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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 commutative2012-11-09
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    @Amr Yes; and to be clear I've also written that $R$ should contain a unit element.2012-11-09
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    Why is the set of divisors of R a subring of R2012-11-09

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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 $t$.2012-11-09
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    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.2012-11-09
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    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.2012-11-09
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The set of divisors of R does not contain 0, thus it is not a subring of R

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    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.2012-11-09
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    Even if this is the case, one can it guarantee that {zero divisors} U {0} is closed under addition.2012-11-09
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    You cannot guarantee it but you can suppose it. As A Walker specifically stated.2012-11-09
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    I didnt undertand that he imposed this condition when I read his question. I will think about it again.2012-11-09
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    As a side comment, some authors allow 0 as a zero divisor.2012-11-09
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    @GiovanniDeGaetano You are correct; I was working with the notion that $0$ is a zero divisor (and a nilpotent element, but this second assertion is common).2012-11-09