Prove that $C(a) $ ={$r \in R : ra = ar$} is a subring of $R$ containing $a$. For a fixed element $a \in R$ define $C(a)$ ={$r \in R : ra = ar$}. Prove that $C(a)$ is a subring of $R$ containing $a$.
attempt: Recall by definition ,  $B$ is a subring of $A$ if and only if $B$ is closed under subtraction and multiplication.
Then Suppose $r,c \in C(a)$. Then its closed under the subtraction (that is if and only if $C(a)$ is closed with respect to both addition and negatives.
Closed under +
$(r+c)a = ra + ca = ar + ac = a(r + c)$
Closed under 
Likewise under multiplication: 
$(rc)a = r(ca) = r(ca) = (ra)c = (ar)c = a(rc)$.
Then from the closure of addition we have $ra + ca = ar + ac$. So 
$ra = ar$ and $ca = ac$, thus $a \in C(a)$ for all $r\in R$, and $-c \in C(a)$. so $C(a)$ is a subring of $R$.
Can someone please verify this? Any help or better approach would be really appreciated it ! thanks.

Let $c\in C (a)$. Then $ac=ca$. Now we have $-ca=-(ca)=-(ac)=-ac=a (-c)$. This shows that $-c\in C (a)$, and the above argument completes the proof.
