How can the height of a non zero prime ideal be $0$?

Question

In my exercise sheet I am supposed to prove that the only prime ideal of height $0$ in an integral domain domain is $(0)$, and to compute the prime ideals of height $0$ in $\mathbb R[x,y]/(xy)$. I seem to be missing something: we defined the height of a prime ideal $\mathfrak{p}\subset R$ as $$ \operatorname{ht}\mathfrak{p}:=\sup\{n\in\mathbb{N}_0\ |\ \exists\text{ prime ideals }\mathfrak{p}_0,...,\mathfrak{p}_n\subset R\text{ with }\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq...\subsetneq\mathfrak{p}_n=\mathfrak{p}\} $$ But then if $\mathfrak{p}$ is a nonzero prime ideal, we can set $\mathfrak{p}_0=(0)$ and $\mathfrak{p}_1=\mathfrak{p}$, hence $\operatorname{ht}\mathfrak{p}\geq 1$. So it really confuses me that I have the extra assumption of $R$ being an integral domain to prove this and that I am supposed to compute the prime ideals of height zero in a given ring, as to me they seem to be always $0$. Where am I mistaking?

Answer 1

If the ring is not an integral domain, then $(0)$ is not prime.