The center of a ring is {$z \in R | zr=rz$ for all $r \in R$}. Prove that the center of a ring is a subring that contains the identity. I started like this.\
Proof: Let $C =$ {$z \in R | zr=rz$ for all $r \in R$}. Let $a ,b \in C$ then $ar-br = ra-rb$ which implies that $(a-b)r = r(a-b) \in C$.
This is where my difficulty is, $arbr = rarb$, how do I show that the product $ab$ is in the center? Also how does this help me show that the identity is in the subring?
Best Answer
First, the identity is clearly in the center since part of the definition of an identity element is that it commutes with every other element.
Secondly, if $a,b \in C$ then your task is to show that $ab \in C$. Note that for any $r \in R$:
$rab=(ra)b=(ar)b=a(rb)=a(br)=abr$. Does this help?