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# central idempotent

Let $R$ be a ring. An element $e\in R$ is called a *central idempotent* if it is an idempotent and is in the center $Z(R)$ of $R$.

It is well-known that if $e\in R$ is an idempotent, then $eRe$ has the structure of a ring with unity, with $e$ being the unity. Thus, if $e$ is central, $eRe=eR=Re$ is a ring with unity $e$.

It is easy to see that the operation of ring multiplication preserves central idempotency: if $e,f$ are central idempotents, so is $ef$. In addition, if $R$ has a multiplicative identity $1$, then $f:=1-e$ is also a central idempotent. Furthermore, we may characterize central idempotency in a ring with $1$ as follows:

###### Proposition 1.

An idempotent $e$ in a ring $R$ with $1$ is central iff $eRf=fRe=0$, where $f=1-e$.

###### Proof.

If $e$ is central, then clearly $eRf=fRe=0$. Conversely, for any $r\in R$, we have $er=er-erf=er(1-f)=ere=(1-f)re=re-fre=re$. ∎

Another interesting fact about central idempotents in a ring with unity is the following:

###### Proposition 2.

The set $C$ of all central idempotents of a ring $R$ with $1$ has the structure of a Boolean ring.

###### Proof.

First, note that $0,1\in C$. Next, for $e,f\in C$, we define addition $\oplus$ and multiplication $\odot$ on $C$ as follows:

$e\oplus f:=e+f-ef\qquad\mbox{and}\qquad e\odot f:=ef.$ |

As discussed above, $\oplus$ and $\odot$ are well-defined (as $C$ is closed under these operations). In addition, for any $e,f,g\in C$, we have

1. $(C,1,\odot)$ is a commutative monoid, in which every element is an idempotent (with respect to $\odot$). This fact is clear.

2. $\oplus$ is commutative, since $C\subseteq Z(R)$.

3. $\oplus$ is associative:

$\displaystyle e\oplus(f\oplus g)$ $\displaystyle=$ $\displaystyle e+(f+g-fg)-e(f+g-fg)$ $\displaystyle=$ $\displaystyle e+f+g-ef-fg-eg+efg$ $\displaystyle=$ $\displaystyle(e+f-ef)+g-(e+f-ef)g$ $\displaystyle=$ $\displaystyle(e\oplus f)\oplus g.$ 4. $\odot$ distributes over $\oplus$: we only need to show left distributivity (since $\odot$ is commutative by $1$ above):

$\displaystyle e\odot(f\oplus g)$ $\displaystyle=$ $\displaystyle e(f+g-fg)=ef+eg-efg$ $\displaystyle=$ $\displaystyle ef+eg-eefg=ef+eg-efeg$ $\displaystyle=$ $\displaystyle ef\oplus eg=(e\odot f)\oplus(e\odot g).$

This shows that $(C,0,1,\oplus,\odot)$ is a Boolean ring. ∎

## Mathematics Subject Classification

16U99*no label found*20M99

*no label found*

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