# Proving that a subset of a ring $R$ is a subring

In this example, $R$ is a ring with unity $1$, with $a\in R$ having the property $a^2=a$ (making it a Boolean ring). I know every Boolean ring is of characteristic 2 since: $a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a \implies a+a=0$

The subset is defined as $aRa\subseteq R$ by $aRa=${$ara | r\in R$}.

How would I go about proving, or disproving, that $aRa$ is a subring of $R$ given the subset?

Would $aRa$ contain the same unity element $1$

Excuse the lengthy question, rings are proving to be a particularly pertinent frustration for me in Abstract Algebra.

• Well, you switched between $a$ and $e$, but the question is clear. Another fact that you might find to be relevant here is that boolean rings are actually always commutative. – Dustan Levenstein Dec 17 '15 at 13:02
• Thanks for pointing that out, I changed them to $a$. For some reason a generic identity element $e$ kept creeping into my mind. – Argon14 Dec 17 '15 at 13:09
• @Argon14 Let me know if I need to add anything to my answer for you. Don't let the terminology of whether or not the two rings share identity get in the way of proving that $eRe$ is a ring. – rschwieb Dec 17 '15 at 18:30

Boolean rings are commutative (see below for a proof) so that $ara=a^2r=ar$. So $aRa$ will only contain the identity if $a$ is a unit.

Also $a(1-a)=a-a^2=a-a=0$ showing that $a\neq1$ implies that $a$ is not a unit.

Final conclusion $aRa$ is a subring of $R$ if and only if $a=1$.

proof of commutativity:

$a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$ so that $ab+ba=0=ab+ab$. So $ba=ab$.

## Don't miss out

One can adopt the requirement that a subring shares the same identity as the containing ring or not. Considering that this fact doesn't rely on $R$ being boolean, or on $R$ even having identity, it seems like a better choice to work with rings that may not have identity, and call the rings contained inside them 'subrings.'

Yes, you heard right: for any idempotent element $e$ in any ring $R$, the subset $eRe$ is a ring with identity $e$ (whether you call that a subring or not.)

To prove it's a ring, the easiest way is to follow the subring test. Just convince yourself why $ere-ese\in eRe$ , and why $(ere)(ese)\in eRe$ for any $r,s\in R$, and why $(ere)e=e(ere)=ere$.

For a Boolean ring, every element is idempotent, so you get a lot of such rings in $R$.

## Extra

This type of ring is sometimes called a corner ring for that idempotent. The reason is that $\hom(eR_R,fR_R)\cong eRf$, and then this leads to an isomorphism $R\cong\begin{bmatrix}eRe&eR(1-e)\\(1-e)Re&(1-e)R(1-e)\end{bmatrix}$. I think this is the reason for the name.