Suppose that $R$ is a commutative ring (with unity) of characteristic $0$, and that the set $Z$ of zero divisors in $R$ forms an ideal. Does it follow that the characteristic of $R/Z$ is $0$?
Equivalently (with less verbiage), suppose that the ring homomorphism $\iota:\mathbb{Z} \to R$ injects. Does it follows that $\iota(m)$ is not a zero divisor, for all $m$?
This proposition holds when all zero divisors are nilpotent:
Example: Suppose that $Z = \mathrm{nil}(R)$, the nilradical of $R$. Then $Z$ is an ideal, and if $\iota(m) \in Z$, then $\iota(m^n)=0$. Here $\mathrm{char}(R)=0$ forces $m^n=0$, so $m=0$ and $\mathrm{char}(R/Z)=0$.
If this proposition does not hold under my given hypotheses, I would appreciate a counter-example or a new hypothesis on $Z$; something more restrictive than $Z$ forming an ideal and less restrictive than $Z =\mathrm{nil}(R)$.