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

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

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 $p \equiv 1 \pmod{4}$, $q\equiv 1 \pmod{3}$.

• Jordan and Livne gave a complete description of points of $S$ over local fields.

• Jordan gave some results on points of $S$ over imaginary quadratic fields. For example, if $D = 39$, $k = \mathbb{Q}(\sqrt{-3})$, then $S_k$ does not satisfy the Hasse Principle.

• Skorobogatov and Siksek showed that if $S_k$ does not have a $k$-rational divisor class of degree 1 and $\Sh({\mathrm{Jac}}(S_k))$ is finite, then the failure of the Hasse Principle for $S_k$ is accounted for by the Manin obstruction.

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

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

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

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:

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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