TITLE: An omission in K. Conrad's notes on the conductor ideal
QUESTION [11 upvotes]: 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?

REPLY [15 votes]: 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 $IJ = \mathcal O \not\subseteq \mathfrak p$.