Does there always exist a prime ideal that contains $x$ but not $y$?

Question

Let $A$ be a commutative ring with unity. Say $x, y$ are two distinct elements of $A$, neither of which are nilpotent. I was wondering can we always find a prime ideal $P$ of $A$ such that $x \in P$ and $y \not \in P$? Thank you.

Answer 1

No; if $x$ and $y$ differ by a unit, then they generate the same ideal, and so any prime ideal containing $x$ contains $y$, and vice-versa.

Suppose, though, that all prime ideals containing $x$ contained $y$, and vice-versa. Then $y$ is contained in all prime ideals of $A/(x)$, since the prime ideals of $A/(x)$ are the prime ideals of $A$ containing $x$. Since the intersection of all prime ideals of a ring is its nilradical, we have that $y$ is contained in the nilradical of $A/(x)$, i.e. $y \in \sqrt{(x)}$, the ideal of $A$ which is the radical of the ideal generated by $x$. If we swap $x$ and $y$, we can say that that $\sqrt{(x)} = \sqrt{(y)}$.