Height of a principal ideal generated by a zero divisor.

Question

Let $A$ be a f.g. algebra over a field $k$. Let $x$ be a zero divisor of $A$. Does it follow that $\mathrm{ht}(x)=0$?

I know that by Hauptidealsatz ht(x)$\leq$1.

Answer 1

No. Let $R=k[x,y]/(x^2,xy)$ and let $I=(y)$. Then $I$ has height $1$ since $I^2=(x,y)^2$, but clearly $y$ is a zero divisor in $R$.