Maximal element of $(I : x)$, where $x$ is in $A - I$, is prime belonging to $I$ Given that $I$ is decomposable, I am supposed to prove that any maximal element $P$ of the set {$(I : x) | x \in A - I$} must belong to $I$, i.e., $P$ is prime and for every reduced primary decomposition $I = \cap_i Q_i$, there is some i so that $\sqrt{Q_i} = P$ when $P = (I: x)$ is maximal in this set. 
The trouble is that I'm very unfamiliar with this set, and don't even know how to show that a maximal element of it must be prime. About the most information I can come up with for $(I : x)$ is that it must contain $I$ and if $ab \in (I : x)$, then $ab = xy$ for some y in A, all of which is pretty obvious. Any hints to just get me started would be appreciated. Is there some special property that $x \in A - I$ has whenever $(I : x)$ is maximal in the given set?

To show that a maximal $(I:x)$ is prime: Choose $x$ such that $(I:x)$ is maximal.  If $(I:x)$ is not prime, then we can choose $a$ and $b$ such that $ab \in (I:x)$ but $a \not \in (I:x)$, $b \not \in (I:x)$.  Now what can you say about $(I:ax)$?
