prime and semiprime ring Can someone help me to solve these problems ?

1:Let $R$ be a prime ring. How can we prove that if $a \in R$ commutes with every element from a nonzero left ideal $L $ of $R$, then $a \in Z(R)$? 

Is this still true if $R$ is semiprime?
My other question is :

2:Is $0$ the only idempotent in a domain without unity?

Thanks for your attention.
 A: $1-)$ If $x\in R$, $r\in L$ then $xr\in L$ hence $a(xr)=(xr)a$. But, $ar=ra$; we get that $(ax-xa)r=0$, which is to say, $(ax-xa)L=0$ for all $x\in R$. Since $R$ is prime and $L\neq 0$, we conclude that $ax=xa$ for all $x\in R$, hence $a\in Z(R)$.  I thınk It is is false for semiprime ring.  We can think $M_2(\mathbb{Z_2}).$( I will add more detail  in my free time )$2-)$ IT is true any idempotent $a\neq 1$ in  domain is zero divisior.   You can see more detail in Herstein 1976. 
A: 1) 1ENIGMA1's idea can't be improved upon really. To rephrase it, one can check that $L$ annihilates all commutators $ax-xa$ for $x\in R$. So the ideal generated by these commutators, call it $A$, satisfies $AL=\{0\}$ for a nonzero ideal $L$. In a prime ring, this implies $A=0$, and consequently that $x$ commutes with everything.
It's clearly false for a semiprime ring. Let $R$ be any noncommutative prime ring, and $S$ be any prime ring. Then $R\times S$ is a semiprime ring that isn't commutative. Select an element $r\in R$ that isn't in the center of $R$. Then clearly $(r,0)$ isn't in the center of $R\times S$, but $(r,0)$ commutes with every element in the ideal $\{0\}\times S$.
2) As shown here, a nonzero idempotent in a ring without nonzero zero divisors must be the identity. So yes, the only idempotent in a domain (commutative or not) without identity, is $0$.
