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In our lecture on Algebraic Number Theory, we came up with the following definition of an conductor:

Let $A$ be a Dedekind domain, $K$ the quotient field of $A$, $L/K$ a finite, seperabel extension, $B$ the integral closure of $A$ in $L$ and $\theta \in B$ a primitive element for $L/K$ as well as $A[\theta]$ a subring of $B$. Then the conductor $\mathcal{F}$ of the subring $A[\theta]$ is defined as $$\mathcal{F}= \{ \alpha \in B \vert \alpha \cdot B \subseteq A[\theta]\}.$$

We then state that $\mathcal{F}$ is an ideal of $O_K$ and even $= O_K$ if $A[\theta]=B$ holds.

Now I have a hard time understanding why $\mathcal{F}$ should be an ideal of $O_K$? Could it possibly be that instead of $O_K$ it would have to be $B$ in the above definition?

Moreover, in a following application, we need to then determine whether a prime ideal of $A$ is relative prime to the conductor in $B$. Does any of you may be know any examples on how to calculate this? (preferably for some quadratic number field, e.g. $K= \mathbb{Q}(\sqrt{-29})$). Is there any possible way to work with the adjoint element and its results modulo $4$?

Thank you!

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    I think there might be a confusion with the notation: $\;A\;$ is *the basis* integral domain we begin with, $\;K\;$ its Q.F., and $\;L/K\;$ a finite, separable extension of fields. Then, as far as I can see it, $\;B=\mathcal O_L:=\;$ is the integral closure of $\;A\;$ in $\;L\;$ , meaning: from all the (algebraic over $\;K\;$) elements in $\;L\;,\;\;\mathcal O_L\;$ is the set of the ones that satisfy a monic polynomial *in* $\;A[x]\;$ . What has $\;\mathcal O_K\;$ to do here is beyond my comprehension **unless** we already *had before* that $\;K/\Bbb Q\;$ is a finite extension and...etc. – 2017-01-25
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    @DonAntonio Yes, that was exactly what I was wondering as we did not mention anything about the extension $K/ \mathbb{Q}$ in this chapter. I think it was supposed to be $\mathcal{O}_L$. So thank you for your clarification! – 2017-01-25

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As the notation problem seems to have been solved in the comment by DonAntonio, my answer refers to your second question: Deciding if a prime ideal in $B$ is relatively prime to the conductor $\mathcal{F}$ of $A[\theta]$, when you have a quadratic extension.

So let $\mathbb{Q}(\sqrt d)$ be a quadratic extension, where $d$ is squarefree. (so in this case let $A := \mathbb{Z}$, $K := \mathbb{Q}(\sqrt d)$, $B := \mathcal O_K$). I assume you want to check for coprimality in order to being able to for example compute the prime ideals in the ring of integers which lie above any prime.

Note in this case, that you get

$$\mathcal O_K = \begin{cases} \mathbb{Z}[\displaystyle \frac{1+\sqrt d}{2}], & \text{for } d \equiv 1 \bmod 4\\ \mathbb{Z}[\displaystyle \sqrt d], & \text{for } d \equiv 2,3 \bmod 4\\ \end{cases}$$

So you can choose $\theta$ accordingly to the result modulo 4 of $d$ and get $\mathcal O_K = \mathbb{Z}[\theta]$, thus the conductor $\mathcal F = \{ \alpha \in \mathbb{Z}[\theta] \vert \alpha \cdot \mathbb{Z}[\theta] \subseteq \mathbb{Z}[\theta]\} = \mathbb{Z}[\theta] = \mathcal O_K$.

So for any prime ideal $\mathfrak p \subset \mathbb{Z}$, the ideal $\mathfrak p \cdot \mathcal O_K$ is coprime to the conductor, as $\mathfrak p \cdot \mathcal O_K + \mathcal F = \mathcal O_K$.

If, of course, a different $\theta$ has been given, it depends on the setup.

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    Thank you for your help! :) – 2017-01-25