Let $K \subset L=K(\alpha)$ be a number field extension with rings of integers $\mathcal{O}_K$ and $\mathcal{O}_L=\mathcal{O}_K[\alpha]$ respectively. Let $\pi$ be a prime ideal in $O_K$, and let $F = \mathcal O_K/\pi$.
Finally, let $p$ be the minimum monic polynomial of $\alpha$ over $\mathcal O_K$, and let $p(x) = \prod_i(p_i(x))^{f_i}$ be the factorization of $p$ in $F[x]$.
Then we can show that $$\mathcal{O}_L/(\pi \mathcal{O}_L) = \prod_i (\mathcal{O}_L/(\pi,p_i(\alpha)))^{f_i} .$$ I have seen multiple proofs of this theorem but none of them show that $$\underbrace{\pi \mathcal{O}_{L_\strut}}_{P} = \underbrace{\prod_i (\pi,p_i(\alpha))^{f_i}}_{Q}.$$
It is easy to show that $Q \subset P$ by simply multiplying everything out but can we show that $P \subset Q$ (or even that $P = Q$) directly? By this I mean mostly avoiding the use of the norm function(size of the ideals) to show that $Q|P \Rightarrow P = Q$