Prime ideal and nilpotent elements 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!
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. $\rm\: a^n\! = 0\in P.$
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.