# $x$ not nilpotent implies that there is a prime ideal not containing $x$.

Let $\mathscr{N}(R)$ denote the set of all nilpotent elements in a ring $R$.

I have actually done an exercise which states that if $x \in \mathscr{N}(R)$, then $x$ is contained in every prime ideal of $R$.

The converse of this statement is: if $x \notin \mathscr{N}(R)$, then there is a prime ideal which does not contain $x$.

But I am not able to prove it. Can anyone provide me a proof of this result?

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You've stated the converse wrong. You must mean "then there is a prime ideal which does not contain $x$". (That would be the "inverse", or the contrapositive of the converse.) Equivalently, you want to prove that if $x$ is contained in every prime ideal of $R$, then $x$ is nilpotent. –  Jonas Meyer Jan 19 '11 at 22:47
@Jonas: Thanks for the correction. –  anonymous Jan 20 '11 at 9:10

HINT $\$ If $x$ isn't nilpotent then localize at the monoid generated by $x$ to deduce that that some prime ideal doesn't contain $x$.
Let $f$ not be nilpotent. Consider the set $\Sigma$ of ideals $I$ such that if $n > 0$ then $f^n \notin I$. It is non-empty because $(0) \in \Sigma$. Order $\Sigma$ by inclusion and by Zorn's lemma choose a maximal element $P$. Show that $P$ is prime: let $x,y \notin P$. Then the ideals $P+(x)$ and $P+(y)$ strictly contain $P$, and so er not in $\Sigma$. Thus $f^m \in P+(x)$ and $f^n \in P+(y)$ for some $m,n$ (by def of $\Sigma$). It follows that $f^{m+n} \in P+(xy)$, hence $P+(xy) \notin \Sigma$, that is $xy \notin P$, so $P$ is prime. Hence we have a prime ideal $P$ not containing $f$.
Remark $\$ The maximal element constructed above is simply a maximal ideal in the localization $\rm\ S^{-1}\ R\$mentioned in my post. This contracts to a prime in $\rm\:R\:$ by basic properties of localization, viz. there is a one-one order-preserving correspondence between prime ideals in $\rm\ S^{-1} R\$ and those prime ideals in $\rm\:R\:$ disjoint from $\rm\:S\:.$ –  Bill Dubuque Jan 19 '11 at 23:07