For a ring $R$ and an ideal $I$ of $R$, the primes of the form $\sqrt{(I:a)}$ for some $a\in R$ are called associated prime ideals of $I$ (Here $(I:a)=\{r\in R:ar\in R\}$). Let $\text{Ass}(I)$ denote the set of such associated prime ideals of $I$.      How is it possible to prove that if $R$ is Noetherian, then    $$\text{Ass}(I)=\{P\text{ prime}:P=(I:a)\text{ for some }a\in R\}?$$   (In other words, it is not necessary to take radicals).   I guess that one should use the fact that for Noetherian rings there exists always a minimal primary decomposition of any ideal $I$ of $R$, but so far I have failed to prove the claim.
Maybe this question is too obvious to everyone.      In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I)$, then $P=I:a$ for some $a\in R$ where $\mathrm{Ass}(I)$ denotes associated primes of $I$.   This is an exercise 8.29 in Gathmann's Commutative Algebra notes. I am supposed to find this statement obvious. However, up to now, I still did not find this statement obvious. Can someone explain intuitively why noetherianness here allows me removing radical $\sqrt{I:a}$ without going into details?  If $Q$ is a primary ideal of $I$, then certainly $\sqrt{Q:a}=P$ for $a\not\in P$. It seems taking $(p_1,\dots, p_n)=P$, $p_i^{n_i}a\in Q$. Since $Q$ primary, I must have $p_i\in Q$ and $p_i^{n_i}\in Q$. (This argument seems very wrong as this deduces the same conclusion without noetherian property.)
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