# The set of prime ideals in the set of ideals $\mathrm{rad}(\mathfrak a:x)$ equals the set of prime ideals in the set of ideals $(\mathfrak a:x)$?

I want to find a direct proof that for any Noetherian ring $$R$$ and $$\mathfrak a$$ an ideal of $$R$$, the prime ideals occurring in the set of ideals $$\mathrm{rad}(\mathfrak a:x)$$ is the same as the set of prime ideals occurring in the set of ideals $$(\mathfrak a:x)$$.

$$(a:x)=\{ r \in R : rx \in a \}$$, 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.

Prop. If $$P$$ is a finitely generated prime ideal which equals $$\mathrm{rad}(0:x)$$ (or just minimal over $$(0:x)$$) for some $$x\neq 0$$, then there exists an $$y$$ such that $$P=(0:y).$$
Proof. Suppose $$P=(a_1,...,a_n)$$, of course assuming our $$a_i$$'s are not zero. Let us consider $$A_P$$. Then $$P=\mathrm{rad}(0:x)$$ in $$A_P$$. You can find a minimal number $$e_1$$ such that $$a_1^{e_1}x=0$$ in the local ring. Now replace $$x$$ by $$a_1^{e_1-1}x$$. Then you may assume that $$a_ix=0$$ in the local ring for all $$i$$. That is $$(0:x)A_P=PA_P$$.
The meaning of last equation is that you can find $$s_i\notin P$$ such that $$a_is_ix=0$$ in $$A$$. Now replace $$x$$ by $$xs_1s_2\cdots s_n$$, and then $$P=(0:x)$$.