I want to find a direct proof that for any Noethern ring $R$ with $a$ an ideal of $R$ the prime ideals occurring in the set of ideals $rad(a:x)$ is the same as the set of prime ideals occurring in the set of ideals $(a:x)$
$(a:x)=\{ r \in R : rx \in a \}$, and $rad$ is the radical of an ideal.
The reason for me believing this to be true is that for $ a = \bigcap_{i=1}^{n}q_i$ being a minimal primary decomposition of $a$ with $rad(q_i)=p_i$ we have that $ass(R/a)=\{p_1,p_2,...,p_n\}$ (1st uniqueness theorem of primary decomposition as stated in Reid undergraduate commutative algebra) And we also have that the prime ideals occuring in the set of ideals of $rad(a:x)$ $x\in R$ is the same set of primes. (1st uniqueness theorem as stated in McDonald, Atiyah) If we now note that $ass(R/a)=\{r \in R : rx = 0, x \in R/a \} = \{ r \in R : rx \in a \}= (a:x) $ we have to conclude that the set of prime ideals in $rad(a:x)$ has to be the same as the prime ideals of $(a:x)$ for any ideal $a$ of $R$
For any prime ideal of $(a:x)$ of course $rad(a:x)$ is prime so all prime ideals in $(a:x)$ is prime in $rad(a:x)$ but what about the other direction? Let p be prime in $rad(a:x) =rad(ann(x))$ $x\in R/a$ since the radical of $ann(x)$ is the intersection of all minimal prime ideals $w_i$ containing $ann(x)$ we may write $\bigcap_{i=1}^{n}w_i = rad(ann(x))$ Now assuming that this intersection is prime we must have that $rad(ann(x))=w_i$ for some $i$. ... But this doesn't show $ann(x)$ to be prime?
If the stated proposition is false please adress how the two different versions of the 1st uniqueness theorem may both be true.