Characteristic of a Quotient Ring

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 $Z =\mathrm{nil}(R)$.

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Is R commutative –  Amr Nov 9 '12 at 9:27
@Amr Yes; and to be clear I've also written that $R$ should contain a unit element. –  A Walker Nov 9 '12 at 9:31
Why is the set of divisors of R a subring of R –  Amr Nov 9 '12 at 9:44

No. Consider $R = \mathbb{Z}[t]/(m t)$ for any integer $m > 1$ together with the natural inclusion $\mathbb{Z} \hookrightarrow R$.

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Unfortuantely, it seems to me that the image of $m+t$ is not a zero divisor of $R$, so that $R$ does not satisfy the hypothesis that its set of zero divisors forms an ideal (even if you throw in $0$). It looks like this can be repaired by replacing $R$ by its quotient $R = \mathbb{Z}[t]/(mt,t^2)$; now the image of $(km+lt)t$ for $k,l\in\Bbb Z$ is zero, and the zero divisors are the ideal generated by the images of $m$ and of $t$. –  Marc van Leeuwen Nov 9 '12 at 11:59
For $R=\mathbb{Z}[t]/(4t,t^2)$, the image of $2$ is a zero divisor, as $2 \cdot 2t =0$. So here shouldn't $Z=(2,t)$? In general, one should expect $Z$ to have generators with divisors of your $m$. Of course, this too can create a problem: Consider $R=\mathbb{Z}[t]/(6t,t^2)$. Both $2$ and $3$ are zero divisors, but $1$ is not. It seems to me that your example should work exactly when $m$ is a prime power. –  A Walker Nov 9 '12 at 17:23
I see your point, Marc. Thanks for mentioning it. A Walker is also right that the ideal generated by $m$ and $t$ does not contain all zero divisors in general. I am very tired now, so I cannot tell if one can fix this example to make it work. –  Lennart Nov 9 '12 at 21:25
It looks to me that you can improve your answer. What about if $Z:= \{\text{zero divisors}\}\cup \{0\}$? What about the "equivalent" statement suggested by A Walker? Moreover you forgot a "zero" in your answer. –  Giovanni De Gaetano Nov 9 '12 at 9:52