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I want to show that having only the zero as a nilpotent element is a local property for a Ring $R$.

Assume $R$ only has the zero element as a nilpotent element and there exists a prime ideal $p$ such that $R_p$ has a nonzero nilpotent element. Call it $t$. Then $t^n=0$ for same $n \in \mathbb{N} $. Thus there exist an $s \in R\setminus p$ such that $R \ni st^n=0$. So we conclude that $(st)^n = 0$ in $R$. It follows that $st=0$ since $R$ has only the zero element as a nilpotent element. Now $st=0$ is equivalent to $t=0 \in R_p$. So that gives us a contradiction to the choice of $t$.

Now assume $R_p$ has no nonzero nilpotent elements for all $p$ prime. Assume there exists a nonzero nilpotent element in $R$. Call it $t$ and define the $R$-module $M := (t)$. Since $R_p$ has no nonzero nilpotent elements $M_p=0$ for all $p$ prime ideals. Since being zero is a local property we know that $M$ must be zero and therefore $t=0$.

Is this proof correct I am not sure here especially I am not confident that the following holds: Since $R_p$ has no nonzero nilpotent elements $M_p=0$ for all $p$ prime ideals.

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  • $\begingroup$ Allright but from my lecture I already know that the nilradical that contains all nilpotent elements is the intersection of all prime ideals. so with this knowledge the proof is correct? $\endgroup$ – CookieMaster Jul 14 '15 at 19:56
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    $\begingroup$ It's correct.${}$ $\endgroup$ – user26857 Jul 14 '15 at 20:19

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