Conductor coprime to an ideal of a ring of integers in a number field Let $L/K$ be an extension of number fields. Suppose that $\alpha \in \mathcal O_L$ is a primitive element, i.e., $L = K(\alpha)$. Of course, we have $\mathcal O_K[\alpha] \subset \mathcal O_L$. The conductor with respect to $\alpha$ is
$$\mathfrak f := \{ \gamma \in \mathcal O_L ~ | ~ \gamma \mathcal O_L \subset \mathcal O_K[\alpha]\}.$$
Clearly, $\mathfrak f \subset \mathcal O_K[\alpha]$, and $\mathfrak f$ is an ideal.
Consider now a non-zero prime ideal $\mathfrak p \subset \mathcal O_K$. My question is: how to show that the implication
$$\mathfrak p \mathcal O_L + \mathfrak f = \mathcal O_L \implies \mathfrak p + (\mathfrak f \cap \mathcal O_K) = \mathcal O_K \qquad (*)$$
holds? Neukirch's Algebraic Number Theory (Theorem 8.3) just uses it, but the implication seems not to be justified.
Alternatively, I would also be interested in a proof of the fact that the natural morphism
$$\mathcal O_K[\alpha]/\mathfrak p \mathcal O_K[\alpha] \to \mathcal O_L/\mathfrak p \mathcal O_L$$
is an isomorphism, if $\mathfrak p \mathcal O_L + \mathfrak f = \mathcal O_L$, which does not rely on $(*)$. Surjectivity is rather clear, since $\mathfrak p \mathcal O_L + \mathfrak f = \mathcal O_L$ implies $\mathfrak p \mathcal O_L + \mathcal O_K[\alpha] = \mathcal O_L$, so the natural morphism
$$\mathcal O_K[\alpha] \to \mathcal O_L/\mathfrak p \mathcal O_L$$
is onto. Furthermore, it is clear that $\mathfrak p \mathcal O_K[\alpha]$ is contained in its kernel $\mathfrak p \mathcal O_L \cap \mathcal O_K[\alpha]$ -- the other inclusion is the problem, and I don't know how to show it without using $(*)$. Taking $(*)$ for granted, we can show the converse inclusion as follows: write $x + f = 1$, where $x \in \mathfrak p$ and $f \in \mathfrak f \cap \mathcal O_K$. Then, for every $y \in \mathfrak p \mathcal O_L \cap \mathcal O_K[\alpha]$, we obtain $y = xy + fy$. Finally, $xy \in \mathfrak p \mathcal O_K[\alpha]$ and the definition of $\mathfrak f$ implies that $fy \in \mathfrak p \mathcal O_K[\alpha]$, too.
I however have no clue about the proof of $(*)$, so every piece of help is appreciated! I am probably missing something.
 A: Here is an argument that the kernel $\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L$ of $\mathcal O_K[\alpha] \to \mathcal O_L/\mathfrak p \mathcal O_L$ is equal to $\mathfrak p\mathcal O_K[\alpha]$ without using $(*)$.
Only $\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L \subset \mathfrak p\mathcal O_K[\alpha]$ needs a proof.
Claim: $\mathfrak f + \mathfrak p \mathcal O_L = \mathcal O_L$ implies that $\mathfrak f + \mathfrak p \mathcal O_K[\alpha] = \mathcal O_K[\alpha]$.
Proof of the Claim: If $\mathfrak f + \mathfrak p \mathcal O_K[\alpha]$ is a proper ideal of $\mathcal O_K[\alpha]$, then there is a maximal ideal $\mathfrak m \subset \mathcal O_K[\alpha]$ containing $\mathfrak f + \mathfrak p \mathcal O_K[\alpha]$. In particular,
$$\mathfrak f \subset \mathfrak m \qquad \text{and}\qquad \mathfrak p  \mathcal O_K[\alpha] \subset \mathfrak m.$$
We obtain that
$$\mathfrak f \mathcal O_L = \mathfrak f \subset \mathfrak m \mathcal O_L \qquad \text{and}\qquad \mathfrak p  \mathcal O_L \subset \mathfrak m\mathcal O_L,$$
contradicting the hypothesis that $\mathfrak f$ and $\mathfrak p \mathcal O_L$ are coprime. $\Box_{\text{Claim}}$
Now, from the Claim, we obtain that
\begin{align*}\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L = \mathcal O_K[\alpha] \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) &= (\mathfrak f + \mathfrak p \mathcal O_K[\alpha])(\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) \\
&\stackrel{(1)}{\subset} \mathfrak f \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) + \mathfrak p \mathcal O_K[\alpha] \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) \\
&\stackrel{(2)}{\subset} \mathfrak f \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) + \mathfrak p \mathcal O_K[\alpha] \\
&\stackrel{(3)}{\subset} \mathfrak f \mathfrak p \mathcal O_L + \mathfrak p \mathcal O_K[\alpha] \\
&\stackrel{(4)}{=} \mathfrak f \mathfrak p +  \mathfrak p \mathcal O_K[\alpha] \\
&\stackrel{(5)}{\subset} \mathfrak p \mathcal O_K[\alpha].
\end{align*}
Let me justify the steps:
$(1)$ The ideal $(\mathfrak f + \mathfrak p \mathcal O_K[\alpha])(\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L)$ is generated by all elements of the form $(f+x) \cdot y$, where $f \in \mathfrak f$, $x \in \mathfrak p \mathcal O_K[\alpha]$ and $y \in \mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L$.
Similarly, $\mathfrak f \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) + \mathfrak p \mathcal O_K[\alpha] \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L)$ is generated by all elements of the form $f \cdot y_1 + x \cdot y_2$, where $f \in \mathfrak f$, $x \in \mathfrak p \mathcal O_K[\alpha]$, and $y_1,y_2 \in \mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L$. The inclusion follows.
$(2)$ is clear, since $\mathfrak p \mathcal O_K[\alpha] \cdot (\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L) \subset \mathfrak p \mathcal O_K[\alpha]$.
$(3)$ follows immediately, since $\mathcal O_K[\alpha] \cap \mathfrak p \mathcal O_L \subset \mathfrak p \mathcal O_L$.
$(4)$ The conductor $\mathfrak f$ is an ideal in $\mathcal O_L$, and hence $\mathfrak f \mathcal O_L = \mathfrak f$.
$(5)$ From $\mathfrak f \subset \mathcal O_K[\alpha]$.
