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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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    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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    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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    @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