I am trying to prove that if $R$ is a commutative Northerian ring with $1$, for all ideals $I$, $E$, then $E \cap I^n \subseteq EI$ for some $n\in \mathbb{N}$.
I have tried to go in this direction, but I have not obtained it:
Let $I= Q_{1}\cap \cdots \cap Q_{m}$ be the primary decomposition of $I$ with $P_{i}=\text{rad}(Q_{i})$, for all $i\in \{1,\cdots,n\}$ and $E=S_{1}\cap \cdots \cap S_{t}$ a primary decomposition of $I$.
Since $R$ is Noetherian, for each $i\in \{1,\cdots, m\}, $ $Q_{i}^{n_{i}}\subseteq P_{i}$ for some $n_{i}\in \mathbb{N}$.
Let $t=\max\{n_{i} \mid i\in \{1,\cdots,m\}\}$. Then, $Q_{i}^t \subseteq P_{i}$ for all $i\in \{1,\cdots,m\}$. Therefore, $$(Q_{1}\cap \cdots \cap Q_{m})^t\subseteq P_{i}\subseteq Q_{i},$$ for $i\in \{1,\cdots, m\}$.
This is useless because I only obtain that $I^t\subseteq I$. However, I do believe that I should use the existence of primary decomposition in Noetherian rings. Can someone give me a clue, please?
Suppose that $EI= Q_{1}\cap \cdots \cap Q_{n}$ where $Q_{i}$ is primary and $P_{i}=\text{rad}(Q_{i})$ for $i\in \{1,\cdots,n\}$.
Suppose that $E$ is not contained in $Q_{i}$. Then, there exists $x$ in $E$ such that $x$ is not in $Q_{i}$. Let $z\in I$. Then, $xz\in EI=Q_{1}\cap \cdots \cap Q_{n}$, so $xz\in Q_{i}$ and therefore, $z\in P_{i}$.
Since $R$ is a commutative Noetherian ring, there exists $k_{i}\in \mathbb{N}$ such that $P_{i}^{k_{i}}\subseteq Q_{i}$.
Let $k=\max\{k_{i}\mid i\in\{1,\cdots,n\}\}$. Then,
$E \subseteq \bigcap_{E\subseteq Q_{i}} Q_{i}$ and $I^k\subseteq \bigcap_{I^k\subseteq Q_{i}} Q_{i}$, so $E\cap I^k\subseteq EI$.