Let $A = K[X,Y,Z]/(XZ-Y^2)$, $P = (\bar{X}, \bar{Y}) = (X,Y)/(XZ-Y^2) \subseteq A$ ideal. I know that $P^{2} = (\bar{X}) \cap (\bar{X},\bar{Y},\bar{Z})^{2}$ and I want to show that this is a primary decomposition (minimal) of $P^{2}$. It's clear that $(\bar{X},\bar{Y},\bar{Z})^{2}$ is primary because its radical is a maximal ideal, but I'm getting stuck with $(\bar{X}) = (X,Y^2)/(XZ-Y^2)$.
So, is $(\bar{X})$ primary?
I think this works...
It is equivalent to showing that in $R=(K[X,Y,Z]/(XZ-Y^2))/((X, Y^2)/(XZ-Y^2))\cong K[Y,Z]/(Y^2)$, the zero divisors are all nilpotent.
Let $a$ be a nonzero zero divisor in $R$, that is, there exists $a,b\in K[Y,Z]\setminus (Y^2)$ such that $ab\in(Y^2)$. Since $Y$ is a prime in the UFD $K[Y,Z]$, and $Y^2|ab$, we have that two copies of $Y$ must appear in the factorization of $ab$.
But they can't both appear in $a$ or both appear in $b$ because of the assumption that $a,b\notin (Y^2)$. So $a$ has at least one copy of $Y$ in its factorization, hence $a^2\in (Y^2)$, and the image of $a$ is nilpotent in $R$.