Let $R$ be an arbitrary (commutative) ring. Let $I \subseteq R$ be an ideal such that $I$ has a minimal primary decomposition $$I =\bigcap_{i=1}^s Q_i$$ with $Q_i$ $P_i$-primary ideal, $i=1, \dots,s$.
We know that $\sqrt I = \bigcap_{P \in \mathrm{Min}(I)}P$. If $I$ is radical we can say: $$\bigcap_{i=1}^s Q_i = \bigcap_{i=1}^s P_i=\bigcap_{P \in \mathrm{Min}(I)}P.$$ Can we conclude that $Q_i$ must be equal to $P_i$ for all $i=1, \dots,s$?
I see that there is a similar question about embedded prime ideals, but it only explains that, in the same hypothesis, the $P_i$'s must be minimal primes. I understood this statement, but I can't understand why in this case holds the equality $Q_i =P_i$.
$Q_i$ contains the intersection of all associated primes. In particular, $Q_i$ contains an associated prime (why?), say $P_j$. From $P_j\subseteq Q_i$ we get $P_j\subseteq P_i$ (why?). Then $P_j=P_i$ (why?). We get $P_i\subseteq Q_i$, so $P_i=Q_i$ (why?).