Prime Ideal Squared (9.1.12)

Dummit and Foote Abstract Algebra, section 9.1, exercise 12:

Let the overbar denote passage from $\mathbb{Q}[x,y,z]$ into $\mathbb{Q}/(xy-z^2)$. Prove that $\overline{P}=(\overline{x},\overline{z})$ is a prime ideal. Show that $\overline{xy}∈\overline{P}^2$ but that no power of $\overline{y}$ lies in $\overline{P}^2$, showing that $\overline{P}$ is a prime ideal whose square is not a primary ideal.

Proof: It is easy to check that for a surjective homomorphism $φ$, we have $(φ(a_1),φ(a_2),...)=φ(a_1,a_2,...)$, so that $(\overline{x},\overline{z})=(x,z)/(xy-z^2)$. Evidently $(xy-z^2) \subseteq (x,z)$ so that $\overline{\mathbb{Q}[x,y,z]}/\overline{(x,z)}≅\mathbb{Q}[x,y,z]/(x,z)≅\mathbb{Q}[y]$ is an integral domain, so that $\overline{P}$ is a prime ideal. Furthermore, $\overline{xy-z^2}=\overline{0}$ so that $\overline{xy}=\overline{z}^2∈\overline{P}^2$ as claimed. Now, if $\overline{y}^n∈\overline{P}^2 \subseteq \overline{P}$ for some $n∈\mathbb{Z}^+$ then we have the natural image of $\overline{y}^n$ in the previously mentioned quotient isomorphic to $\mathbb{Q}[y]$ is zero, an impossibility by the commutativity of natural projections.$~\square$