“…Thus for n ≥ 2 in the intersection of (I + J) n as in Lemma 2.5 we cannot omit any component involving P 2 . We certainly cannot omit the minimal component P 1 + Q 1 , and we cannot omit the component for P 1 + Q 2 because n i=0 p i1 q n−i,1 ∩ n i=0 p i2 q n−i,1 ∩ n i=0 p i2 q n−i,2 = n i=0 (x 1 , x 2 ) 4i q n−i,1 ∩ n i=0 p i2 J n−i = n i=0 (x 1 , x 2 ) 4i q n−i,1 ∩ J n + (x 4 1 , x 3 1 x 2 , x 1 x 3 2 , x 4 2 , x 3 ) contains x 3 q n1 and is thus not a subset of (I + J) n . This proves that for all n ≥ 2, (I + J) n has four associated primes.…”