# Ideal maximal with respect to certain property is prime

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.

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

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

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

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.

• I follow up until the line 'If (0) were not a prime ideal of R/P, x would be in every prime ideal, and thus would be nilpotent'. I'm not sure I understand why x would be in every prime ideal – David Warren Katz Dec 26 '16 at 17:45
• $I$ is chosen to be an arbitrary nonzero prime ideal of $R/P$, and then we show that $x \in I$. Thus, $x$ is in every nonzero prime ideal of $R/P$, and if the zero ideal is not prime, then it is in fact in every prime ideal. – Starfall Dec 26 '16 at 17:48
• This is an excellent answer. Loved it. +1 – DonAntonio Dec 26 '16 at 19:09

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 $R$.