# Nilpotent elements lie in some prime ideal in a commutative ring with $1$

$R$ is a ring with $1$. We call $r\in R$ nilpotent if $\exists n\in \mathbb{N}$ such that $r^n=0$. Show every nilpotent element lies in some prime ideal.

The fact that $1\in R$ may not be needed here since this problem has multiple parts.

I know that "every nilpotent element lies in every prime ideal," but I don't assume $R$ has prime ideals. So I would like to construct one (possibly in terms of a given nilpotent element $r$). Suppose $r^n=0$ with $n$ minimal.

My attempts at constructing a prime containing $r$ have included $(r),$ the collection of nilpotent elements, the collection of zero divisors, the collection of all $s$ with $s^n=0$ and $n$ minimal. But I cannot show that either of these are prime ideals mostly due to the fact that $R$ may not be an integral domain. How can I construct such a prime ideal?

Let $S$ be the set of all proper ideals containing $r$ (which is non-empty obviously). With inclusion this set is partially ordered. Take any chain of ideals $r\in \mathfrak{a}_1\subset \mathfrak{a}_2\subset \cdots$. Then $\bigcup \mathfrak{a}_i$ is an ideal too, and it is in $S$. This is an upper bound for you chain. So the hypothesis of Zorn lemma holds. Hence $S$ has a maximal element. This maximal element is a maximal ideal containing $r$. Every maximal ideal is prime. Hence there exists a prime ideal containing $r$.