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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.

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    $\begingroup$ $B$ should also be nonempty and I'm sure you've verified that $a\in C(a)$. Otherwise, this looks fine. $\endgroup$ – David Hill Oct 21 '15 at 22:15
  • $\begingroup$ This don't hold in a ring without $1$. $\endgroup$ – TokenToucan Oct 21 '15 at 22:17
  • $\begingroup$ so I also need to check it has the identity? $\endgroup$ – Mahidevran Oct 21 '15 at 22:19
  • $\begingroup$ I think many books let you assume that rings have $1$, which is important to your proof. A counterexample is $C(2)$ in the ring of even integers mod $12$ (you can check that $2\not\in C(2)$). $\endgroup$ – TokenToucan Oct 21 '15 at 22:24
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    $\begingroup$ @CuddlyCuttlefish $a\in C(a)$ holds in any ring, unital or not, because $a$ commutes with itself. $\endgroup$ – Arthur Oct 21 '15 at 22:28
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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.

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