It seems like "a prime ring has no nontrivial central idempotent" is a well known fact since the Book of Rowen and Goodearl and lots of other papers use this fact freely. However, I cannot find the proof of this fact.
Can anyone give me at least a hint to prove : a prime ring has no nontrivial central idempotent?
On
Suppose $\;R\;$ is a prime ring and $\;0,1\neq r\in R\;$ is central and idempotent, so for any $\;x\in R\;$ :
$$(1-r)rx=(r-r^2)x=(r-r)x=0\cdot x=0\implies\,1-r=0\;\;\text{or}\;\;x=0$$
In the first case we get $\;r=1\;$ and thus $\;r\;$ isn't trivial, and in the second we get $\;R=\{0\}\;$
On
A ring is prime iff the product of two nonzero ideals is nonzero.
If $e$ is a nontrivial central idempotent, then $A=eR$ and $B=(1-e)R$ are two ideals, and $AB=\{0\}$.
Actually in this case, $A\oplus B=R$ is a ring decomposition into two other nonzero rings, which is never a prime ring. This is sort of like being prime in the sense it can't be factorized into smaller rings.
Factorizations of a ring into smaller nonzero rings correspond to central idempotents. If a ring has no nontrivial central idempotents, you can think of it as "irreducible." Prime rings are always irreducible, but the converse is not always true. (This closely parallels the standard situation of prime and irreducible elements in integral domains.
A ring $R$ is prime if, from $axb=0$, for all $x\in R$, it follows $a=0$ or $b=0$.
If $e$ is a central idempotent, $ex(1-e)=e(1-e)x=0$, so, by definition, either $e=0$ or $1-e=0$.