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.