Conductor coprime to an ideal of a ring of integers in a number field

Question

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.

Answer 1

It is probably too late for you but i will write a solution for other students, cause I wasted a few hours on this proof. I will use a notation similar to the one in Neukirchs book.

We want to show that $\mathfrak{p} \mathcal{O} \cap \mathcal{O}' = \mathfrak{p}\mathcal{O}'$. The one side is obvious by using ideal properties, $\mathfrak{p}\mathcal{O}' \subset \mathfrak{p} \mathcal{O} \cap \mathcal{O}'$. So lets show the other property by showing that $\mathfrak{p} + (\mathfrak{F} \cap \mathcal{O}_K) = \mathcal{O}_K$ using @reuns tip. Suppose this would not be the case. Then $\mathfrak{p} + (\mathfrak{F} \cap \mathcal{O}_K) = \mathfrak{p}$ (since $\mathfrak{p}$ maximal and the sum of ideals contains the ideals). But then $\mathfrak{F}$ is an ideal of $\mathcal{O}$ s.t. $\mathfrak{F}\cap \mathcal{O}_K \subset \mathfrak{p}$. If we decompose $p\mathcal{O} = \prod_{i=1}^n \mathfrak{P}_i^{e_i}$ in prime ideals $\mathfrak{P}_i \subset \mathcal{O}$, then these are exactly the prime ideals of $\mathcal{O}$ s.t. $\mathfrak{P}_i \cap \mathcal{O}_K = \mathfrak{p}$, so there exists an $i$ with $\mathfrak{F} \subset \mathfrak{P}_i$. But then $\mathfrak{p}\mathcal{O} + \mathfrak{F} \subset \mathfrak{P}_i \neq \mathcal{O}$, which is a contradiction to the assumption that they are relatively prime. Hence $\mathfrak{p} + (\mathfrak{F} \cap \mathcal{O}_K) = \mathcal{O}_K$.

Now $\mathfrak{p} \mathcal{O} \cap \mathcal{O}' \subset \mathcal{O}_K \cdot (\mathfrak{p} \mathcal{O} \cap \mathcal{O}') = (\mathfrak{p} + (\mathfrak{F} \cap \mathcal{O}_K))\cdot (\mathfrak{p} \mathcal{O} \cap \mathcal{O}')$
$= \mathfrak{p}\cdot (\mathfrak{p} \mathcal{O} \cap \mathcal{O}') + (\mathfrak{F} \cap \mathcal{O}_K)\cdot (\mathfrak{p} \mathcal{O} \cap \mathcal{O}') \subset \mathfrak{p}\mathcal{O}'$, where we used $\mathfrak{p} \mathfrak{F} \mathcal{O} \subset \mathfrak{p}\mathcal{O}'$ in the last inclusion.

Answer 2

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 $\mathfrak f \subset \mathcal O_K[\alpha]$.