In a comment to this answer it is mentioned that all the associated primes of $(XY,(X-Y)Z)$ are minimal using some Cohen-Macaulay ring.
Since I don't know anything about Cohen-Macaulay rings, is there any other way to show that all the associated primes are minimal?
Clearly $V(XY,(X-Y)Z)=V(X,Y) \cup V(X,Z) \cup(Y,Z)$ So we know that the minimal primes are $\{(X,Y),(X,Z),(Y,Z)\}$. But how to show that there is no embedded prime?
Added: Is this true that all the embedded primes of $I$ occurs in the decomposition (not the irreducible decomposition, of course, in the irreducible decomposition only minimal primes occurs) of the variety $V(I)$? For example, if $I=(x^2,xy) \subset k[x,y]$ then $V(I)= V(x) \cup V(x,y)$, so here the embedded prime $(x,y)$ also occurs.