Refined class number formulas for elliptic units

Abstract

We generalize the notions of Kolyvagin and pre-Kolyvagin systems to prove “refined class number formulas” for quadratic extensions of a quadratic imaginary fields $K$ of class number one. Our main result generalises the results and conjectures of Darmon (Canad. J. Math. 47:302–317, 1995), by replacing circular units in abelian extensions of $\mathbb {Q}$ by elliptic units in abelian extensions of K.

Résumé

Nous généralisons ici les notions de systèmes et de pré-systèmes de Kolyvagin pour prouver des “formules de nombres de classes raffinées” pour les extensions quadratiques de corps quadratiques imaginaires $K$ de groupe de classes trivial. Notre résultat principal généralise les résultats et les conjectures de Darmon (Canad. J. Math. 47:302–317, 1995) en remplaçant les unités circulaires dans les extensions abéliennes de $\mathbb {Q}$ par les unités elliptiques dans les extensions abéliennes de $K$.

Appendices

Appendix A: About the augmentation quotient

Let $\mathcal {I}_\mathfrak {m}^{new}$ be the subgroup generated by the elements of the form $\prod _{\mathfrak {l}|\mathfrak {m}^+}(g_\mathfrak {l}-1)$ where $g_\mathfrak {l} \in \Gamma _\mathfrak {l}$. Denote also $P_\mathfrak {n}$, the composition $\mathbb {Z}[\Gamma _\mathfrak {m}]\rightarrow \mathbb {Z} [\Gamma _\mathfrak {n}]\rightarrow \mathbb {Z}[\Gamma _\mathfrak {m}]$. To simplify the notation denote $r:=r(\mathfrak {m}^+)$. We have the following statement.

Proposition 10.4

For any abelian group $A$, if $x\in A \otimes I^r/I^{r+1}$ is such that $P_{\mathfrak {m}/\mathfrak {l}}(x) = 0$ for all $\mathfrak {l}|\mathfrak {m}^+$, then $x\in A\otimes \mathcal {I}_\mathfrak {m}^{new}$.

Proof

As a first remark, since $P_{\mathfrak {m}/\mathfrak {l}_1}\circ P_{\mathfrak {m}/\mathfrak {l}_2}=P_{\mathfrak {m}/(\mathfrak {l}_1 \mathfrak {l}_2)}$, we have

$$\begin{aligned} P_{\mathfrak {m}^-\mathfrak {d}}(x) = 0 \forall \mathfrak {d}|\mathfrak {m}^+ . \end{aligned}$$

For any $\mathfrak {l}|\mathfrak {m}$ choose a generator $\gamma _\mathfrak {l}$ of $\Gamma _{\mathfrak {l}}$. Then the elements of the form $\prod (\gamma _\mathfrak {l}-1)$ generate $I^r/I^{r+1}$. Indeed for any $\prod _j (g_j-1) \in I^r/I^{r+1}$, we have

$$\begin{aligned} \prod _j (g_j-1)&= \prod _j \left( \prod _\mathfrak {l} \gamma _\mathfrak {l}^{n_{\mathfrak {l},j}} -1\right) \\&= \prod _j \left( \sum _\mathfrak {l} n_{\mathfrak {l},j}(\gamma _\mathfrak {l} -1\right) \quad \text { since } g_1g_2-1= g_1-1 + g_2-1 \in I_\mathfrak {m}/I_\mathfrak {m}^2\\&=\sum _{I} n_I \prod _{\mathfrak {l}\in I} (\gamma _\mathfrak {l} -1). \end{aligned}$$

Using the previous fact and using the notations

$$\begin{aligned} \mathfrak {m}^+=\prod _{i=1}^r \mathfrak {l}_i\quad \text{ and } \quad \mathcal {J}^-_r:=\left\{ \lbrace \mathfrak {l}_1,\dots ,\mathfrak {l}_r \rbrace \text { such that } \forall i, \mathfrak {l}_i | \mathfrak {m}^- \right\} , \end{aligned}$$

