Decompose your ring using idempotents

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Decompose your ring using idempotents

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[QUICK DESCRIPTION]

If $e$ is an idempotent in a ring $R$ (not necessarily with identity), then we can decompose $R$ as the direct sum of four components (subrings), each of them related to $e$. Concretely,
[math]
R=eRe \oplus eR(1-e) \oplus (1-e)Re \oplus (1-e)R(1-e);
[/math]
note that here $(1-e)R$ is just ''notation'' that means $\{x-ex|x\in R\}$,
and similarly for $R (1-e)$.

This decomposition is known as the ''Peirce decomposition'' of $R$ with respect to $e$.

An important feature of this decomposition is that each of
$e R e$ and $(1-e) R (1-e)$ is a subring of $R$, and the former even
has an identity, namely, the element $e$[add]{{, which does indeed lie
in $e R e$, since $e = e e e$; also, $e (e r e) = e^2 r e = e r e $
for all $e r e \in e R e $.}}[/add]

If $R$ ''does'' contain an identity $1$, then $1 - e$ is defined,
and is again an idempotent, and one has $e + (1 - e) = 1$ (obviously),
and $e ( 1 - e) = (1-e) e = 0$ [add]{{since $e( 1 - e) = 
(1-e) e = e - e^2 = e - e = 0,$
as $e$ is idempotent.}}[/add]  One says that $e$ and $1-e$ form a pair
of [[w:idempotent | orthogonal idempotents]].
Furthermore, in this case, the subring $(1-e)R(1-e),$ as defined above,
coincides with the set of products  $\{(1-e)r(1-e) \, | \, r \in R\}$.
(A similar remark applies to each of $e R (1-e)$ and $(1-e) R e$.)

More generally, continuing to assume that $R$ contains an identity,
if $\{e_\alpha\}_{\alpha \in A}$ is a set of orthogonal idempotents for $R$
with the property that $1 = \sum_{\alpha \in A} e_{\alpha},$
then [math] R= \bigoplus_{\alpha,\beta \in A} e_{\alpha}Re_{\beta}. [/math]

[PREREQUISITES]
The basic idea of decomposing a ring via idempotents should be accessible to anyone knowing the basics of ring theory.  Several of the examples
below refer to contexts which require a more specialized degree of knowledge;
in such cases, this is indicated at the beginning of the example.

[EXAMPLE central]  If $e$ lies in the centre of $R$ (in particular,
if $R$ is commutative), then $eRe = eR$,
$(1-e)R(1-e) = (1-e)R,$ and $eR(1-e) = (1-e)R e = 0.$  Thus we obtain
the simpler decomposition $R = eR \times (1-e)R.$

[EXAMPLE Spec] This example requires
a basic knowledge of the costruction of the [[w:spectrum of a ring| spectrum]]
$\mathrm{Spec}\,R$ of a commutative ring $R$ with identity.

If $R$ is commutative with $1$ and $e \in R$ is an idempotent, then
$R = e R \times (1-e)R$, as noted in [ref Example #central].  As remarked above,
we may form the idempotent $1 - e \in R$, and so
in particular, both $e R$ and $(1-e)R$ are commutative rings with identity (namely
$e$ and $(1-e)$ respectively).
Thus $\mathrm{Spec}\, R,$ $\mathrm{Spec}\, e R$, and $\mathrm{Spec}\, (1-e)R$
are all defined, and the factorization $R = e R \times (1-e) R$ induces a 
decomposition
[maths]
\mathrm{Spec} \,R = \mathrm{Spec}\, e R \coprod \mathrm{Spec}\, (1-e)R
[/maths]
of $\mathrm{Spec} \,R$ into a union of two open subsets.  Conversely,
any such decomposition of $\mathrm{Spec} \,R$ arises from an idempotent $e \in R$
in this way.

In short, if we think of $R$ as being the ring of 
[[w: regular function | regular functions]] on $\mathrm{Spec} \,R$,
then  idempotents in $R$ serve precisely as the 
[[Turn sets into functions | indicator (or characteristic) functions]] of simultaneously 
open and closed
subsets of $\mathrm{Spec} \,R$.  (In particular, $1 \in R$ is the indicator
function of $\mathrm{Spec}\, R$ itself, while $0 \in R$ is the indicator function
of
the empty subset.)

Note that in the case that $R$ is a [[w:boolearn ring | Boolean ring]],
so that every element is idempotent (by definition), the space
$\mathrm{Spec} R$ is totally disconnected, and the above discussion 
specializes to the 
[[w:Stone's representation theorem for Boolean algebras | Stone representation theorem]].

[EXAMPLE]
This example requires some familiarity with
[[w:topological group | topological group theory]], including the notion of
[[w:Haar measure | Haar measure]]
 on a locally compact topological group, and various related notions.

Suppose that $G$ is a topological group which admits a
[[w:neighbourhodd basis | neighbourhood basis]]  of
the identity consisting of compact open subgroups.  (A basic example of such a group is $\mathrm{GL}_n(\Q_p)$, the general linear group of invertible $n\times n$-matrices over the field $\Q_p$ of [[w:p-adic number| $p$-adic numbers]].)
One consequence of this assumption is that $G$ is locally compact, and so we can choose a [[w:Haar measure | Haar measure]] $\mu$ on $G$.

Let $\mathcal H$ denote the $\C$-vector space of all compactly supported
locally constant $\C$-valued functions on $G$.  We can make $\mathcal H$ into
a $\C$-algebra by defining a convolution product on its elements as follows:
[maths]
(f_1 * f_2)(g) := \int_{G} f_1(g')f_2(g^{\prime -1} g) \, d \mu(g') ,
[/maths]
for any two functions $f_1, f_2 \in \mathcal H$.
The $\C$-algebra $\mathcal H$ is then referred to as the
[[w:Hecke algebra | Hecke algebra]]of the group $G$.

If $K$ is any compact open subgroup of $G$, then we may define an idempotent
$e_K \in \mathcal H$ via the formula $e_K := \dfrac{1}{\mu(K)} \chi_K,$
where $\chi_K$ is the
[[Turn sets into functions | indicator (or characteristic) function]] of $K$. [add]
{{Since $K$ is open, the function $\chi_K$ is
locally constant and compactly supported, and thus
lies in $\mathcal H$, and hence so does $e_K$.}} [/add]

In this situation, for any $f \in \mathcal H$, the convolution
$e_K *f$ is the function obtained by averaging $f$ on the
left via the action of the compact open subgroup $K$, and simililarly
$f* e_K$ is the is the function obtained by averaging $f$ via the action of
$K$ on the right.
In particular
$e_K * f = f$ if and only
if $f(k g) = f(g)$ for all $k \in K$ and $g \in G$, and simililarly,
$f * e_K = f$ if and only if $f(g k) = f(g)$ for all $k \in $K and $g \in G$.
Thus the subalgebra $e_K \mathcal H e_K$ of $\mathcal H$
consists of those elements of $\mathcal H$ that are bi-invariant under
the action of $K$, i.e. such that $f(k_1 g k_2) = f(g)$ for all
$k_1, k_2 \in K$ and $g \in G$.

Since any $f \in \mathcal H$ is compactly supported and locally constant,
by virtue of our assumption on $G$ we may find some sufficiently small
compact open subgroup $K$ such that $f$ is bi-$K$-invariant.  Thus
we find that
[maths]
\mathcal H = \bigcup_{K} e_K \mathcal H e_K,
[/maths]
where the union is indexed by the collection of all compact open subgroups of $G$.

[GENERAL DISCUSSION]

Decompose your ring using idempotents

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