Let *R* be a general ring, not necessarily commutative. An element *x*∈*R* is said to be **idempotent** if *x*^{2} = *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*) –*f*^{2}(*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*) =*f*^{2}(*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 {*e _{i}*} is said to be

**orthogonal**if

*e*= 0 for all

_{i}e_{j}*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 *x _{i}* is idempotent since and each since

*I*is a left module. Since

_{j}*R*is the direct sum of

*I*‘s we have for all

_{j}*j*≠

*i*and for all

*i*. Hence the are orthogonal idempotents.

Conversely given {*e _{i}*}, let us define

*I*:=

_{i}*Re*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

_{i}*e*‘s are orthogonal. Similarly, for all

_{i}*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 *e _{i}*‘s summing to 1 and let Clearly with and this is the unique representation. ♦

Note that under the above correspondence,

*I*= 0 if and only if_{i}*e*=0;_{i}- 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 *I _{i}*. Such modules correspond to primitive idempotents.

Definition. A non-zero idempotent e is said to beprimitiveif 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,
*Rf*_{1}∩*Rf*_{2}= 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* = *S*_{3}, 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 *g*α = α*g* 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*])/π*Z*(*R*[*G*]) and

Next, we look at idempotents.

Definition. An idempotent e of ring R is said to becentralif 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 iscentrally primitiveif 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, ifwhere 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