# 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 $(a:x)$?

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.

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It has been proved in chapter8 or 7 in Atiyah's book, I have not this book in hand right now.

However, I want to try to give a proof.

Prop. If $P$ is a finitely generated prime ideal which equals $rad(0:x)$ (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's are not zero. Let us consider $A_P$. Then $P=rad(0:x)$ in $A_P$ in both case. 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 then 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 $xs_1s_2...s_n$, then $P=(0:x)$..

It is so complicated texting in iPad.

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