I am self studying commutative algebra from a class notes based on atiyah and macdonald and I am struck on this proof.
Statement; Let $I=\cap_{i=1}^n q_i$ be a minimal primary decomposition of I. Then $\sqrt{q_i}=p_i's $ are precisely the prime ideals in $S=${$ \sqrt{I:x} : x\in A$}
Proof: If $\sqrt{I:x } =p$ is a prime ideal $=> \sqrt{\cap_{i=1}^n q_i :x} =p= \sqrt{ \cap_{i=1}^n (q_i:x)} =\cap_{i=1}^n \sqrt{ q_i:x}= \cap_{ x\notin q_i}p_i$.
I am not able to understand the last step how author wrote $\cap_{i=1}^n \sqrt{ q_i:x}= \cap_{ x\notin q_i}p_i$.
Can you please help me prove this assertion?