How does this calculation of Siegel make sense?

Question

I am reading Siegel's paper Zum Beweise des Starkschen Satzes. Let $K$ be an imaginary quadratic field with $d_K=-p$, $p=4k+3$ a prime, and such that $h_K=1$.

Let $f=4m+1$ be a prime inert in $K$, and consider the order $\mathcal{O}=\mathbb Z+f\mathcal O_K$ with conductor $f$. Let $$\omega=\frac{fp+\sqrt{-p}}{2}.$$ Siegel defines the lattices $$\mathfrak c_k=[f,k-\omega]\quad\text{for } -\frac{f-1}{2}\leq k\leq \frac{f-1}{2}$$ and $$\mathfrak{c}_\infty =[1,-f\omega].$$ These are proper fractional ideals of $\mathcal O$, mutually non-homothetic. By a well-known formula relating the class numbers $h_K$ and $h(\mathcal O)$ we know that $h(\mathcal O)=f+1$, so the ideals above represent all ideal classes of $\mathcal O$. A little calculation shows that $N(\mathfrak{c}_k)=N(\mathfrak{c}_\infty)=1$. Indeed

$$N(\mathfrak c_k)=\frac{N([f^2,fk-f\omega])}{f^2}=\frac{1}{f^2}\frac{\text{disc}([f^2,fk-f\omega])^{1/2}}{\text{disc}(\mathcal O)^{1/2}}=\frac{1}{f^2}\frac{\begin{vmatrix}f^2 & fk-f\omega \\ f^2 & fk -f\overline \omega\end{vmatrix}}{\begin{vmatrix}1 & -f\omega \\ 1& -f\overline \omega\end{vmatrix}}=1.$$

Then Siegel proceeds to calculate the values of the character defined by $$\chi(\mathfrak a)=\left(\frac{f}{N(\mathfrak a)}\right)=\left(\frac{fd_K}{N(\mathfrak a)}\right).$$

This does not make sense to me, because the norms are equal to $1$, but Siegel gets different values. See the referenced paper, beginning of the section 2., p. 183.

Update

We have $\mathfrak c_k \not \subset \mathcal O$ but $f\mathfrak c_k\subset \mathcal O$, so we can use the relation $N(f)N(\mathfrak c_k)=N(f\mathfrak c_k)$:

$$ N(\mathfrak c_k) =\frac{N(f\mathfrak c_k)}{N(f)} = \frac{N(f\mathfrak c_k)}{f^2} .$$ To compute $N(f\mathfrak c_k)$ we use the following fact: if $M\subset L$ are free modules of the same rank $n$, $(e_i)$ and $(u_i)$ bases for $L,M$ respectively, $u_i=\sum c_{ij}e_j$, then $(L:M)=\lvert \det(c_{ij})\rvert.$ Therefore $$\mathcal O=[1,-f\omega],\qquad f\mathfrak c_k=[f^2,fk-f\omega],\qquad N(f\mathfrak c_k)=\begin{vmatrix}f^2 & 0\\ fk &1\end{vmatrix}=f^2.$$

Consequently, $N(\mathfrak c_k)=1$.

Update 2

The $\mathfrak c_k$ are not ideals of $\mathcal O_K$. Let $m$ be a rational integer. We prove that if $m\omega \mathfrak c_k\subset \mathfrak c_k$ then $m$ is a multiple of $f$.

Suppose that $m\omega \mathfrak c_k\subset \mathfrak c_k$ and that $(f,m)=1$. Then $$m\omega(k-\omega)=xf+y(k-\omega),\qquad x,y\in \mathbb Z,$$ $$(m\omega-y)(k-\omega)\in f\mathcal O_K.$$ But $f$ was assumed to be inert in $K$, so $f\mathcal O_K$ is a prime ideal, and $\omega\not\in \mathcal O$. Therefore $\omega\equiv y/m$ modulo $f\mathcal O_K$, because $m$ is invertible. On the other hand, from the definition of $\omega$ we have $4\omega^2\equiv p$ modulo $f\mathcal O_K$. Therefore $(p|f)=1$. But since $f=4m+1$ is inert $-1=(-p|f)=(p|f)$, a contradiction.

This shows that $\mathfrak c_k$ is a proper ideal of $\mathcal O$.

Answer 1

Here's what I found out so far.

Let $K$ be a complex quadratic number field with discriminant $\Delta < -4$. The ring class group modulo $f$ is a special case of a ray class group: Two ideals (coprime to $f$, as everything below) are equivalent in the ring class group modulo $f$ if $\alpha {\mathfrak a} = \beta {\mathfrak b}$ for elements $\alpha, \beta \in {\mathcal O}_K$ congruent to a rational integer modulo $f$. The different classes can be represented by ideals in ${\mathcal O}_K$ (as I just did), as ${\mathbb Z}$-modules, or as ideals in the order ${\mathcal O}_f$. There are a lot of isomorphisms floating around, and the underlying sets of these objects are, in general, not the same.

Let me give an example. Consider $K = {\mathbb Q}(\sqrt{-7})$ and $f = 5$. The formula for the number of ring classes (see Cox, Primes of the form $x^2 + ny^2$ or, better yet, Cohn's Advanced number theory) gives $h = 6$. The corresponding ring classes are represented by the ideals $(1)$ (the principal class) and the ideals $(k+\alpha)$ for $k = 0, 1, \ldots, 4$, where $\alpha = \frac{1 + \sqrt{-7}}2$. This does not contain the number theoretic information we are interested in.

We therefore consider the ${\mathbb Z}$-modules $M_k = [5, k-\omega]$ and $M_\infty = [1, -5\omega]$, where $\omega= \frac{35 + \sqrt{-7}}2$. To these modules $M_k = [\alpha, \beta]$ we associate quadratic forms $Q_k = N(\alpha x + \beta y)$. Here's what we get: $$ \begin{array}{c|cc} k & Q_k & \text{reduced form} \\ \hline 1 & 25x^2 - 165xy + 274y^2 & (4, -1, 11) \\ 2 & 25x^2 - 155xy + 242y^2 & (2, 1, 22) \\ 3 & 25x^2 - 145xy + 212y^2 & (2, -1, 22) \\ 4 & 25x^2 - 135xy + 184y^2 & (4, 1, 11) \\ 5 & 25x^2 - 125xy + 158y^2 & (7, 7, 8) \\ \infty & x^2 - 175xy + 7700y^2 & (1, 1, 44) \end{array} $$ These are the six form classes of binary quadratic forms with discriminant $-5^2 \cdot 7$. These form classes contain all the information we need for computing class fields using complex multiplication.

The only nontrivial quadratic character $\chi$ on the ring class group is the one with values $-1$ on the nonsquare classes. Since the forms $Q_1$, $Q_4$ and $Q_\infty$ obviously represent squares, we have $\chi(Q_1) = \chi(Q_4) = \chi(Q_\infty) = 1$ and $\chi(Q_2) = \chi(Q_3) = \chi(Q_5) = -1$.

We can also attach ideals in the rings ${\mathcal O}_f$ representing the six equivalence classes by simply associating the ideal $(a, \frac{b - f\sqrt{\Delta}}2)$ to the form $(a, b, c)$. I have not yet checked how the evaluation of the genus character works using these ring ideals.