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.
 A: 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 $ba=ab$.
A: 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.
