Ideal maximal with respect to certain property is prime

Question

Let $R$ be a ring, and $x\in R$ is a non-nilpotent element of $R$. Let $P$ be an ideal of $R$ that is maximal with respect to the property that $x^n \notin P $, $\forall n$. Show that $P$ is prime.

Note that $P$ may not be a maximal ideal, only maximal with respect to this property. I know that a non-nilpotent element must note be contained in some prime ideal, since the nilradical is the intersection of the prime ideals. However, I am not really sure how to approach this problem.

Answer 1

Suppose $\;rs\in P\;$ , and nevertheless $\;r,s\notin P\;$ , then

$$P\lneq P+rR,\,\,P\lneq P+sR\implies\;\exists\,n,m\in\Bbb N\;\;s.t.\;\;x^n\in P+rR,\,\,x^m\in P+sR\implies$$

$$x^nx^m=x^{n+m}\in(P+rR)(P+sR)\le P\;\;\ldots\;\text{contradiction}$$

Answer 2

Let $ I $ be a nonzero prime ideal of $ R/P $. It corresponds to an ideal $ I_R $ of $ R $ properly containing $ P $, and by maximality of $ P $ with respect to the given property, it follows that $ x^n \in I_R $ for some $ n \in \mathbb N $, and thus $ x \in I $ in $ R/P $ by primality of $ I $. Thus, every nonzero prime ideal of $ R/P $ contains $ x $. If $ (0) $ were not a prime ideal of $ R/P $, $ x $ would be in every prime ideal, and thus would be nilpotent; which is impossible by the given condition. Thus, the ideal $ (0) $ is prime in $ R/P $, i.e $ R/P $ is a domain.

Answer 3

Consider the set $S=\{x^n:x\ge0\}$ and the ring of quotients $A=S^{-1}R$. The ideals in this ring are in bijective monotonic correspondence with the ideals of $R$ that do not intersect $S$ via extension and contraction.

Pick a maximal ideal of $A$; then its contraction is a prime ideal of $R$.