We generalize L. J. Mordell’s construction of cubic surfaces for which the Hasse principle fails.

Jordan, Rotger and de Vera-Piquero proved that Shimura curves have no points rational over imaginary quadratic fields under a certain assumption. In this article, we extend their results to the case of number fields of higher degree. We also give counterexamples to the Hasse principle on Shimura curves.

Let X be a nice variety over a number field k. We characterise in pure “descent-type” terms some inequivalent obstruction sets refining the inclusion $X{{(}_k)}^{ét,Br}\subset X{{(}_k)}^{Br₁}$. In the first part, we apply ideas from the proof of $X{{(}_k)}^{ét,Br}=X{{(}_k)}^{{}_k}$ by Skorobogatov and Demarche to new cases, by proving a comparison theorem for obstruction sets. In the second part, we show that if $\subset {\subset }_k$ are such that $\subset Ext\left({,}_k\right)$, then $X{{(}_k)}^{}=X{{(}_k)}^{}$. This allows us to conclude, among other things, that $X{{(}_k)}^{ét,Br}=X{{(}_k)}^{{}_k}$ and $X{{(}_k)}^{Sol,Br₁}=X{{(}_k)}^{So{l}_k}$.

Si un système d’équations polynomiales à coefficients entiers admet une solution dans ${𝐐}^n$, il en admet sur tout complété $p$-adique ou réel de $𝐐$. La réciproque a été démontrée par Hasse pour les quadriques, mais elle est fausse en général. Une grande partie des contre-exemples connus peuvent être expliqués à l’aide de l’obstruction de Brauer-Manin, basée sur la théorie du corps de classe. Il est donc naturel de se demander si, pour certaines classes de variétés, cette obstruction est la seule. Le but...

Let $C_m:y^2=x^3-m^{2}x+p^2{q}^2$ be a family of elliptic curves over $\mathbb{Q}$, where $m$ is a positive integer and $p$, $q$ are distinct odd primes. We study the torsion part and the rank of $C_m\left(\mathbb{Q}\right)$. More specifically, we prove that the torsion subgroup of $C_m\left(\mathbb{Q}\right)$ is trivial and the $\mathbb{Q}$-rank of this family is at least 2, whenever $m\neg \equiv 0(mod3)$, $m\neg \equiv 0(mod4)$ and $m\equiv 2(mod64)$ with neither $p$ nor $q$ dividing $m$.