Why is $(x,y)\cap(x,z)\cap(x,y,z)^2$ a minimal primary decomposition of $(x,y)(x,z)$?
I understand that the ideals are primary and also that one has $$(x,y)\cap(x,z)\cap(x,y,z)^2=(x,y)(x,z).$$ But I don't know what I have to show for the decomposition to be minimal.
Is it enough to show that omitting any ideal from the decomposition makes the equality a proper inclusion? Where do I have to use associated primes? My lecture notes are quite inconclusive this.
Let's work with the primary decomposition $I=(x,y)\cap(x,z)\cap(x,y,z)^2$ you gave and call $Q_1 = (x,y),Q_2 = (x,z), Q_3 = (x,y,z)^2$.
$Q_1,Q_2$ are primary because they are prime (and so rad($Q_1$)=$Q_1$ and similarly for $Q_2$). Now $Q_3$ is primary because its radical ideal is $(x,y,z)$ which is maximal.
To show this is minimal you need to show the radicals (say, $P_1,P_2,P_3$, respectively) of the three ideals $Q_1,Q_2,Q_3,$ respectively, are distinct, and as you said you must show that if you omit $Q_i$ then you get a strictly bigger ideal (and do this for each $i=1,2,3$).
Looking at the radicals we see they are all different.
Finally, we must show that we can't omit any $Q_i$'s. $z^2\in Q_2 \cap Q_3$, but $z^2 \notin Q_1$ so $z^2 \notin I$ (and $Q_2 \cap Q_3\nsubseteq Q_1$); $y^2\in Q_1 \cap Q_3$, but $y^2 \notin Q_2$ so $y^2 \notin I$; and $x\in Q_1 \cap Q_2$, but $x\notin Q_3$ so $x\notin I$. Hence, we cannot omit any $Q_i$, $i=1,2,3$ from the decomposition and it is thus minimal.