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Let $R$ be a ring. The center of $R$ is the set $\{x\in \!\,R\mid ax=xa \text{ for all }a \in R\}$. Prove that the center of a ring is a sub-ring.

I am not sure how to start off this problem.

Michael Hardy
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KGTW
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    you have to start this problem by just writing down all properties that a subset of a ring should satisfy to be a sub ring and check if your subset satisfy that... –  Dec 04 '13 at 16:39

2 Answers2

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A subset is a subring if and only if it's nonempty and closed under addition, subtraction, and multiplication, and in some cases one might write the definition so as to require it to contain the unit element $1$.

Does $1$ belong to this set? I.e. is it true that $a1=1a$ for all $a\in R$?

If $ax=xa$ and $ay=ya$ for all $a\in R$, then is it true that $a(x\pm y)=(x\pm y)a$ for all $a\in R$?

If $ax=xa$ and $ay=ya$ for all $a\in R$, then is it true that $a(xy)=(xy)a$ for all $a\in R$?

darij grinberg
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Michael Hardy
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In my book . There is a subring Test to check the properties as Michael Hardy stated above to see if a subset $S$ of a ring $R$ is a subring of $R$ if $S$ is itself a ring with the operations of $R$. So we can say let $a$ and $b$ belong to the center. right? because we need to know if the center of $R$ is a subring.

According to the Subring Test " A nonempty subet $s$ of a ring $R$ is a subring if $S$ is closed under subration and multiplication - that is, if $a-b$ and $ab$ are in $S$ whenever $a$ and $b$ are in $S$.

let $a$ and $b$ belong to the center for every $x\in R$. then we will get $ax=xa$ and $bx=xb$. Then $(a-b)x=ax=bx=xa=xb=x(a-b)$ and $(ab)x=a(bx)=a(xb)=(ax)b=x(ab)$. So this satisfies the subring test. We can conclude that the center of the ring $R$ is a subring which the ring $R$.

KGTW
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