TITLE: Let $R$ be a ring and $H$ be $Z(R)$'s abelian group. If $R/H$ is cyclic, then why can we say $R$ is commutative ring?
QUESTION [0 upvotes]: Let $R$ be a ring and $H$ be $Z(R)$'s underlying abelian group($Z(R)$ is center of $R$ as a ring).
If $R/H$ is cyclic, then why can we say $R$ is commutative ring ?
My try
Let $R/H$ is generated by $x$.
Let $X$ be lift of $x$ to $R$. I want to know why I can write $R＝Z(R)[x]$.
If I could prove this equation, I gain the result.

REPLY [1 votes]: Let $x$ be a cyclic generator of the abelian group $R/Z(R)$, and $X$ be a lift of $x$ to $R$. Just by considering the cosets of $Z(R)$ in $R$, we find that if $r\in R$ then $r = nX + z$ for some $n\in\mathbb{Z}$ and $z\in Z(R)$. You can check that if $r = nX+z$ and $r' = n'X+z'$ with $n,n'\in\mathbb{Z}$ and $z,z'\in Z(R)$, then $rr' = r'r$, and this proves that $R$ is commutative.
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