Prime ideals corresponding to primary ideals in decompositon Let $R$ be a commutative ring with unity different from $0$, and $I$ be a decomposable ideal of $R$, i.e. 
$$I=\mathfrak{q}_1\cap \mathfrak{q}_2 \cap \cdots \cap \mathfrak{q}_r $$
where (1) $\mathfrak{q}_i$ are primary ideals; (2) $rad(\mathfrak{q}_i)$ are all distinct; (3) No $\mathfrak{q}_i$ contains intersection of other $\mathfrak{q}_j$'s. 
In this situation, it is well-known that 
(1) $\mathfrak{p}_i$ are uniquely determined (independent of whether $R$ is noetherian or not).
(2) $\mathfrak{p}_i$ are precisely the prime ideals appearing in the set $\{ rad(I:x) \,\, | \,\, x\in R\}$.
If $R$ is noetherian, we can say more (see Prop.2.10(iii), page 21 for proof): 
(2') $\mathfrak{p}_i$ are precisely the prime ideals appearing in the set $\{ (I:x) \,\, | \,\, x\in R\}$.
Q. Can't we conclude (2') if $R$ is not  Noetherian?

Nice question!
I'm following the suggestion made by John Brevik in a comment.
The answer is No.
Here is a ring $R$ which does not satisfy (2'):
Let $K$ be a field, let $X_1,X_2,\dots$ be indeterminates, and set 
$$
R:=K[X_1,X_2,X_3,\dots]/(X_1,X_2^2,X_3^3,\dots).
$$
Let $x_i$ be the image of $X_i$ in $R$. Clearly $\mathfrak m:=(x_1,x_2,\dots)$ is a maximal ideal of $R$, and is its nilradical. In particular the zero ideal of $R$ is $\mathfrak m$-primary, and it suffices to show that $\mathfrak m$ is not the annihilator of $r$ for any $r$ in $R$.
Let $B$ be the set of those monomials in $x_1,x_2,\dots$ which are of degree less than $i$ in $x_i$ for all $i$. I leave it to the reader to check that $B$ is a $K$-basis of $R$.
Let $r$ be a nonzero element of $R$. It suffices to show that we have $rx_i\ne0$ for some $i$. 
If $i\ge2$ is such that $x_i$ does not occur in $r$ when $r$ is written as a $K$-linear combination of elements of $B$, then we have $rx_i\ne0$ as desired.
