Factoring ideals in algebraic number rings using Dedekind's theorem 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 $Q|P \Rightarrow P = Q$
 A: I don't know if this is what you mean by norms (you mean the absolute norm?). If this is what you already knew I apologize in advance!
$$\begin{aligned}\mathcal{O}_L/\pi\mathcal{O}_L &\cong \mathcal{O}_K[T]/(p(T))/(\pi\mathcal{O}_K[T]/(p(T))\\ &\cong \mathcal{O}_K[T]/(p(t),\pi)\\ &\cong \left(\mathcal{O}_K/\pi\right)[T]/(p(T))\\ &= \prod_i \left(\mathcal{O}_K/\pi\right)[T]/(p_i(T)^{f_i})\end{aligned}$$
But, just by counting dimensions:
$$\#\left(\mathcal{O}_K/\pi)\right)[T]/(p_i(T)^{f_i})=\left(\#(\mathcal{O}_K/\pi)[T]/(p_i(T))\right)^{f_i}$$
So, $\mathcal{O}_L/P$ and $\mathcal{O}_L/Q$ have the same cardinality. So, containment implies equality.
A: There is a slight generalization of what you call Dedekind's theorem, where one only assumes that p does not divide the index of O_K [a] in O_L, p denoting the prime of Z lying under Pi). See Daniel A. Marcus, "Number Fields", Springer Universitex , chapter 3, theorem 27. The proof is clear and easy to follow, and - as far as I remember - doesn't use norms. 
