# Prime ideal and nilpotent elements [closed]

If $\mathfrak p \subset R$ is a prime ideal, prove that for every nilpotent $r \in R$ it follows that $r \in \mathfrak p$.

The only hint that my tutor gave me was to use induction. Can someone explain what he means by this?

Thanks for the help!

## closed as off-topic by Carl Mummert, Xander Henderson, mrtaurho, ancientmathematician, Parcly TaxelMar 10 at 3:57

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• He probably means induction on the smallest power of $r$ that equals $0$. – Tobias Kildetoft Apr 29 '13 at 18:49

Hint $\$ Induction on $\rm\,n\,$ shows $\rm\ b = a_1\cdots\, a_n\in P\,$ prime $\rm\: \Rightarrow\:$ some $\rm\ a_i\in P.\:$

Yours is the special case $\rm\: b=0,\ \ a_i = a,\$ i.e. $\rm\: a^n\! = 0\in P.$

Ok, since $P$ is prime, if $ab \in P$, either $a \in P$, or $b \in P$, right?

Assume that $x$ is nilpotent, then there exists some positive integer $n$, such that $x^n = 0$, i.e $\underbrace{x.x.x....x}_{n \mbox{ times}} = x^n = 0 \in P$, what can you say about $x$?

You can also use Proof by Contradiction by noticing that if $a \notin P$, and $b \notin P$, then $ab \notin P$.

$r^n=0$ and $0\in P$, but $P$ is prime, hence contains at least one factor, if it contains the product.

You may prove the the nilradical$=\{x\in R:x^{n}=0,n\in\mathbb{N}\}=\cap P$ where $P$ is a prime ideal of $R$. That should complete the proff, hint use Zorns lemma.