Conductor of an order

Let $\mathcal O\subseteq\mathcal O_K$ be an order in a number field $K$ , with ring of integers $\mathcal O_K$ . Its conductor is

\mathfrak f=\{x\in\mathcal O_K:x\mathcal O_K\subseteq\mathcal O\}.

It is the largest $\mathcal O_K$ -ideal contained in $\mathcal O$ , and is also an ideal of $\mathcal O$ . Since $1\in\mathcal O_K$ , every element of $\mathfrak f$ already belongs to $\mathcal O$ .

Why it is nonzero

The additive quotient $\mathcal O_K/\mathcal O$ is finite. If its order is $m$ , then $m\mathcal O_K\subseteq\mathfrak f$ . Thus the conductor records a finite obstruction to the order being maximal; it equals $\mathcal O_K$ exactly when $\mathcal O=\mathcal O_K$ .

Quadratic orders

If $K$ is quadratic and $\mathcal O=\mathbb Z+f\mathcal O_K$ with $f\ge1$ , then $\mathfrak f=f\mathcal O_K$ . The integer $f$ is also called the conductor, whereas $\mathfrak f$ is the conductor ideal.

For $K=\mathbb Q(\sqrt{-3})$ and $\mathcal O=\mathbb Z[\sqrt{-3}]$ , this gives

\mathfrak f=2\mathcal O_K=(2,1+\sqrt{-3}).

It is the noninvertible ideal from the invertibility example.

Away from the conductor

For an integral ideal $I\subseteq\mathcal O$ , being prime to the conductor means $I+\mathfrak f=\mathcal O$ . Such nonzero ideals are invertible. Localization outside the primes containing $\mathfrak f$ removes the difference between $\mathcal O$ and $\mathcal O_K$ .

References

Andrew Sutherland, MIT 18.785 Lecture 7, §7.1, Lemma 7.1, Theorem 7.8, and Corollary 7.10.