Proving identity In a ring $R$ with identity$,$if every idempotent is central$,$then prove that $a$$b$ $=$ $1$$,$$($$a$$,$$b$$\in$$R$$)$$,$implies that $b$$a$ $=$ $1$.please help .I am stucked.
 A: Step 1: $ba$ is idempotent.
Step 2: $1=abab=baab=ba$.
QED
A: $(ba)^2=b(ab)a=ba$, so $ba$ is idempotent, so is central.
Thus $$\forall x\in R, bax=xba$$
If $x=a$ we have $ba^2=aba\Rightarrow ba^2=a$ because $ab=1$.
If $x=b$ we have $bab=b^2a\Rightarrow b=b^2a$ because $ab=1$.
Combining thes two equalities we get :
$$ab=(ba^2)(b^2a)=ba(ab)ba=b(ab)a=ba$$
Since $ab=1$, we have $ba=1$.
A: Haven't done a lot of ring theory lately, but lmk if this works:
Your ring $R$ has all idempotent elements also central.
AND
you have $ab = id$.
take that and multiply $ab*ab = ab*id = id$
so now we know that ab is idempotent.
now let's use the fact that it is central
$ab*b^{-1} = b^{-1}*ab$ because $ab$ is idempotent and thus central, so $ab*x = x*ab \forall x \in R$
$ab*b^{-1} = a$ by $b*b^{-1} = id$
but also 
$ab*b^{-1}= b^{-1}$ by $ab = id$
we know now, then, that $a = b^{-1}$.
Thus, since there are unique multiplicative inverses in ring theory, $ab = ba = 1$ ($ab = b^{-1}*b$, and $ba = b*b^{-1}$, which by definition are $1$)
