Kummer-Dedekind theorem for number fields via valuation theory

I'm following these notes https://websites.math.leidenuniv.nl/algebra/localfields.pdf and i'm struggling with exercise $3.10$ regarding the valuation theory proof of Kummer-Dedekind. The statement of the problem is as follows:

Let $L/K$ be an extension of number fields and $\alpha \in \mathcal{O}_L$ an element that generates $L$ over $K$. Suppose that $\mathfrak{p}$ is a prime in $\mathcal{O}_K$ that does not divide $[\mathcal{O}_L : \mathcal{O}_K[\alpha]]\mathcal{O}_K$. If $m_{\alpha,K}(X)$ factors over $\mathcal{O}_L/\mathfrak{p}$ as $\overline{m}_{\alpha,K}(X) = \prod_{i} \overline{g}_i(X)^{e_i}$, then $\mathfrak{p}$ factors in $\mathcal{O}_L$ as $\mathfrak{p}\mathcal{O}_L = \prod_{i} \mathfrak{q}_i^{e_i}$ with $\mathfrak{q}_i$ the prime ideal generated by $\mathfrak{p}_i$ and $g_i(\alpha)$ for any lift $g_i \in \mathcal{O}_K[X]$ of $\overline{g}_i$. Hint: we have $\overline{m}_{\alpha,K}(X) = \prod_i f_i \in K_{\frak{p}}[X]$ by Hensel’s lemma, and $L_{\frak{q}_i} = K_{\frak{p}}[X]/(f_i)$ has residue class field $K[X]/(g_i)$.

I've worked out the classical proof (using only ring theory and ideal norms), but i would like to find the intended solution, which i suppose should use as little ring theory as possible. I know there is a one to one correspondence between prime ideals and non-archimedean valuations over number field, and that $$ L \otimes_K K_{\mathfrak{p}} \cong \prod_{\mathfrak{q} \mid \mathfrak{p}} L_{\mathfrak{q}} $$

On the other hand, by hint, $$ L \otimes_K K_{\mathfrak{p}} \cong \prod_{i} K_{\mathfrak{p}}[X] / \left< f_i \right> $$

How do we know that each $f_i$ is irreducible?

Exercise $3.8$ might be useful:

Let $\phi$ be a valuation on a field $K$ and $L$ a finite extension of $K$. Show that the natural homomorphism $K_\phi \otimes_K L \rightarrow \prod_{\psi \mid \phi} L_\psi$ is surjective, and that the maximal ideals of $K_\phi \otimes_K L$ correspond bijectively to the extensions $\psi$ of $\varphi$ to $L$. Show also that the image of $L = 1 \otimes L$ is dense in $\prod_{\psi \mid \phi} L_\psi$

By exercise, using index condition, we know that $A_{\mathfrak{p}}[\alpha] = \prod_{\mathfrak{q} \mid \mathfrak{p}} A_{\mathfrak{q}}$, since the image of $\mathcal{O}_K[\alpha]$ is dense (by index condition) and $A_{\mathfrak{p}}[\alpha]$ is compact with $\mathcal{O}_K[\alpha]$ dense, so we have the desired equality.

