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R is a ring. $W=\left\{ x\in R|\text{x is a nilpotent}\right\}$ and $P\left( R\right) =\cap \{ P|P\unlhd A, \text{P is prime}\}$ .How to proof $W\subseteq P\left( R\right)$

FMath
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  • By exponential zero, do you mean nilpotent? – Tobias Kildetoft May 19 '17 at 19:09
  • What is $A$ in $P(R)$? – Wuestenfux May 19 '17 at 19:15
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    Your last three questions have been poorly asked, and moreover they were poorly asked duplicates. Please make efforts to stop doing that. Search for answers before you ask, and when you ask, put some effort into the answers. See the FAQ for tips on writing better questions. – rschwieb May 19 '17 at 20:13

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Suppose $x\in W$, so that $x^n =0$ for some $n\geq 1$. Let $m$ be the least natural number such that $x^m \in P$ (we know that at least one such power exists, namely $n$, as $x^n = 0 \in P$). If $m>1$, then $xx^{m-1}\in P$. Since $P$ is prime, either $x\in P$ or $x^{m-1}\in P$. Either way, this contradicts the minimality of $m$, and so it must be the case that $m=1$, and so $x\in P$. This works for all prime ideals $P$, and so $x\in P(R)$.

Therefore, $W\subseteq P(R)$.

florence
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