any element $x\in A \otimes I^{r}/I^{r+1}$ can be written as

$$\begin{aligned} x&= x_{new} \otimes \prod _{i=1}^r(\gamma _{\mathfrak {l}_i}-1) + \sum _{J\in \mathcal {J}^-_r} x_J \otimes \prod _{\mathfrak {l} \in J} (\gamma _{\mathfrak {l}}-1)\\&\quad + \sum _{a_1=1}^r\left( \sum _{J\in \mathcal {J}_{r-a_1}} x_{J,1}\otimes (\gamma _{\mathfrak {l}_1}-1)^{a_1} \prod _{\mathfrak {l} \in J} (\gamma _{\mathfrak {l}}-1)\right) +\cdots \\&\quad \cdots +\sum _{a_r=1}^r\left( \sum _{J\in \mathcal {J}_{r-a_r}} x_{J,r}\otimes (\gamma _{\mathfrak {l}_r}-1)^{a_r} \prod _{\mathfrak {l} \in J} (\gamma _{\mathfrak {l}}-1)\right) \\&\quad +\sum _{a_1+a_2=2,a_1,a_2\ne 0}^r \left( \sum _{J\in \mathcal {J}_{r-a_1-a_2}} x_{J,1,2}\otimes (\gamma _{\mathfrak {l}_1}-1)^{a_1}(\gamma _{\mathfrak {l}_2}-1)^{a_2} \prod _{\mathfrak {l} \in J}(\gamma _{\mathfrak {l}}-1)\right) +\cdots \\&\qquad \qquad \qquad \qquad \qquad \quad \vdots \qquad \qquad \qquad \qquad \qquad \vdots \\&\quad + \sum _{\mathfrak {l}|\mathfrak {m}^-} x_{\mathfrak {l},\mathfrak {l}_r} \otimes (\gamma _{\mathfrak {l}_1}-1)\cdots (\gamma _{\mathfrak {l}_{r-1}}-1) (\gamma _\mathfrak {l}-1)+\cdots \\&\quad \dots +\sum _{\mathfrak {l}|\mathfrak {m}^-} x_{\mathfrak {l},\mathfrak {l}_1}\otimes (\gamma _{\mathfrak {l}_2}-1)\cdots (\gamma _{\mathfrak {l}_r}-1)(\gamma _\mathfrak {l}-1). \end{aligned}$$

To finish the proof, we have to show that $x = x_{new} \otimes \prod _{\mathfrak {l}|\mathfrak {m}^+}(\gamma _\mathfrak {l}-1)$ and we prove it by induction. By applying $P_{\mathfrak {m}^-}$, we see that

$$\begin{aligned} \sum _{J\in \mathcal {J}^-_r} x_J \otimes \prod _{\mathfrak {l} \in J} (\gamma _{\mathfrak {l}}-1)=0 \end{aligned}$$

and the second row is zero. Then applying $P_{\mathfrak {m}^-\mathfrak {l}_i}$, we have

$$\begin{aligned} \sum _{a_i=1}^r\left( \sum _{J\in \mathcal {J}_{r-a_i}} x_{J,i}\otimes (\gamma _{\mathfrak {l}_i}-1)^{a_i} \prod _{\mathfrak {l} \in J} (\gamma _{\mathfrak {l}}-1)\right) =0. \end{aligned}$$

Applying it for all $i$ we see that the third row is zero. Then, by induction on the rows, we apply $P_{\mathfrak {m}^-\mathfrak {l}_{i_1}\cdots \mathfrak {l}_{i_n}}$ for all $n-$uple $(i_1,...i_n)$ and we see that the $(n+2)$-th row is zero. $\square$

Appendix B: On a technical hypothesis in [3]

We finally end this paper with a small erratum in [3]. Stated as they are, Lemma 2.1.4 and Lemma 3.5.2 are wrong and we state here a “good version” of Lemma 3.5.2. It does not change anything in [3], since Lemma 3.5.2 is only used in a special case where it is actually true. We suggest here a new formulation of this lemma with its proof. We follow the notation of [3].

Lemma 10.5

Suppose that (H0), (H1) and (H3) hold. Then for any ideal $J$ of $R$ and any submodule $S$ of $T$ and $S'$ of $T^*$, we have

$$\begin{aligned} (S/J T)^{G_\mathbb {Q}}=(S'/J T^*)^{G_\mathbb {Q}}=0. \end{aligned}$$

Proof

By (H1), if $(T/\mathfrak {m} T)^{G_\mathbb {Q}} \ne 0$, then $G_\mathbb {Q}$ acts trivially on $T$. In this case we can find a nonzero element in $H^1(\mathbb {Q}(T,\mu _{p^\infty })/\mathbb {Q},T/\mathfrak {m} T)$ since $Gal(\mathbb {Q}(T,\mu _{p^\infty })/\mathbb {Q})$ has a quotient isomorphic to $Z/pZ$ and $T/\mathfrak {m} T \simeq Z/pZ$. By (H3), this shows $(T/\mathfrak {m} T)^{G_\mathbb {Q}} = 0$.

For any submodule $S$ of $T$, and any ideal $J\subseteq R$, $(S/J T)^{G_\mathbb {Q}}$ injects in $(T/J T)^{G_\mathbb {Q}}$, so we have to show that $(T/J T)^{G_\mathbb {Q}}=0$. Suppose $\mathfrak {m}^{i+1} \subsetneq J \subseteq \mathfrak {m}^{i}$ and let $\overline{x} \in (T/J T)^{G_\mathbb {Q}}$ with antecedent $x$ in $T$. By definition, $x\in (T/\mathfrak {m} T)^{G_\mathbb {Q}}=0$ and $x \in \mathfrak {m}T$. Write $x=a_1 x_1$. We have $a_1 x_1 \in \in (T/J T)^{G_\mathbb {Q}}$. If $a_1 \in J$, $\overline{x}=0$ and we are done. If not, by (H0), $x_1 \in \mathfrak {m}T$ and we can write $x_1=a_2 x_2$ with $a_2 \in \mathfrak {m}$. By induction, $x=a_1\cdots a_{i+1} x_{i+1}$ with $a_k \in \mathfrak {m}$ and $\overline{x}=0$. The proof for $T^*$ is similar. $\square$