under the above correspondence,
- Ii = 0 if and only if ei =0;
- any idempotent e∈R gives orthogonal {e, 1-e}, and so R = Re ⊕ R(1-e).
Indecomposable Left Ideals
Now suppose R is an artinian ring (and hence noetherian by the Hopkins–Levitzki theorem). The Krull-Schmidt theorem says that R is a direct sum of indecomposable projective modules Ii. Such modules correspond to primitive idempotents.
Definition. A non-zero idempotent e is said to be primitive if it cannot be written as a sum of non-zero orthogonal idempotents
Proposition. If e is an idempotent, then Re is indecomposable if and only if e is primitive.
Proof
Indeed, if then
is the direct sum of two non-zero left modules.
- First, Rf1 ∩ Rf2 = 0: indeed, if
then
so we do have a direct sum on the RHS.
- Clearly
- Finally for
in the RHS, we have
Conversely, if Re = I ⊕ J, express e = x+y and we have a sum of two orthogonal idempotents (proof left to the reader). ♦
In writing R as a direct sum of indecomposable projective modules, the terms are unique up to isomorphism and permutation, but this does not mean the corresponding idempotents are unique. Specifically, we can have
where
as R-modules but they are distinct left ideals of R.
Examples
Suppose R is the ring of 2×2 upper-triangular matrices with real entries. We have the following decomposition:
Correspondingly, we have the (orthogonal) idempotents:
For an example in group algebras, let K be a field whose characteristic is not 2 or 3. Then is semisimple and we have the following orthogonal idempotents:
Central Idempotents
Recall that the centre of a ring, denoted Z(R), is the set of all x∈R, such that xy = yx for all y∈R. This is a commutative subring of R. The centre of the group algebra is easy to describe.
Lemma. If R is commutative, then Z(R[G]) is the free R-module with basis:
where C ⊆ G is a conjugancy class.
For example if G = S3, then Z(R[G]) is the free R-module with basis e, (1,2)+(2,3)+(1,2) and (1,2,3)+(1,3,2).
Proof
Let where
; this lies in the centre iff gα = αg for all g∈G. Multiplying gives
It follows that α lies in Z(R[G]) iff for all g, x in G, or equivalently for all g, x,
♦
In particular, this means if R is a complete discrete valuation ring with uniformizer π, residue field k and field of fractions K, then Z(k[G]) = Z(R[G])/πZ(R[G]) and
Next, we look at idempotents.
Definition. An idempotent e of ring R is said to be central if it lies in Z(R).
Correspondingly we have the following
Theorem. There is a bijection between:
- an isomorphism
as a product of rings;
- a decomposition
as a direct sum of (two-sided) ideals;
- an expression
as a sum of orthogonal central idempotents.
Proof
First we prove the correspondence between the second and third sets.
By the earlier correspondence gives us
where
is a collection of n orthogonal idempotents. Let us show that
commutes with all x in R: indeed,
since
is a two-sided ideal. Now
and
and since
is a direct sum, matching components gives us
Conversely, suppose where
are orthogonal central idempotents. The prior correspondence gives
which is a two-sided ideal since
commutes with everything.
The correspondence between the first and second collections is left as an exercise. ♦
As before, let us consider the case where the decomposition is maximal.
Definition. Let e be a non-zero central idempotent (in Z(R)). We say e is centrally primitive if we cannot write e as a sum of two non-zero orthogonal central idempotents.
Note
If R is artinian, then we can write as a finite direct sum of ideals, where each
cannot be decomposed further as a direct sum of two non-zero ideals. This corresponds to writing
as a sum of orthogonal central idempotents which are centrally primitive.
A central idempotent e can be centrally primitive without being primitive, i.e. e can be written as a sum of two non-zero orthogonal idempotents, but neither of these is central. We will see an explicit example later.
Unlike the case of general idempotents, we have:
Proposition. For any ring R, if
where each of
and
is a set of centrally primitive central idempotents which are orthogonal, then r=s and there is a permutation σ of {1,…,r} such that
for all i.
Proof
First note that are distinct: indeed if e is orthogonal to itself, then
Same goes for
Next, consider Since
is centrally primitive and
are orthogonal central idempotent:
it follows that we must have for some unique i and all remaining terms are zero. Likewise, for this i, we have
for some unique k. So we have
and since
we must have j=k and so
Since
are distinct, as are
the result follows. ♦
Example
Let us find all central idempotents of Note that its centre is spanned by 1, a := (1,2)+(1,3)+(2,3) and b := (1,2,3)+(1,3,2). These satisfy
Now we can write:
which is a sum of orthogonal central idempotents, which gives an isomorphism of rings. Since this decomposition is clearly maximal, the above three idempotents are all centrally primitive. Note that the first term is centrally primitive but not primitive since