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Question: $R$ a commutative ring with identity and $x\in R$ is non-nilpotent. Prove there exists a prime ideal $P$ of $R$ such that $x$ is not in $P$.

Thoughts: I was thinking about trying to play with a maximal ideal. Since, if $P$ is maximal then $R/P$ is a field, so $R/P$ is an integral domain and thus $P$ is prime. Now, since $x$ is a non-nilpotent element, we have that $x^n\neq 0$ for $n\geq 1$. So we can take the multiplicative set $\{x,x^2,\dots, x^n,\dots\}$ which doesn't contain $0$, and then I want to say that there must be a maximal ideal of $R$, not intersecting the above multiplicative set, but that would require showing that if, say, $M$ is maximal among non-principal ideals of $R$, then $M$ is a prime ideal, which I am not sure how to do. So, I was wondering if there was an alternative method of proving this, or, if not, is there a nice way to prove the "M maximal among non-principal ideals...." statement? Thank you!

User7238
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    See the proof of Proposition 1.8 (page 5) in Introduction to Commutative Algebra by Atiyah & Macdonald. – azif00 Oct 06 '21 at 01:19
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    are you assuming $R$ is commutative? You should say so; note that for a non-commutative ring, you can have maximal ideals $P$ such that $R/P$ is not a field. E.g., $2\times 2$ matrices over a field. I note that you don't mention commutativity anywhere, even in the tags. – Arturo Magidin Oct 06 '21 at 01:38
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    @ArturoMagidin edited. Thank you for pointing that out! – User7238 Oct 06 '21 at 02:24

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This follows from the fact that the nilradical of a commutative ring coincides with the intersection of all its prime ideals.

Thus if $x$ is in every prime ideal, then $x$ is nilpotent.

A proof is available on this wiki page. The basic idea is to use Zorn's lemma, more or less in the same way of showing the existence of maximal ideals.

As an alternative point of view, consider $R_x$, the localization of $R$ with respective to the multiplicative set $\{x^i: i \geq 0\}$.

Since $x$ is not nilpotent, we have $\frac 0 1 \neq \frac 1 1$ in $R_x$, i.e. $R_x$ is not the zero ring. Thus it has a maximal ideal $m_x$, i.e. there exists a homomorphism $\pi_x: R_x \rightarrow k$ for some field $k$.

Composing $\pi_x$ with the natural homomorphism $\iota: R \rightarrow R_x$, we get a homomorphism $\pi: R \rightarrow k$.

The kernel of $\pi$ does not contain $x$, as $\iota(x)$ is a unit in $R_x$ and cannot be mapped to $0$ via $\pi_x$.

The image of $\pi$, being a subring of $k$, is an integral domain.

Therefore the kernel of $\pi$ is a prime ideal of $R$ not containing $x$.

WhatsUp
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    I wouldn't say follows: it is precisely that statement, more or less! – Pedro Oct 06 '21 at 02:33
  • @PedroTamaroff The other direction (i.e. if $x$ is nilpotent, then it is in every prime ideal) is easy, but still needs a proof, right? – WhatsUp Oct 06 '21 at 02:34
  • Sure. ${}{}{}{}$ – Pedro Oct 06 '21 at 02:36
  • I do not think it is a good idea to reason with "This follows from the fact that the nilradical of a commutative ring coincides with the intersection of all its prime ideals.", because the question being asked is almost always used to prove the fact in the quoted sentence. So, if the quoted sentence is given as rationale, odds are good that circular reasoning will have occurred. (I hedge my statement this way because the probability of a proof of "the fact" avoiding the asked question seems low.) But everything that follows is just the ticket, reasoning with the multiplicative set. – rschwieb Oct 06 '21 at 13:37