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.

  • 1
    $\begingroup$ $B$ 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
  • 2
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.