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 $\mathcal O$.

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