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I am referring to the very useful K. Conrad's notes on the conductor ideal of an order in a Dedekind domain: https://kconrad.math.uconn.edu/blurbs/gradnumthy/conductor.pdf

$\DeclareMathOperator\Cl{Cl}$On page 13 it is claimed that every ideal class of the order contains a representative coprime with the conductor, and this would be crucially needed in a paper I am writing. Unfortunately, the proof is omitted (Theorem 5.2). By Lemma 5.1, the statement is valid in the Dedekind domain, and the document suggests that this could be extended to the order by some work. For sure, one of the classes in the preimage via the extension map from $\Cl(\mathcal{O})$ to $\Cl(\mathcal O_{K})$ of an element of $\Cl(\mathcal{O}_{K})$ has the property, but I could not prove why this should also occur with the remaining ones.

Has anyone succeeded in completing this proof?

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    $\begingroup$ Oops! I did not intend to post the file with that incomplete argument. A proof that would be overkill is to identify the ideal class group (Picard group) of the order with a generalized ideal class group and thus Galois group of a finite abelian extension (Theorems 3.8, 3.11, and 4.2 in arxiv.org/abs/1405.5776), so each ideal class should be represented by infinitely many prime ideals. That gives ideals relatively prime to the conductor. I wanted to find a much simpler argument, but I then forgot about it. $\endgroup$
    – KConrad
    Commented Apr 16, 2022 at 6:10
  • $\begingroup$ Thank you very much for your very useful answer, this solves the problem I was having $\endgroup$
    – Hair80
    Commented Apr 16, 2022 at 16:36
  • $\begingroup$ You said "I could prove why this should also occur …", but the question strongly suggested you meant that you could not prove it. I edited accordingly. I hope that this was correct. $\endgroup$
    – LSpice
    Commented Apr 16, 2022 at 19:43
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    $\begingroup$ I updated the file using Will's argument below, so now there is no longer an omission. After adding an example at the start of Section 5, the labels Lemma 5.1 and Theorem 5.2 have turned into Lemma 5.2 and Theorem 5.3 (and the goal of the lemma has been changed). The claim that you wanted is now the first part of Theorem 5.3 and it holds in an arbitrary one-dimensional Noetherian domain, not just in an order in a number field (see footnote 4). $\endgroup$
    – KConrad
    Commented Apr 16, 2022 at 23:19
  • $\begingroup$ The "footnote 4" in my previous comment should be "footnote 5". $\endgroup$
    – KConrad
    Commented Aug 18, 2022 at 3:24

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I can give a non-overkill argument.

Let $I$ be an invertible ideal of $\mathcal O$, and $J$ its inverse. Then for each $a \in J$ we obtain an ideal $aI$ in the ideal class of $I$. The ideal $aI$ is prime to $\mathfrak p$ if and only if $ a b \notin \mathfrak p$ for at least one $b\in I$, and $aI$ is prime to the conductor if and only if this is satisfied for each prime $\mathfrak p$ dividing the conductor.

For each $\mathfrak p$, this condition clearly only depends on the congruence class of $a$ in $J / \mathfrak p J$. Now by the Chinese remainder theorem for $J$, the map $J \to \prod_{\mathfrak p \mid \mathfrak c} J/ \mathfrak p J$ is a surjection (its proof is the same as the usual Chinese remainder theorem).

So it suffices to check for each $\mathfrak p$ that there is $a \in I$ and $b \in J$ with $ab \notin \mathfrak p$. But this is obvious since $IJ = \mathcal O \not\subseteq \mathfrak p$.

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  • $\begingroup$ Thank you very much for your very useful answer, it is really helpful $\endgroup$
    – Hair80
    Commented Apr 16, 2022 at 16:38

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