Given a ring with unity and a central idempotent element e, prove some isomorphic relations Given a ring $R$ with  $1\neq  0$, and an element $e$ that is idempotent which is central in $R$, I want to prove that $R/Re \cong R(1-e)$, $R/R(1-e)\cong Re$, and subsequently, $R\cong Re\times R(1-e)$. My intuition is pointing me to the isomorphism theorems, and the last should follow from the Chinese remainder theorem, but I'm at a loss as to how to connect the dots.
 A: You can prove directly that $R\cong Re\times R(1-e)$. The natural map works:

Let $f=1-e$. Then $e+f=1$ and $ef=0=fe$. Consider $\phi: R\to Re\times R(1-e)$ given by $\phi(x)=(xe,xf)$. it is easy to see that $\phi$ is injective. To prove that it is surjective, prove that $\phi(ye+zf)=(ye,zf)$.

A: Your ideas are exactly right.
First note that since $e$ is central, $Re$ and $R(1-e)$ are two sided ideals. Since $1\in Re+R(1-e)$. 
To show that $R/Re \cong R(1-e)$, note that $\varphi(x)=x(1-e)$ is a homomorphism of rings (since $\varphi(xy)=x(1-e)y(1-e)=xy(1-e)$). Note that $Re \subset \ker \varphi$ since $\varphi(xe)=xe(1-e)=x(e-e)=x0=0$.
Then we want to use the first isomorphism theorem (since $\varphi$ is clearly surjective), so we just need $\ker\varphi= Re$. Suppose $x\in\ker\varphi$, then $x(1-e)=0$, or $x-xe=0$, but then $x=xe$, so $x\in Re$. Therefore $\ker\varphi=Re$ and by the first isomorphism theorem $R/Re\cong R(1-e)$. The other proof is identical.
Now we just need the last statement.
Note that we want the intersection of the ideals to be trivial to use the Chinese remainder theorem to get what we want. To see this, note that multiplying by $e$ is the identity on $Re$ since $e$ is idempotent, however $x(1-e)e=x(e-e)=x0=0$, so no element of $R(1-e)$ can be in $Re$ except 0. Thus the intersection is trivial. We also have $Re+R(1-e)=R$, so by the Chinese remainder theorem $R\cong R/(0)=R/(I\cap J) \cong R/Re \times R/R(1-e)\cong R(1-e)\times Re$.
