I'm thinking about this problem:
Let $k$ be a field and $A:=k[x,y,z]∕(xyz-z^2)$.
$1)$ Show that $(\overline{x})$ and $(\overline{y})$ are primary and that $(\overline{z})$ is prime in $A$.
$2)$ Find a minimal primary decomposition of $(\overline{0})$ in $A$.
I already know how to solve $1)$, but I'm stuck with $2)$.
From the way the question is posed, it looks like the decomposition is supposed to be something like $(\overline{0})=(\overline{x})\cap(\overline{y})\cap(\overline{z})\cap...$
But my first thought was to try $(\overline{0})=(\overline{xy-z})\cap(\overline{z})$.
First, we have:
$\overline{f}\overline{g}\in(\overline{xy-z})\implies fg=s(xy-z)+t(xyz-z^2)\in(xy-z)\implies fg\in(xy-z)\overset{(*)}{\implies}$ $f\in(xy-z)$ or $g\in(xy-z)\implies \overline{f}\in(\overline{xy-z})$ or $\overline{g}\in(\overline{xy-z})$.
therefore $(\overline{xy-z})$ is prime in $A$ (where $^{(*)}$ is because $xy-z$ is irreducible in $k[x,y,z]$).
Second, if $\overline{p}\in(\overline{xy-z})\cap(\overline{z})$, then $p=q(xy-z)+r(xyz-z^2)$ and $p=rz+s(xyz-z^2)$, therefore
$$q(xy-z)=rz+s(xyz-z^2)-r(xyz-z^2)\in(z)$$ $$\Rightarrow q\in(z)\Rightarrow p\in(xyz-z^2)\Rightarrow\overline{p}=\overline{0}$$
So, indeed, $(\overline{0})=(\overline{xy-z})\cap(\overline{z})$, and I didn't need either $(\overline{x})$ or $(\overline{y})$.
Am I missing something?