Primary decomposition of $(XY,(X-Y)Z)$ in $k[X,Y,Z]$

Solution 1:

Let $I=(XY,(X-Y)Z)$. As you observed, $I$ has three minimal primes, one of them being $(X,Z)$. We have $I:(X,Z)=(XY,Y^2,(X-Y)Z)$. Using this we get $$I=(XY,Y^2,(X-Y)Z)\cap(X,Z).$$ Set $J=(XY,Y^2,(X-Y)Z)$. Note that $J$ has two minimal primes, namely $(Y,Z)$ and $(X,Y)$. We have $J:(Y,Z)=(X-Y,XY,Y^2)=(X-Y,Y^2)$, and therefore $J=(Y,Z)\cap(X-Y,Y^2)$. Finally we have got the primary decomposition, $$I=(Y,Z)\cap(X-Y,Y^2)\cap(X,Z).$$

Edit. Let $I$ be an ideal in a noetherian ring $R$ having a unique primary decomposition, $I=Q_1\cap\cdots\cap Q_m$. (This happens, for example, when all the associated primes of $I$ are minimal.) Set $P_i=\sqrt{Q_i}$. Suppose that $(I:P_1)$ has only $m-1$ associated primes. Then $I=P_1\cap (I:P_1)$.

We have $(I:P_1)=(Q_1:P_1)\cap(Q_2:P_2)\cap\cdots\cap(Q_m:P_m)$. Since $P_1\nsubseteq P_i$ for $i\ne 1$ we get $(Q_i:P_1)=Q_i$ for $i\ne 1$. Moreover, if $P_1\neq Q_1$, then $(Q_1:P_1)$ is $P_1$-primary (here I have used that $P_1$ is finitely generated, so some power of $P_1$ is lying in $Q_1$). But we assumed that $(I:P_1)$ has only $m-1$ associated primes, so $P_1=Q_1$. Thus we get $(I:P_1)=Q_2\cap\cdots\cap Q_m$ and hence $I=P_1\cap(I:P_1)$.