MHB How can we prove that the center of a ring is a subring?

[cbarker1]

Dear Everyone, I am stuck on a portion of the sub-ring criterion. The Problem states:
"The center of a ring $R$ is $\{z\in R| zr=rz \ \forall r\in R\}$. Prove that the center of the ring is a subring that contains the identity as well as the center of a division ring is a field."

I am doing the subring first, then the identity portion second. So here is my attempt:

Let $S$ be the center of the ring. We know that the $0\in S$ since $0\in R$ by the definition of a ring. So $S\ne\emptyset$.
Let $a,b\in S$. Then $ar=rb$. $r(a-b)=0$. Thus $a-b \in S$. Here is where I am stuck as well as the next step in the criterion.

What am I doing correctly or wrongly?

Thanks
Cbarker1

 

[Olinguito]

Hello Cbarker1.

Let $S$ be the center of the ring. We know that the $0\in S$ since $0\in R$ by the definition of a ring.

Nope. The reason we know $0\in S$ is because $0\cdot r=0=r\cdot $ for all $r\in R$, i.e. $0$ satisfies the condition for membership of $S$.

Similarly $1\in S$. Can you show this?

Let $a,b\in S$. Then $ar=rb$. $r(a-b)=0$. Thus $a-b \in S$. Here is where I am stuck as well as the next step in the criterion.

Concentrate on the given criterion! An element $x\in R$ is in $S$ iff $x\cdot r=r\cdot x$ for all $x\in R$. You want to show that $S$ is a subring of $R$. It was already established above that $0,1\in S$. It remains to show that $S$ is an additive subgroup of $R$ and that it is closed under multiplication.

Let $a,b\in S$. Then $ar=ra$ and $br=rb$ for all $r\in R$ (why?). We want to show that $a-b\in S$ and $ab\in S$. That is to say, we want to show that for all $r\in R$, $(a-b)r=r(a-b)$ and $(ab)r=r(ab)$. Can you do that? Hints: distributive law for the first one, associativity of multiplication for the second.

The last part of the question asks to show that if $R$ is a division ring then $S$ is a subfield. After all the work done in the first part, the only thing left to do is showing that multiplication in $S$ is commutative. (Note that it need not be commutative in $R$ itself.) Again, let $a,b\in S$. We want to show that $ab=ba$. But this is trivial because $b\in S$ $\implies$ $b\in R$. Hence, given $a\in S$ and $b\in R$, what can you conclude?

 

[cbarker1]

I figured out the 1st idea (proving $1 \in S$) as well as the second idea. But I am having trouble with how we have this $ar=ra$ and $br=rb$ to get to this $(ab)r=r(ab)$.

Thanks
Cbarker1