Yafaev: Descent on certain Shimura curves

Joint work with A. Skorobogatov.

Let $ B$ be an indefinite division quaternion algebra over $ \mathbb{Q}$ (indefinite means that $ \mathbb{R}\tensor B = M_2(\mathbb{R})$). Let $ \mathcal{O}_B$ be a maximal order in $ B$. Let $ D$ be the discriminant of $ B$ (the product of primes that do not split $ B$), and suppose $ D >
1$. Let $ \mathbb{H}^{\pm}$ be the union of the upper and lower half planes and consider the Shimura curve

$\displaystyle S = \mathcal{O}_B^\times \backslash \mathbb{H}^{\pm}.$    

Then $ S$ is a compact Riemann surface $ S$ with a canonical model over $ \mathbb{Q}$.

Let $ N$ be a prime, $ (N, D) = 1$. Let $ \Gamma_0(N)$, $ \Gamma_1(N)$ be the inverse images in $ \mathcal{O}_B^\times$ of the usual subgroups of $ GL_2(\mathbb{Z}/N)$. Let $ X =
\Gamma_1(N)\backslash\mathbb{H}^{\pm}$, $ Y =
\Gamma_0(N)\backslash\mathbb{H}^{\pm}$. Then $ X \rightarrow Y$ is a Galois covering, with Galois group $ \mathbb{Z}/N\mathbb{Z}^\times$, which is unramified if $ \exists p, q \vert D$ such that $ p \equiv 1
\pmod{4}$, $ q\equiv 1 \pmod{3}$.

We have $S(\mathbb{R}) = \emptyset$, and if $k$\ is im...
... M_2(k)$, and $\vert Cl(k)\vert = 1$, then $S(k) \neq

Consider now the covering $ X \rightarrow Y$ of curves over $ \mathbb{Q}$, as defined above. We want to find an imaginary quadratic field $ k$ and $ X$ and $ Y$ such that for any twist of $ X$ by a character $ {\mathrm{Gal}}(\overline{k}/k) \xrightarrow{\sigma}{}
(\mathbb{Z}/N)^\times/\pm 1$, $ X^\sigma(\mathbb{A}_k) = \emptyset$ but $ Y(\mathbb{A}_k) \neq \emptyset$. Since, as in Skorobogatov's talk, $ Y(k) = \bigcup_\sigma X^\sigma(k)$, this shows that the Hasse principle for $ Y$ fails and provides a cohomological obstruction that explains this failure.

Suppose first that $ v \vert p$ for a prime $ p$ dividing $ D$. Then $ Y$ has bad reduction at $ v$. Furthermore, it is known (Jordan-Livne-Varshavsky) that $ Y(k_v) \neq \emptyset$ if $ p$ is inert. So we suppose that all primes dividing $ D$ are inert in $ k$.

Now consider what happens at $ N$. We have the usual model of $ Y_{\mathbb{F}_N}$ with two components $ S_{\mathbb{F}_N}$, intersecting at supersingular points defined over $ \mathbb{F}_{N^2}$. We assume again that $ N$ is inert. It can be shown that $ \vert S(\mathbb{F}_{N^2}) \setminus \{$Supersingular Points$ \}\vert > 0$. This then shows that if $ N$ is inert, $ Y(k_v) \neq \emptyset$ for $ v\vert N$.

Finally, we consider places $ v$ which do not divide $ ND$. If $ \sigma:
{\mathrm{Gal}}(\overline{k}/k) \rightarrow (\mathbb{Z}/N)^\times/\pm 1$ is ramified at such a place $ v$, $ X^\sigma(k_v) = \emptyset$. So we need only consider characters unramified outside $ ND$. If we suppose further

  1. that $ (N-1)/2$ is prime to $ p(p^2-1)$ for any $ p\vert D$
  2. $ \vert Cl(k)\vert$ is prime to $ (N-1)/2$
then we are left with characters corresponding to $ \mathbb{Q}(\zeta_N)^+k/k$.

Now, $ Y$ and $ X^\sigma$ have good reduction outside $ ND$. To deal with places $ v\notdivides ND$, we count points to show that the curve has points over $ \mathbb{F}_v$ (which is $ \mathbb{F}_{p}$ or $ \mathbb{F}_{p^2}$ according as $ p$ is split or inert in $ k$), then lift them using Hensels Lemma.

The point counts make use of the following trace formulas: $ p \notdivides ND$ then

$\displaystyle {\mathrm{Tr}}(F_p^r\vert H^1_{\text{\'et}}(Y_{\overline{\mathbb{F}}_p},
\mathbb{Q}_\ell))$ $\displaystyle =$ $\displaystyle {\mathrm{Tr}}(pT_{p^{r-2}} - T_{p^r}\vert H^0(Y_{\overline{\mathbb{F}}_p}, \Omega^1))$  
$\displaystyle {\mathrm{Tr}}(F_p^r\vert H^1_{\text{\'et}}(X^\sigma_{\overline{\mathbb{F}}_p},
\mathbb{Q}_\ell))$ $\displaystyle =$ $\displaystyle {\mathrm{Tr}}(\gamma^r pT_{p^{r-2}} - \gamma^r T_{p^r}\vert H^0(X_{\overline{F}_p}, \Omega^1))$  

where $ F_p$ is Frobenius and $ \gamma = \sigma(F_p)$. We get $ Y(\mathbb{F}_{p^2})\vert > 0$, and $ Y(\mathbb{F}_p)\vert > 0$ if $ \exists t
\in \mathbb{Z}$, $ \vert t\vert < 2\sqrt{p}$ such that
  1. all primes dividing $ D$ are split in $ \mathbb{Q}(\sqrt{t^2-4p})$
  2. $ p \notdivides t $ or $ p$ is not split in $ \mathbb{Q}(\sqrt{t^2-4p})$
  3. $ N$ is not inert in $ \mathbb{Q}(\sqrt{t^2-4p})$.

Putting these together we get the following proposition:
\begin{proposition}[Local points on $X^\sigma$]
Suppose that for all $m \in \{ 0...
...acters $\sigma$\ we have $X^\sigma(\mathbb{A}_k) =
We can now find counterexamples to the Hasse principle as follows:

  1. Choose $ D = q_1 q_2$, with $ q_1 \equiv 1 \pmod{4}$ and $ q_2 \equiv 1 \pmod{3}$.
  2. Choose $ N$ such that $ (N-1)/2$ is coprime to $ q_1(q_2^2-1)$ $ q_2(q_2^2
- 1)$.
  3. Let $ p$ be a prime such that $ p \notdivides ND$. Call $ p$ good if there exists $ t$ such that condition [*] above is satisfied, bad otherwise. (The set of bad primes is finite.) Now find a good $ p$ such that ([*]) is satisfied.
  4. Now choose a field $ k$ imaginary quadratic such that $ q_1, q_2, N$ are inert in $ k$, bad primes are inert, primes from Step [*] are split or ramified, and $ \vert Cl(k)\vert$ is prime to $ (N-1)/2$. Ono can prove that if there is one $ k$satisfying these conditions there are infinitely many.

Example: $ D = 35$, $ N = 23$, $ k = \mathbb{Q}(\sqrt{-127})$.

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