Zerodivisors in a ring whose center is an integral domain Suppose that $R$ is a ring whose center $Z(R)$ is an integral domain. Is it still possible for $R$ to have a central zerodivisor? That is can there exist nonzero elements $c \in Z(R)$ and $r\in R$, such that $cr=0$?
Simple rings can not give rise to such examples since the center is then a field. But prime rings are good candidates, because then the center is an integral domain. I believe that an example should exist but I am not able to find one.
 A: The ring $\Bbb{Z}\langle x,y\rangle / (2x,2y)$ should work.
It will end up being the graded ring 
$$R=\Bbb{Z}\oplus \bigoplus_{n=1}^\infty \Bbb{F}_2^{2^n}, $$
with basis for the homogeneous component $R_i$ being the set of words of length $i$ in $x$ and $y$.
To show that the center is just $\Bbb{Z}$, observe that if $p\in R$, then for each word in $p$ which contains $y$, define the left $x$-degree of the word to be the number of $x$s preceding the first $y$ in the word. Consider the word $w$ (containing $y$) in $p$ with the highest left $x$-degree (as long as there are words containing $y$). Since $w$ contains a $y$, $wx$ has the same left $x$-degree, but $xw$ has left $x$-degree one higher than $w$. Thus the highest left $x$-degree of a word in $xp$ containing $y$ is higher than the highest left $x$-degree of a word in $px$. Thus if $p$ contains a $y$ somewhere, it is not in the center. By symmetry if $p$ contains an $x$, it is also not in the center. Thus the center can only consist of the constants, $\Bbb{Z}$.
Then $2$ is a central zero divisor.
Edit:
I find reuns's comment in many ways a clearer explanation than my own, so I'm reproducing it here, since comments are ephemeral.

So you meant $R=\left\{n+\sum_{j=1}^J w_{e_j}\right\}$, where the $w_i$ are the non-empty words of $x,y$ and the product is defined by concatenation $w_iw_l$ and $2w_i=0$. Then for $p\in R−\Bbb{Z}$ containing a word with some $y$ and $k$ larger than the maximal length of its words $x^kp\ne px^k$. So $Z(R)=\Bbb{Z}$. And $2x=0$.

