Let $L$ and $K$ be number fields such that $L/K$ is a finite extension. Suppose $\mathfrak{a},\mathfrak{b}$ are ideals in $\mathcal{O}_K$ and $\mathfrak{a}\mathcal{O}_L|\mathfrak{b}\mathcal{O}_L$. Prove that $\mathfrak{a}|\mathfrak{b}$ and that $\mathfrak{a}=\mathfrak{a}\mathcal{O}_L \cap \mathcal{O}_K$.
My thought: Consider the prime factorization $\mathfrak{a}\mathcal{O}_L= \prod_i \mathfrak{P}_i^{e_i}, \mathfrak{b}\mathcal{O}_L= \prod_i \mathfrak{P}_i^{f_i}$. Then we have $e_i \leq f_i$ since $\mathfrak{a}\mathcal{O}_L|\mathfrak{b}\mathcal{O}_L$. But then I dont know how to move on ,please helps.
I am assuming you have uniqueness of prime ideal factorization for number rings already at your disposal. Suppose an ideal ${\frak a}\triangleleft{\frak O}_K$ has the prime factorization ${\frak a}={\frak p}_1^{v_1}\cdots{\frak p}_r^{v_r}$. Then
$${\frak aO}_L=\prod_{i=1}^{r}({\frak p}_i{\frak O}_L)^{v_{\large i}}=\prod_{i=1}^r\left(\prod_{{\frak P\mid p}_i}{\frak P}^{e({\frak P\mid p})}\right)^{\large v_i}=\prod_{{\frak P\mid p}_i\mid{\frak a}}{\frak P}^{e({\frak P\mid p})v_{\large i}}.$$
This gives $v_{\frak P}({\frak aO}_L)=e({\frak P\mid p})v_{\frak p}({\frak a})$. Use it to rewrite $v_{\frak P}({\frak aO}_L)\le v_{\frak P}({\frak bO}_L)$ ($\Leftarrow{\frak aO}_L\mid{\frak bO}_L$).