If $A$ is Noetherian, and $x$ is a non-unit and non zero-divisor, then can $(xr)=(r)$ for any non-unit $r$? Problem
I am attempting to prove that if $A$ is a Noetherian commutative ring with unit, and if $x$ is a non-unit and not a zero divisor, then any minimal ideal of $(x)$ has height $1$.
Attempt
I have shown that if $Q\subseteq(x)$ is a prime ideal with $Q\ni q=xr_1\neq0$, then $x\in Q$ or $r_1\in Q$. If $x\in Q$, then $Q=(x)$ and so we consider later. Otherwise we get a chain of ideals
$$(r_1)\subseteq(r_2)\subseteq(r_3)\subseteq\cdots$$
of ideals with $q=x^kr_k$ for each $k$ by the non zero-divisor property. By the Noetherian property, the chain has $(r_k)=(r_{k+1})=\cdots$ eventually. Then, we have $(xr_{k+1})=(r_{k+1})$. I want to conclude that this means $x$ is a unit, a contradiction, or $r_{k+1}$ is a unit, thus giving $x^{k+1}\in Q$ and so $x\in Q$, but maybe that's the wrong conclusion.
I'm not sure where to go from here. What am I missing?
 A: The answer to your title question is no. Indeed, let $A=F[x,y]\big/\langle xy-y\rangle$, where $F$ is your favorite field, and let $a,b$ denote the images of $x$ and $y$ in $A$. We have $ab=b$, so of course $\langle ab\rangle=\langle b\rangle$, but I claim that $a$ is neither a unit nor a zero-divisor.
To see it is not a unit, note that $\langle x\rangle+\langle xy-y\rangle=\langle x,y\rangle$ is a proper ideal of $F[x,y]$, so that $\langle a\rangle$ is a proper ideal of $A$. To see it is not a zero-divisor, suppose otherwise; then $px\in\langle xy-y\rangle$ for some $p\notin\langle xy-y\rangle$. Say $px=q(xy-y)$ for a given $q\in F[x,y]$. Since $x$ does not divide $xy-y$, and $F[x,y]$ is a UFD, thus $x$ divides $q$; say $q=xq'$.  But now $xp=xq'(xy-y)$, so $p=q'(xy-y)$, contradicting that $p\notin\langle xy-y\rangle$.

Regarding your main question, we can prove it as follows. By Krull's principal ideal theorem, the height of any minimal prime over $a$ is at most $1$, so we need only show there is no minimal prime over $a$ of height $0$. If not, then $a\in P$ for some minimal prime $P$ of $A$, contradicting that $a$ is not a zero-divisor.
