Let \(\OO\) be the ring of integers in a quadratic number field \(\QQ(\sqrt{-d})\). Show that any subring \(R \subseteq \OO\), \(R \neq \ZZ\), is of the form \(R=\ZZ+f \cdot \OO\), for a uniquely determined integer \(f \geq 1\). This integer \(f\) is called the conductor of the ring \(R\).
Solution
Put \[ K=\QQ(\sqrt{-d}). \] Since \([K:\QQ]=2\), the ring of integers has an integral basis \[ \OO=\ZZ\oplus\ZZ\omega \] with \(1,\omega\) a \(\ZZ\)-basis.
1. There is a unique integer \(f\ge1\) such that\[ \{b\in\ZZ:b\omega\in R\}=f\ZZ. \]
Because \(R\) is a subring with identity, \(\ZZ\subseteq R\). Since \(R\ne\ZZ\), choose \[ \alpha=a+b\omega\in R\setminus\ZZ, \qquad a,b\in\ZZ. \] Then \(b\ne0\), and \[ b\omega=\alpha-a\in R. \] Thus \[ I=\{b\in\ZZ:b\omega\in R\} \] is a nonzero additive subgroup of \(\ZZ\). Every nonzero subgroup of \(\ZZ\) has the form \(f\ZZ\) for a unique positive integer \(f\).
2. For the integer \(f\) of step 1,\[ \boxed{R=\ZZ+f\OO.} \]
Since \[ \OO=\ZZ\oplus\ZZ\omega, \] we have \[ \ZZ+f\OO = \ZZ+f(\ZZ\oplus\ZZ\omega) = \ZZ\oplus f\ZZ\omega. \] Step 1 gives \(f\omega\in R\), while \(\ZZ\subseteq R\). Hence \[ \ZZ\oplus f\ZZ\omega\subseteq R. \]
Conversely, let \[ \alpha=a+b\omega\in R. \] Since \(a\in\ZZ\subseteq R\), we have \[ b\omega=\alpha-a\in R. \] By the definition of \(f\) in step 1, this means \(b\in f\ZZ\). Therefore \[ \alpha\in\ZZ\oplus f\ZZ\omega. \] Thus the reverse inclusion also holds.
3. The integer \(f\) is uniquely determined by \(R\); indeed\[ \boxed{f=[\OO:R].} \]
With respect to the basis \(1,\omega\) of \(\OO\), step 2 gives the basis \[ 1,f\omega \] of \(R\). Therefore \[ [\OO:R] = \left| \det \begin{pmatrix} 1&0\\ 0&f \end{pmatrix} \right| =f. \] The index depends only on \(R\), so no other positive integer can give the same subring.
4. Q.E.D.
Steps <1>1–<1>2 prove existence of the required representation, and step 3 proves uniqueness of its conductor.