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!

Solution 1:

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. $\rm\: a^n\! = 0\in P.$

Solution 2:

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 $ab \notin P$.

Solution 3:

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