$a^n = 0 \implies a \in P$ (where $P$ is a prime ideal)

Question

Is the above true? (I think it is!) if so, please can somebody explain why? I don't see it!

Answer 1

By definition, $P$ is a prime ideal when $xy\in P \implies x\in P$ or $y\in P$. By induction, $x_1 x_2 \cdots x_n \in P \implies x_i \in P$, for some $i$.

In particular, since $0 \in P$, if $a^n=0$, then $a^n \in P$, and so $a\in P$.

Answer 2

It can be proved by induction.

$a.a^{n-1}=a^n=0\in P$ implies that $a\in P\vee a^{n-1}\in P$ (characteristic for prime ideals)

hence (inductionstep) $a\in P$.

Note that the stronger $a^n\in P\Rightarrow a\in P$ is also true. It can be proved exactly the same way.

Answer 3

Perhaps you want to know the following fact (the ring $R$ is assumed to be commutative with identity). Let $$N= \{a \in R\mid a^n=0 \text{ for some integer } n\gt 0\}.$$ This set equals all the so-called nilpotent elements. One can show that actually $$N=\bigcap\{P \mid P \text{ is a prime ideal of }R\}.$$ To prove this: one inclusion ($\subseteq$) is done by the answers above. The reverse inclusion is less trivial, and usually uses the following fact: every ideal maximal with respect to being disjoint from a multiplicative subset of $R$ is a prime ideal of $R$. With this, one can show that for each non-nilpotent element, there is a prime ideal disjoint from the powers of that element, and hence the prime does not contain the element. This shows how non-nilpotents are excluded from the intersection, so that everything there is nilpotent.