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.


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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    $\begingroup$ Your ideals are integral ideals, and their norms seem to be $f$, not $1$. $\endgroup$ – Franz Lemmermeyer Mar 25 '20 at 16:56
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    $\begingroup$ The norm of an ideal with basis $[m, a+n\omega]$ is $mn$. $\endgroup$ – Franz Lemmermeyer Mar 26 '20 at 7:48
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    $\begingroup$ Anyway, using the definition of the norm of an ideal as the cardinality of its residue class ring, use the element $k - \omega$ to reduce any algebraic integer to an ordinary integer; using the element $f$ of the ideal we find that the residue class ring is simply ${\mathbb Z}/f{\mathbb Z}$, so the norm if the ideal is $f$. $\endgroup$ – Franz Lemmermeyer Mar 26 '20 at 14:29
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    $\begingroup$ You're right. But they are ${\mathcal O}_k$-modules, and these have norm $f$. Siegel's "Ringideal" is defined as in Hecke's lecture "Analysis und Zahlentheorie" (edited by Roquette), p. 125ff. $\endgroup$ – Franz Lemmermeyer Mar 26 '20 at 21:08
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    $\begingroup$ ${\mathbb Z}-modules$ in ${\mathcal O}_k$ of course - sorry. I'll go through the stuff carefully until tomorrow. $\endgroup$ – Franz Lemmermeyer Mar 27 '20 at 9:21

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.

  • $\begingroup$ Thank you for your answer. However, it is still not clear to me what is wrong with my calculation of the norm. According to the question above, every ideal class of $\text{Cl}(\mathcal O_f)$ contains a fractional ideal of norm equal to $1$. But by this question/answer this happens if and only if the class number $h_f$ is odd. But $h_f=f+1$ above is even. $\endgroup$ – Shimrod Mar 28 '20 at 20:08
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    $\begingroup$ Neither do I. For the moment, I think that Siegel is a little bit sloppy with his language. $\endgroup$ – Franz Lemmermeyer Mar 28 '20 at 23:28

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