Noetherian ring and primary decomposition result I'm struggling with the following problem and I would appreciate some help if possible
Let $R$ be Noetherian and let $I,J$ be ideals. Define $(I:J^{\infty}) = \bigcup_{n}(I:J^{n})$.
(a) If $Q$ is primary, prove that $(Q:J^{\infty}) = Q$ for any $J \subset R$ with $J$ not contained in the radical of $Q$.
(b) If $P$ is prime and $I = Q_{1} \cap \ldots \cap Q_{k}$ is a finite intersection of primary ideals, then show that $(I : P^{\infty})$ is the intersections of the $Q_{i}$ for which $P \not \subset P_{i}$. 
Thank you!
 A: I think the basic observation key to a) is as user26857 says, the ideal version of the definition of a primary ideal

Let $I,J \subset R$ be finitely generated ideals, and $Q \subset R$ a primary ideal.  Suppose that $IJ \subseteq Q$.  Then either $I \subseteq Q$ or for some $m$ we have $J^m \subseteq Q$. 

The equivalence follows easily from the elementwise definition of primary ideals.  
Now let $Q$ be $P$-primary (i.e. $\sqrt{Q} = P$) with $J \nsubseteq P$. We want to show that $(Q:J^{\infty}) = Q$.
One direction is immediate: $(Q:J^{\infty}) = \bigcup_{n}(Q:J^{n}) \supseteq (Q : R) = Q$. So we just need $(Q : J^{\infty}) \subseteq Q$.  We have $q \in (Q : J^n) \implies q J^n \subseteq Q \implies q \in Q$ or $J^{mn} \subseteq Q$ (using the above observation and that $J$ is f.g. because $R$ assumed Noetherian).  In the second case, however, taking radicals gives $J \subseteq \sqrt{J} = \sqrt{J^{mn}} \subseteq P$, a contradiction.  Hence $q \in Q$, and (a) is done.
For (b), we suppose that $P$ is prime and $I = Q_{1} \cap \ldots \cap Q_{k}$ where $Q_i$ is $P_i$-primary.  We want to show that $(I : P^{\infty})$ is equal to the intersections of the $Q_i$ for which $P \nsubseteq P_i$. 
First note that $(I : P^{\infty}) = (Q_{1} \cap \ldots \cap Q_{k} : P^{\infty}) = (Q_1 : P^{\infty})\cap \dots \cap (Q_k : P^{\infty})$. 
Consider from part (a) that if $P \nsubseteq P_i$ then $(Q_i : P^{\infty}) = Q_i$. 
On the other hand if $P \subseteq P_i = \sqrt{Q_i}$ then for some $m$ we have $P^m \subseteq Q_i$ (using that $P$ is f.g.) so $R = (Q_i : P^m) \subseteq (Q_i : P^{\infty})$ and b) is done.  
Note that the Noetherian assumption in this problem is very shallow.  We just needed for the ideal $J$ in part a) to be f.g. and for the prime ideal $P$ in part b) to be f.g. (and of course b) is more natural to exist in settings which admit primary decompositions).  
