If $x$ is not nilpotent, how to prove there exists a prime ideal does't contain $x$

Question

Actually, I find some explanation using Zorn's lemma and localization. However, our class doesn't include these until now. So can someone prove it in an easier way?

Answer 1

I don't think you can avoid Zorn's lemma (or some weaker assumption thereof), but you can avoid localization.

Suppose $x$ is not nilpotent and consider $S=\{x^n:n\ge0\}$. Consider the family of all ideals $I$ such that $I\cap S=\emptyset$. Such a family is not empty because it contains $\{0\}$.

Also, the union of a chain of ideals in this family is again an ideal in this family. By Zorn's lemma, we can choose a maximal member $P$. Let's prove $P$ is prime.

Suppose $ab\in P$, but neither $a$ nor $b$ belong to $P$. Then $aR+P$ and $bR+P$ are ideals properly containing $P$ and, by maximality, they intersect $S$. Suppose $x^m\in aR+P$ and $x^n\in bR+P$. Then $x^m=ar+u$ and $x^n=bs+v$, with $r,s\in R$ and $u,v\in P$. Now $$ x^{m+n}=abrs+(bsu+arv+uv)\in P $$ which is a contradiction.