Proving Division of Local Ring in Ring with Idempotents

[xixi]

Let $R$ be a ring . Suppose that $e$ and $f=1-e$ are two idempotent elements of $R$ and we have $R=eRe \oplus fRf$ (direct sum ) and $R$ doesn't have any non-trivial nilpotent element . Set $R_1=eRe$ and $R_2=fRf$ . If $R_1=\{0,e\}$ and $R_2$ is a local ring , then prove that $R_2$ is a division ring . (note that $e$ and $f$ are central idempotents and therefore $fRf=fR$ )

 

[fresh_42]

I.) $ef=e(1-e)=e-e^2=e-e=0$ and $fe=(1-e)e=e-e^2=e-e=0$

II.) With $R=\{\,0,e\,\}\oplus fRf$ we have for elements $p=frf$ by (I) that $ep=0=pe$ and for elements $p=e+frf$ we get $ep=e^2+efrf=e^2=e=(e+frf)e=pe$, hence $e \in R$ is central. But $fr=(1-e)r=r-er=r-re=r(1-e)=rf$ so $f\in R$ is also central. Thus $R=\{\,0,e\,\}\oplus fR$.

III.) If $f$ is no unit in $R_2$ then $1-f=e \in R_2$ is a unit by locality. Now $e\in R$ is also a unit, but $ef=0$. So $f=0$ and $e=1$.
If $f$ is a unit, then by $ef=0$ we get $e=0$ and $f=1$.
Thus the only possibilities are $(e,f)\in\{\,(1,0)\, , \,(0,1)\,\}$.

IV.) If $f=0$, then $R=R_1\oplus R_2=\{\,0,1\,\}\oplus \{\,0\,\} = \mathbb{Z}_2$ is a field, and $R_2=\{\,0\,\}$. Hence we have $f=1$ and $e=0$, i.e. $R=R_1\oplus R_2=\{\,0\,\}\oplus R_2=R_2$.
The nilradical of $R$ is zero, so the intersection of all prime ideals of $R$ is zero.
The Jacobson radical $J(R)$, the intersection of all maximal ideals of $R$, is zero if $R$ is a division ring. For a local ring $R$ we have that $R/J(R)$ is a division ring. So all comes down to show that $J(R)=\{\,0\,\}$.

However, as I see it, this needs additional information which we do not have, e.g. $R$ could be left-Artinian.