Classical proof: The maps $$ \mathcal{O}_K(\alpha) \longrightarrow \mathcal{O}_L \longrightarrow \mathcal{O}_L / \mathfrak{q_i} $$ induces a map $$ \varphi_i: \mathcal{O}_K(\alpha) / \left(\mathfrak{p }\mathcal{O}_K(\alpha) + g_i(\alpha) \mathcal{O}_K(\alpha) \right) \longrightarrow \mathcal{O}_L / \mathfrak{q_i} $$ By definition, $[\mathcal{O}_L : \mathcal{O}_K[\alpha]]\mathcal{O}_L \subseteq \mathcal{O}_K[\alpha]$. But \begin{gather*} \frac{\mathcal{O}_K(\alpha)}{\mathfrak{p }\mathcal{O}_K(\alpha) + g_i(\alpha) \mathcal{O}_K(\alpha)} \cong \frac{\frac{\mathcal{O}_K[X]}{\left<m_{\alpha,K}(X)\right>}}{ \mathfrak{p }\frac{\mathcal{O}_K[X]}{\left<m_{\alpha,K}(X)\right>} + \left(g_i + \left<m_{\alpha,K}(X)\right>\right) \frac{\mathcal{O}_K[X]}{\left<m_{\alpha,K}(X)\right>}} \cong \\ \cong \frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X] + g_i\mathcal{O}_K[X] \mathcal{O}_K[X]} \cong \frac{\frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X]}}{g_i\frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X]}} \cong \frac{\frac{\mathcal{O}_K}{\mathfrak{p}}[X]}{\overline{g}_i\frac{\mathcal{O}_K}{\mathfrak{p}}[X]} \end{gather*} Thus, $\varphi$ is injective or $\mathfrak{q}_i = \mathcal{O}_L$. The index conditon states that $$ [\mathcal{O}_L : \mathcal{O}_K[\alpha]] \mathcal{O}_L / \mathfrak{q_i} \subseteq \varphi\left( \mathcal{O}_K(\alpha) / \left(\mathfrak{p }\mathcal{O}_K(\alpha) + g_i(\alpha) \mathcal{O}_K(\alpha) \right) \right) $$ But $[\mathcal{O}_L : \mathcal{O}_K[\alpha]] \in \left(\mathcal{O}_K(\alpha) / \mathfrak{p }\mathcal{O}_K(\alpha)\right)^\times$ since $[\mathcal{O}_L : \mathcal{O}_K[\alpha]] \in \left(\mathcal{O}_K / \mathfrak{p }\right)^\times$, so $\varphi$ is surjective. Thus, $\mathfrak{q}_i = \mathcal{O}_L$ or $\mathfrak{q}_i$ is a maximal ideal. The map $$ \mathcal{O}_K(\alpha) \longrightarrow \mathcal{O}_L \longrightarrow \mathcal{O}_L / \mathfrak{p}\mathcal{O}_L $$ induce a map $$ \varphi: \mathcal{O}_K(\alpha) / \mathfrak{p }\mathcal{O}_K(\alpha) \longrightarrow \mathcal{O}_L / \mathfrak{p}\mathcal{O}_L $$ But $$ \frac{\mathcal{O}_K(\alpha)}{\mathfrak{p }\mathcal{O}_K(\alpha)} \cong \frac{\frac{\mathcal{O}_K[X]}{\left<m_{\alpha,K}\right>}}{ \mathfrak{p }\frac{\mathcal{O}_K[X]}{\left<m_{\alpha,K}\right>}} \cong \frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X] + m_{\alpha,K}\mathcal{O}_K[X]} \cong \\ \cong \frac{\frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X]}}{m_{\alpha,K}\frac{\mathcal{O}_K[X]}{\mathfrak{p}\mathcal{O}_K[X]}} \cong \frac{\frac{\mathcal{O}_K}{\mathfrak{p}}[X]}{\overline{m}_{\alpha,K}\frac{\mathcal{O}_K}{\mathfrak{p}}[X]} \cong \prod \frac{\frac{\mathcal{O}_K}{\mathfrak{p}}[X]}{\overline{g}_{i}^{e_i}\frac{\mathcal{O}_K}{\mathfrak{p}}[X]} $$ By definition, $[\mathcal{O}_L : \mathcal{O}_K[\alpha]]\mathcal{O}_L \subseteq \mathcal{O}_K[\alpha]$. The index conditon states that $$ [\mathcal{O}_L : \mathcal{O}_K[\alpha]] \mathcal{O}_L / \mathfrak{q_i} \subseteq \varphi\left( \mathcal{O}_K(\alpha) / \left(\mathfrak{p }\mathcal{O}_K(\alpha) + g_i(\alpha) \mathcal{O}_K(\alpha) \right) \right) $$ But $[\mathcal{O}_L : \mathcal{O}_K[\alpha]] \in \left(\mathcal{O}_K(\alpha) / \mathfrak{p }\mathcal{O}_K(\alpha)\right)^\times$ since $[\mathcal{O}_L : \mathcal{O}_K[\alpha]] \in \left(\mathcal{O}_K / \mathfrak{p }\right)^\times$, so the above inclusion is in fact an equality, so $\varphi$ is surjective. We can restate this as $$ \mathcal{O}_L = \mathfrak{p}\mathcal{O}_L + \mathcal{O}_K[\alpha] $$ Since the ideal and subring are comaximal, we conclude that $$ \mathfrak{p}\mathcal{O}_L \cap \mathcal{O}_K[\alpha] = \mathfrak{p}\mathcal{O}_K[\alpha] $$ Thus, $\varphi$ is injective, so it is indeed and isomorphism. Finally, $$ \mathfrak{p}\mathcal{O}_L \mid \prod_i \mathfrak{q}_i^{e_i} $$ and by considering norms, $$ N(\mathfrak{p}\mathcal{O}_L) = \left| \frac{\mathcal{O}_L}{\mathfrak{p}\mathcal{O}_L} \right| = \left| \frac{\mathcal{O}_K(\alpha)}{\mathfrak{p }\mathcal{O}_K(\alpha)} \right| = \prod_i \left| \frac{\mathcal{O}_K}{\mathfrak{p}} \right|^{f_i e_i} \leq \\ \prod_i \left| \frac{\mathcal{O}_K}{\mathfrak{q}_i} \right|^{e_i} = \prod_i N(\mathfrak{q}_i) $$ So indeed we have and equality, and thus $\mathfrak{p}\mathcal{O}_L = \prod_i \mathfrak{q}_i^{e_i}$ and each $\mathfrak{q}_i$ is prime, so each $\varphi_i$ is and isomorphism.