In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I),P=I:a$ for some $a\in R$. 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.)
 A: Since $R$ is Noetherian, let $I=Q_{1}\cap \cdots\cap Q_{k}$ be a minimal primary decomposition. Let $P_{i}=\operatorname{rad} Q_{i}$. Then $\mathrm{Ass}(I)=\{P_{1},\dots,P_{k}\}$.
Claim: each $P_{i}$ is of form $I:a$.
Assuming the claim, we prove that $\mathrm{Ass}(I)=\{P: P \textrm{ prime}, P=I:a, \textrm{ for some } a\in R\}$.
"$\subset$" is true by the claim.
"$\supset$": let $I:a$ be a prime, then $\mathrm{rad}(I:a)=I:a$ is prime, so $I:a\in \mathrm{Ass}(I)$ by the definition of $\mathrm{Ass}(I)$.
Proof of the claim: let $L_{i}=\cap_{j\neq i} Q_{j}$, let $K_{i}=L_{i}\setminus Q_{i}$. Then $K_{i}$ is non-empty by definition of minimal primary decomposition. Let $S_{i}=\{(I:\mu):\mu\in K_{i}\}$. Since $R$ is Noetherian, $S_{i}$ has a maximal element $I:m_{i}$. Our goal is to show $P_{i}=I:m_{i}$.
Step 1: $(I:m_{i})\subset P_{i}$.
$$ \begin{align*}
I:m_{i}
&=(Q_{1}\cap\cdots\cap Q_{k}):m_{i}\\
&=(Q_{1}:m_{i})\cap\cdots\cap(Q_{k}:m_{i})\\
&=Q_{i}:m_{i} \quad \textrm{since } m_{i}\in K_{i}\\
&\subset P_{i}, \end{align*}
$$
the proof for the last line: let $a\in Q_{i}:m_{i}$, then $am_{i}\in Q_{i}$. Since $Q_{i}$ primary and $m_{i}\notin Q_{i}$, $a^{n}\in Q_{i}$, so $a\in P_{i}$.
Step 2: $(I:m_{i})$ is prime.
It is not $R$ by step 1. Let $ab\in (I:m_{i})$ and $a\notin (I:m_{i})$. Then $abm_{i}\in I$, so $b\in I:m_{i}a$. We have $I:m_{i}\subset I:m_{i}a$. We show $m_{i}a\in K_{i}$. It is in $L_{i}$. Since $a\notin I:m_{i}=Q_{i}:m_{i}$, $m_{i}a\notin Q_{i}$, so $m_{i}a\in K_{i}$. By maximality, $I:m_{i}=I:m_{i}a$. So $b\in I:m_{i}$.
Step 3: $(I:m_{i})=P_{i}$.
Since $Q_{i}\subset Q_{i}:m_{i}=I:m_{i}$, $\operatorname{rad} Q_{i}\subset \mathrm{rad}(I:m_{i})$, so $P_{i}\subset I:m_{i}$. Therefore, $P_{i}=I:m_{i}$.
